ARS MATHEMATICA CONTEMPORANEA Volume 17, Number 2, Fall/Winter 2019, Pages 349-698 Covered by: Mathematical Reviews zbMATH (formerly Zentralblatt MATH) COBISS SCOPUS Science Citation Index-Expanded (SCIE) Web of Science ISI Alerting Service Current Contents/Physical, Chemical & Earth Sciences (CC/PC & ES) dblp computer science bibliography The University of Primorska The Society of Mathematicians, Physicists and Astronomers of Slovenia The Institute of Mathematics, Physics and Mechanics The Slovenian Discrete and Applied Mathematics Society The publication is partially supported by the Slovenian Research Agency from the Call for co-financing of scientific periodical publications. ARS MATHEMATICA CONTEMPORANEA Where Do We Stand? This volume completes the 11th year of publication of AMC. A very natural question for our readers, authors, reviewers and editors is this: Where do we stand? Although we do not believe in one-dimensional ranking, we would like to present some figures that tell how SCImago sees our journal. SCImago uses Scopus data to rank a total of 31971 scientific journals worldwide, according to their SJR index. Among these, AMC ranks 3612th, which puts AMC in the top 12%. Similarly, among the 2011 mathematics journals in the Scopus database, AMC ranks 246th, putting it in the top 13%. At first sight this might not seem all that impressive. But among the 926 scientific journals published in Eastern Europe, AMC ranks 7th, and among the 152 of those that are in mathematics, AMC ranks first! Moreover, according to SCImago, Ars Mathematica Contemporanea covers papers from Algebra and Number Theory, Discrete Mathematics and Combinatorics, Geometry and Topology, and Theoretical Computer Science, and in each of these categories, AMC is in Q1 (the first quartile). It is also in the Scopus first quartile in the field of Mathematics for the year 2018. Still, these are just numbers, and are not the most important for our community. What matters is the quality of the papers that we publish. Also we are aware of the fact that some authors are not happy with our large backlog, and for their career, other rankings may be more important. For AMC itself, however, these numbers matter quite a lot. In particular, it is notable that among the 78 scientific journals covered by Scopus that are published in Slovenia, our journal also ranks first. This is of crucial importance when we seek financial support from our government. Klavdija Kutnar, Dragan Marušic and Tomaž Pisanski Editors in Chief xvii ARS MATHEMATICA CONTEMPORANEA Contents Girth-regular graphs Primož Potočnik, Janoš Vidali........................349 The existence of square non-integer Heffter arrays Nicholas J. Cavenagh, Jeffery H. Dinitz, Diane M. Donovan, Emine §ule Yazici...............................369 Semigroups with fixed multiplicity and embedding dimension Juan Ignacio García-García, Daniel Marín-Aragón, María Ángeles Moreno-Frías, José Carlos Rosales, Alberto Vigneron-Tenorio.......397 Grundy domination and zero forcing in Kneser graphs Boštjan Brešar, Tim Kos, Pablo Daniel Torres................419 S12 and P12-colorings of cubic graphs Anush Hakobyan, Vahan Mkrtchyan.....................431 The complement of a subspace in a classical polar space Krzysztof Petelczyc, Mariusz Zynel.....................447 Unicyclic graphs with the maximal value of Graovac-Pisanski index Martin Knor, Jozef Komorník, Riste Škrekovski, Aleksandra Tepeh.....455 On separable abelian p-groups Grigory Ryabov ................................ 467 Strong geodetic problem on complete multipartite graphs Vesna Iršic, Matjaž Konvalinka........................481 Archimedean toroidal maps and their minimal almost regular covers Kostiantyn Drach, Yurii Haidamaka, Mark Mixer, Maksym Skoryk.....493 Symplectic semifield spreads of PG(5,qt), q even Valentina Pepe ................................ 515 On the Hamilton-Waterloo problem: the case of two cycles sizes of different parity Melissa S. Keranen, Adrián Pastine......................525 The finite embeddability property for IP loops and local embeddability of groups into finite IP loops Martin Vodicka, Pavol Zlatoš.........................535 On graphs with the smallest eigenvalue at least — 1 - a/2, part III Sho Kubota, Tetsuji Taniguchi, Kiyoto Yoshino...............555 Lobe, edge, and arc transitivity of graphs of connectivity 1 Jack E. Graver, Mark E. Watkins.......................581 xvii ARS MATHEMATICA CONTEMPORANEA Bipartite edge-transitive bi-p-metacirculants Yan-Quan Feng, Yi Wang...........................591 On flag-transitive automorphism groups of symmetric designs Seyed Hassan Alavi, Ashraf Daneshkhah, Narges Okhovat.........617 The symmetric genus spectrum of abelian groups Coy L. May, Jay Zimmerman.........................627 Extremal embedded graphs Qi Yan, Xian'an Jin..............................637 Reconfiguring vertex colourings of 2-trees Michael Cavers, Karen Seyffarth.......................653 Volume 17, Number 2, Fall/Winter 2019, Pages 349-698 xvii ARS MATHEMATICA CONTEMPORANEA ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 17 (2019) 349-368 https://doi.org/10.26493/1855-3974.1684.b0d (Also available at http://amc-journal.eu) Girth-regular graphs * Primož Potočnik Faculty of Mathematics and Physics, University of Ljubljana, Jadranska 19, SI-1000 Ljubljana, Slovenia, and Institute of Mathematics, Physics and Mechanics, Jadranska 19, SI-1000 Ljubljana, Slovenia Janoš Vidali Faculty of Mathematics and Physics, University of Ljubljana, Jadranska 19, SI-1000 Ljubljana, Slovenia Received 23 April 2018, accepted 4 March 2019, published online 4 November 2019 We introduce a notion of a girth-regular graph as a k-regular graph for which there exists a non-descending sequence (ai, a2,..., ak) (called the signature) giving, for every vertex u of the graph, the number of girth cycles the edges with end-vertex u lie on. Girth-regularity generalises two very different aspects of symmetry in graph theory: that of vertex transitivity and that of distance-regularity. For general girth-regular graphs, we give some results on the extremal cases of signatures. We then focus on the cubic case and provide a characterisation of cubic girth-regular graphs of girth up to 5. Keywords: Graph, girth-regular, cubic, girth. Math. Subj. Class.: 05C38 1 Introduction This paper stems from our research of finite connected vertex-transitive graphs of small girth. The girth (the length of a shortest cycle in the graph) is an important graph theoretical invariant that is often studied in connection with the symmetry properties of graphs. For example, cubic arc-transitive graphs (a graph is called arc-transitive if its automorphism group acts transitively on its arcs, where an arc is an ordered pair of adjacent vertices) and cubic semisymmetric (regular, edge-transitive but not vertex-transitive) graphs of girth up * Supported in part by the Slovenian Research Agency, projects J1-5433, J1-6720, and P1-0294. E-mail addresses: primoz.potocnik@fmf.uni-lj.si (Primož Potočnik), janos.vidali@fmf.uni-lj.si (Janoš Vidali) Abstract ©® This work is licensed under https://creativecommons.org/licenses/by/4.0/ 370 Ars Math. Contemp. 17 (2019) 185-202 to 9 and 10 have been studied in [5,11] and [6], respectively, and tetravalent edge-transitive graphs of girths 3 and 4 have been considered in [14]. Recently, a classification of all cubic vertex-transitive graphs of girth up to 5 was obtained in [8]. In our investigation of vertex-transitive graphs of small girth, it became apparent to us that the condition of vertex-transitivity is almost never used in its full strength. What was needed in most of the arguments was only a particular form of uniformity of the distribution of girth cycles throughout the graph. Let us make this more precise. For an edge e of a graph r, let e(e) denote the number of girth cycles containing the edge e. Let v be a vertex of r and let jei,..., ek } be the set of edges incident to v ordered in such a way that e(e1) < e(e2) < • • • < e(ek). Then the k-tuple (e(e1), e(e2),..., e(ek)) is called the signature of v. A graph r is called girth-regular provided all of its vertices have the same signature. The signature of a vertex is then called the signature of the graph. We should like to point out that girth-regular graphs of signature (a, a,..., a) for some a have been introduced under the name edge-girth-regular graphs in [10], where the authors focused on the families of cubic and tetravalent edge-girth-regular graphs. By definition, every girth-regular graph is regular (in the sense that all its vertices have the same valence). Further, it is clear that every vertex-transitive as well as every semisym-metric graph is girth-regular. Slightly less obvious is the fact that every distance-regular graph is also girth-regular. The notion of girth-regularity is thus a natural generalisation of all these notions. On the other hand, examples of girth-regular graphs exist that are neither vertex-transitive, nor semisymmetric nor distance-regular (for example, the truncation of a 3-prism is such a graph; see Section 3.2). The central question we would like to propose and address in this paper is the following: Question 1.1. Given integers k and g, for which tuples a = (a1, a2,..., ak) G Zk does a girth-regular graph of girth g and signature a exist? The above question seems to be very difficult if considered in its full generality. We begin by stating three theorems proved in Section 2, which give an upper bound on the entries a4 of the signature in terms of the valence k and the girth g, and consider the case where this upper bound is attained. Theorem 1.2. If r is a girth-regular graph of valence k, girth g, and signature (a1,... ,ak), then ak < (k — 1)d, where d = |_g/2j. Theorem 1.3. If r is a connected girth-regular graph of valence k, girth 2d for some integer d, and signature (a1,..., ak) such that ak = (k — 1)d, then a1 = a2 = • • • = ak and r is the incidence graph of a generalised d-gon of order (k — 1,k — 1). In particular, if k = 3, then g G {4, 6, 8,12} and r is isomorphic to K3,3 (if g = 4), the Heawood graph (if g = 6), the Tutte-Coxeter graph (if g = 8) or to the Tutte 12-cage (if g = 12). For a description of the graphs mentioned in the above theorem, see Note 2.2. Theorem 1.4. If r is a connected 3-valent girth-regular graph of girth 2d +1 for some integer d and signature (a1, a2 ,a3) such that a3 = 2d, then r is isomorphic to K4 or the Petersen graph. In the second part of the paper, we focus on 3-valent graphs (also called cubic graphs) and obtain a complete classification of cubic girth-regular graphs of girth at most 5 (see P. Potočnik and J. Vidali: Girth-regular graphs 351 Theorem 1.5 below). Prisms and Möbius ladders are defined in Section 4, the notion of a dihedral scheme and truncation is defined in Section 3.2, and graphs arising from maps are discussed in Section 3.3. Theorem 1.5. Let r be a connected cubic girth-regular of girth g with g < 5. Then either the signature of r is (0,1,1) and r is a truncation of a dihedral scheme on some g-regular graph (possibly with parallel edges), or one of the following occurs: 1. g = 3 and r = K4 with signature (2,2, 2); 2. g = 4 and r is isomorphic to a prism or to a Möbius ladder, with signature (4,4,4) if r = K3,3, signature (2, 2,2) if r is isomorphic to the cube Q3, and signature (1,1,2) otherwise; 3. g = 5 and r is isomorphic to the Petersen graph with signature (4,4,4), or to the dodecahedron with signature (2, 2, 2). Since every vertex-transitive graph is girth-regular, the above result can be viewed as a partial generalisation of the classification [5] of arc-transitive cubic graphs of girth at most 9 and also a recent classification [8] of vertex-transitive cubic graphs of girth at most 5. Unless explicitly stated otherwise, by a graph, we will always mean a finite simple graph, defined as a pair (V, where V is the vertex-set and ~ an irreflexive symmetric adjacency relation on V. However, in Section 3.2 it will be convenient to allow graphs possessing parallel edges; details will be explained there. Finally, in Section 3.3, when considering embeddings of graphs onto surfaces, we will intuitively think of a graph in a topological context as a 1-dimensional CW complex. See that section for details. 2 An upper bound on the signature This section is devoted to the proof of Theorems 1.2, 1.3 and 1.4 that give an upper bound on the number of girth cycles through an edge in a girth-regular graph and in some cases characterise the graphs attaining this bound. 2.1 Moore graphs and generalised n-gons We begin by a well-known result that sets a lower bound on the number of vertices for a k-regular graph of finite girth g. Proposition 2.1 (Tutte [16, 8.39], cf. Brouwer, Cohen & Neumaier [2, §6.7]). Let r be a k-regular graph with n vertices and finite girth g > 2. Let d = \g/2\. Then ' i + k £?=-o3)/2(* - i)j = if g is odd, .2£?=o2)/2(* - i)j = ifg is even. ( ^ Note 2.2. Let r be a k-regular graph of girth g for which equality holds in (2.1). If g is odd, then such an extremal graph is called a Moore graph. It is well known (see [7] or [1], for example) that a Moore graph is either a complete graph, an odd cycle, or has girth 5 and valence k G {3,7, 57}. Of the latter, the first two cases uniquely determine the Petersen graph and the Hoffman-Singleton graph, respectively, while no example is known for k = 57. 372 Ars Math. Contemp. 17 (2019) 185-202 If the girth g is even, then r is an incidence graph of a generalised (g/2)-gon of order (k-1, k-1) (see [15] or [2, §6.5], for example). For k = 2, we have ordinary polygons, and their incidence graphs are even cycles. For k > 3, such generalised (g/2)-gons only exist if g/2 G {2,3,4, 6} (see [9, Theorem 1]). In particular, if k = 3, then r is the incidence graph of a generalised d-gon of order (2, 2), where d G {2,3,4, 6}. For d =2, this is a geometry with three points incident to three lines, so its incidence graph is K3 3. For d = 3, we get the Fano plane, and its incidence graph is the Heawood graph, which is the unique cubic arc-transitive graph on 14 vertices. For d = 4, there is a unique generalised quadrangle of order (2,2), cf. Payne & Thas [12, 5.2.3], and its incidence graph is the Tutte-Coxeter graph, also known as the Tutte 8-cage, which is the unique connected cubic arc-transitive graph on 30 vertices. For d =6, there is a unique dual pair of generalised hexagons of order (2,2), cf. Cohen & Tits [3], and their incidence graph (on 126 vertices), also known as the Tutte 12-cage, is not vertex-transitive. However, the latter graph is edge-transitive, making it semisymmetric - in fact, it is the unique cubic semisymmetric graph on 126 vertices (see [4], where this graph is denoted by S126). 2.2 Proof of Theorems 1.2,1.3 and 1.4 Equipped with these facts, we are now ready to prove Theorems 1.2, 1.3 and 1.4. Let us thus assume that r is a simple connected girth-regular graph of valence k > 3, let g be its girth and let (ai, a2,..., ak) be its signature. Set d = [g/2j. In order to prove Theorem 1.2, we need to show that ak < (k - 1)d, or equivalently, that e(e) < (k - 1)d for every edge e of r. For an integer i and a vertex v of r, let S*(v) denote the set of vertices of r that are at distance i from v, and for an edge uv of r, let Dj(u, v) = Sj(u) n Sj(v). If i and j are integers such that |i - j| > 2, then clearly Dj(u, v) = 0. Now let uv be an arbitrary edge of r and let i G {2,..., d}. For simplicity, let Dj = Dj(u, v). If A and B are two sets of vertices of r, let E(A, B) be the set of edges with one end-vertex in A and the other in B. Since g > 2d, the following facts can be easily deduced: (1) D* = 0 if i < d - 1; (2) each of D*-1 and D|_ i is an independent set; (3) each vertex in D*-1 has precisely one neighbour in D*__2, and if i < d - 1, precisely k - 1 neighbours in D*+1; (3') each vertex in D*_1 has precisely one neighbour in D*-1, and if i < d - 1, precisely k - 1 neighbours in D*+1; (4) |D*_i| = |Di-1| = (k - 1)i-i; (5) if g is even, then e(uv) = |E(Dd_1,Dd_1 )|; (6) if g is odd, then every vertex in D^ has precisely one neighbour in each of the sets Dd-1, Dd_1 and e(uv) = D|. Henceforth, let uv be an arbitrary edge of r such that e(uv) = ak and let Dj = Dj (u, v). The structure of r with respect to the sets Dj is then depicted in Figure 1. P. Potočnik and J. Vidali: Girth-regular graphs 351 Figure 1: The partitions of the vertices of r of girth g corresponding to an edge uv lying on (k - 1)d girth cycles, where d = [g/2j. (a) shows the case when g is even, while (b) shows the case when r is cubic and g is odd. The sets Dj with i + j < 2d are independent sets, while the set D^ in the odd case induces a perfect matching. Suppose first that g is even. Let d D = J (DU U Di-1) -i i=i and observe that all of the vertices in D, except possibly those in Dd-i and Dd-i, have all of their neighbours contained in D. By (4), we see that |D| = 2(1 + (k - 1) + ••• + (k - 1)d-i)=2(k -1. Moreover, it follows from (1) - (5) that ak = e(uv) = |E(Dd-i,Dd-i)| < (k - 1)|Dd-i| = (k - 1)d. This proves Theorem 1.2 in the case when g is even. (The case when g is odd will be considered later.) To prove Theorem 1.3, assume that ak = (k - 1)d. Then equality holds in the above equation, implying that |E(Dd-i, Ddf-i)| = (k - 1)|Dd-i|, which means that each vertex in Dd-i has k - 1 neighbours within Dd-i. This implies that every vertex from the set D has all of its neighbours contained in D, and by connectivity of r, we see that V(r) = D. But then by Proposition 2.1 and Note 2.2, the graph r is the incidence graph of a generalised g/2-gon of order (k - 1, k - 1). If, in addition, k = 3 holds, then r is one of the graphs mentioned in the statement of Theorem 1.3. This proves Theorem 1.3. Let us now move to the case where g is odd, prove Theorem 1.4 and finish the proof of Theorem 1.2. Suppose henceforth that g is odd. Even though Theorem 1.4 is only about cubic graphs, we will try to continue the proof without this assumption for as long as we can. Let d D = Dd U J(Di-i U Di-i) i=i and observe that |D| < (k - 1)d + 2(1 + (k - 1) + • • • + (k - 1)d-i) = k(k - 1)d - 2. k2 368 374 Ars Math. Contemp. 17 (2019) 349-368 If we prove that every vertex in D has all of its neighbours contained in D, the connectivity of r will imply that V(T) = D. But then Proposition 2.1 will imply that r is a Moore graph. Since the only cubic Moore graphs are K4 and the Petersen graph, this will then imply Theorem 1.4. Note that by (2), (3) and (3'), it follows that the neighbourhoods of all vertices, except possibly those contained in D^-1, D'd_l or Dd, are contained in D. By (3), (3') and (6), it follows that |Dd| < (k - 1)|D^-1| = (k - 1)d, and by (6) we see that ak = e(uv) = |Dd| < (k - 1)d, thus proving Theorem 1.2 also for the case when g is odd. Assume now that ak = (k - 1)d. Then, by (6), |Dd| = (k - 1)d, implying that every vertex in D^-1 (as well as in D^_1) has k - 1 neighbours in D^, and thus none outside the set D. To prove Theorem 1.4, it thus suffices to show that every vertex from Dd has all of its neighbours in D. Since every vertex in D^+1 or D^+1 has to have at least one neighbour in D^-1 or Dd_1, respectively, and since all of the neighbours of vertices in the latter two sets lie in Dd, D^-2 and it follows that the sets D^+1 and D^+1 are empty. By consequence, the sets D]+1(u, v) and D|+1(u, v) for i > d are also empty. Let us summarise that in Lemma 2.3. Lemma 2.3. Let r be a girth-regular graph of girth 2d +1 and signature (a1;..., ak) such that ak = (k - 1)d. If uv is an edge of r such that e(uv) = ak, then for i > d the sets D|+1(u, v) and D|+1(u, v) are empty. Suppose now that V(r) = D. Then a vertex y G Dd has a neighbour w outside D. Since the girth of r is 2d + 1, there exists a unique path of length d from y to u. Let v' be the neighbour of u through which this path passes, and let u' be a neighbour of v' other than u such that e(v'u') = e(uv). Let Ej = Dj (u',v') and observe that by Lemma 2.3, the sets Ed+1 and Ed+1 are empty. Furthermore, since w is not in D but has a neighbour y in D, we see that d(w, u) — d +1, implying that w G Dj^+J. We shall now partition the set Dj^-1 with respect to the distance to the vertices v' and u'. In particular, we will show that Dj-1 is a disjoint union of the sets X = Dd-1 n Ed-3, Y = Dd-1 n Ed-1, Z = Dd-1 n Ed. To prove this, note first that a vertex in Dj-1 is at distance d -1 from u and thus by (1), it is either at distance d - 2 or d from v'. Furthermore, those vertices that are at distance d - 2 from v' are either at distance d - 3 or d - 1 from u', and therefore belong to X or Y. Now let x be an element of Dj 1 that is at distance d from v '. Since Ed+1 = 0, this implies that x is either in Ej-1 or in Ed. If x G Ej-1, then there exist two distinct paths of length d from x to v', one passing through u and one passing through u', yielding a cycle of length at most 2d, which is a contradiction. Hence x G Ed, and therefore x G Z. P. Potočnik and J. Vidali: Girth-regular graphs 355 We will now determine the sizes of X, Y and Z. In particular, we will show that: |X| = (k - 1)d-3, |Y| = (k - 2)(k - 1)d-3, |Z| = (k - 2)(k - 1)d-2. To prove the first equality, observe that X consists of all the ends of paths of length d - 2 that start with v'u'. The equality for |X| then follows from the fact that there are (k - 1)d-3 such paths. Further, note that Y consists of all the ends of paths of length d - 2 that start in v' but do not pass through u' or u. There are (k - 2)(k - 1)d-3 such paths, proving the equality for | Y| . Finally, to prove the equality for | Z| , observe that Z consists of all the ends of paths of length d - 1 that start in u but do not pass through v or v'; there are clearly (k - 2)(k - 1)d-2 such paths. We will now partition the set Dd into sets X', Y' and Z' defined as follows. Let x be a vertex of Dd and observe that there is a unique path from x to u of length d. If this path passes through X, then we let x G X', if it passes through Y, then we let x G Y', and if it passes through Z, we let x G Z'. Since each vertex in Dd-1 has k - 1 neighbours in Dd and each vertex in Dd has precisely one neighbour in D^1, we see that |X'| = (k - 1)|X| = (k - 1)d-2, |Y'| = (k - 1)|Y| =(k - 2)(k - 1)d-2, |Z'| = (k - 1)|Z| =(k - 2)(k - 1)d-1. Observe furthermore that a vertex x in X', having a neighbour in X, is at distance at most d - 2 from u', but since it is at distance d from u, it is at distance exactly d - 2 from u'. Similarly, d(x, v') < d - 1 and since d(x, u) = d, we see that d(x, v') = d - 1. In particular, x G ^ and thus X' = Dd n Ed-2 = Edd-2. A similar argument shows that Y' = Dd n Ed-1. Let us now consider the set Z', and in particular the intersection A = Z' n Ej-1. Note that each vertex in Z must have at least one neighbour in A, for otherwise it could not be at distance d from u'. This implies that |A| > |Z| = (k - 2)(k - 1)d-2. On the other hand, for a similar reason, each vertex in A must have a neighbour in X'. By comparing the sizes of A and X', we may thus conclude that every vertex in X' has k - 2 neighbours in A and each vertex in A has precisely one neighbour in X'. In particular, every vertex in X' has all of its neighbours in D, and consequently, the vertex w has no neighbours in X'. Therefore, we have y G Y'. Now recall that w G implying that d(w, v') > d. On the other hand, w has a neighbour in Y', which is a subset of Ed-1, implying that d(w, v') = d. Since Ed+1 = 0, it follows that w G Ed, and hence there exists a path wz1z2... zd-1u' of length d from w to u'. By considering possibilities for such a path, one can now easily see that z1 g Z' and z2 g X'. But then z1 has at least four neighbours: z2, w, a neighbour in Z, and a neighbour in Dd-1, see Figure 2. This contradicts our assumption that the valence k is 3. This contradiction shows that V(r) = D, and thus completes the proof of Theorem 1.4. 356 Ars Math. Contemp. 17 (2019) 349-368 Figure 2: The partitions of the vertices of r of girth g, where g is odd, corresponding to the edges uv and u'v', both lying on 2d girth cycles, where d = [g/2\. Assuming there is a vertex w e D c, and 3. if a > 1 and c = a + b, then g is even. Proof. Let u be a vertex of r and let and e1,e2 and e3 be the three edges incident to u, lying on a, b and c g-cycles, respectively. Further, let x, y, z be the number of g-cycles the 2-paths eie2, e2e3 and e3ei lie on, respectively. Clearly, we have a = x + z, b = x + y and c = y + z. Then a + b + c = 2(x + y + z), showing that this sum is even. Further we may express x = (a + b - c)/2, y = (-a + b + c)/2 and z = (a — b + c)/2. Since these numbers are nonnegative, it follows that a + b > c. Now suppose that a > 1 and c = a + b. Let us call an edge e with e(e) = c saturated and others unsaturated. Note that c > b, implying that e1 and e2 are unsaturated while e3 is saturated. Since y + z = c = a + b = 2x + y + z, we see that x = 0. Since u was an arbitrary vertex of r, this shows that a 2-path in r consisting of two unsaturated edges belongs to no g-cycles. In particular, when traversing a g-cycle in r, saturated and unsaturated edges must alternate, implying that g is even. □ Lemma 3.2. If the signature of a cubic girth-regular graph is (0, b, c), then b = c = 1. Proof. Let r be a cubic girth-regular graph with signature (0, b, c) and let g be its girth. By part (2) of Lemma 3.1, it follows that b = c. Suppose that b > 1. Let e be an edge of P. Potočnik and J. Vidali: Girth-regular graphs 357 u 2 o- 1 1 o- w 2 \D> D. 12 (a) u 2 o— 1 1 O- w 2 KDD2- 12 (b) Figure 3: The partitions of the vertices of r of girth g corresponding to a 2-path uvw lying on 2d-1 girth cycles, where d = [g/2j. (a) shows the case when g is even, while (b) shows the case when g is odd. The sets Dj with i + j < 2d are independent sets, while the set D^ may contain edges. Note that no vertex of D* (i G {d - 1, d}) with a neighbour in Di-can have a neighbour in D*-1 or D*_ 1. r lying on b g-cycles, and let C, C' be two distinct g-cycles containing e. Since C = C', there exists a vertex u such that one of the edges incident to u lies on both C and C', while each of the remaining two edges incident to u belongs to exactly one of C and C'. However, this contradicts a = 0. □ Corollary 3.3. If r is a cubic girth-regular graph with signature (a, b, c) and girth g, where g is odd, then a =1. Proof. Suppose that a = 1. By part (2) of Lemma 3.1, c = b or c = b + 1. If b = c, then a + b + c is odd, contradicting part (1) of Lemma 3.1. Hence c = b +1 = a + b, and by part (3) of Lemma 3.1, g is even, contradicting our assumptions. □ Lemma 3.4. Let r be a cubic girth-regular graph of girth g with signature (a, b, c). Let m = 2Lg/2J 1. Then a > c — m and b < a — c + 2m. 2 D 1 1 1 1 v v 1 1 2 D 1 Proof. Let us first show that any 2-path in r lies on at most m girth cycles. Let uvw be a 2-path in r, and let Dj be the set of vertices at distance i from u and at distance j from w. Set d = |_g/2j. Similarly as in the proof of Theorem 1.2, we can see that the number of girth cycles containing the 2-path uvw equals the number of common neighbours of vertices in the sets D^-2 and Dj_2 if g is even, and the number of edges between the vertices in the sets Ddd-1 and D^_1 if g is odd, see Figure 3. In the even case, |D^-21 = D-21 = 2d-2, and each of the vertices from D^-2 or D^_2 may have at most two common neighbours with vertices of the other set, so uvw can lie on at most 2d-1 = m girth cycles. In the odd case, we have |D^-1|, |D^-1| < 2d-1, and each vertex from D^-1 or D^-1 may have at most one neighbour in the other set, as otherwise we would have a cycle of length 2d < g. Therefore, uvw can lie on at most m girth cycles also in this case. 358 Ars Math. Contemp. 17 (2019) 349-368 As each of a, b, c is the sum of the number of girth cycles two distinct 2-paths sharing the central vertex lie on, the quantity c - a equals the difference between the numbers of girth cycles two such 2-paths lie on, and is therefore at most m, from which a > c - m follows. Also, the quantity —a + b + c equals twice the number of girth cycles a 2-path in r lies on, and is therefore at most 2m. From this, b < a — c + 2m follows. □ 3.2 Dihedral schemes, truncations and signature (0,1,1) In this section we will allow graphs to have parallel edges and loops. A graph with parallel edges and loops is defined as a triple (V, E, d) where V and E are the vertex-set and the edge-set of the graph and d: E ^ {X : X C V, |X | < 2} is a mapping that maps an edge to the set of its end-vertices. If |d(e) | = 1, then e is a loop. Further, we let each edge consist of two mutually inverse arcs, each of the two arcs having one of the end-vertices as its tail. If the graph has no loops, we may identify an arc with tail v underlying edge e with the pair (v, e). The set of arcs of a graph r is denoted by A(r) and the set of the arcs with their tail being a specific vertex u by outr (u). The valence of a vertex u is defined as the cardinality of outr(u). A dihedral scheme on a graph r (possibly with parallel edges and loops) is an irreflexive symmetric relation ^ on the arc-set A(r) such that the simple graph (A(r), is a 2-regular graph each of whose connected components is the set outr (u) for some u G V(r). (Intuitively, we may think of a dihedral scheme as a collection of circles drawn around each vertex u of r intersecting each of the arcs in outr(u) once.) Note that, according to this definition, the minimum valence of a graph admitting a dihedral scheme is at least 3. The group of all automorphisms of r that preserve the relation ^ will be denoted by Aut(r, and the dihedral scheme ^ is said to be arc-transitive if Aut(r, acts transitively on A(r). Given a dihedral scheme ^ on a graph r, let Tr(r, be the simple graph whose vertices are the arcs of r and two arcs s, t G r are adjacent in r if either t ^ s or t and s are inverse to each other. The graph Tr(r, is then called the truncation of r with respect to the dihedral scheme Note that Tr(r, is a cubic graph which is connected whenever r is connected. As we shall see in Section 3.3, a natural source of arc-transitive dihedral schemes are arc-transitive maps (either orientable or non-orientable). However, not all dihedral schemes arise in this way. Clearly, the automorphism group Aut(r, acts naturally as a group of automorphisms of Tr(r, implying that Tr(r, is vertex-transitive whenever the dihedral scheme ^ is arc-transitive. The following result gives a characterisation of arc-transitive dihedral schemes in group theoretical terms. Here, the symbol Dd denotes the dihedral group of order 2d acting naturally on d points, while Zd is the cyclic group acting transitively on d points. Lemma 3.5. Let r be an arc-transitive graph (possibly with parallel edges) of valence d for some d > 3. Then r admits an arc-transitive dihedral scheme if and only if there exists an arc-transitive subgroup G < Aut(r) such that the group GUutr(u) inducedby the action of the vertex stabiliser Gu on the set outr(u) is permutation isomorphic to the transitive action of Dd, Zd or (when d is even) D d on d vertices. Proof. Suppose that ^ is a dihedral scheme on r and that G = Aut(r, Then GUutr(u) P. Potočnik and J. Vidali: Girth-regular graphs 359 preserves the restriction of the relation ^ onto outr(u), and thus acts as a vertex-transitive group of automorphisms on the simple graph (outr(u), ^u). Since the latter graph is a cycle of length d, we thus see that GUutr (u) is a transitive subgroup of Dd and thus permutation isomorphic to one of the transitive actions mentioned in the statement of the lemma. Conversely, suppose that for some vertex u, the group GUutr(u) is permutation isomorphic to the transitive action of Dd, Zd, or (if d is even) Dd/2 on d vertices. In all three cases, we may choose an adjacency relation on outr (u) preserved by GUutr (u) in such a way that (outr (u), is a cycle. For every v e V(r), choose an element gv e G such that vgv = u, and let be the relation on outr(v) defined by s t if and only if sgv ^u tgv. Then clearly (outr(v), ) is a cycle, implying that the union ^ of all for u G V(r) is a dihedral scheme. Moreover, it is a matter of straightforward computation to show that ^ is invariant under G. □ We are now ready to prove the following characterisation of cubic girth-regular graphs of signature (0,1,1). Theorem 3.6. If r is a simple cubic girth-regular graph of girth g with signature (0,1,1), then r = Tr(A, where ^ is a dihedral scheme on a g-regular graph A (possibly with parallel edges). Moreover, if r is vertex-transitive, then the dihedral scheme is arc-transitive. Proof. Let V be the vertex-set of r, let T be the set of girth cycles in r, let M be the set of edges that belong to no girth cycle in r, and let G = Aut(r). Note that since the signature of r is (0,1,1), each vertex v e V is incident to exactly one edge in M and to exactly one girth cycle in T. For an edge v'v e M, let C and C' be the girth cycles that pass through v and v', respectively, and let d(v'v) = {C, C'}. This allows us to define a graph A = (T, M, d). Note that since C, C' e V(A) are girth cycles of r, we have C = C', and so A has no loops. This allows us to view an arc of A as a pair (C, e) where e e M and C is a girth cycle of r passing through one of the two end-vertices of e. For two such pairs (Ci, ei) and (C2, e2) we write (C1, e1) ^ (C2, e2) if and only if C1 = C2 and the end-vertices of e1 and e2 that belong to C1 are two consecutive vertices of C1. Then ^ is a dihedral scheme on A. Let r' = Tr(A, We will now show that r' = r. By the definition of truncation, the vertex-set of r' equals the arc-set of A. For an arc (C, e) of A let y(C, e) be the unique end-vertex of e that belongs to C. Since each vertex of r is incident to exactly one edge in M and exactly one cycle in T, it follows that ^ is a bijection between V(r') and V(r). If (C1, e1) and (C2, e2) are adjacent in r', then either (C1, e1) ^ (C2, e2) or (C1, e1) and (C2, e2) are inverse arcs in r'. In the first case, C1 = C2 and the vertices ^(C1, e1) and y(C2, e2) are adjacent on C1. In the second case, e1 = e2 and the vertices ^(C1, e1) and ^(C2, e2) are the two end-vertices of e1. In both cases ^(C1, e1) and ^(C2, e2) are adjacent in r. By a similar argument we see that whenever ^(C1, e1) and y(C2, e2) are adjacent in r, (C1, e1) and (C2, e2) are adjacent in r'. Since both r and r' are simple graphs (one by assumption, the other by definition), this shows that ^ is a graph isomorphism. Suppose now that G is transitive on the vertices of r. Since both sets T and M are invariant under the action of G, there exists a natural action of G on A that preserves the dihedral scheme that is, G < Aut(A, Now let (C1, e1) and (C2, e2) be two arcs 360 Ars Math. Contemp. 17 (2019) 349-368 of A, and for i e {1, 2}, let vj be the unique end-vertex of ej that lies on Q. Since G is vertex-transitive on r, there exists g e G mapping vi to v2. Since Cj is the unique girth-cycle through vj for i e {1, 2}, it follows that Cg = C2. Similarly, since ej is the unique edge in M incident with vj for i e {1,2}, it follows that eg = e2. This shows that G acts transitively on the arcs of A. □ Note 3.7. Parallel edges occur in the graph A as in Theorem 3.6 whenever there exist two girth cycles in r such that there are at least two edges with an end-vertex in each of the two girth cycles. In fact, it can be easily seen that in a girth-regular graph r with signature (0,1,1), there are at most two such edges between any two girth cycles, leading to at most two parallel edges between each two vertices, with the exception of the case when r is the 3-prism (see Section 4) and A is the graph with two vertices and three parallel edges between them. Note 3.8. No nontrivial bound on the girth of the graph A as in Theorem 3.6 can be given. In fact, we can construct a family of graphs of constant girth such that their truncations with respect to appropriate dihedral schemes are cubic girth-regular graphs with signature (0,1,1) and unbounded girth. Let A be a graph obtained by doubling all edges in a k-regular graph of girth at least k^1 - the girth of A is then 2. Equip A with a dihedral scheme ^ such that each two arcs with a common tail belonging to two parallel edges are antipodal in the connected component of the graph defined by ^ they belong to. Then Tr(A, is a cubic girth-regular graph of girth k and signature (0,1,1). 3.3 Maps and signatures (2, 2, 2) and (1,1, 2) In this section, it will be convenient to think of a graph (possibly with parallel edges) as a topological space having the structure of a regular 1-dimensional CW complex with the vertices of the graph corresponding to the 0-cells of the complex and the edges corresponding to the 1-cells. A simple closed walk (that is, a closed walk that traverses each edge at most once) in the graph then corresponds to a closed curve in the corresponding topologi-cal space which may intersect itself only in the points that correspond to the vertices of the graph. Given a graph r (viewed as a CW complex) and a set of simple closed walks T in r, one can construct a 2-dimensional CW complex in the following way. First, take a collection D of topological disks, one for each walk in T. Then choose a surjective continuous mapping from the boundary of each disk to the closed curve in r representing the corresponding walk in T, such that the preimage of each point that is not a vertex of the graph is a singleton. Finally, identify each point of the boundary of the disk with its image under that continuous mapping. Note that the resulting topological space is independent of the choice of the homeomorphisms D and thus depends only on the choice of the graph and the set of closed walks T. When r is connected and the resulting topological space is a closed surface (either orientable or non-orientable), the CW complex is also called a map. Its open 2-cells are then called the faces of the map, the closed walks in T are called the face-cycles and the graph r is the skeleton of the map. A map whose skeleton is a k-regular graph and all of whose face cycles are of length m is called an {m, k}-map. The following lemma provides a sufficient condition on the set of cycles T under which the resulting 2-dimensional CW complex is indeed a map. P. Potočnik and J. Vidali: Girth-regular graphs 361 Lemma 3.9. Let r be a graph and T a set of simple closed walks in r such that every edge of r belongs to precisely two walks in T. For two arcs s and t with a common tail, write s ^ t if and only if the underlying edges of s and t are two consecutive edges on a walk in T. If ^ is a dihedral scheme, then r is the skeleton of a map whose face cycles are precisely the walks in T. Proof. Let us think of r as a 1-dimensional CW complex and let us turn it into a 2-dimensional CW complex by adding to it one 2-cell for each walk in T as described above. Let us now prove that the resulting topological space M is a closed surface. It is clear that the internal vertices of the 2-cells have a regular neighbourhood. Further, since each edge of r lies on precisely two walks in T, the internal points of edges also have a regular neighbourhood, made up from two half-disks, each contained in the 2-cell glued to one of the walks in T passing through that edge. Finally, let u be a vertex of r, let k be the valence of u, and let jsj : i e Zk} be the set of arcs with the initial vertex u such that s0 ^ si ^ • • • ^ sk-1 ^ s0. By the definition of each pair of arcs (sj, si+1) (i e Zk) lies on a unique walk Cj in T. Note that C = Ci+1, for otherwise the edge underlying si+1 would lie on only one walk in T. This implies that a regular neighbourhood of u in M can be built by taking appropriate half-disks from the 2-cells corresponding to the cycles C (i e Zk), and gluing them together in the order suggested by the relation This shows that M is a 2-manifold without a boundary. Finally, since r is finite, M is compact, and thus a closed surface. Hence, M is a map with r as its skeleton. □ Each face of a map can be decomposed further into flags, that is, triangles with one vertex in the centre of a face, one vertex in the centre of an edge on the boundary of that face and one in a vertex incident with that edge. In most cases, a flag can be viewed as a triple consisting of a vertex, an edge incident to that vertex, and a face incident to both the vertex and the edge. An automorphism of a map is then defined as a permutation of the flags induced by a homeomorphism of the surface that preserves the embedded graph. A map is said to be vertex-transitive or arc-transitive provided that its automorphism group induces a vertex-transitive or arc-transitive group on the skeleton of the map, respectively. Note 3.10. If a map is built from a graph r and a set of simple closed walks T as in Lemma 3.9, then each automorphism of r that preserves the set of walks T clearly extends to an automorphism of the map. If M is a map on a surface S, then the sets V, E and F of the vertices, edges and faces, respectively, satisfy the Euler formula |V | — |E| + |F | = x(S) where x(S) is the Euler characteristic of the surface S. It is well known that x(S) < 2 with equality holding if and only if S is homeomorphic to a sphere. Moreover, if x(S) is odd, then S is non-orientable. As the following two results show, skeletons of maps arise naturally when analysing cubic vertex-transitive graphs of signature (2, 2, 2) or (1,1, 2). Theorem 3.11. Let r be a simple connected cubic girth-regular graph of girth g and order n with signature (2,2, 2). Then g divides 3n and r is the skeleton of a {g, 3}-map 362 Ars Math. Contemp. 17 (2019) 349-368 embedded on a surface with Euler characteristic (3 1 x = nU-2 Moreover, every automorphism of r extends to an automorphism of the map. In particular, if r is vertex-transitive, so is the map. Proof. Let T be the set of girth cycles of r. Since the valence of r is 3, it follows easily that the relation ^ from Lemma 3.9 satisfies the conditions stated in the lemma; that is, ^ is a dihedral scheme. Lemma 3.9 thus yields a map M whose skeleton is r and whose face-cycles are precisely the walks in T; in particular, M is a {g, 3}-map, as claimed. Since r is a cubic graph with n vertices, it has 3n/2 edges, and since each vertex lies on three face-cycles and since each face-cycle contains g vertices, the map M has 3n/g faces (showing that g must divide 3n). The Euler characteristic of M thus equals n - f + t = n(f - 2). Since every automorphism of r preserves T, it extends to an automorphism of M (see Note 3.10). □ Theorem 3.11 has the following interesting consequence. Corollary 3.12. There exists only finitely many connected cubic girth-regular graphs with signature (2,2,2) of girth at most 5. Proof. Suppose that r is a connected cubic girth-regular graph with signature (2,2,2) of girth g and order n. By Theorem 3.11, r is a skeleton of a map on a surface of Euler characteristic x = n(3/g - 1/2). Hence, if g < 5, then x > n/10, and since x < 2, it follows that n < 20. □ Note 3.13. For each g > 6, there are infinitely many girth-regular graphs of girth g with signature (2, 2, 2). If M is a map and r is its skeleton, then one can define a dihedral scheme ^ on r by letting s ^ t whenever the arcs s and t have a common tail and the underlying edges of s and t are two consecutive edges on some face-cycle of M. The truncation Tr(r, is then simply referred to as the truncation of the map M and denoted Tr(M). Note that this construction in some sense complements Lemma 3.9. We are now equipped for a characterisation of cubic girth-regular graphs with signature (1,1,2). Theorem 3.14. Let r be a simple connected cubic girth-regular graph of girth g with n vertices and signature (1,1, 2). Then g is even and r is the truncation of some map M with face cycles of length g/2. In particular, g/2 divides n. Moreover, if r is vertex-transitive, M is an arc-transitive {g/2, £}-mapfor some £ > g. Proof. By part (3) of Lemma 3.1 we know that g is even and in particular, g > 4. Let X be the set of edges of r that belong to exactly one girth cycle and let Y be the set of edges that belong to two girth cycles. Since the signature of r is (1,1,2), every vertex of r is incident to two edges in X and one edge in Y. Consequently, the edges in Y form a perfect matching of r and the subgraph induced by the edges in X is a union of vertex-disjoint cycles of r that cover all the vertices of r. Let us denote the set of these cycles by C. P. Potočnik and J. Vidali: Girth-regular graphs 363 Observe also that two edges in X sharing a common end-vertex, say v, cannot be two consecutive edges on the same girth cycle, for otherwise that would be a unique girth cycle through v, contradicting the fact that the third edge incident with v belongs to two girth cycles. Since the edges in Y form a complete matching of r, the same holds for the edges in Y, implying that the edges on any girth cycle alternate between the sets X and Y. For an edge e in Y with end-vertices u and v, let Cu and Cv be the unique cycles in C that pass through u and v, respectively, and define d(e) to be the pair {Cu, Cv}. Let A = (C, Y, d). Note that since the edges of A are precisely those edges of r that belong to Y, we may think of the arc-set A(A) as being the set of arcs of r that underlie edges in Y. Note also that it may happen that for some e G Y, we may have Cu = Cv and then the graph A has loops. If D is a girth cycle of r, then the edges of D that belong to Y induce a simple closed walk in the graph A of length g/2, which we denote D. Let T be the set of walks D where D runs through the set of girth cycles of r. Since edges of A correspond to the edges of r that pass through two girth cycles of r, each edge of A belongs to two walks in T. As |Y| = n/2, it follows that g/2 divides n. Let ^ be the relation on the arcs of A defined by T as explained in Lemma 3.9. It is easy to see that ^ is a dihedral scheme. Indeed, let C g C be a vertex of A viewed as a cycle in r and let v0, vi,..., vfc_i g V(r) be its vertices listed in a cyclical order as they appear on C. Further, for each i G Zk, let sj be the arc of r with tail v» that underlies an edge contained in Y. The arc sj can thus also be viewed as an arc of A. Observe that out a (C) = {s0, s1,..., sk-1} and that s0 ^ s1 ^ • • • ^ sk-1 ^ s0. In particular, ^ is a dihedral scheme. By Lemma 3.9, there exists a map M with skeleton A in which T is the set of face-cycles. Moreover, ^ equals the dihedral scheme arising from that map. Let r' = Tr(M) and let s be a vertex of r'. Then s is an arc of A and thus also an arc of r underlying an edge in Y. By letting y(s) be the tail of s (viewed as an arc of r), we define a mapping y: V(r') ^ V(r). Note that the mapping which assigns to a vertex v G V(r) the unique arc of r with tail v that underlies an edge in Y is the inverse of y, showing that y is a bijection. Furthermore, note that two vertices s and t of r' are adjacent in r' if and only if one of the following happens: (1) they are inverse to each other as arcs of A; or (2) they have a common tail and s ^ t. In case (1), y(s) and y(t) are adjacent in r via an edge in Y, while in case (2), y(s) and y(t) are adjacent in r via an edge in X. Conversely, if for some s, t G V(r'), the images y(s) and y(t) are adjacent in r, then either s and t are inverse to each other as arcs of A (this happens if y(s) and y(t) form an edge in Y), or s and t have a common tail and s ^ t (this happens if y(s) and y(t) form an edge in X). In both cases, s and t are adjacent in r'. This implies that y is an isomorphism of graphs and thus r = Tr(M), as claimed. Since every automorphism of r preserves each of the sets Y and X (and thus also C), it clearly induces an automorphism of the graph A which preserves the set T. In particular, every automorphism of r induces an automorphism of the map M. Finally, suppose that r is vertex-transitive. Then all cycles of C have the same length I > g. As each vertex of a cycle of C is incident to precisely one edge of Y, it follows that A is an ¿'-regular graph, and M is then a {g/2, ¿}-map. Let G be a group of automorphisms of r acting transitively on V(r). Note that every vertex of r is the tail of precisely one arc of r that underlies an edge of Y. In view of our identification of the arcs of r' with the arcs of r that underlie an edge in Y, we thus see that the transitivity of the action of G on V(r) implies the transitivity of the action of G on the arcs of M. □ 364 Ars Math. Contemp. 17 (2019) 349-368 4 Cubic girth-regular graphs of girths 3 and 4 Before stating the theorem about girth-regular cubic graphs of girth 3, let us point out that every cubic graph admits a unique dihedral scheme, which is preserved by every automorphism of the graph. This allows us to talk about truncations of cubic graphs without specifying the dihedral scheme. Theorem 4.1. Let r be a connected cubic girth-regular graph of girth 3. Then one of the following holds: (a) r is isomorphic to the complete graph K4; (b) r has signature (0,1,1) and is isomorphic to the truncation of a cubic graph. Proof. Let (a, b, c) be the signature of r. By Theorem 1.2 it follows that c < 2. If c =2, then Theorem 1.4 implies that r is isomorphic to K4. On the other hand, if c = 1, then Lemmas 3.1 and 3.2 imply that the signature of r is (0,1,1), and by Theorem 3.6, itfollows that r is the truncation of a cubic graph. □ Let us now move our attention to graphs of girth 4. Before stating the classification theorem, let us define two families of cubic vertex-transitive graphs. For n > 3, let the n-Mobius ladder Mn be the Cayley graph Cay(Z2n, { — 1,1, n}). Note that such a graph has girth 4. The graph Mn has signature (4,4,4) if n = 3 (in this case it is isomorphic to the complete bipartite graph K3,3), and (1,1, 2) if n > 4. An n-Mobius ladder can also be seen as the skeleton of the truncation of the {2, 2n}-map with a single vertex embedded on a projective plane. For n > 3, the n-prism Yn is defined as the Cartesian product Cn □ K2 or, alternatively, as the Cayley graph Cay(Zn x Z2, {( — 1,0), (1,0), (0,1)}). The girth of Y3 is 3, while the girth of Yn for n > 4 is 4. The graph Yn has signature (2,2,2) if n = 4 (in this case it is isomorphic to the cube Q3), and (1,1, 2) if n > 5. An n-prism can also be seen as the skeleton of the truncation of the {2, n}-map with two vertices embedded on a sphere, i.e., an n-gonal hosohedron. Theorem 4.2. Let r be a connected cubic girth-regular graph of girth 4. Then r is iso-morphic to one of the following graphs: (a) the n-Mobius ladder Mn for some n > 3; (b) the n-prism Yn for some n > 4; (c) Tr(A, for some tetravalent graph A and a dihedral scheme ^ on A. Proof. Let (a, b, c) be the signature of r. By Theorem 1.2, we see that c < 4, and by Theorem 1.3, if c = 4, then the signature of r is (4,4,4) and r = K3,3 = M3. Suppose now that c =3. Then, by Lemma 3.1, a + b is odd, and by Lemma 3.2, a > 1. Hence either a = 1 and then b = 2, or a = 2 and then b = 3. The possible signatures in this case are thus (1,2,3) and (2,3,3). Let us show that neither can occur. Let uv be an edge of r lying on three 4-cycles, and u0,u1 and v0,v1 be the remaining neighbours of u and v, respectively. There must be three edges with one end-vertex in {u0, u1} and the other in {v0, v1}; without loss of generality, these edges are u0v0, u0v1 and u1v1 (see Figure 4(c)). Then the edges uu0 and vv1 already lie on three 4-cycles, P. Potočnik and J. Vidali: Girth-regular graphs 365 (a) (b) (c) Figure 4: (a) The graph K4 of girth 3 with signature (2,2, 2). (b) The graph Y3 of girth 3 with signature (1,1,2). (c) Constructing a graph of girth 4 with c =3. The dashed edges should lie on two 4-cycles, however the doubled edge already lies on three 4-cycles. u v so we have b = 3, and thus a = 2. In particular, e(uui) = e(vv0) = 2. Then the edge u1v1, being incident to both u1 and v1, belongs to precisely three 4-cycles, that is e(u1v1) = 3. Similarly, e(w0v0) = 3. It follows that the edge u0v1 lies on precisely two 4-cycles. However, we have already determined three 4-cycles on which u0v1 lies; these are uu0v1v, v0u0v1v, and uu0v1 u1. This contradiction shows that the case c = 3 is not possible. Suppose now that c = 2. By Lemma 3.1, a + b is even, and by Lemma 3.2, a > 1. Hence the signature of r is either (1,1, 2) or (2,2,2). If (a, b, c) = (1,1,2), then, by Theorem 3.14, r is the skeleton of the truncation of a connected map M with face cycles of length 2. Since every edge belongs to two faces and every face is surrounded by two edges, the number of faces equals the number of edges. The Euler characteristics x(S) of the underlying surface S thus equals |V(M)|. Since x(S) < 2, it follows that M has one or two vertices, depending on whether S is the projective plane or the sphere - in particular, the skeleton of M is an ¿'-regular graph for some I > 4. If M has one vertex only, then it consists of ¿/2 loops embedded onto the projective plane in such a way that its truncation is the Mobius ladder Mn with n = ¿/2 > 4, see Figure 5(a) (note that M3 = K3,3 has signature (4,4,4)). On the other hand, if M has two vertices, then M is the map with two vertices and I parallel edges embedded onto the sphere. The graph r is then isomorphic to the n-prism Yn with n = I > 5. If (a, b, c) = (2, 2,2), then, by Theorem 3.11, r is the skeleton of a {4,3}-map embedded on a surface of Euler characteristic x = n/4 > 0. As above, x < 2 and thus x =1 or 2. For x = 1, we get the hemicube on the projective plane (see Figure 5(b)), and its skeleton is isomorphic to K4 of girth 3. For x = 2, we get the cube on a sphere, and its skeleton is isomorphic to Y4 with signature (2,2,2). This completes the case c = 2. If c = 1, then since a + b + c is even, we see that a = 0 and b =1, and then by Theorem 3.6, r is the truncation of a 4-regular graph with respect to some dihedral scheme. □ 366 Ars Math. Contemp. 17 (2019) 349-368 Figure 5: (a) A {2, 8}-map with a single vertex, four edges and four labelled faces embedded on the projective plane. Its truncation has the graph M4 with signature (1,1, 2) as its skeleton. (b) The hemicube on the projective plane with labelled faces. Its skeleton is the graph K4. 5 Cubic girth-regular graphs of girth 5 Theorem 5.1. Let r be a connected cubic girth-regular graph of girth 5. Then either the signature of r is (0,1,1) and r is the truncation of a 5-regular graph with respect to some dihedral scheme, or r is isomorphic to the Petersen graph or to the dodecahedron graph. Proof. Let (a, b, c) be the signature of r. By Theorem 1.2, we see that c < 4. If c = 4, then by Theorem 1.3 and Theorem 1.4, the signature of r is (4,4,4) and r is isomorphic to the Petersen graph. We may thus assume that c < 3. If a = 0, then by Lemma 3.2 the signature of r is (0,1,1), and then by Theorem 3.6, r is the truncation of a 5-regular graph with respect to some dihedral scheme. Moreover, by Corollary 3.3, a = 1. We may thus assume that a > 2. If c =2, then the signature of r is (2,2,2) and by Theorem 3.11, r is the skeleton of a {5, 3}-map embedded on a surface of Euler characteristic x = n/10, where n is the order of the graph r. In particular, x G {1,2}. If x =1, then n =10 and since the girth of r is 5, Proposition 2.1 and Note 2.2 imply that r is the Petersen graph (whose signature is in fact (4,4,4)). If x = 2, then n = 20 and r is the skeleton of a {5, 3}-map on the sphere. It is well known that there is only one such map, namely the dodecahedron. Finally, suppose that c = 3. Then, by Lemma 3.1, a + b is odd, and since a > 2, the signature of r is (2,3,3). We will now show that this possibility does not occur. Let uv be an edge of r lying on three 5-cycles, and u0, ui, v0, v1 be vertices of r with adjacencies u0 ~ u ~ u1 and v0 ~ v ~ v1. Then there should be three vertices adjacent to one of u0, u1 and one of v0, v1. Without loss of generality, let w00, w10, w11 be vertices such that u0 ~ w00 ~ v0 ~ w10 ~ u1 ~ w11 ~ v1. Further, let x be the neighbour of u0 other than u and w00, and let y be the neighbour of v1 other than v and w11, see Figure 6(a). Observe that x = y, for otherwise the edge uv would belong to four 5-cycles. Note also that x is not adjacent to any of the three neighbours of v, for otherwise the girth of r would be at most 4. The signature implies that for each vertex, two edges incident to it lie on three 5-cycles. Suppose that uu0 lies on three 5-cycles. As x and v have no common neighbours, w00 and x must have a common neighbour with u1, so we have w00 ~ w11 and x ~ w10, see Figure 6(b). But then the edge uu1 lies on four 5-cycles, contradiction. Therefore, the edge P. Potočnik and J. Vidali: Girth-regular graphs 367 u V (a) (b) (c) Figure 6: Constructing a graph of girth 5 with c = 3. The bold edges lie on three 5-cycles, and the dashed edges lie on two 5-cycles. The general setting is shown in (a). In (b), the arc (u, u0) is assumed to lie on three 5-cycles, but the doubled edge then lies in four 5-cycles. In (c), the obtained distribution of edges among cycles is shown, which, however, cannot be completed. uu0 must lie on two 5-cycles, and a similar argument shows the same for vv1. Thus, the arcs uui, vv0, u0x, u0w0o, v1w11 and v1y must lie on three 5-cycles, see Figure 6(c). Since the edge u0w00 lies on three 5-cycles, there should be three vertices adjacent to one of u, x and one of v0 and the remaining neighbour of w00. Similarly, v1w11 lying on three 5-cycles implies that there should be three vertices adjacent to one of v, y and one of u1 and the remaining neighbour of w11. As w10 is the only potential common neighbour for x, v0 and for y, u1 , it follows that at least one of these pairs does not have a common neighbour. Without loss of generality, we may assume that x and v0 do not have a common neighbour. The vertex u already has a common neighbour with v0, and it must also have a common neighbour with the remaining neighbour of w00. Then the remaining neighbour of w00 must be w11, which however has no common neighbour with x, contradiction. Therefore, (a, b, c) = (2, 3,3) is not possible. □ 6 Concluding remarks Theorem 1.5 gives a complete classification of simple connected cubic girth-regular graphs of girths up to 5. While extending the classification to non-simple graphs (i.e., girths 1 and 2) is straightforward, increasing the girth leads to exponentially many more possible signatures. For example, the census of connected cubic vertex-transitive graphs on at most 1280 vertices by Potocnik, Spiga and Verret [13] shows that 9 distinct signatures appear among graphs of girth 6, while many more signatures are allowed by the results in Sections 1, 2 and Subsection 3.1. A classification of connected cubic vertex-transitive graphs of girth 6 will thus be given in a follow-up paper. References [1] E. Bannai and T. Ito, On finite Moore graphs, J. Fac. Sci. Univ. Tokyo Sect. IA Math. 20 (1973), 191-208, doi:10.15083/00039786. [2] A. E. Brouwer, A. M. Cohen and A. Neumaier, Distance-Regular Graphs, volume 18 of 368 Ars Math. Contemp. 17 (2019) 349-368 Ergebnisse der Mathematik und ihrer Grenzgebiete, Springer-Verlag, Berlin, 1989, doi: 10.1007/978-3-642-74341-2. [3] A. M. Cohen and J. Tits, On generalized hexagons and a near octagon whose lines have three points, European J. Combin. 6 (1985), 13-27, doi:10.1016/s0195-6698(85)80017-2. [4] M. Conder, A. Malnic, D. Marusic and P. Potocnik, A census of semisymmetric cubic graphs on up to 768 vertices, J. Algebraic Combin. 23 (2006), 255-294, doi:10.1007/s10801-006-7397-3. [5] M. Conder and R. Nedela, Symmetric cubic graphs of small girth, J. Comb. Theory Ser. B 97 (2007), 757-768, doi:10.1016/j.jctb.2007.01.001. [6] M. Conder and S. S. Zemljic, private communication with M. Conder, 2017. [7] R. M. Damerell, On Moore graphs, Proc. Cambridge Philos. Soc. 74 (1973), 227-236, doi: 10.1017/s0305004100048015. [8] E. Eiben, R. Jajcay and P. Sparl, Symmetry properties of generalized graph truncations, J. Comb. Theory Ser. B 137 (2019), 291-315, doi:10.1016/j.jctb.2019.01.002. [9] W. Feit and G. Higman, The nonexistence of certain generalized polygons, J. Algebra 1 (1964), 114-131, doi:10.1016/0021-8693(64)90028-6. [10] R. Jajcay, G. Kiss and S. Miklavic, Edge-girth-regular graphs, European J. Combin. 72 (2018), 70-82, doi:10.1016/j.ejc.2018.04.006. [11] K. Kutnar and D. Marusic, A complete classification of cubic symmetric graphs of girth 6, J. Comb. Theory Ser. B 99 (2009), 162-184, doi:10.1016/j.jctb.2008.06.001. [12] S. E. Payne and J. A. Thas, Finite Generalized Quadrangles, EMS Series of Lectures in Mathematics, European Mathematical Society (EMS), Zürich, 2nd edition, 2009, doi:10.4171/066. [13] P. Potocnik, P. Spiga and G. Verret, Cubic vertex-transitive graphs on up to 1280 vertices, J. Symbolic Comput. 50 (2013), 465-477, doi:10.1016/j.jsc.2012.09.002. [14] P. Potocnik and S.Wilson, Tetravalent edge-transitive graphs of girth at most 4, J. Comb. Theory Ser. B 97 (2007), 217-236, doi:10.1016/j.jctb.2006.03.007. [15] J. Tits, Sur la trialité et certains groupes qui s'en déduisent, Inst. Hautes Études Sci. Publ. Math. 2 (1959), 13-60, http://www.numdam.org/item?id=PMIHES_195 9_2_13_0. [16] W. T. Tutte, Connectivity in Graphs, number 15 in Mathematical Expositions, University of Toronto Press, Toronto, 1966, http://www.jstor.org/stable/10.3138/j. ctvfrxhc8. ARS MATHEMATICA CONTEMPORANEA ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 17 (2019) 369-395 https://doi.org/10.26493/1855-3974.1817.b97 (Also available at http://amc-journal.eu) The existence of square non-integer Heffter arrays Nicholas J. Cavenagh Department of Mathematics, The University of Waikato, Private Bag 3105, Hamilton 3240, New Zealand Jeffery H. Dinitz Department of Mathematics and Statistics, University of Vermont, Burlington, VT 05405, USA Diane M. Donovan School of Mathematics and Physics, The University of Queensland, Queensland 4072, Australia Emine §ule Yazici Department of Mathematics, Koc University, Istanbul, Turkey Received 8 October 2018, accepted 13 April 2019, published online 4 November 2019 A Heffter array H(n; k) is an n x n matrix such that each row and column contains k filled cells, each row and column sum is divisible by 2nk + 1 and either x or —x appears in the array for each integer 1 < x < nk. Heffter arrays are useful for embedding the graph K2nk+1 on an orientable surface. An integer Heffter array is one in which each row and column sum is 0. Necessary and sufficient conditions (on n and k) for the existence of an integer Heffter array H(n; k) were verified by Archdeacon, Dinitz, Donovan and Yazici (2015) and Dinitz and Wanless (2017). In this paper we consider square Heffter arrays that are not necessarily integer. We show that such Heffter arrays exist whenever 3 < k < n. Keywords: Heffter arrays, biembedding cycle systems. Math. Subj. Class.: 05B30 E-mail addresses: nickc@waikato.ac.nz (Nicholas J. Cavenagh), jeff.dinitz@uvm.edu (Jeffery H. Dinitz), dmd@maths.uq.edu.au (Diane M. Donovan), eyazici@ku.edu.tr (Emine Sule Yazici) Abstract ©® This work is licensed under https://creativecommons.org/licenses/by/4.G/ 370 Ars Math. Contemp. 17 (2019) 349-368 1 Introduction A Heffter array H(m, n; s, t) is an m x n matrix of integers such that: (1) each row contains s filled cells and each column contains t filled cells; (2) the elements in every row and column sum to 0 in Z2ms+1; and (3) for each integer 1 < x < ms, either x or — x appears in the array. If the Heffter array is square, then m = n and necessarily s = t. We denote such Heffter arrays by H(n; k), where each row and each column contains k filled cells. A Heffter array is called an integer Heffter array if Condition (2) in the definition of a Heffter array above is strengthened so that the elements in every row and every column sum to zero in Z. Archdeacon, in [1], was the first to define and study a Heffter array H(m, n; s, t). He showed that a Heffter array with a pair of special orderings can be used to construct an embedding of the complete graph K2ms+1 on a surface. This connection is formalised in the following theorem. For definitions of simple and compatible orderings refer to [1]. Theorem 1.1 ([1]). Given a Heffter array H(m,n; s,t) with compatible orderings ur of the symbols in the rows of the array and on the symbols in the columns of the array, then there exists an embedding of K2ms+1 such that every edge is on a face of size s and a face of size t. Moreover, if wr and wc are both simple, then all faces are simple cycles. The embedding of K2ms+1 given in Theorem 1.1 provides a connection with the embedding of cycle systems. A t-cycle system on n points is a decomposition of the edges of Kn into t-cycles. A t-cycle system C on Kn is cyclic if there is a labeling of the vertex set of Kn with the elements of Zn such that the permutation x ^ x + 1 preserves the cycles of C. A biembedding of an s-cycle system and a t-cycle system is a face 2-colorable topological embedding of the complete graph K2ms+1 in which one color class is comprised of the cycles in the s-cycle system and the other class contains the cycles in the t-cycle system, see for instance [4, 5, 6, 8, 9, 10, 11] for further details. A number of papers have appeared on the construction of Heffter arrays, H(m, n; s, t). The case where the array contained no empty cells was studied in [2], with results summarised in Theorem 1.2. Theorem 1.2 ([2]). There is an H(m, n; n, m) for all m,n > 3 and an integer Heffter array H (m, n; n, m) exists if and only if m, n ^ 3 and mn = 0, 3 (mod 4). The papers [3, 7] focused on square integer Heffter arrays H(n; k) and verified their existence for all admissible orders. This result is summarized in the following theorem. Theorem 1.3 ([3, 7]). There exists an integer H(n; k) if and only if 3 < k < n and nk = 0, 3 (mod 4). Table 1 lists the possible cases and cites the article which verifies existence of square integer Heffter arrays, where DNE represents a value that does not exist. In these cases we will verify existence for the non-integer Heffter arrays H(n; k). The main result of this paper is the following. Theorem 1.4. There exists an H (n; k) if and only if 3 < k < n. N. J. Cavenagh et al.: The existence of square non-integer Heffter arrays 371 Table 1: Existence results for square integer Heffter arrays H(n; k). n \ k 0 1 2 3 0 [3] [3,7] [3] [3] 1 [3] DNE DNE [3] 2 [3] DNE [3] DNE 3 [3] [3,7] DNE DNE Table 2: Cases for non-integer Heffter arrays H(n; k). Case A Case B CaseC CaseD Case E k 2 (mod 4) 3 (mod 4) 3 (mod 4) 1 (mod 4) 1 (mod 4) n 1, 3 (mod 4) 3 (mod 4) 2 (mod 4) 1 (mod 4) 2 (mod 4) From Theorem 1.2 above, the case n = k has been solved, so we henceforth assume that n > k. The cases that need to be addressed are set out in Table 2. Cases A, B, C, D and E are solved by Theorems 3.2,4.2, 5.2,6.2 and 7.2, respectively, thus proving Theorem 1.4. In this paper the rows and columns of a square n x n array are always indexed by the elements of {1, 2,..., n}. Unless otherwise stated, when working modulo n, replace 0 by n, so we use the symbols 1, 2,..., n instead of 0,1,...,n -1. While rows and columns are calculated modulo an integer, entries are always expressed as non-zero integers. Throughout this paper A[r, c] = x denotes the occurrence of symbol x in cell (r, c) of array A. By A±z we refer to the array obtained by replacing A[r, c] by A[r, c]+z (if A[r, c] > 0) and A[r, c] -z (if A[r, c] < 0). If each row and each column of A contains the same number of positive and negative numbers, then A ± z has the same row and column sums as A. In this case we say A is shiftable. The support of an array A is defined to be the set containing the absolute value of the elements contained in A. If A is an array with support S and z a nonnegative integer, then A ± z has support S + z. 2 Increasing k from base cases For each of the cases set out in Table 2 our overall strategy is to generate a base case H(n; k) where k takes the smallest possible value and then increase k by multiples of 4, adjoining 4 additional entries to each row and column. In this section we outline various tools to enable this process. To this end, we introduce the following definitions. We associate the cells of an n x n array with the complete bipartite graph Kn,n where partite sets are denoted {aj | i = 1, 2,..., n} and {bj | j = 1,2,..., n} and the edge {aj, bj} corresponds to the cell (i, j). We say that in an n x n array a set of cells S forms a 2-factor if the corresponding set of edges in the graph Kn,n forms a spanning 2-regular graph and forms a Hamilton cycle if the corresponding set of edges forms a single cycle of length 2n. For each d G {0, 1,...,n - 1},we define the diagonal Dd to be the set of cells of the form (r + d, r), 1 < r < n (evaluated modulo n). Observe that the cells Dj U Dj form a Hamilton cycle whenever j - i is coprime to n. Lemma 2.1. Let Si and S2 be two disjoint sets of cells in an n x n array which each form 372 Ars Math. Contemp. 17 (2019) 349-368 Hamilton cycles. The cells of Si U S2 can be filled with the elements of {1,2,..., 4n} so that each row and column sum is equal to 8n + 2. Proof. Let the cells of Si and S2 be {ei | 1 < i < 2n} and {fi | 1 < i < 2n}, respectively, where: • Cells ei and ei+i are in the same row (column) whenever i is odd (respectively, even); • Cells fi and fi+i are in the same row (column) whenever i is odd (respectively, even); • Cells ei and fi are in the same row. Place 1 in cell ei, 4n in cell fi and: • 2n — 2i + 1 in cell e2i+i, where 1 < i < n — 1; 2n + 2i — 1 in cell e2i where 1 ^ i ^ n; • 2n + 2i in cell f2i+i where 1 < i < n — 1; 2n — 2i + 2 in cell f2i where 1 < i < n. The entries in cells ei, e2, fi and f2 add to 1 + (2n + 1) + 4n + 2n = 8n + 2. For every other row, there are two cells from Si with entries adding to 4n + 2 and two cells from S2 with entries adding to 4n. For every column, there are two cells from Si adding to 4n and two cells from S2 adding to 4n + 2. See the example below. □ We demonstrate Lemma 2.1 below when n = 9. The elements of Si are underlined. Si U S2 1 19 18 36 34 17 21 2 15 23 10 26 4 13 25 32 14 11 27 22 20 9 29 16 30 7 31 6 24 12 5 33 35 8 28 3 The following theorem will be crucial in Cases A and D. Theorem 2.2. Let H(n; k) be a Heffter array such that each row and column sums to 2nk + 1. Suppose there exist Hamilton cycles Hi and H2 disjoint to each other and to the filled cells of H(n; k). Then there exists an H(n; k + 4) Heffter array with row and column sums equal to 2n(k + 4) + 1, where the filled cells are precisely the filled cells of H (n; k), Hi and H2. Proof. Let A0 represent the H(n; k) and negate each element so that each row and column has sum equal to —(2nk + 1). From Lemma 2.1, there exists an array Ai on the cells of Hi and H2 such that each row and column sum is equal to 8n + 2; add n(k + 4) — (4n) = nk to N. J. Cavenagh et al.: The existence of square non-integer Heffter arrays 373 each element of Ai to create a new array A i that has support { nk +1, nk+2,..., n (k+4)}. Note that in Ai each row and column sum is equal to 8n + 2 + 4nk. Let A be the union of A0 with Ai. The resulting array A has support {1, 2,..., n(k + 4)}, with k + 4 filled cells in each row and column. Finally, each row and column sum of A is — (2nk + 1) + (8n + 2) + 4(nk) = 2n(k + 4) + 1, as desired. □ The following lemma generalizes Theorem 2.2 from [7], and is used in Cases B and C. Lemma 2.3. Let Si and S2 be two disjoint sets of cells in an n x n array which each form Hamilton cycles. Then for any positive integers t and s > t + 2n, the cells of Si U S2 can be filled with elements to make a shiftable array with support {s + i, t + i | 1 < i < 2n} so that the four elements in each row and each column sum to 0. Proof. Let the sets of cells of Si and S2 be {ej | 1 < i < 2n} and {/ | 1 < i < 2n}, respectively, where: • Cells ei and ei+i are in the same row (column) whenever i is odd (respectively, even); • Cells fi and fi+i are in the same row (column) whenever i is odd (respectively, even); • Cells ei and fi are in the same row. Place: • s + 2n in cell ei and — (t + 2n) in cell fi, with sum s — t; • s + 2i in cell e2i+i and —(t + 2i) in cell f2i+i, with sum s — t, for 1 < i < n — 1, • —(s + 2i — 1) in cell e2i and t + 2i — 1 in cell f2i, with sum t — s, for 1 < i < n. It now follows that the row sums are 0. Using similar arguments it can be seen that the columns also sum to 0. Observe that there are two positive and two negative integers in each row and column; thus the array is shiftable. □ The proof of the following lemma is similar to the proof of Lemma 2.3; we use this in Case E. Lemma 2.4. Let n be even. Let Si and S2 be two disjoint sets of cells in an n x n array which each form 2-factors that are the union of two n-cycles. Then for any positive integers s, t, u and v where s>t + n, t > u + n and u > v + n, the cells of Si U S2 can be filled with elements to make a shiftable array with support {s + i,t + i, u + i,v + i | 1 < i < n} so that the four elements in each row and column sum to 0. Proof. Let Ci, Cj be the cycles of the 2-factor Sj, i € {1, 2}, where Ci and Ci share a row and C2 and C2 share a row. Let the sets of cells of Ci, C{, C2 and C2 be {ei | 1 < i < n}, {/i | 1 ^ i ^ n}, {gi | 1 < i < n} and {hi | 1 < i < n}, respectively, where: • Cells ei and ei+i are in the same row (column) whenever i is odd (respectively, even); 374 Ars Math. Contemp. 17 (2019) 349-368 • Cells fi and fi+1 are in the same row (column) whenever i is odd (respectively, even); • Cells gi and gi+1 are in the same row (column) whenever i is odd (respectively, even); • Cells hi and hi+1 are in the same row (column) whenever i is odd (respectively, even); • Cells e1 and g1 are in the same row; cells f1 and h1 are in the same row. Place: • s + n in cell e1, — (t + n) in cell g1 and u + n in cell f1, — (v + n) in cell h1; • s + 2i in cell e2i+1 and — (t + 2i) in cell g2i+1, for 1 < i < n/2 — 1; • —(s + 2i — 1) in cell e2i and t + 2i — 1 in cell g2i, for 1 < i < n/2; • u + 2i in cell f2i+1 and — (v + 2i) in cell h2i+1, for 1 < i < n/2 — 1; • —(u + 2i — 1) in cell f2i and v + 2i — 1 in cell h2i, for 1 < i < n/2. It now follows that the row sums are 0. Using similar arguments it can be seen that the columns also sum to 0. □ 3 Case A: k = 2 (mod 4) In this section we construct a Heffter array H(n; k), for n = 1,3 (mod 4) and k = 2 (mod 4), where k < n. Row and column sums will always equal 2nk + 1. We start with an example of our construction. H(15; 6) 6 —4 89 81 1 8 12 —88 83 87 85 2 86 18 —82 77 3 79 73 80 24 —76 71 9 15 67 74 30 —70 65 59 21 61 68 36 —64 —58 53 27 55 62 42 —52 47 33 49 56 48 —46 41 39 43 50 54 —40 35 45 37 44 60 —34 29 51 31 38 66 —28 23 57 25 32 72 —22 17 63 19 26 78 — 16 11 69 13 20 84 — 10 5 75 7 14 90 N. J. Cavenagh et al.: The existence of square non-integer Heffter arrays 375 Lemma 3.1. For n = 1, 3 (mod 4), n ^ 7 and k = 6 there exists a Heffter array H(n; 6). Proof. We remind the reader that rows and columns are calculated modulo n but the array entries are not. The array A = A[r, c] is defined as follows, where 1 < i < n: A[i,i] = 6i, A[i + 2, i] = 6n + 2 - 6i, A[i + 1, n - 2 + i] = 6n +1 - 6i, A[i + 2, n - 2 + i] = 6i - 3, A[i, n - 5 + i] = 6n + 5 - 6i, A[i + 1, n - 5 + i] = -6n - 4 + 6i. Then the support of A is {1, 2,..., 6n}. The sets of elements in rows 1, 2 and i, 3 < i < n, are, respectively: {6, 8,1, 6n - 9, 6n - 1, -4}, {12, 2, 6n - 5, 6n - 3, 6n - 7, -(6n - 2)}, {6i, 6n + 2 - 6(i - 2), 6n +1 - 6(i - 1), 6(i - 2) - 3, 6n + 5 - 6i, -6n - 4 + 6(i - 1)}. Thus in each case the sum of elements in a row is 12n +1. The set of elements in column i, 1 < i < n - 5 is: {6i, 6n + 2 - 6i, 6n +1 - 6(i + 2), 6(i + 2) - 3, 6n + 5 - 6(i + 5), -6n - 4 + 6(i + 5)}. The set of elements in columns n - 4, n - 3, n - 2, n - 1 and n are, respectively: {6n - 24, 26,13, 6n - 15, 6n - 1, -(6n - 2)}, {6n - 18, 20, 7, 6n - 9, 6n - 7, -(6n - 8)}, {6n - 12,14,1, 6n - 3, 6n - 13, -(6n - 14)}, {6n - 6, 8, 6n - 5, 3, 6n - 19, -(6n - 20)}, {6n, 2, 6n - 11, 9, 6n - 25, -(6n - 26)}. Thus in each case the sum of elements in a column is 12n +1. □ Theorem 3.2. There exists a Heffter array H(n; k) for all n = 1, 3 (mod 4) and k = 2 (mod 4), where n > k ^ 6. Proof. Let k = 4p + 6. Then 4p + 6 < n - 1 so p < (n - 7)/4. We have solved the case p = 0 in Lemma 3.1 so we may assume p > 1. Observe that the Heffter array given in the proof of that lemma uses only elements in diagonals D0, D2, D3, D4, D5 and D6 and so does not intersect the diagonals D7, D8,..., Dn-1. We can apply Theorem 2.2 recursively, where the diagonals D7, D8,..., D6+2p-1, D6+2p can be paired to give sets of cells S1 and D6+2p+1, D6+2p+2,..., D6+4p-1, D6+4p paired to give sets of cells S2. The result is a Heffter array H(n; k) with constant row and column sum 2nk + 1 whenever k = 2 (mod 4), n = 1, 3 (mod 4) and n > k > 6. □ 4 Case B: k = 3 (mod 4) and n = 3 (mod 4) In this section we construct a Heffter array H(n; k) where n = 4m + 3, k = 4p + 3 and k < n. 376 Ars Math. Contemp. 17 (2019) 349-368 We will begin with k = 3. We first assume that m > 4 and construct an n x n array which is the concatenation of three smaller arrays, AO = AO[r, c], of dimension (4m — 7) x (4m — 7), Ai = Ai[r, c] of dimension 7 x 7 and C of dimension 3 x 3, each containing 3 filled cells per row and column. So we see that n = 4m + 3. The sum of the rows and columns in A0 and A1 will be 0, while the sum of the rows and columns in C will be 2nk + 1. We begin with an example of the main construction of this section. H (19; 3) 16 -48 32 17 27 -44 -33 -14 47 21 15 -36 -18 -13 31 29 -9 -20 -34 -12 46 45 -10 -35 -19 30 -11 -25 24 1 22 -50 28 3 37 -40 26 23 -49 -38 42 -4 -51 8 43 -2 41 -39 5 57 53 54 6 55 56 52 7 Lemma 4.1. For n = 3 (mod 4) and n ^ 7 there exists a Heffter array H(n; 3). Proof. Let n = 4m + 3. We first assume that m > 4 and so n > 19. The small cases will be dealt with at the end of the proof. Let AO be: A0[2i — 1, 2i] = 8m + 1 — i, 1 < i < m, AO[2i, 2i — 1] = —(8m + i), 1 < i < m, AO [2i, 2i + 1] = 12m — i, 1 < i < m — 1, AO[2i + 1, 2i] = —(4m + 1 + i), 1 < i < m — 1, AO [2m — 2 + 2i, 2m — 1 + 2i] = 5m + i, 1 < i < m — 3, AO[2m — 1 + 2i, 2m — 2 + 2i] = —(11m + 1 — i), 1 < i < m — 3, N. J. Cavenagh et al.: The existence of square non-integer Heffter arrays 377 A0 [2m - 1 + 2«, 2m + 2i] = 9m + i, 1 < i < m - 4, A0[2m + 2i, 2m - 1 + 2i] = -(7m +1 - i), 1 < i < m - 4, A0[i + 1,i + 1] = -(4m - 1 - i), 1 < i < 2m - 2, A0 [2m + i, 2m + i] = 2m - i, 1 < i < 2m - 8, A0[2m, 2m] = 4m - 1, A0[1,4m - 7] = -12m, A0[1,1] = 4m, A0 [4m - 7,4m - 7] = 6m + 3. A0[4m - 7,1] = 4m + 1, We illustrate Ai, in the case m — 4: A0(m = 4) 16 32 -48 -33 -14 47 -18 -13 31 -34 -12 46 -19 -11 30 -35 -10 45 -20 -9 29 -36 15 21 17 -44 27 First observe that the array A0 is a (4m - 7) x (4m - 7) array that has 3 filled cells in each row and column. To confirm that the row and columns sums are 0, note that this array was constructed by taking the first 4m - 8 rows and columns of the integer Heffter array H(4m; 3) given in [3], then placing entry -12m in cell (1,4m - 7), entry 4m +1 in cell (4m - 7,1) and entry 6m + 3 in cell (4m - 7,4m - 7). Thus we need only check the sum of row 4m - 7 which is (4m + 1) + (6m + 3) - (10m + 4) = 0 and the sum of column 4m - 7 which is -12m + (6m - 3) + (6m + 3) = 0. Hence each row and column in the array A0 sums to zero. Although not necessary for the case k = 3, for larger values of k (see the following theorem) we map the rows and columns of A0 so that the filled cells are a subset of the union of diagonals D0 U D1 U D2 U Dn-2 U Dn-1. This can be done by applying the mapping P 2i - 1, when 1 < i < 2m - 3, i 1 8m - 2i - 12, when 2m - 2 < i < 4m - 7, to the rows and column of A0. This does not change the row and column sum and the support is still {1, 2,.. ., 4m + 3} \ {1, 2, .. ., 7, 2m, 6m - 2,. .., 6m + 2, 6m + 4, 10m - 3, .. ., 10m + 3, 12m + 1,12m + 2,. .., 12m + 9}. We call this rearranged array A0; see H(19; 3) above for A0 when m = 4. 378 Ars Math. Contemp. 17 (2019) 349-368 Next, let Ai be: A i — (6m + 1) 6m 1 6m — 2 — (12m + 2) 6m + 4 3 10m — 3 — 10m 6m + 2 6m — 1 — (12m + 1) — (10m — 2) 10m + 2 —4 — (12m + 3) 2m 10m + 3 —2 10m + 1 — (10m — 1) It is easy to check that this array has row and column sum 0 and support {1, 2, 3,4, 2m, 6m - 2,..., 6m + 2, 6m + 4, 10m - 3, .. ., 10m + 3, 12m + 1,12m + 2,12m + 3}. We place A1 on the intersection of row and column sets {4m — 6,4m — 5,..., 4m}. Finally, place the block C on the intersection of the row and column sets {4m + 1, 4m + 2,4m + 3}. Observe that the rows and columns sum to 2nk + 1. It is convenient to express C in terms of n and k, as it will be part of more general constructions in the next theorems. However, if k = 3, observe that {12m + 4,12m + 5,..., 12m + 9} = {nk — 5, nk — 4,. .., nk}. C 5 nk nk — 4 nk — 3 6 nk — 2 nk — 1 nk — 5 7 Let A be the concatenation of A0, A1 and C to obtain an H(4m + 3; 3) Heffter array for all m > 4. An H(7; 3) is given in the Appendix. When n = 11 or 15, concatenating the array B(n) given in the Appendix for these size with C will produce an H(n; 3) with similar properties and support. □ We now consider the case when k > 3; we apply the techniques developed in Section 2. Theorem 4.2. There exists a Heffter array H(n; k) for all n = 3 (mod 4) and k = 3 (mod 4), where n > k ^ 3. Proof. Given Lemma 4.1 we only need to address the case k = 4p + 3 where 1 < p < m. Since k < n — 4, p < (n — 7)/4. First observe that the filled cells of H(n; 3) given in the previous lemma are a subset the union of diagonals D0, D1, D2, Dn-2, Dn-1 with support {1, 2,..., 12m+3}U{nk—5, nk—4,..., nk}. We next identify p disjoint Hamilton cycles, that are also disjoint from diagonals D0, D1, D2, Dn-2, Dn-1, by pairing the remaining (n —5) diagonals. Then we apply Lemma 2.3, contributing {12m+4,12m+5,... ,nk — 6} to the support. The result is a Heffter array H(n; k) with each row and column sum equal to 0 except for the final three rows and columns which sum to 2nk + 1. □ N. J. Cavenagh et al.: The existence of square non-integer Heffter arrays 379 5 Case C: k = 3 (mod 4) and n = 2 (mod 4) We construct a Heffter array H(n; k) where n = 4m + 6 and k = 4p + 3 for k < n. As in the previous section, we will begin with k = 3. Firstly we will assume that m > 7 and construct an n x n array which is the concatenation of several smaller arrays. Here we use four smaller arrays, AO = AO[r, c] of size (4m - 13) x (4m - 13), A1 = A^r, c] of size 13 x 13, A2 = A2 [r, c] of size 3 x 3 and A3 = A3[r, c] of size 3 x 3, each containing 3 filled cells per row and column. The sum of the rows and columns in A0, A1 and A2 is 0, while the sum of the rows and columns in A3 is 2nk + 1. Lemma 5.1. For n = 2 (mod 4) and n ^ 34 there exists a Heffter array H(n; 3). Proof. Let n = 4m + 6 and m > 7. Let AO be: AO [2i - 1, 2i] = 8m + 1 - i, AO[2i, 2i - 1] = -(8m + i), AO [2i, 2i +1] = 12m - i, AO[2i + 1, 2i] = -(4m +1 + i), AO [2m - 2 + 2i, 2m - 1 + 2i] = 5m + i, AO[2m - 1 + 2i, 2m - 2 + 2i] = -(11m + 1 - i), AO [2m - 1 + 2i, 2m + 2i] = 9m + i, AO[2m + 2i, 2m - 1 + 2i] = -(7m +1 - i), AO[i + 1, i + 1] = -(4m - 1 - i), AO [2m + i, 2m + i] = 2m - i, AO[2m, 2m] = 4m - 1, AO[1, 4m - 13] = -(12m), AO[1,1] = 4m, AO [4m - 13,1] = 4m + 1, AO [4m - 13,4m - 13] = 6m + 6. We illustrate AO when m = 7: AO(m = 7) 28 56 -84 -57 -26 83 -30 -25 55 -58 -24 82 -31 -23 54 -59 -22 81 -32 -21 53 -60 -20 80 -33 -19 52 -61 -18 79 -34 -17 51 -62 -16 78 -35 -15 50 -63 27 36 29 -77 48 1 /A i /A m, 1 /A i /A m, 1 /A i /A m- 1, 1 /A i /A m- 1, 1 /A i /A m- 6, 1 /A i /A m- 6, 1 /A i /A m- 7, 1 /A i /A m- 7, 1 /A i /A to 2, 1 /A i /A to 14, 380 Ars Math. Contemp. 17 (2019) 349-368 Observe that the array A0 is a (4m - 13) x (4m - 13) array that has 3 filled cells in each row and column. Similarly to the previous case, this array was constructed by taking the first 4m — 14 rows and columns of the integer Heffter array H(4m; 3) given in [3], then placing entry —12m in cell (1,4m — 13), entry 4m + 1 in cell (4m — 13,1) and entry 6m + 6 in cell (4m — 13,4m — 13). Thus we need only check the sum of row 4m — 13 which is (4m + 1) + (6m + 6) — (10m + 7) = 0 and the sum of column 4m — 13 which is — 12m + (6m — 6) + (6m + 6) = 0. Hence all rows and columns in the array A0 sum to zero. We apply the same mapping as in the proof of Lemma 4.1 to the rows and columns so that all non-empty cells are a subset of the diagonals D0, Di, D2, D„_2,D„-i. Let A0 be the resultant array. The support for A0 is {1, 2, .. ., 12m + 8} \ {1, 2,. .., 13, 2m, 6m — 5, 6m — 4, .. ., 6m + 5, 6m + 7, 10m — 6, 10m — 5, ..., 10m + 6,12m + 1,12m + 2,.. ., 12m + 18}. We now arrange the 57 missing symbols into a 13 x 13 array Ai and two 3 x 3 arrays A2 and A3, where A2 and A3 ares shown below and Ai is given at the end of the Appendix. A A3 —8 nk — 2 — (nk — 10) — (nk — 9) —9 nk nk — 1 — (nk — 11) — 10 11 nk — 3 nk — 7 nk — 6 12 nk — 5 nk — 4 nk — 8 13 The support for Ai is {1, 2, .. ., 7, 2m, 6m — 5, 6m — 4, .. ., 6m + 5, 6m + 7, 10m — 6,10m — 5,. .., 10m + 6,12m + 1,12m + 2, .. ., 12m + 6}. Finally we give A2 and A3 with support {8, 9,10,11,12,13} U {nk — 11, nk — 10,..., nk}. In the case k = 3, observe that {nk — 11, nk — 10,..., nk} = {12m + 7,12m + 8, ..., 12m + 18}. The concatenation of A0, Ai, A2 and A3 gives a H(4m+6; 3) Heffter array for m > 7. □ Theorem 5.2. There exists a Heffter array H(n; k) for all n = 2 (mod 4) and k = 3 (mod 4), where n > k ^ 3. Proof. An H(6; 3) is given in the Appendix. Otherwise let n = 4m + 6 where m > 1. When 1 < m < 5 concatenate the B(4m+6) given in the Appendix with the array C given in Case B to get an H(4m + 6; 3). Observe that when 1 < m < 4 the entries are only on diagonals D0, Di, D2, Dn-2, Dn-i as before. When n = 26 the entries are on diagonals D0, Di, D2, D24, D25 and D9. We pair up the rest of the diagonals as {D2i, D2i+i} for 5 < i < 11, and {D3, D4}, {D5, D6}, {D7, D8} to get the required Hamilton cycles. When m = 6, n = 30; an H(30; 3) is given in the Appendix. Two Hamilton cycles H and K are also given as a reference on the array. These Hamilton cycles together with H(30; 3) only have entries on diagonals D0, Di, D2, D3, D27, D28, D29, Di8 and Dii. For the rest of the diagonals, pair them up as {D4, D5}, {D6, D7}, {D8, D9}, {Di2, Di3}, N. J. Cavenagh et al.: The existence of square non-integer Heffter arrays 381 {D14,D15}, {D16,D17} and {D2i-1,D2i} for 10 < i < 13 to get the necessary Hamilton cycles. When m > 7, a H(n; 3) exists by Lemma 5.1, then the proof follows as in Lemma 4.1. □ 6 Case D: k = 1 (mod 4) and n = 1 (mod 4) In this section we construct a Heffter array H(n; k) where n = 4m + 1 and k = 4p +1, with k < n and hence m > 2. The case H(9; 5) is given in the Appendix and so henceforth we assume m > 3. We begin with k = 5 construct an n x n array for which the sum of each row and column is 2nk + 1 = 40m + 11. First we give an example H(17; 5) of our construction. (Note a Hamilton cycle H has been included for the case k > 5.) H(17; 5) 85 -27 11 50 H H 52 68 84 -28 12 35 H H 67 83 -29 13 H H 37 66 82 -30 14 39 H H 65 81 -31 15 41 H H 64 80 -32 43 H 16 H 63 79 -33 17 45 H H H H 62 78 -34 18 47 -20 H H 61 60 21 49 H H 77 59 -19 3 51 36 H H 76 58 -22 23 38 H H 75 57 -4 5 40 H H 25 74 56 -24 42 H H 73 55 -6 7 2 44 H H 72 54 -1 9 46 H 53 71 H -8 H 10 48 -26 H 70 69 Lemma 6.1. For n = 1 (mod 4), n ^ 13 and k = 5 there exists a Heffter array H(n; 5). Proof. Let n = 4m +1 for m > 3. In every case here row and column sums will equal 40m + 11. We give the general construction of A below. A [4 m — 1,4m] = —1, A[4m — 1,1] = 2, A[4m, 2] = 2m + 1, A[4m + 1, 3] = 2m + 2, A[2m — 2,4m] = 4m, A[2m — 1, 2m + 2] = 4m + 1, A[2m, 2m + 3] = 4m + 2, A[2m + 2, 2m + 3] = —(4m + 3), A[2m +1,1] = —(4m + 4), A[1, 2m] = 12m + 2, 382 Ars Math. Contemp. 17 (2019) 349-368 A [4m +1, 2m + 1] = -(6m + 2), A[4m - 3, 2m + 1] = 6m + 1, A[2m, 2m + 2] = -(8m + 2), A[2, 2m + 1] = 8m + 3, A [3,4m] = 8m + 5, A[1,4m + 1] = 12m + 4, A[4m, 2m + 2] = 12m + 5, A[4m + 1,4m + 1] = 16m + 5. A[i, i + 3] = 2m + i + 2, 1 < i ^ 2m 3, (6.1) A[i + 1,i] = 16m + 5 - i, 1 < i ^ 2m, (6.2) A[i, i] = 20m + 6 - i, 1 < i ^ 2m, (6.3) A[2m - i, 2m + 1 - i] = -(8m + 2 - i), 1 < i ^ 2m 1, (6.4) A[2m + 3 - i, 4m + 2 - i] = 12m + 5 - 2i, 1 < i ^ 2m 1, (6.5) A[2m + i, 2m + i] = 14m + 5 - i, 1 < i ^ 2m 1, (6.6) A[2m + 1 + i, 2m + i] = 18m + 6 - i, 1 < i ^ 2m, (6.7) A[4m + 2 - i, 2m - i] = 12m + 2 - 2i, 1 < i ^ 2m 1, (6.8) A[2m + 2i, 2m + 2i + 3] = 2i + 1, 1 < i ^ m - 1, (6.9) A[2m + 2i + 2, 2m + 2i + 3] = -(2i + 2), 1 < i ^ m - 1, (6.10) A [2m + 2i - 1, 2m + 2i + 2] = 4m + 2i + 3, 1 < i ^ m - 2, (6.11) A [2m + 2i + 1, 2m + 2i + 2] = -(4m + 2i + 4), 1 < i ^ m - 2. (6.12) We note that the support of A contains: • 3,4,..., 2m by (6.9) and (6.10), • 2m + 3, 2m + 4,..., 4m - 1 by (6.1), • 4m + 5, 4m + 6,..., 6m by (6.11) and (6.12), • 6m + 3, 6m + 4,..., 8m +1 by (6.4), • 8m + 4, 8m + 6, 8m + 7,..., 12m, 12m + 1,12m + 3 by (6.5) and (6.8), • 12m + 6,12m + 7,..., 16m + 4 by (6.6) and (6.2), • 16m + 6,16m + 7,..., 20m + 5 by (6.7) and (6.3). It follows that the support of A is {1, 2,..., 20m + 5} as required. To verify the sum of each row and column is 40m +11 we begin by noting that, respectively, (6.1), (6.2), (6.3), (6.4) and (6.5) give the sum for row r (where 4 < r < 2m - 3) and (6.1), (6.2), (6.3), (6.4) and (6.8) give the sum for column c (where 4 < c < 2m - 1) as: (2m + r + 2) +(16m + 6 - r) + (20m + 6 - r) - (6m + r + 2) +(8m - 1 + 2r) = 40m + 11 and (2m + c - 1) + (16m + 5 - c) + (20m + 6 - c) - (6m + 1 + c) + (8m + 2 + 2c) = 40m +11. For row 2m + r, where 3 < r < 2m - 4 (or r = 2m - 2) and column 2m + c, where 4 < c < 2m - 1, we argue as follows. N. J. Cavenagh et al.: The existence of square non-integer Heffter arrays 383 Respectively (6.6), (6.7) and (6.8) give a partial sum of 40m + 10 for row r and (6.5), (6.6) and (6.7) give a partial sum of 40m + 12 for column c: (14m + 5 — r) + (18m + 7 — r) + (8m — 2 + 2r) = 40m + 10, (6.13) (8m + 1 + 2c) + (14m + 5 — c) + (18m + 6 — c) = 40m + 12. (6.14) Next (6.9) and (6.10) imply that if r is even, row 2m + r contain the entries r + 1 and —r giving a partial sum of 1. If r is odd, (6.11) and (6.12) imply that row 2m + r contains the entries 4m + r + 4 and — (4m + r + 3) also giving a partial sum of 1. Adding this to the partial sum in (6.13) we get an overall row sum of 40m + 11. Then (6.9) and (6.10) imply that if c is odd, column 2m + c contains the entries c — 2 and — (c — 1) giving a partial sum of —1. If c is even, (6.11) and (6.12) imply that column 2m + c contains the entries 4m + c +1 and —(4m + c + 2) also giving a partial sum of —1. Adding this to the partial sum in (6.14) we get a column sum of 40m +11. The remaining rows and columns can be checked individually to complete the proof that all rows and columns sum to 40m +11. Thus we have the required H (4m +1;5). □ Next, with care, we add up to 2(m — 2) Hamilton cycles to obtain an H(4m + 1; 4p + 5) where p < m — 2. Theorem 6.2. For n = 1 (mod 4) and k = 1 (mod 4) there exists a Heffter array H(n; k), where k < n. Proof. Let n = 4m + 1. The Heffter array H(9; 5) is given in the Appendix. Otherwise, an H(n; 5) exists by Lemma 6.1. This was labeled A in the proof of that lemma. Observe that the occupied cells of A are a subset of the union of diagonals D := D„-3 U D„-2 U Dn_1 U Do U D1 U D4 U D2m—4 u D2m-2 U D2m-1 U D2m U D2m+2. For each m > 3, the following set of cells is a subset of D, that does not intersect A and forms a Hamilton cycle H: {(i, 2m + 1 + i), (i, 2m + 2 + i) | 1 < i < 2m — 3} U {(2m — 2, 4m — 1), (2m — 2, 4m + 1)} U {(2m — 1, 4m + 1), (2m — 1, 4m), (4m, 4m), (4m, 2m + 1)} U {(i, i — 2m +1), (i, i — 2m + 2) | 2m < i < 4m — 1} U {(4m + 1,1), (4m +1, 2m + 2)}. (See the array H(17; 5) above for an example.) Thus there exists 4m+1 —11 = 4(m—2)—2 diagonals that do not intersect A U H and so it is possible to construct 2m — 5 disjoint Hamilton cycles by pairing empty diagonals that are either distance 1 or 2 apart. For m > 6 a possible pairing of diagonals is: {D2, D3}; {D2m-5, D2 m— {D5+2i, D6+2i}, 0 < i < m — 6; {D2m+1, D2m+3}; {D2m+4+2i, D2m+5+2i}, 0 < i < m — 4. 384 Ars Math. Contemp. 17 (2019) 349-368 When m = 4 we pair the diagonals as {D2,D3}; {D9,Du}; {D12,D13} and when m = 5 we pair the diagonals as {D2,D3}; {D5,D7}; {Dn,D13}; {D14,D15}; {Die,D17}. Together with H this gives a total of 2m - 4 Hamilton cycles. Thus applying Theorem 2.2 recursively, we can form a Heffter array for each k such that k = 1 (mod 4) and k < n - 4. □ 7 Case E: k = 1 (mod 4) and n = 2 (mod 4) In this section we construct a Heffter array H(n; k) where n = 4m + 2 and k = 4p + 1, where k < n. We demonstrate the following construction in the case m = 4 with an example of an H(18; 5); the cycles H and K will be needed later for the case k > 5. H (18; 5) 2 51 20 -38 -35 -40 4 53 19 -36 -42 -28 6 37 27 39 -44 -29 8 26 41 25 -46 -30 10 43 24 -48 -31 12 45 23 -50 -32 14 47 22 -52 -33 16 49 21 -54 -34 18 1 81 K H 69 -60 K 90 H H 3 80 K H 70 -61 K 89 88 H 5 79 K H 71 -62 K 72 87 H 7 78 K H K -63 -55 K 86 H 9 77 K H 64 K -56 K 85 H 11 76 65 H 75 66 -57 K 84 H 13 H K K H 67 -58 K 83 H 15 74 H K H 68 -59 K 82 73 17 Lemma 7.1. For n = 2 (mod 4), n ^ 6 there exists a Heffter array H(n; 5). Proof. Let n = 4m + 2. The case H(n; k) = H(6; 5) is given in the Appendix and so henceforth we assume m > 2. Our Heffter array will be the concatenation of an array A0 in the first 2m + 1 rows and columns and an array A1 in the last 2m +1 rows and columns. The row and column sums of A0 will be 0 and the row and column sums of A1 will be 2nk + 1. In our definition of A0, rows and columns are calculated modulo 2m + 1 rather than n. We begin by defining a (2m +1) x (2m + 1) array AO which has support {2, 4, 5, , 4m + 2} U {4m + 3, 4m + 4, , 12m + 6} N. J. Cavenagh et al.: The existence of square non-integer Heffter arrays 385 and for which each row sums to 0 and each column sums to 0 except for columns 1 and 2m + 1, which sum to -(2m + 1) and 2m + 1, respectively. Then we swap the entry -(8m + 4) in cell (2,1) with the entry -(8m + 8) in cell (2,2m + 1) and swap the entry 6m + 2 in cell (4,1) with the entry 8m + 7 in cell (4, 2m +1). The result will be an array A0 defined on row and column set {1,2,..., 2m + 1} with row and column sums equal to 0. To this end, for 1 < i < 2m + 1 let A0[i, i] = 2i; A0[2 + i, 1 + i] = -(6m + 3 + i); AO[i, i - 2] = -(8m + 4 + 2i). Now let Ao [r, c] = < AO [3 - i, 2m + 1 - i] = 4m + 2 + i; AO [2 + i,i - 2] = 8m + 3 + 2i; -(8m + 8), (r, c) = (2,1), -(8m + 4), (r, c) = (2, 2m + 1), 8m + 7, (r, c) = (4,1), 6m + 2, (r, c) = (4, 2m + 1), _ AO [r, c], otherwise. The case m = 4 is illustrated in the example H(18; 5) given above. It will be useful to note that the non-empty cells of this (2m +1) x (2m + 1) array A0 are a subset of the union of diagonals D0 := D0 U D1 U D2 U D3 U D4. Next, we define a (2m + 1) x (2m + 1) array A1, on row and column set {2m + 2, 2m + 3,..., 4m + 2} with support {1, 3,4,..., 4m + 1} U {12m + 7,12m + 8,..., 16m + 8} U {kn - 4m - 1, kn - 4m,..., kn}. For the case k = 5, observe that {kn - 4m - 1, kn - 4m,..., kn} = {16m + 9,16m + 10,..., 20m + 10}. Thus when k = 5 the support of A0 U A1 is {1,2,..., 20m + 10} as required. Each row and column of A1 will sum to 2nk + 1. We first give A1 for the cases m = 2 and m = 3 separately; then we present the general formula. A1(m = 2,n = 10) 1 10k - 4 -31 10k - 2 37 39 3 10k -34 10k - 7 -33 40 5 10k - 8 10k - 3 10k - 1 10k - 6 36 7 -35 10k - 5 -32 10k - 9 38 9 A1 (m = 3, n = 14) 1 14k -43 14k - 7 50 14k - 1 3 51 14k - 8 -44 -45 5 14k - 2 55 14k - 12 14k - 3 7 14k - 11 54 -46 56 -47 9 14k - 13 14k - 4 14k - 6 53 -48 14k - 9 11 -49 14k - 10 14k - 5 52 13 386 Ars Math. Contemp. 17 (2019) 349-368 Otherwise m > 4 and we define A1 as follows. The rows and columns are defined modulo 2m + 1 rather than modulo n. To construct the overall Heffter array, the array Ai is then shifted by adding 2m + 1 (as an integer) to each row and column. A1[4,1] = 14m + 8; A1[5, 2m + 1] = 14m + 9; (7.1) A1[6, 2m] = 16m + 8; A1[2m - 1,1] = kn - 4m + 1; (7.2) Ai[2m, 2m + 1] = kn - 4m; A1[2m + 1, 2m] = kn - 4m - 1; (7.3) A1[i,i] = 2i - 1, 1 < i < 2m + 1; (7.4) A1[i, 2m - 1 + i] = kn + 1 - i, 1 < i < 2m + 1; (7.5) A1[i, i + 1] = kn - 2m - i, 1 < i < 2m - 2; (7.6) A1[4 + i,i] = -(12m + 6 + i), 1 < i < 2m + 1; (7.7) A1 [6 + i, 1 + i] = 14m + 9 + i, 1 < i < 2m - 2. (7.8) We note that the support for A1 is the union of the sets: • {1, 3, 5, 6,..., 4m + 1} (by (7.4)), • {12m + 7,12m + 8,..., 14m + 7} (by (7.7)), • {14m + 8,14m + 9} (by (7.1)), • {14m + 10,..., 16m + 7} (by (7.8)), • {16m + 8} (by (7.2)), • {kn - 4m - 1, kn - 4m, kn - 4m + 1} (by (7.2), (7.3)), • {kn - 4m + 2,..., kn - 2m - 1} (by (7.6)) and • {kn - 2m,..., kn} (by (7.5)). We next check the row and column sums. For row r in the range 7 to 2m + 1, (7.5) and (7.6) give a partial sum of (kn +1 - r) + (kn - 2m - r) = 2kn - 2m + 1 - 2r, while (7.7) and (7.8) give a partial sum of (-12m - 6 - (r - 4)) + (14m + 9 + (r - 6)) = 2m + 1. Now combined with (7.4) the sum of these rows is (2nk - 2m +1 - 2r) + (2m + 1) + (2r - 1) = 2nk + 1, as required. For column c in the range 2 to 2m - 1, (7.5) and (7.6) give a partial sum of (kn +1 - (c + 2)) + (kn - 2m - (c - 1)) = 2kn - 2m - 2c, while (7.7) and (7.8) give a partial sum of (-12m - 6 - c) + (14m + 9 + (c - 1)) = 2m + 2. N. J. Cavenagh et al.: The existence of square non-integer Heffter arrays 387 Now combined with (7.4) the sum of these columns is (2nk — 2m — 2c) + (2m + 2) + (2c — 1) = 2nk + 1. The sum of the remaining rows and columns can be calculated individually and overall the rows and columns of A1 sum to 2nk +1 as required. Thus the concatenation of A0 with A1 gives an H(4m + 2; 5). □ Theorem 7.2. For n = 2 (mod 4), n ^ 6 and k = 1 (mod 4) there exists a Heffter array H(n; k), where k < n. Proof. Let n = 4m + 2 and k = 4p +1. A H(n; 5) exists by Lemma 7.1. Otherwise, k > 9 and m > 2. We take the array A = A0 U A1 from the previous lemma. We will construct m — 2 cycles of length n (that is, on 2(2m + 1) = n cells) in the upper left-hand (A0) and lower right-hand (A1) quadrants, and m further cycles of length n in each of the remaining quadrants. Together these form 2m — 2 disjoint 2-factors. From the proof of Lemma 7.1, within A0 there are 2m +1 — 5 = 2(m — 2) empty diagonals, which we take in pairs to obtain m — 2 cycles of length n. Next take the intersection of the last 2m + 1 rows and columns, this is the (2m +1) x (2m +1) subarray that contains A1. We will refer to diagonals within that subarray only, recalling that the rows and columns are calculated modulo 2m + 1. We aim to find m — 2 cycles of length n from this subarray. The case m = 2 is trivial and for the case m = 3, observe that the empty cells of A1 in the previous lemma form a cycle of length 14. Otherwise m > 4 and the array A1 occupies diagonals D0, D1, D2, D3, D4, D5, D7, D2m—2 and D2m. Next take the following cells H = (j(i + 1, i), (2m — 2 + i, i) | 1 < i < 2m + 1} \ {(2m — 1,1), (2m + 1, 2m)}) U {(2m — 1, 2m), (2m + 1,1)} K = ({(3 + i, i), (7 + i, i) | 1 < i < 2m + 1} \ {(4,1), (6, 2m)}) U {(4, 2m), (6,1)}. In the example H(18; 5) above these cycles are shown in cells marked by H and K, respectively. Observe that H and K form two cycles of length 4m + 2 disjoint from A1 but are a subset of D1 U D2m-2 U D2m U D3 U D7 U D5. Thus there exists 2m +1 — 9 = 2(m — 4) diagonals that do not intersect A1 U H U K. For m > 6, we can thus form m — 4 cycles of length n by taking pairs of diagonals: {De ,D8}; {D2 m — 3, D2 m— 1}; {Dg+2i, D10+2i}, 0 < i < m — 7, and when m = 5, 2m +1 = 11 so we get one cycle of length n by taking the diagonals De and D9. Thus with H and K we have m — 2 cycles of length n that are disjoint from A1 and each other; together these form m — 2 2-factors, each consisting of two cycles of length n. We can create a further m cycles of length n in each of the remaining quadrants, as these cells are all empty. Altogether we have 2m — 2 disjoint 2-factors. Thus by Lemma 2.4, 388 Ars Math. Contemp. 17 (2019) 349-368 we can fill 4(p - 1) cells in each row and column with support {16m + 9,16m + 10,..., kn - 4m - 2} without changing the row and column sums, where k = 4p +1. Thus there exists an H(4m + 2; 4p +1) Heffter array for each m > 2 and p < m. □ References [1] D. Archdeacon, Heffter arrays and biembedding graphs on surfaces, Electron. J. Combin. 22 (2015), #P1.74 (14 pages), https://www.combinatorics.org/ojs/index.php/ eljc/article/view/v22i1p7 4. [2] D. S. Archdeacon, T. Boothby and J. H. Dinitz, Tight Heffter arrays exist for all possible values, J. Combin. Des. 25 (2017), 5-35, doi:10.1002/jcd.21520. [3] D. S. Archdeacon, J. H. Dinitz, D. M. Donovan and E. S. Yazici, Square integer Heffter arrays with empty cells, Des. Codes Cryptogr. 77 (2015), 409-426, doi:10.1007/s10623-015-0076-4. [4] M. Buratti and A. Del Fra, Existence of cyclic k-cycle systems of the complete graph, Discrete Math. 261 (2003), 113-125, doi:10.1016/s0012-365x(02)00463-6. [5] S. Costa, F. Morini, A. Pasotti and M. A. Pellegrini, Globally simple Heffter arrays and orthogonal cyclic cycle decompositions, Australas. J. Combin. 72 (2018), 549-593, https: //ajc.maths.uq.edu.au/pdf/72/ajc_v72_p54 9.pdf. [6] J. H. Dinitz and A. R. W. Mattern, Biembedding Steiner triple systems and n-cycle systems on orientable surfaces, Australas. J. Combin. 67 (2017), 327-344, https://ajc.maths.uq. edu.au/pdf/67/ajc_v67_p327.pdf. [7] J. H. Dinitz and I. M. Wanless, The existence of square integer Heffter arrays, Ars Math. Con-temp. 13 (2017), 81-93, doi:10.26493/1855-3974.1121.fbf. [8] M. J. Grannell and T. S. Griggs, Designs and topology, in: A. Hilton and J. Talbot (eds.), Surveys in Combinatorics 2007, Cambridge University Press, Cambridge, volume 346 of London Mathematical Society Lecture Note Series, 2007 pp. 121-174, doi:10.1017/ cbo9780511666209.006, papers from the 21st Biennial British Combinatorial Conference held in Reading, July 8 - 13, 2007. [9] T. S. Griggs and T. A. McCourt, Biembeddings of symmetric n-cycle systems, Graphs Combin. 32 (2016), 147-160, doi:10.1007/s00373-015-1538-1. [10] T. A. McCourt, Biembedding a Steiner triple system with a Hamilton cycle decomposition of a complete graph, J. Graph Theory 77 (2014), 68-87, doi:10.1002/jgt.21774. [11] A. Vietri, Cyclic fc-cycle systems of order 2kn+k: a solution of the last open cases, J. Combin. Des. 12 (2004), 299-310, doi:10.1002/jcd.20002. N. J. Cavenagh et al.: The existence of square non-integer Heffter arrays 389 Appendix Case B H(7; 3) 15 -13 -2 -11 14 -3 -4 -8 12 -1 10 -9 5 21 17 18 6 19 20 16 7 B(11) -1 18 -17 24 -2 -22 -23 -3 26 -16 20 -4 19 -8 -11 -9 21 -12 -10 25 -15 -13 -14 27 B(15) 1 -36 35 -34 -3 37 33 -22 -11 39 -21 -18 -14 -12 26 -15 -17 32 28 -19 -9 30 -10 -20 -16 24 -8 -13 38 -25 -29 31 -2 -4 -23 27 390 Ars Math. Contemp. 17 (2019) 349-368 CaseC H(6; 3) -1 -16 17 -11 -4 15 12 -9 -3 -2 10 -8 13 -7 -6 18 14 5 B(10) 1 22 -23 17 2 -19 -18 15 3 -24 14 10 9 11 -20 4 -16 12 13 -21 8 B(14) -34 -1 35 -2 24 -22 36 -32 -4 -23 -3 26 -20 28 -8 30 -9 -21 -10 -19 29 -11 27 -16 25 -12 -13 -14 -17 31 -15 33 -18 B (18) 1 21 -22 -36 -4 40 35 -12 -23 -17 33 -16 -11 42 -31 -28 41 -13 -18 43 -25 -26 45 -19 -14 34 -20 -30 -2 32 27 -24 -3 -15 44 -29 -8 46 -38 -39 -9 48 47 -37 -10 N. J. Cavenagh et al.: The existence of square non-integer Heffter arrays 391 o lO 1 o CO o 1—1 1 oo lO OS 1 Oi 1 oo 1 1—1 lO 1 Oi m 1 b- lO oo i—i 1 lO lO CM i—l 1 oo 1 1—1 1—1 1 lO 1 CO lO CO oo CM 1 oo i—i 1 i—i 1 b- CM 1 i—i b- CM 1 oo CM 1 boo 1 lO i—l 1 CM lO o CM 1 OS oo OS i—l 1 b- i—l 1 CM to CM 1 i—l CM 1 oo to CM oo 1 oo co 1 lO CO i—i 1 CO CM 1 oo CM CM 1 co co 1 i—i oo 1 iO co CM 1 oo oo 1 CO co 1 1 O i—i o co 1 OS CM 392 Ars Math. Contemp. 17 (2019) 349-368 Iß CO Iß 1 i—1 i—1 iß iß œ 1—1 1 CO oo 1 boo 1 CM i—1 iß CM b- lß 1 oo i—i iß oo 1 CO CM 1 iß oo 1 CO 1 oo oo 1 i—1 CM oo iß 1 iß i—i oo o CO oo oo 1 b- CM 1 CM 1 oo CM i—i CM 1 CM oo 1 CM b- O 1 i—i CO O oo 1 i—1 oo 1 œ CM 1 CM 1 i—i b- CM CO oo 1 oo CM 1 i—i 1 O 1—1 1 i—1 LO O CM 1 œ Iß oo 1 œ CO CM iß 1 i> 1—1 1 o b- CM CM 1 oo 1 o Iß 1 oo CO oo i—i 1 œ 1 oo Iß 1 oo CO CO 1—1 1 b- 1 i—i Iß CO CO CO 1 CO 1 oo 1 b- CO N. J. Cavenagh et al.: The existence of square non-integer Heffter arrays 393 1 s -a cs 1 s -a s -a CD lO 1 s -a ^ lO CO 1 S -a 1 S -a X t- to lO 1 CO 1 CD CD 1 co CO 7 o CO 1 cs 1 CD X 00 to lO 1 CO lO 1 lO to 1 ^ o ^ lO lO X lO ^ CD 1 cs 1 ■sf 1 ^ ^ co to t-7 ^ cs lO 1 CD 1 00 to lO X to ^ 00 1 ^ 00 cs 1 cs t-1 co ^ co co X o t- 7 lO 1 X cs t-cs 1 lO (N co t- CD CO 1 CO 1 iu n1 then {n1,..., ne-1, ne — n1} is also a minimal system of generators of a numerical semigroup. Proof. In other case, there exists k € {1,..., e — 1} such that nk € {ne — ni} + (ni,. .. ,nfc_i,nfc+i,. .., ,ne_i,n — ni). But it is not possible because ne — n1 + n1 = ne > nk. □ Let S be a numerical semigroup. We denote by M(S) the maximum of msg(S). If S € L(m, e), we define the following sequence of elements of L(m, e): • So = S, • Sn+1 = ((msg(Sn) \ {M(Sn)}) U {M(S„) — m}) if M(S„) — m > m. Because of Lemma 3.1, there exists a sequence: S = S0 C S1 C • • • C Sk = ^(S) € C(m, e). 402 Ars Math. Contemp. 17 (2019) 349-368 Example 3.2. Let S G L(5,3) be the semigroup minimally generated by {5,13,21}. Then, we have the following sequence of elements of L(5,3): S0 = (5,13, 21) C Si = (5,13,16) C S2 = (5,11,13) C S3 = (5, 8,11) C S4 = (5, 6, 8) = ^(S) G C(5, 3). Let S be in C(m, e). We define the graph G([S]) as follows: [S] is the set of vertices and (A, B) g [S] x [S] is an edge if msg(B) = (msg(A) \ {M(A)}) U {M(A) - m}. Theorem 3.3. If S G C(m, e) then G([S]) is a tree with root S. Moreover, if P G [S] and msg(P) = {n1 < n2 < • • • < ne} then the children of P in G([S]) are the numerical semigroups of the form (({n1,... , ne} \ {nk}) U {nk + n1}) such that k G {2,..., e}, nk + ni > ne and nk + ni G ({ni, ...,ne}\ {nk}). Proof. From the definition and the comment after Lemma 3.1, we have that G ([S]) is a tree with root S. Let k be in {2,..., e} such that nk + n1 > ne and nk + n1 G ({n1,..., ne} \ {nk}). If H = (({n1,..., ne} \ {nk}) U {nk + n1}) is clear that msg(H) = ({n1, ...,ne}\ {nk }) U {nk + nj and msg(P) = (msg(H) \ {M(H)}) U {M(H) - m}. Therefore H is a child of P. Conversely, if H is a child of P then (H, P) is an edge of G([S]) and we obtain that H is as the theorem describes. □ The previous theorem provides us with a method to recursively build the elements of [S] as it is shown in the next example. Example 3.4. Figure 1 shows some levels of the tree G([(5,6,8)]). Note that the cardinality of [S] is infinity, so it is impossible to compute all the elements of [S]. However, in the next section, we show that it is possible to compute all the elements of [S] with a fixed Frobenius number or genus. 4 Frobenius number and genus Let P be a numerical semigroup with minimal generating set {n1 < n2 < • • • < ne}, k G {2,..., e} and H be the numerical semigroup generated by ({n1,..., ne }\{nk}) U {nk + n1}. Then H c P, F(P) < F(H) and g(P) < g(H). We can formulate the following result. Proposition 4.1. If S G C(m, e), P G [S] and (H, P) is an edge of G([S]) then F(P) < F(H) and g(P) < g(H). As a consequence of the previous proposition, we have that if we go along through the branches of the tree G([S]), the numerical semigroups that we are finding have greater or equal Frobenius number, and also a greater genus. This fact enables us to formulate the Algorithms 1 and 3 in order to compute all the elements in [S] with Frobenius number less than or equal to a given integer, and genus less than or equal to another given integer, respectively. J. I. Garcia-Garcia et al.: Semigroups with fixed multiplicity and embedding dimension 403 (5, 6, 8) (5, 8, 11) (5, 11, 13) (5, 6, 13) (5, 13, 16) (5, 16, 18) (5, 11, 18) (5, 13, 21) (5, 11, 23) (5, 18, 21) (5, 16, 23) (5, 11, 28) (5, 21, 23) (5, 18, 26) (5, 16, 28) (5, 23, 26) (5, 21, 28) (5, 18, 31) (5, 16, 33) Figure 1: Seven levels of the tree of the packed numerical semigroup (5, 6, 8}. Algorithm 1 An algorithm to determinate the elements T G [S] such that F (T) < F for a fixed integer F. INPUT: (S, F) where S is a packed numerical semigroup and F is a positive integer. OUTPUT: {T G [S] | F(T) < F}. 1: if F(S) > F then 2: return 0 3: while true do 4: A = {S} and B = {S}. 5: C = {h | H is a child of an element of B, F(H) < F}. 6: if C = 0 then 7: return A 8: A = A U C, B = C. The following example illustrates how the previous algorithm works. Example 4.2. We compute all the elements of [(5,6, 8}] with Frobenius number less than or equal to 25. • A = {(5,6,8}}, B = {(5, 6,8}} and C = {(5,8,11}, (5,6,13}}. • A = {(5,6,8}, (5,8,11}, (5, 6,13}}, B = {(5,8,11}, (5, 6,13}} and C = {(5,11,13}}. 404 Ars Math. Contemp. 17 (2019) 349-368 • A = {(5,6, 8), (5,8,11), (5, 6,13), (5,11,13)}, B = {(5,11,13)} and C = {(5,11,18)}. • A = {(5,6, 8), (5,8,11), (5, 6,13), (5,11,13), (5,11,18)}, B = {(5,11,18)} and C = 0. Therefore, the set {T G [(5, 6,8)] | F(T) < 25} is equal to {(5,6,8), (5, 8,11), (5, 6,13), (5,11,13), (5,11,18)}. The next algorithm allows us to compute all the numerical semigroups with multiplicity m, embedding dimension e and Frobenius number less than or equal to F. Note that if S is a numerical semigroup, such that S = N then m(S) - 1 G S and then m(S) - 1 < F(S). Algorithm 2 An algorithm to determinate the numerical semigroups with a fixed embedding dimension and multiplicity, and bounded Frobenius number. INPUT: m, e, and F positive integers such that 2 < e < m < F +1. OUTPUT: {S | S numerical semigroup, m(S) = m, e(S) = e and F (S) < F}. 1: compute C(m, e), using Proposition 2.4. 2: for all S G C(m, e) do 3: compute A(S) = {T g [S] | F(T) < F}, using Algorithm 1. 4: return USec(m,e)A(S). Now, we change Frobenius number by the genus in Algorithm 1 and Algorithm 2. Algorithm 3 An algorithm to determinate the elements T G [S] such that g(T) < g for a fixed integer g. INPUT: (S, g) where S is a packed numerical semigroup and g is a positive integer. OUTPUT: {T G [S] | g(T) < g}. 1: if g(S) > g then 2: return 0 3: A = {S} and B = {S}. 4: while true do 5: C = {H | H is a child of an element of B, g(H) < g}. 6: if C = 0 then 7: return A 8: A = A U C, B = C. We illustrate now the above algorithm. Example 4.3. We compute all the elements of [(5,6, 8)] with genus less than or equal to 15. • A = {(5,6, 8)}, B = {(5,6,8)} and C = {(5, 8,11), (5,6,13)}. • A = {(5,6, 8), (5,8,11), (5, 6,13)}, B = {(5,8,11), (5, 6,13)} and C = {(5,11,13)}. • A = {(5,6, 8), (5,8,11), (5, 6,13), (5,11,13)}, B = {(5,11,13)} and C = {(5,11,18)}. J. I. Garcia-Garcia et al.: Semigroups with fixed multiplicity and embedding dimension 405 • A = {(5,6,8), (5,8,11), (5, 6,13), (5,11,13), (5,11,18)}, B = {(5,11,18)} and C = 0. Algorithm 3 returns {(5,6,8), (5,8,11), (5, 6,13), (5,11,13), (5,11,18)}. Note that if S is a numerical semigroup such that S = N then {1,..., m(S) - 1} C N \ S and then m(S) - 1 < g(S). Combining the above results, we obtain Algorithm 4. Algorithm 4 An algorithm to compute numerical semigroups with fixed multiplicity, embedding dimension and bounded genus. INPUT: m, e, and g positive integers such that 2 < e < m < g + 1. OUTPUT: {S | S numerical semigroup, m(S) = m, e(S) = e and g(S) < g}. 1: compute C(m, e), using Proposition 2.4. 2: for all S G C (m, e) do 3: compute A(S) = {T g [S] | g(T) < g}, using Algorithm 3. 4: return USec(m,e)A(S). Note that applying Algorithms 1 and 2 we have to compute the Frobenius number and the genus, respectively, of the numerical semigroups we recursively obtain when we build [S]. Results of the next section enable us to easily compute the Frobenius number and the genus of every semigroup of [S]. 5 The Apery set of the elements of [S] Let S be a numerical semigroup and n G S \ {0}. The Apery set (named by [1]) of n in S is Ap(S, n) = {s G S | s - n G S}. The next result is a consequence of Lemma 2.4 from [11]. Lemma 5.1. Let S be a numerical semigroup and n G S \ {0}. Then Ap(S, n) has cardinality n. Moreover, Ap(S, n) = {w(0) = 0, w(1),..., w(n — 1)} where w(i) is the least element in S congruent with i modulo n. The set Ap(S, n) gives us a lot of information of S. The following result is found in [13]. Lemma 5.2. Let S be a numerical semigroup and n G S \ {0}. Then: • F (S) = max(Ap(S, n)) — n. • g(S) = nn (EwEAp(s,n) w) — ^. The following result is a consequence of Lemma 5.1. Lemma 5.3. Let S be a numerical semigroup with minimal system of generators {ni, n2, ..., ne} and Ap(S, n1) = {0, w(1),..., w(n1 — 1)}. Then w(i) = min{a2n2 + • • • + aene | (a2,..., ae) G Ne-i and a2n2 + • • • + aene = i (mod ni)}. 406 Ars Math. Contemp. 17 (2019) 349-368 Note that the set j(a2,..., ae) G Ne-1 | a2n2 + • • • + aene = i (mod ni)} has a finite number of minimal elements (using the usual ordering in Ne-1) by Dickson's Lemma (Theorem 5.1 from [10]). We denote the set of these minimal elements by M((n1,..., ne), i). The following result is obtained from Lemma 5.3. Proposition 5.4. Let S be a numerical semigroup with minimal system of generators {n1, n2,..., ne} and Ap(S, n1) = {0, w(1),..., w(n1 — 1)}. Then w(i) = min{a2n2 + • • • + aene | (a2,. .., ae) G M((n1, .. ., ne), i)}. We illustrate the above proposition with an example. Example 5.5. In this example we try to compute the Apery set of the numerical semigroups of [(5, 6,8}] that we obtained in Example 3.4. For every i G {1, 2, 3,4} let A(i) be the set {(a2, a3) G N2 | a2 • 1 + a3 • 3 = i (mod 5)}, and let M(i) be the set of the minimal elements of A(i). Then, M(1) = {(1,0), (0, 2)}, M(2) = {(2, 0), (0, 4), (1, 2)}, M(3) = {(3,0), (0,1)} and M(4) = {(4,0), (0, 3), (1,1)}. Now, if we take an element from [(5,6, 8}], for example S = (5, 21,13}, and we want to compute Ap(S, 5) = {0, w(1), w(2), w(3), w(4)}, by applying Proposition 5.4 we have that w(1) = min{21, 26} = 21, w(2) = min{42, 52, 47} = 42, w(3) = min{63,13} = 13 and w(4) = min{84, 39, 34} = 34. Note that in the previous example it was easy to compute M (i) for every i G {1, 2,3,4}. Now, our next goal is to give an algorithm for computing M((n1,..., ne), i). In order to do it, we introduce the following sets: C(1) = {(x2, .. ., Xe) G Ne-1 | n2X2 +-----+ neXe = i (mod n1)}, C (2) = {(x1, X2, ...,Xe) G Ne | (—n1)x1 + n2X2 +-----+ neXe = i}, C(3) = {(X1,X2,...,Xe,Xe+1) G Ne+1 | (—n1)X1 + n2X2 +----+ neXe + (—i)Xe+1 = 0}. Lemma 5.6. If (a2,..., ae) G C(1) then there exists a1 G N such that (a1, a2,..., ae) G C(2). Proof. It is enough to note that if n2a2 + • • • + neae = i (mod n1) then, there exist a1 G N such that n2a2 + • • • + neae = i + a1 n1. □ Thanks to [12] we know that C(3) is a finitely generated submonoid of Ne+1. The next result can be deduced from Lemma 2 of [12]. Lemma 5.7. Let A be the set {a1,..., at} with Oj = (ai1, ai2,..., aie, aie+1) a system of generators of C(3). If we suppose that a1,..., ad are the elements in A with the last coordinate equal to zero and ad+1,..., aq are the elements of S with the last coordinate equal to 1, then C(2) = { F(S)}. Let m be a positive integer. We denote by L(m) the set {S | S is a numerical semigroup with m(S) = m}. Clearly L(m) is a Frobenius pseudo-variety and max(L(m)) = {0, m, = (m, m +1, ..., 2m - 1). So, as a consequence of Proposition 7.1, we have the following result which is fundamental in this work. Theorem 7.2. The graph G(L(m)) is a tree rooted in (m, m + 1,..., 2m — 1). Moreover, the set formed by the children of a vertex S G L(m) is {S \ {x} | x G msg(S), x > F(S) and x = m}. The previous theorem allows us to recursively construct L(m) from its root by recursively adding its children to the computed vertices. We illustrate this with an example. Example 7.3. We show some levels of the tree G(L(4)) giving its vertices and edges, and the minimal removed generators for obtaining the children. (4, 5, 6, 7) (4,9, 10, 11) (4,7, 10, 13) (4,7, 9) (4,6, 11, 13) (4,6, 9) (4, 5) J. I. Garcia-Garcia et al.: Semigroups with fixed multiplicity and embedding dimension 411 If G is a tree with root r, the level of a vertex x is the length of the only path which connect x and r. The height of a tree is the value of its maximum level. If k G N, we denote by N(k, G) = {v G G | v has level k}. So in Example 7.3 we have: N(0, L(4)) = {(4, 5, 6, 7)}, N(1, L(4)) = {(4, 6, 7, 9), (4, 5, 7), (4, 5, 6)}, N(2, L(4)) = {(4, 7, 9,10), (4, 6, 9,11), (4, 6, 7), (4, 5,11)}, N(3, L(4)) = {(4, 9,10,11), (4, 7,10,13), (4, 7, 9), (4, 6,11,13), (4, 6, 9), (4, 5)}. 8 Elements of L(m, e) with minimum genus Our aim in this section is to give an algorithm that allows us to compute g(m, e) and {S | S G L(m, e) and g(S) = g(m, e)}. The following result is a consequence of Theorem 7.2. Proposition 8.1. If m is a positive integer and (S, T) an edge of G(L(m)), then g(S) = g(T ) + 1. As a direct consequence of the previous proposition we have the following result. Corollary 8.2. Let us fix m, e G N. If P = min{k G N | N(k, G(L(m))) n L(m, e) = 0} then {S G L(m, e) | g(S) = g(m, e)} = N(P, G(L(m))) n L(m, e). Moreover, g(m, e) = m — 1 + P. It is clear that if m > e > 2 then (m, m + 1,..., m + e — 1) G L(m, e). In this way, we have the following result. Proposition 8.3. Let m and e be positive integers. 1. If m < e then L(m, e) = 0. 2. If e = 1 and L(m, e) = 0 then m =1 and L(m, e) = {N}. 3. If m > e > 2 then L(m, e) = 0. We now give an algorithm to compute g(m, e) and {S G L(m, e) | g(S) = g(m, e)}. Algorithm 5 An algorithm to compute g(m, e) and the set of semigroups with a fixed multiplicity and embedding dimension such that its genus is g(m, e). INPUT: m and e positive integers such that m > e > 2. OUTPUT: g(m, e) and {S | S G L(m, e) and g(S) = g(m, e)}. 1: Set k = 0 and A = {(m, m + 1,..., 2m — 1)}. 2: while True do 3: if A n L(m, e) = 0 then 4: return m — 1 + k and A n L(m, e) 5: for S G A do 6: C(S) = {T | T is a child of S}. 7: A = USeA C(S), k = k + 1. 412 Ars Math. Contemp. 17 (2019) 349-368 We illustrate the above algorithm in the following example. Example 8.4. We compute g(5,3) and {S G L(5,3) | g(S) = g(5,3)} using Algorithm 5. • k = 0 and A = {(5,6,7,8, 9)}. • k =1 and A = {(5, 7, 8, 9,11), (5, 6, 8, 9), (5, 6, 7, 9), (5, 6, 7, 8)}. • k = 2 and A = {(5, 8, 9,11,12), (5, 7, 9,11,13), (5, 7, 8,11), (5, 7, 8, 9), (5, 6, 9,13), (5, 6, 8), (5, 6, 7)}. It returns g(5,3) = 6 and {S G L(5,3) | g(S) = 6} = {(5,6,8), (5,6,7)}. In the package FrobeniusNumberAndGenus ([5]), we can run the command ComputeMinimumGenusLme[5,3] to obtain this result. If S is a numerical semigroup, n G S \ {0} and Ap(S, n) = {w(0) = 0, w(1),..., w(n — 1)} (see [11, Lemma 2.4, Lemma 2.6]), then w(i) = kjn + i for some k G N and kn + i G S if and only if k > kj. Therefore, using Lemma 5.2, we have the following (see the proof of [11, Proposition 2.12]). Lemma 8.5. Let S be a numerical semigroup, n G S \ {0} and Ap(S, n) = {0, kin + 1, ..., kn-in + n — 1}. Then g(S) = ki + • • • + kn-i. The next result is easily deduced from Corollary 4 of [6]. Lemma 8.6. Let m, e, q, r be integers such that m > e > 2, S = (m, m+1,..., m+e—1), and m — 1 = q(e — 1) + r, with q, r G N and r < e — 2. Then Ap(S, m) = {0, m + 1,.. ., m + e — 1, 2m + (e — 1) + 1,. .., 2m + 2(e — 1),. .., qm + (q — 1)(e — 1) + 1,. .., qm + q(e — 1), (q + 1)m + q(e — 1) + 1, .. ., (q + 1)m + q(e — 1) + r}. If a, b G N and b = 0 we denote by a mod b the remainder of dividing a by b. If q is a rational number we denote by |_qj = max{z G Z | z < q}. Note that a = Lf J b + (a mod b). From Lemma 8.5 and Lemma 8.6 we have the following result. Proposition 8.7. Let m and e be integers such that m > e > 2 and S = (m, m +1, ..., m + e — 1). Then, g(S) = m—1 e - 1 +1 m— i e-1 (e — 1) + (m — 1) mod (e — 1) Clearly (m, m +1, ...,m + e — 1) G L(m, e) and therefore we have the following result. Corollary 8.8. If m and e are integers such that m > e > 2 then g(m,e) < m— 1 e - 1 +1 m— 1 e-1 (e — 1) + (m — 1) mod (e — 1) J. I. Garcia-Garcia et al.: Semigroups with fixed multiplicity and embedding dimension 413 For many examples the equality holds. However, there are some cases where the semigroup (m, m + 1,..., m + e - 1} does not have minimum genus in the set L(m, e) as we show in the next example. Example 8.9. S = (8, 9,10} is a numerical semigroup and g(S) = 16. S = (8,9,11} is a numerical semigroup and g(5) = 14. Therefore, in this case g((8, 9,10}) = g(8, 3). The following result is a consequence from Proposition 2.4. Proposition 8.10. If S G L(m, e) then S = ({m} + {x mod m | x G msg(S )}} G C (m, e) and g(<§) < g(S). Moreover, if S G C(m, e) then g(S) < g(S ). We illustrate the previous proposition with an example. Example 8.11. If S = (5,11,17} G L(5,3) then S = ({5} + {0,1, 2}} = (5,6, 7} G C(5,3). Therefore, g(S) < g(S). Moreover, S G C(5,3), so g(Š) < g(S). The next result is a consequence of Proposition 8.10. Corollary 8.12. Let m and e be integers such that m > e > 2. Then 1. g(m, e) = min{g(S) | S G C (m, e)}. 2. {S G L(m, e) | g(S) = g(m, e)} = {S G C(m, e) | g(S) = g(m, e)}. Note that C (m, e) is finite and therefore the previous corollary gives us another algorithm for computing g(m, e) and {S G L | g(S) = g(m, e)}. We give more details about this method using Proposition 2.4 and the calculations which appear in Example 2.5. Example 8.13. From the calculations of Example 2.5, we have C(6, 3) = {(6, 7, 8}, (6, 7, 9}, (6, 7,10}, (6, 7,11}, (6, 8, 9}, (6, 8,11}, (6, 9,10}, (6, 9,11}, (6,10,11}}. A simple computation shows us g((6, 7, 8}) = 9, g((6, 7, 9}) = 9, g((6, 7,10}) = 9, g((6, 7,11}) = 10, g((6, 8, 9}) = 10, g((6, 8,11}) = 11, g((6,9,10}) = 12, g((6,9,11}) = 13 and g((6,10,11}) = 13. Therefore, g(6, 3) = 9 and the set {S G L(6, 3) | g(S) = 9} is equal to {(6,7, 8}, (6,7,9}, (6,7,10}}. 9 Elements of L(m, e) with minimum Frobenius number Our aim in this section is to obtain algorithmic methods for computing F (m, e) and {S G L(m, e) | F (S) = F (m, e)}. The next result is a consequence of Theorem 7.2. Proposition 9.1. If m is a positive integer and (S, T) is an edge of G(L(m)), then F (T ) < F (S). 414 Ars Math. Contemp. 17 (2019) 349-368 The following result can be deduced from [9]. Proposition 9.2. If m is a positive integer and (S, T) is an edge of G(L(m)), then e(S) < e(T). Clearly F(m, m) = m — 1 and {S e L(m, m) | F(S) = m — 1} = {(m, m + 1,.. ., 2m — 1)}. It is well known (see [14] for example) that if S = (ni, n2) is a numerical semigroup, then F(S) = nin2 — ni — n2. Therefore, we obtain the following result. Proposition 9.3. Let m be an integer such that m > 2. 1. F(m, m) = m — 1 and {S e L(m, m) | F(S) = m — 1} = {(m, m + 1,..., 2m — 1)}. 2. F(m, 2) = m2 — m — 1 and {S e L(m, 2) | F(S) = m2 — m — 1} = {(m, m +1)}. If q is a rational number we denote by [q] = min{z e Z | q < z}. The next result is deduced from Corollary 5 of [6]. Proposition 9.4. If m and e are integers such that m > e > 2, then m — 1 e — 1 As a consequence of the previous proposition we get the following result. Corollary 9.5. If m and e are integers such that m > e > 2, then m1 F((m, m + 1, .. ., m + e — 1)) m — 1. F(m, e) < m — 1. e — 1 In the previous corollary, equality often holds, but in some cases F((m, m + 1,..., m + e — 1)) = min{F(S) | S e L(m, e)}. For example, F((4,5,6)) = 7 and F((4,5,7)) = 6. From the above results, we obtain the following algorithm where the projections from the cartesian product L(m) x N are denoted by ni and n2. Algorithm 6 An algorithm to compute F(m, e) and the set of semigroups with a fixed multiplicity and embedding dimension such that its Frobenius number is F(m, e). INPUT: m and e integers such that m > e > 2. OUTPUT: F(m, e) and {S e L(m, e) | F(S) = F(m, e)}. 1: A = {(m, m + 1,..., 2m — 1)}, I = 0 and a = [^-T]m — 1. 2: while True do 3: C = {(S, F(S)) | S is child of some element of A and F(S) < a}. 4: K = {S e ni(C) | e(S) > e}. 5: if K = 0 then 6: return F(m, e) = ^(Z) and {S e L(m,e) | F(S) = F(m,e)} = ni(1) 7: A = K, B = {(S, F(S)) | S e K and e(S) = e}. 8: a = min(n2(B) U {a}), I = {(S, F(S)) e I U B | F(S) = a}. We illustrate how this algorithm works with an example. J. I. Garcia-Garcia et al.: Semigroups with fixed multiplicity and embedding dimension 415 Example 9.6. We compute F(4, 3) and {S G L(4,3) | F(S) = F(4,3)} using Algorithm 6. • A = {(4,5,6, 7)}, I = 0 and a = \§]4 - 1=7. • C = {((4, 6,7,9), 5), ((4,5, 7), 6), ((4, 5,6), 7)} and K = {(4, 6, 7, 9), (4, 5, 7), (4, 5, 6)}. • A = {(4, 6, 7, 9), (4, 5, 7), (4, 5, 6)}, B = {((4, 5, 7), 6), ((4, 5, 6), 7)}, a = min{6, 7} = 6 and I = {((4, 5, 7), 6)}. • C = {((4, 7,9,10), 6)} and K = {(4,7, 9,10)}. • A = {(4,7,9,10)}, B = 0, a = 6 and I = {((4, 5,7), 6)}. • C = 0 and K = 0. Therefore, F(4,3) = 6 and {S G L(4,3) | F(S) = 6} = {(4,5,7)}. Using the Math-ematica package [5], we obtain 6 and (4, 5, 7), running the commands MinFrob[4,3] and FrobeniusEmbeddingDimensionMultiplicity[6,3,4], respectively. Our next goal is to give an alternative algorithm for computing F(m, e) and {S G L(m, e) | F(S) = F(m, e)}. The next result is deduced from Proposition 2.4. Proposition 9.7. If S G L(m, e) then S = ({m} + {x mod m | x G msg(S)}) G C(m, e) and F(S) < F(S). As a consequence of the previous proposition we get the following result. Corollary 9.8. If m and e are integers such that m > e > 2, then F(m, e) = min{F(S) | S G C(m, e)}. The set C(m, e) is finite, so previous corollary give us an algorithmic method for computing F(m, e). Example 9.9. We compute F(6, 5). First, we calculate C(6, 5) by using Proposition 2.4 and then we apply Corollary 9.8. So, C(6, 5) = {(6, 7, 8, 9,10), (6, 7, 8, 9,11), (6, 7, 8,10,11), (6, 7, 9,10,11), (6, 8, 9,10,11)} and therefore F(6, 5) = min{F((6, 7, 8, 9,10)) = 11, F((6, 7, 8, 9,11)) = 10, F((6, 7, 8,10,11)) = 9, F((6, 7, 9,10,11)) = 8, F((6, 8, 9,10,11)) = 13} = 8. We are now interested in giving a method for computing {S G L(m, e) | F(S) = F(m, e)}. The next example shows us that there exist semigroups S G L(m, e) such that S G C(m, e) and F(S) = F(m, e). Example 9.10. The numerical semigroups Si = (7, 9,10,15) and S2 = (7,8,10,19) verify that Si,S2 G L(7,4) \C(7,4) and F(Si) = F(S2) = 13 = F(7,4). 416 Ars Math. Contemp. 17 (2019) 349-368 If S G L(m, e) we denote by 0(S) the numerical semigroup generated by {m} + {x mod m | x G msg(S)}. Clearly, 0(S) G C(m, e). Using the partition given in Section 2 and Theorem 2.3, the following two steps are sufficient for computing {S G L(m, e) | F(S) = F(m, e)}. 1. Compute A = {S G C(m, e) | F(S) = F(m, e)}. 2. For every S G A, compute {T G [S] | F(T) = F(S)}. We already know how to compute 1. We now focus on giving an algorithm that allows us to compute 2. Using Algorithm 1, for S G C(m, e) and F G N we get the set {T G [S] | F(T) < F}. Clearly if S G C(m,e) then {T G [S] | F(T) = F(S)} = {T G [S] | F(T) < F(S)}. We are going to adapt Algorithm 1 to our needs for computing 2. We now recall some definitions of Section 3. If S is a numerical semigroup, M(S) = max(msg(S)). If S G C(m, e) the graph G([S]) was defined as follows: [S] is its set of vertices and (A, B) g [S] x [S] is an edge if msg(B) = (msg(A) \ {M(A)}) U {M(A) - m}. Now, using the Theorem 3.3, we give an algorithm which for a semigroup S G C(m, e) computes the set {T g [S] | F(T) = F(S)}. Algorithm 7 An algorithm to compute the semigroups of each equivalence class such that their Frobenius number is minimum._ INPUT: S G C(m,e). OUTPUT: {T G [S] | F(T) = F(S)}. 1: A = {S} and B = {S}. 2: while True do 3: C = {H | H is child of an element of B and F(H) = F(S)}. 4: if C = 0 then 5: return A 6: A = A U C, B = C. We finish this section with an example to illustrate the above algorithm. Example 9.11. We use now Algorithm 7 for computing {T G [S] | F(T) = F(S) = 10} where S = (6,7, 8, 9,11} G C(6, 5). • A = {(6,7, 8, 9,11}} and B = {(6,7,8,9,11}}. • C = {(6, 8, 9,11,13}, (6, 8,11,13,15}}. • A = {(6,7, 8, 9,11}, (6,8, 9,11,13}, (6,8,11,13,15}} and B = {(6, 8, 9,11,13}, (6, 8,11,13,15}}. • C = 0. Thus, {T G [S] | F(T) = 10} = {(6, 7, 8, 9,11}, (6, 8, 9,11,13}, (6, 8,11,13,15}}. This result is also obtained with the package FrobeniusNumberAndGenus ([5]) by executing the command ComputeEquivalenceClass[{6,7,8,9,11}]. J. I. Garcia-Garcia et al.: Semigroups with fixed multiplicity and embedding dimension 417 References [1] R. Apéry, Sur les branches superlinéaires des courbes algébriques, C. R. Acad. Sci. Paris 222 (1946), 1198-1200. [2] V. Barucci, D. E. Dobbs and M. Fontana, Maximality Properties in Numerical Semigroups and Applications to One-Dimensional Analytically Irreducible Local Domains, Memoirs of the American Mathematical Society, American Mathematical Society, Providence, RI, 1997, doi:10.1090/memo/0598. [3] E. Contejean and H. Devie, An efficient incremental algorithm for solving systems of linear Diophantine equations, Inform. and Comput. 113 (1994), 143-172, doi:10.1006/inco.1994. 1067. [4] J. Fromentin and F. 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[16] Wolfram Research, Inc., Mathematica, Version 11.2, Champaign, Illinois, 2017, https:// www.wolfram.com/mathematica. /^creative ^commor ARS MATHEMATICA CONTEMPORANEA ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 17 (2019) 419-430 https://doi.org/10.26493/1855-3974.1881.384 (Also available at http://amc-journal.eu) Grundy domination and zero forcing in Kneser graphs* Boštjan Brešar t Faculty of Natural Sciences and Mathematics, University of Maribor, Slovenia Institute of Mathematics, Physics and Mechanics, Ljubljana, Slovenia Tim Kos * Institute of Mathematics, Physics and Mechanics, Ljubljana, Slovenia Pablo Daniel Torres § Depto. de Matemática, Universidad Nacional de Rosario and Consejo Nacional de Investigaciones Científicas y Técnicas, Argentina Received 17 December 2018, accepted 20 May 2019, published online 6 November 2019 Abstract In this paper, we continue the investigation of different types of (Grundy) dominating sequences. We consider four different types of Grundy domination numbers and the related zero forcing numbers, focusing on these numbers in the well-known class of Kneser graphs Kn,r. In particular, we establish that the Grundy total domination number Ygr (Kn,r) equals (2rr) for any r > 2 and n > 2r +1. For the Grundy domination number of Kneser graphs we get ygr (Kn,r) = a(Knr) whenever n is sufficiently larger than r. On the other hand, the zero forcing number Z(Kn,r) is proved to be - (2rr) when n > 3r + 1 and r > 2, while lower and upper bounds are provided for Z(Kn,r) when 2r + 1 < n < 3r. Some lower bounds for different types of minimum ranks of Kneser graphs are also obtained along the way. Keywords: Grundy domination number, Grundy total domination number, Kneser graph, zero forcing number, minimum rank. Math. Subj. Class.: 05C69, 05C76, 05D05 *This work was supported in part by the bilateral project MINCYT-MHEST SLO 1409. t Supported by the Slovenian Research Agency (research core funding No. P1-0297 and the project grant No. J1-9109). i Supported by the Slovenian Research Agency (research core funding No. P1-0297). § Author thanks the financial support from PICT 2016-0410 and PID-UNR ING 538-539. He also thanks University of Maribor for the financial support and hospitality. E-mail addresses: bostjan.bresar@um.si (Boštjan Brešar), timkos89@gmail.com (Tim Kos), ptorres@fceia.unr.edu.ar (Pablo Daniel Torres) ©® This work is licensed under https://creativecommons.org/licenses/by/4.0/ 420 Ars Math. Contemp. 17 (2019) 349-368 1 Introduction The Kneser graph, Kn,r, where n, r are positive integers such that n > 2r, has the r-subsets of the n-set as its vertices, and two r-subsets are adjacent in Kn,r if they are disjoint. The class of graphs became well-known by the celebrated Erdos-Ko-Rado theorem [11], which determined the independence number a(Knr) of the Kneser graph Kn r to be equal to (n. Another famous result is Lovasz's proof of Kneser's conjecture, which determines the chromatic number of Kneser graphs [22], see also Matousek for a combinatorial proof of this result [23]. Many other invariants were later considered in Kneser graphs by a number of authors. In particular, the domination number of Kneser graphs was studied in several papers [16, 19, 24], but there is no such complete solution for domination number of Kneser graphs as is the case with the chromatic and the independence number. Various Grundy domination invariants have been introduced in recent years [5, 7, 8], arising from two standard domination invariants, the domination number and the total domination number. In the context of domination, we say that a vertex x totally dominates vertices in its (open) neighborhood, N(x) = {y | xy e E(G)}, and that x dominates vertices in its closed neigborhood, N[x] = N(x) U {x}. A set D in a graph G is a dominating set (resp., a total dominating set) of G if every vertex in V(G) is dominated (resp., totally dominated) by a vertex in D. (Clearly, G must not have an isolated vertex to have a total dominating set.) The minimum cardinality of a dominating set (resp., total dominating set) is the domination number (resp., the total domination number) of G, and is denoted by Y(G) (resp., 7t(G)). Now, Grundy (total) domination number is introduced while applying a greedy algorithm to obtain a (total) dominating set D as follows. Vertices are added to the set D one by one, requiring that a vertex x that was added to D (totally) dominates at least one vertex that was not (totally) dominated before this vertex was added. The longest length of such a sequence in a graph G is the Grundy (total) domination number, Ygr (G) (resp., Ygr (G)). By imposing an additional condition on such dominating sequences, one gets the so-called Z-dominating sequences and Z-Grundy domination number [4], denoted by yZ (G). See Section 2 for formal definitions, and [6, 9] for more results on Grundy domination and Grundy total domination numbers. In [5] a strong connection between the Z-Grundy domination number and the zero forcing number of a graph was established. The latter graph invariant has been intensively studied in recent years; it is closely related to another well-known domination concept called power domination, cf. [13, 14]. Moreover, the zero forcing number is very useful in determining the minimum rank of a graph; see the seminal paper about this connection [1] and some further studies [2, 3, 10]. Minimum rank mr(G) of a graph G is defined as the minimum rank over all symmetric matrices that have non-zero (real) values in the non-zero entries of the adjacency matrix A(G) of G, arbitrary real values in the diagonal, and zero values in all other entries. (See the survey on minimum rank, where many applications and other interesting results on this parameter can be found [12].) In addition, the skew zero forcing number (denoted by Z_(G)) was introduced in [18], and studied in the context of the invariant mr0, which is a version of the minimum rank in which matrices are in addition required to have empty diagonals. Motivated by the results of [5], Lin [20] noticed a similar connection between the Grundy total domination number and the skew zero forcing number of graphs, and also between the Grundy domination number and another version of the minimum rank parameter, denoted by mr^. As shown by Lin [20], the following bounds hold: Ygr(G) < mr,(G), (G) < mro(G), YgZ(G) < mr(G). B. Bresar et al.: Grundy domination and zero forcing in Kneser graphs 421 Consequently, any lower bound on a Grundy domination parameter gives a lower bound on the corresponding minimum rank parameter. In this paper, we study four types of Grundy domination parameters (beside the mentioned ones also the L-Grundy domination number) in Kneser graphs, and apply the obtained results to give some bounds or exact results about zero forcing parameters and minimum rank parameters in Kneser graphs. In the next section, we give all the necessary definitions, establish the notation and present some preliminary results. In Section 3 we prove that Ygr(Kn,2) = a(Kn,2) = n - 1 for n > 6, while Ygr(K5,2) = 5. More generally, for Kneser graphs Kn,r when r > 2 we establish an asymptotic result, which states that there exists an n0 G N dependent on r such that for all n, n > n0, we have Ygr(Kn,r) = a(Kn,r) = . The most complete result is obtained for the Grundy total domination number, where we show that for r > 2 and n > 2r + 1, Ygr (Kn,r) = (2rr). This result is proven in Section 4, using a set theoretic result due to Gyarfas and Hubenko [15], which is the ordered version of Lovasz's result for set-pair collections [21]. Section 5 is about Z-Grundy domination number of Kneser graphs. We prove that for any r > 2 and n > 3r + 1, we have yZ (Kn,r) = (2rr). This immediately implies the result for Kneser grapht with r > 2 and n > 3r +1 about their zero forcing number, notably Z(Kn r) = - (2rr); and a lower bound for the minimum rank, which reads as mr(K„,r) > (2rr). On the other hand, when 2r + 1 < n < 3r and r > 2 we only prove the lower bound for Z-Grundy domination number: y^ (Kn,r) > (2rr) - {Ar3ir1-nn), which is dependent on both r and n. In Section 6, we investigate the L-Grundy domination number of Kneser graphs Kn r, provide the exact result when r = 2, and when r is bigger than 2 we prove a lower and an upper bound, which are not far apart. Finally, in the concluding section, we rephrase the Z-Grundy domination number and the Grundy domination number as set-theoretic problems. 2 Preliminaries and notation Let [n] = {1,2,..., n}, where n G N. In [7] the first type of Grundy dominating sequences was introduced. Let S = (vi,..., vk) be a sequence of distinct vertices of a graph G. The corresponding set {vi,..., vk} of vertices from the sequence S will be denoted by ¡3. The initial segment (vi,..., vj) of S will be denoted by Sj. Given a sequence S' = (ui,..., um) of vertices in G such that S n S' = 0, S © S' is the concatenation of S and S', i.e., S © S' = (vi,..., vk, ui,..., um). A sequence S = (vi,..., vk), where Vj G V(G), is called a closed neighborhood sequence if, for each i j-i N[vj] \ J N[vj] = 0. (2.1) If for a closed neighborhood sequence S, the set S3 is a dominating set of G, then S is called a dominating sequence of G. We will say that vj footprints the vertices from N[vj] \ ji N[vj], and that vj is the footprinter of any u G N[vj] ^ j=i N[vj]. For a dominating sequence S, any vertex in V(G) has a unique footprinter in S. Clearly, the length k of a dominating sequence S = (vi,..., vk) is bounded from below by the domination number Y (G) of a graph G. We call the maximum length of a dominating sequence in G the Grundy 422 Ars Math. Contemp. 17 (2019) 349-368 domination number of a graph G and denote it by Ygr (G). The corresponding sequence is called a Grundy dominating sequence of G or Ygr -sequence of G. In a similar way total dominating sequences were introduced in [8], when G is a graph without isolated vertices. Using the same notation as in the previous paragraph, we say that a sequence S = (vi,...,vk), where vi g V (G), is called an open neighborhood sequence if, for each i i-i N(vi) \ J N(vj) = 0. (2.2) j=i We also speak of total footprinters of which meaning should be clear. It is easy to see that an open neighborhood sequence S in G of maximum length yields S to be a total dominating set; the sequence S is then called a Grundy total dominating sequence or ygr -sequence, and the corresponding invariant the Grundy total domination number of G, denoted Yfgr (G). Any open neighborhood sequence S, where S is a total dominating set is called a total dominating sequence. Two additional variants of the Grundy (total) domination number were introduced in [5]. Let G be a graph without isolated vertices. A sequence S = (vi;..., vk), where vi g V(G), is called a Z-sequence if, for each i i-i N(vi) \ J N[vj] = 0. (2.3) j=i Then the Z-Grundy domination number yZ (G) of the graph G is the length of a longest Z-sequence. Note that such a sequence is also a closed neighborhood sequence and hence yZ (G) < Ygr (G). Given a Z-sequence S, the corresponding set S of vertices will be called a Z-set. Note that yfr(G) = Ygr (G) if and only if there exists a Grundy dominating sequence for G each vertex of which footprints some of its neighbors. mrL mr i Igr Y Z Figure 1: Connections between different Grundy domination and minimum rank parameters. B. Bresar et al.: Grundy domination and zero forcing in Kneser graphs 423 While observing the defining conditions of the three concepts in (2.1), (2.2) and (2.3), it is natural to consider also the fourth concept. It gives the longest sequences among all four versions, and we call it L-Grundy domination. Given a graph G, a sequence S = (vi,..., vk), where vi are distinct vertices from G, is called an L-sequence if, for each i i-1 N[vi] \ U N(vj) = 0. (2.4) j=i Then the L-Grundy domination number Ygr (G) of the graph G is the length of a longest L-sequence. Given an L-sequence S, the corresponding set S of vertices will be called an L-set (the requirement that all vertices in S are distinct, indeed makes S to be a set, and prevents the creation of an infinite sequence by repetition of one and the same vertex). Note that it is possible that vi L-footprints only itself. We will make use of the following result. Proposition 2.1 ([5, Proposition 3.1]). If G is a graph with no isolated vertices, then (i) YZr (G) < Ygr(G) < Ygr(G) - 1, (ii) yZ (G) < Ygr(G) < Ygr(G) and all the bounds are sharp. As mentioned earlier, Lin established the connections between different Grundy domination numbers, zero forcing numbers and minimum rank invariants. (For definitions of different zero forcing and minimum rank parameters we refer to [20].) Theorem 2.2 ([20]). If G is a graph without isolated vertices, then (J) |V (G)| - Z,(G) = = Ygr(G) < mr,(G), (2) |V (G)| - Z-(G) = Ygr (G) < mro(G) (3) |V (G)| - Z(G) = = Ygr(G) < mr(G). (4) |V (G)| - Zl(G) = YgLr (G) < mrL(G) The connections between different Grundy domination and minimum rank parameters that follow from Proposition 2.1 and Theorem 2.2 are presented in the Hasse diagram in Figure 1. 3 Grundy domination number It is easy to check (we did this by computer) that the Grundy domination number of the Petersen graph, K5,2, equals 5. Since this is bigger (by 1) than the independence number of the Petersen graph, it is somewhat surprising that for all n bigger than 5, the Grundy domination number equals the independence number of Kn,2. Proposition 3.1. For n > 6, ygr (Kn,2) = a(Kn,2) = n - 1. 424 Ars Math. Contemp. 17 (2019) 349-368 Proof. Clearly, as a sequence of vertices forming a maximum independent set is a Grundy dominating sequence, we have Ygr(Kn,2) > a(Kn,2) = n - 1. In addition, we have checked by computer small cases, establishing Ygr (Kn,2) = n - 1, for 6 < n < 8. For the proof of the reversed inequality for n > 8, let S = (vi,..., vk) be a Grundy dominating sequence of Suppose k > n - 1. Hence, S is not a stable set. We may assume that Si is a stable set for some i > 1, while vi+1 is adjacent to some vj e Si. Note that, since n > 6, we may assume without loss of generality that vj = {1, j + 1} for 1 < j < i (if i = 3 and Sj = {{1,2}, {1,3}, {2,3}}, then all vertices are already dominated by S^ a contradiction). First, if i < 2, then after S3 = (v1, v2, v3), or S2 = (v1, v2) if i = 1, at most four vertices remain undominated, notably vertices from {{1, 3}, {1,4}, {2, 3}, {2,4}}. Hence, counting the largest possible number of steps, we get k < 7. As n > 8, this is only possible when n = 8, in which case we have Ygr (Kg,2) = 7, already noticed above. Hence, we may assume that i > 2. Therefore, after the segment Si is included, only the vertices in {{1, i + 2},..., {1, n}} remain undominated. Since each vertex in the sequence S has to dominate some undominated vertex k < i + n - (i + 1) = n - 1, a contradiction. □ While we could not find exact values of Grundy domination number for all Kneser graphs Kn,r, we prove in some sense similar result as we have for Kn,2. Namely, as soon as n is large enough with respect to a given r (e.g., when r = 2, this was n > 6), we can prove that Ygr (Kn,r) = a(Kn,r). Theorem 3.2. For any r > 2, there exists n0 e N such that for all n, n > no, we have n1 Ygr (Kn,r) = a(Kn r) . . r - 1 Proof. The result is proven for r = 2 by Proposition 3.1. Fix r > 3, and consider a dominating sequence S = (v1,..., vm) of Kn r, assuming that S is not a stable set. In the first case, suppose that already v1 and v2 are adjacent, and without loss of generality we may assume that v1 = {1,..., r} and v2 = {r + 1,..., 2r}. Let V' be the set of vertices, not dominated by {v1, v2}. Note that every element in V' is an r-set that contains at least one element from v1 and at least one element from v2. Denoting by c = {2r +1,..., n}, any element in V' consists of i elements from c, where 0 < i < r - 2, j elements from v1 , where 1 < j < r - i - 1, and consequently, r - i - j elements from v2 (note that r - i - j > 1 ). Hence IV '1 = g(" - 2r) feXr - r - i Note that Xj=i 1 (j) L-'j—i) is not dependent on n, hence for fixed r this is a constant, while J2r=0 (n=i20 is a polynomial in n of degree r - 2. Hence | V' | = O(nr=2). On the other hand, a(Kn,r) = (n=1), hence the resulting dominating sequence is of size Q(nr-1). Note that the length of the sequence S is at most 2 + | V'|. Hence, if n is big enough, S is n 1 not a Grundy dominating sequence, because its length is less than r 1 . In the second case, the initial segment of S, Sk, is a stable set, for some k > 1, while vfc+1 is adjacent to some vertex in Sk. Without loss of generality, we may assume that B. Bresar et al.: Grundy domination and zero forcing in Kneser graphs 425 vk = {1,..., r} and vfc+1 = {r + 1,..., 2r}. Note as above that the size of the set V' of vertices in Kn,r, which are not in N(vk) U N(vk+1) is O(nr-2). To complete the proof of this case, consider the subgraph G' = Kn r - (UveSfc N[v]). Note that G' must have an edge, for otherwise the proof is already done (indeed, no edges in this graph means, that an optimal sequence S consists of the stable set Sk U V(G')). Let ab be an edge in V(G'). Clearly, for all vertices v in Sk, v is not adjacent to the endvertices of ab, i.e., we have v n a = 0 and v n b = 0. In the same way as above we get that the longest such sequence Sfc has at most 02 (n-2r) Sr—-1 (j)(r-rj-i) vertices, and so |Sfc | = k is of the order O(nr-2). Note that the set of vertices not footprinted by vertices in Sk+1 is contained in V'. Then |Sk+1| + |V'| = k + 1 + |V'| is an upper bound for |S| and remains of the order O(nr-2). Therefore, for n big enough, |S| will not be greater that (n-1), which is of order Q(nr-1). □ For the zero forcing parameter Z^ and minimum rank parameter mr^ Theorem 3.2 gives the following observation. Corollary 3.3. For any r > 2, there exists n0 G N such that for all n, n > n0, we have Note that for 2r +1 < n < n0 a lower bound could be improved by improving the values of ygr(Kn,r). It would be even more interesting to find an upper bound for mr 3. This is an interesting issue yet to be investigated. We could only check by computer that Ygr (K7 3) = 20, while clearly a(K7 3) = 15. 4 Grundy total domination number Unlike the Grundy domination number, we prove that the Grundy total domination number of Kn,r does not depend on n. Moreover, we provide the exact value of jfgr (Kn,r) for all cases. To this end, we use an ordered version of Lovasz's result for set-pair collections [21] provided by Gyarfas and Hubenko [15]. Lemma 4.1 ([15, Lemma 1]). Let T = {(A, Bi) | 1 < i < k} be a set-pair collection with |Ai| = a, ^^ = b satisfying the following conditions: 1. Ai n Bi = 0 for 1 < i < k; 2. Ai n Bj = 0 for 1 < i < j < k. Then k < (°+6). Proposition 4.2. For r > 2 and n > 2r +1, (K„,r) = (2rr). Proof. In order to obtain the lower bound, it is enough to note that any sequence S such that S = {A c [2r] | |A| = r} is a total dominating sequence of Kn r. and 426 Ars Math. Contemp. 17 (2019) 349-368 To prove the upper bound, let S = (vi,..., ) be a total dominating sequence of Kn,r. For 1 < i < k, let w be a vertex totally footprinted by vj. It is not hard to see that the set-pair collection T = {(vj, Wj) | 1 < i < k} satisfies both conditions in Lemma 4.1. Since |vj| = |uj| = r for 1 < i < k, then k < (2rr) and the result follows. □ For the skew zero forcing number (G) and the minimum rank parameter mr0(G) we get the following consequences. Corollary 4.3. If r > 2,n > 2r + 1, then Z_(K„,r) = ) - (2rr) and mro(K„,r) > (2rr). 5 Z-Grundy domination number It is easy to check that the Z-Grundy domination number of the Petersen graph K5 2 and of K6,2 equal to 5. Note that Proposition 4.2 provided a general upper bound, i.e., for r > 2 and n > 2r + 1, yZ (Kn,r ) < (2rr). We can prove that this bound is reached in the following cases. Proposition 5.1. For r > 2 and n > 3r +1, yZ* (K„,r ) = (2rr). Proof. Let us consider the following sets 51 = {A | A C [2r], 1 G A, |A| = r}, 52 = {A | A C {r + 2, r + 3,..., 3r}, |A| = r} - {2r + 1,..., 3r}. Let Si be any sequence of Si and let S2 be any sequence of S2. We claim that S = Si © ({2r + 1,..., 3r}) ©S2 is aZ-dominating sequence. Indeed, each u G Si Z-footprints at least f = [2r] - u, {2r + 1,..., 3r} Z-footprints [r - 1] U {3r + 1} and each v G S2 Z-footprints at least f = {1} U ()r + 2, r + 3,..., 3r} - v). Hence, yZ(Kn,r ) > |S| = |Si| + 1 + |S2| = 2(27i) = (2rr). ' As we have mentioned, the equality follows directly from Proposition 2.1 and 4.2. □ Proposition 5.2. For r > 2 and 2r +1 < n < 3r, yJT (K„,r ) > (2rr) - (^i;"). Proof. Let us consider the following sets 51 = {A | A C [2r], 1 G A, |A| = r}, 52 = {A | A C {n - 2r + 2, n - 2r + 3,..., n}, {2r + 1,..., n} C A, |A| = r}. Note, that '2r - 1\ /2r - 1 - (n - 2r)\ /2r - 1\ /4r - 1 - n^ — — — I 2I 1 1 1 r — (n — 2r) J \ r J ^ 3r — n Let Si be any sequence of Si and let S2 be any sequence of S2. We claim that S = Si © S2 is a Z-dominating sequence. Indeed, each u g Si Z-footprints at least f = [2r] — u and each v g S2 Z-footprints at least f = {1} U ({n — 2r + 2, n — 2r + 3,..., n} — v). Hence, yZ) > |S| = |Si| + |S21 = 2-V) — -V;") = -2;) — ft;1;-). □ Perhaps the most interesting case is that of odd graphs K(2r + 1, r): B. Bresar et al.: Grundy domination and zero forcing in Kneser graphs 427 Corollary 5.3. For r > 2, (K2r+1,r) > (2;) - ft-"2) = ^-H2;-2). For the zero forcing numbers of Kneser graphs we get the following. Corollary 5.4. (i) For r > 2 and 2r + 1 < n < 3r, Z(Kn,r) < (nr) - (2rr) + (4r3;1;n). (ii) For r > 2 and n > 3r +1, Z(Kn,r) = (, - (2rr). Propositions 5.1 and 5.2 yield the following consequences for the minimum ranks, mr(Kn,r). Corollary 5.5. (i) For r > 2 and 2r + 1 < n < 3r, mr(Kn,r) > (2rr) - (4r3;1nn). (ii) For r > 2 and n > 3r +1, mr(Kn,r) > (2rr). 6 L-Grundy domination number Proposition 6.1. For n > 5, 7^,(Kn,2) = n + 2. Proof. First, we prove that S = ({1,2}, {1,3}, {1,4},..., {1, n}, {2,3}, {2,4}, {3,4}) is aL-sequence of Kn,2, where n > 5. Note that S is an L-sequence, since each {1, i}, where i € [n] - {1}, L-footprints itself, {2, 3} L-footprints {1,4}, {2,4} L-footprints {1,3}, and {3,4} L-footprints {1, 2}. Hence, (Kn,2) > n + 2. For the proof of the reversed inequality for n > 5, let S = (v1,..., ) be a maximal L-sequence of Kn,2. We may assume without loss of generality that v1 = {1,2}. Clearly, v1 L-footprints itself and vertices of the form {i, j}, where i, j € [n] - {1, 2} and i = j. Next, assume v2 is a neighbor of v1. Without loss of generality we may assume v2 = {3,4}. Thus, v2 L-footprints itself and vertices of the form {i, j}, where i € {1,2}, j € [n] - {3,4} and i = j. Hence, just the vertices {1,3}, {1,4}, {2,3} and {2,4} remain not L-dominated by S2. The rest can be L-dominated with at most 4 vertices. Follows, |S| < 6 < n + 2. Next, suppose v2 is not a neighbor of v1. Without loss of generality let v2 = {1,3}. In this case, the vertex {2,3} and the vertices of the form {1, i}, where i € [n] - {1}, remain not L-dominated by S2. Next, we distinguish again 4 cases: (i) v3 is a neighbor of v1 and v2 (v3 is of the form {i, j}, where i, j € [n] - {1,2,3} and i = j), (ii) v3 is a neighbor of v1 and not of v2 (v3 is of the form {3,i},where i € [n]-{1, 2,3}), (iii) v3 is a neighbor of v2 and not of v 1 (v3 is of the form { 2, i}, where i € [n] -{1, 2,3}), and (iv) v3 is not a neighbor of v1 or v2 (v3 is {2,3} or is of the form {1, i}, where i € [n] -{1, 2, 3}). In the cases (i), (ii), (iii) or if v3 = {2, 3} in the case (iv), there are at most 3 vertices left, that are not L-dominated by S3. In all cases we can L-dominate the rest with at most 4 vertices. Hence, |S| < 7 < n + 2. 428 Ars Math. Contemp. 17 (2019) 349-368 In the case (iv), where v3 = {1, i} (i G [n] - {1, 2, 3}), the vertex v3 L-footprints just itself and {2,3}. Hence, just the vertices of the form {1, i}, where i G [n] - {1}, remain not L-dominated by S3. Note, that it is possible that v, is first L-footprinted by itself and then later again by another vertex from S. Next, if v4 = {1,4}, then v4 just L-footprints itself and the same vertices stay not L-dominated. Hence, to make the sequence S as large as possible, next in the sequence can be vertices vj_i = {1, i} for i = 4,..., n. Without loss of generality let vn = {2,3} (until now in the sequence are all vertices that contain 1). Hence, just the vertices {1,2} and {1,3} remain not L-dominated by Sn. Follows, |S| < n + 2. □ Proposition 6.2. For r > 2 and n > 2r +1, (K„,r) > (^-1) + (2rr-1). Proof. Let Si = {A | A c [n], 1 G A, |A| = r} and let S2 = {A | A c [2r] - {1}, |A| = r}. Let S1 be any sequence of S1 and let S2 be any sequence of S2. We claim that S = S1 © S2 is an L-dominating sequence. Indeed, each u g S1 L-footprints at least itself and each v g S2 L-footprints at least f = [2r] - v. Hence, yL(Kn,r) > |S| = |S1| + |S2| = (n_1) + (V). □ For the upper bound, we first present a general result bounding the L-Grundy domination number of a graph G with no isolated vertices by using the independence number a(G) and the Grundy total domination number Ygr (G). Proposition 6.3. For a graph G with no isolated vertices, yJT (G) < a(G) + (G) - 1. Proof. Let S = (v1,..., vk) be an L-sequence of G. Let A, B and C be the sets of vertices of S such that every vertex in A only L-footprints itself, every vertex in B L-footprints itself and (at least) one more vertex and every vertex in C does not L-footprint itself. Note that {A, B, C} is a partition of S. Besides, A U B is a stable set of G. Hence, |A| + |B| < a(G). Let S' = (w1,..., wm) be the subsequence of S (respecting the order in S) such that S' = B U C. Clearly, S' is an open neighborhood sequence in G, thus m < 7gr(G). Therefore, |S| = |A| + |B| + |C| < a(G)+ 7gr(G) -|B| < a(G)+ 7gr(G) - 1, since v1 G B. □ Corollary 6.4. For r > 2 and n > 2r + 1, (Kn,r) < (n_1) + (2rr) - 1. Note that the gap between the lower and the upper bound in Proposition 6.2 and Corollary 6.4 is (2r_^ - 1, which is fixed with respect to n. 7 Set-theoretic connections Following the set-theoretic connections as in the case of Grundy total domination number (see Lemma 4.1), we ask the following. Problem 7.1. Let T = {(A,, B,) | (A, U B,) C [n], |A,| = |B,| = r, for all i G [k]} be a set-pair collection satisfying the following conditions: 1. A, n Bj = 0 or A, = Bj for 1 < i < k; 2. A, n Bj = 0 and A, = B^ for 1 < i < j < k. B. Bresar et al.: Grundy domination and zero forcing in Kneser graphs 429 Note that |T| = k. Determine the smallest value f (n, r) for which k < f (n, r) for all such set-pair collections T. From Theorem 3.2, for any r > 2 there exists n0 e N such that for all n, n > n0, we have f (n, r) = Ygr (Kn,r) = a(Kn,r) = . Note that in this case, the cardinality n of the universal set in which Aj and Bj are contained plays an important role in determining f (n, r). As we see in the next case, allowing condition Aj = Bj is also relevant to the problem. In a similar way, we propose the corresponding question for the Z-Grundy domination number. Problem 7.2. Let T = {(Aj, Bj) | (Aj U Bj) C [n], |Aj| = |Bj| = r, for all i e [k]} be a set-pair collection satisfying the following conditions: 1. Aj n Bj = 0 for 1 < i < k; 2. Aj n Bj = 0 and Aj = B^ for 1 < i < j < k. Note that |T| = k. Determine the smallest value fZ (n, r) for which k < fZ (n, r) for all such set-pair collections T. Note that fZ(n, r) = (Kn,r) = (2rr) for n > 3r +1. In this case fZ(n, r) is independent of n, but it is unclear whether this function is dependent on n when 2r +1 < n < 3r. 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ARS MATHEMATICA CONTEMPORANEA 17 (2019) 431-445 https://doi.org/10.26493/1855-3974.1758.410 (Also available at http://amc-journal.eu) Si2 and P12-colorings of cubic graphs * Anush Hakobyan Department of Informatics and Applied Mathematics, Yerevan State University, Yerevan, 0025, Armenia Vahan Mkrtchyan t Dipartimento di Informatica, Universita degli Studi di Verona, Strada le Grazie 15, 37134 Verona, Italy Received 21 July 2018, accepted 1 May 2019, published online 6 November 2019 Abstract If G and H are two cubic graphs, then an H-coloring of G is a proper edge-coloring f with the edges of H, such that for each vertex x of G, there is a vertex y of H with f (dG(x)) = dH(y). If G admits an H-coloring, then we will write H -< G. The Petersen coloring conjecture of Jaeger (P10-conjecture) states that for any bridgeless cubic graph G, one has: P10 -< G. The S10-conjecture states that for any cubic graph G, S10 -< G. In this paper, we introduce two new conjectures that are related to these conjectures. The first of them states that any cubic graph with a perfect matching admits an S12-coloring. The second one states that any cubic graph G whose edge-set can be covered with four perfect matchings, admits a P12-coloring. We call these new conjectures S12-conjecture and P12-conjecture, respectively. Our first results justify the choice of graphs in S12-conjecture and P12-conjecture. Next, we characterize the edges of P12 that may be fictive in a P12-coloring of a cubic graph G. Finally, we relate the new conjectures to the already known conjectures by proving that S12-conjecture implies S10-conjecture, and P12-conjecture and (5,2)-Cycle cover conjecture together imply P10--conjecture. Our main tool for proving the latter statement is a new reformulation of (5, 2)-Cycle cover conjecture, which states that the edge-set of any claw-free bridgeless cubic graph can be covered with four perfect matchings. Keywords: Cubic graph, Petersen graph, Petersen coloring conjecture, S10 -conjecture. Math. Subj. Class.: 05C15, 05C70 * We thank Giuseppe Mazzuoccolo for proving Theorem 2.5. We also thank the anonymous referee for her/his valuable comments that helped us to improve the presentation of the paper. 1 Corresponding author. E-mail addresses: ashunik94@gmail.com (Anush Hakobyan), vahanmkrtchyan2002@ysu.am (Vahan Mkrtchyan) ©® This work is licensed under https://creativecommons.Org/licenses/by/4.0/ 432 Ars Math. Contemp. 17 (2019) 349-368 1 Introduction In this paper, we consider finite, undirected graphs. They do not contain loops, though they may contain parallel edges. We also consider pseudo-graphs, which may contain both loops and parallel edges, and simple graphs, which contain neither loops nor parallel edges. As usual, a loop contributes to the degree of a vertex by two. Within the frames of this paper, we assume that graphs, pseudo-graphs and simple graphs are considered up to isomorphisms. This implies that the equality G = G' means that G and G' are isomorphic. For a graph G, let V(G) and E(G) be the set of vertices and edges of G, respectively. Moreover, let dG(x) be the set of edges of G that are incident to the vertex x of G. A matching of G is a set of edges of G such that any two of them do not share a vertex. A matching of G is perfect, if it contains |V(2G)| edges. A block of G is a maximal 2-connected subgraph of G. An end-block is a block of G containing at most one vertex that is a cut-vertex of G. A subgraph H of G is even, if every vertex of H has even degree in H. A subgraph H is odd, if every vertex of G has odd degree in H. Sometime, we will refer to odd subgraphs as joins. Observe that a perfect matching is a join of a cubic graph. A subgraph H is a parity subgraph if for every vertex v of G dG(v) and dH(v) have the same parity. Observe that H is a parity subgraph of G if G - E(H) is an even subgraph of G. Let G is a cubic graph, and let K be a triangle in G such that each of K is of multiplicity one. For an edge e of K, let f be the edge of G that is incident to a vertex of K and is not adjacent to e. Edges e and f will be called opposite edges. Let G and H be two cubic graphs. An H-coloring of G is a mapping f: E(G) ^ E(H), such that for each vertex x of G there is a vertex y of H, such that f (dG(x)) = dH(y). If G admits an H-coloring, then we will write H -< G. It can be easily seen that if H -< G and K -< H, then K -< G. In other words, -< is a transitive relation defined on the set of cubic graphs. If H -< G and f is an H-coloring of G, then for any adjacent edges e, e' of G, the edges f (e), f (e') of H are adjacent. Moreover, if the graph H contains no triangle, then the converse is also true, that is, if a mapping f: E(G) ^ E(H) has a property that for any two adjacent edges e and e' of G, the edges f (e) and f (e') of H are adjacent, then f is an H-coloring of G (see Lemma 2.1). # Figure 1: The graph P10. Figure 2: The graph S10. Let P10 be the well-known Petersen graph (Figure 1) and let S10 be the graph from Figure 2. The Petersen coloring conjecture of Jaeger states: A. Hakobyan and V Mkrtchyan: S12 and P12 -colorings of cubic graphs 433 Conjecture 1.1 (Jaeger, 1988 [9]). For any bridgeless cubic graph G, P10 < G. Sometimes, we will call this conjecture as P10-conjecture. The conjecture is difficult to prove, since it can be seen that it implies the following classical conjectures: Conjecture 1.2 (Berge-Fulkerson, 1972 [4, 15]). Any bridgeless cubic graph G contains six (not necessarily distinct) perfect matchings F\,... such that any edge of G belongs to exactly two of them. This list of six perfect matchings usually is called a Berge-Fulkerson cover of G. If k(G) is the smallest number of perfect matchings that are needed to cover the edge-set of G, then observe that this conjecture implies that k(G) < 5 for any bridgeless cubic graph. This weaker statement is known as Berge conjecture. Conjecture 1.3 ((5, 2)-even-subgraph-cover conjecture [1, 13]). Any bridgeless graph G (not necessarily cubic) contains five even subgraphs such that any edge of G belongs to exactly two of them. Let us note that some of the even subgraphs stated in this conjecture might be empty. Related with the Jaeger conjecture, the following conjecture has been introduced in [11]: Conjecture 1.4 (V. V. Mkrtchyan, 2012 [11]). For any cubic graph G, S10 -< G. We will call this the Si0-conjecture. A k-edge-coloring is an assignment of colors to edges of a graph from a set of k colors such that adjacent edges receive different colors. The smallest k for which a graph G admits a k-edge-coloring is called a chromatic index of G and is denoted by x'(G). If a is a k-edge-coloring of a cubic graph G, then an edge e = uv is called poor (rich) in a, if the five edges of G incident to u or v are colored with three (five) colors. a is called a normal k-edge-coloring of G if any edge of G is either poor or rich in a. Observe that not all cubic graphs admit a normal k-edge-coloring for some k. An example of such a graph is the graph from Figure 2. On the positive side, all simple cubic graphs admit a normal 7-edge-coloring [10]. The smallest k (if it exists) for which a cubic graph G admits a normal k-edge-coloring is called a normal chromatic index of G and is denoted by xN (G). Normal colorings were introduced by Jaeger in [8], where among other results, he showed that for a cubic graph G, xN (G) < 5 if and only if G admits a P10-coloring. This allowed him to obtain a reformulation of Conjecture 1.1, which states that for any bridgeless cubic graph G, xN (G) < 5. In this paper, we introduce two new conjectures that are related to Conjectures 1.1 and 1.4. In Section 2, we discuss some auxiliary results that will be useful later in the paper. In Section 3, we briefly discuss so-called coloring-hereditary classes of cubic graphs that allowed us to come up with these two new conjectures. Then in Section 4, we present our main results. Finally, in Section 5, we discuss some open problems. Terms and concepts that we do not define in the paper can be found in [17, 18]. 2 Auxiliary results In this section, we present some auxiliary results that will be useful later. Our first two results are lemmas about some properties of H-colorings of cubic graphs. Though all these properties are known before, for the sake of completeness we give complete proofs. 434 Ars Math. Contemp. 17 (2019) 349-368 Lemma 2.1. Assume that G and H are two cubic graphs. Moreover, let H be triangle-free. If a mapping f: E (G) ^ E(H) has a property that for any two adjacent edges e and e' of G, the edges f (e) and f (e') of H are adjacent, then f is an H-coloring of G. Proof. In order to see this, assume that f is not an H-coloring of G. Then G contains a vertex w where the definition of an H-coloring is violated. Let e1, e2 and e3 be the three edges incident to w. Assume that the colors of e1 and e2 in f are the edges xy and yz of H. Observe that z = x, as otherwise we will have f (dG(w)) = dH(x) or f (dG(w)) = dH(y) violating the choice of w. Now, the edge f (e3) of H cannot be incident to y. On the other hand, it must be adjacent to xy and yz. Hence f (e3) connects x and z. Observe that the edges x, y and z form a triangle in H. This contradicts our condition on H. □ Note that the condition H is triangle-free is important in the previous lemma. If G is any 3-edge-colorable cubic graph and H contains a triangle with edges h1, h2 and h3, then consider the 3-edge-coloring of G with colors h1, h2 and h3. Observe that for any two adjacent edges of G, their colors are adjacent edges in H. However, the coloring is not an H-coloring as in every vertex of G its definition is violated. Lemma 2.2. Suppose that G and H are cubic graphs with H -< G, and let f be an H-coloring of G. Then: (a) If M is any matching of H, then f-1(M) is a matching of G; (b) x'(G) < x'(H); (c) If M is a perfect matching of H, then f-1(M) is a perfect matching of G; (d) k(G) < k(H); (e) If H admits a Berge-Fulkerson cover, then G also admits a Berge-Fulkerson cover; (f) For every even subgraph C of H, f-1(C) is an even subgraph of G; (g) For every bridge e of G, the edge f (e) is a bridge of H; (h) If H is bridgeless, then G is bridgeless as well; (i) Xn(G) < Xn (H). Proof. (a) and (c): The proof of (a) follows from the definition of H-coloring: as adjacent edges of G must be colored with adjacent edges of H, then clearly the pre-image of a matching in H must be a matching in G. For the proof of (c) let M be a perfect matching of H. Then by (a), f -1(M) is a matching of G. Let us show that it covers all vertices of G. Let v be a vertex of G. Then the three edges incident to v are colored by a similar three edges of H. Since M is a perfect matching of H, one of these three edges must belong to M, hence f-1(M) n dG(v) = 0. Thus, f-1(M) is a perfect matching of G. (b) and (d): For the proof of (b) assume that x' (H) = s and let M1,...,Ms be the color classes of H in an s-edge-coloring. Consider f -1 (M1),..., f -1(Ms). Observe that due to (a), they are forming s matchings covering the edge-set of G. Thus, x'(G) < s = x'(H). The proof of (d) is similar: let k(H) = s and let M1,...,Ms be the s perfect matchings of H covering E(H). Consider f-1(M1),..., f-1(Ms). Observe that due to (c), they are forming s perfect matchings covering the edge-set of G. Thus, k(G) < s = k(H). (e): Let C = (F1,..., F6) be a Berge-Fulkerson cover of H. Consider the list f-1(C) = (f-1(F1),..., f-1(F6)). Observe that due to (c) they are forming a list of A. Hakobyan and V Mkrtchyan: S12 and P12 -colorings of cubic graphs 435 six perfect matchings of G. Moreover, every edge of G belongs to at least two of these perfect matchings. Hence f-1(C) is a Berge-Fulkerson cover of G. (f): Let C be an even subgraph of H. Let us show that any vertex v of G has even degree in f-1(C). Since H is cubic, C is a disjoint union of cycles. Assume that in f the three edges incident to v are colored with three edges incident to a vertex w of H. Then if w is isolated in C, then clearly v is isolated in f-1(C). On the other hand, if w has degree two in C, then v is of degree two in f-1(C). Thus, v always has even degree in f-1 (C), or f -1(C) is an even subgraph of G. (g): Let e be a bridge of G and let (X, V(G) \ X) be a partition of V(G), such that dG(X) = {e}. Assume that the edge f (e) is not a bridge in H. Then there is a cycle C in H that contains the edge f (e). By (f) f-1 (C) is an even subgraph of G that has non-empty intersection with dG(X). Since the intersection of an even subgraph with dG(X) always contains an even number of edges, we have that dG(X) contains at least two edges which contradicts our assumption. (h): This follows from (g): if H has no bridge, then any edge of G cannot be a bridge, as otherwise its color in f will be a bridge in H. (i): Assume that xN (H) = s, and let g be a normal s-edge-coloring of H. Consider a mapping h defined on the edge-set of G as follows: for any edge e of G, let h(e) = g(f (e)). Let us show that h is a normal s-edge-coloring of G. Let e = vw be any edge of G. Assume that in f the three edges incident to v are colored by the three edges incident to a vertex u1 of H, the three edges incident to w are colored by the three edges incident to a vertex u2 If u1 = u2, then the edge e is poor in h. Thus, we can assume that u1 = u2. Since e G dG(v) n dG(w), we have that m1m2 g E(H) and f (e) = u1u2. Now, observe that since g is a normal edge-coloring, we have that if f (e) is a poor edge in g, then e is a poor edge in h, and if f (e) is a rich edge in g, then e is a rich edge in h. Thus, h is a normal s-edge-coloring of G. Hence xN (G) < s = xN(H). The proof of the lemma is complete. □ We will need some results on claw-free bridgeless cubic graphs. Recall that a graph G is claw-free, if it does not contain 4 vertices, such that the subgraph of G induced on these vertices is isomorphic to K13. In [2], arbitrary claw-free graphs are characterized. In [12], Oum has characterized simple, claw-free bridgeless cubic graphs. In order to formulate Oum's result, we need some definitions. In a claw-free simple cubic graph G any vertex belongs to 1, 2, or 3 triangles. If a vertex v belongs to 3 triangles of G, then the component of G containing v is isomorphic to K4 (Figure 3). An induced subgraph of G that is isomorphic to K4 - e is called a diamond [12]. It can be easily checked that in a claw-free cubic graph no 2 diamonds intersect. A string of diamonds of G is a maximal sequence F1,..., of diamonds, in which Fi has a vertex adjacent to a vertex of Fi+1, 1 < i < k — 1. A string of diamonds has exactly of H. Figure 3: The graph K4. 436 Ars Math. Contemp. 17 (2019) 349-368 2 vertices of degree 2, which are called the head and the tail of the string. Replacing an edge e = uv with a string of diamonds with the head x and the tail y is to remove e and add edges (u, x) and (v, y). If G is a connected claw-free simple cubic graph such that each vertex lies in a diamond, then G is called a ring of diamonds. It can be easily checked that each vertex of a ring of diamonds lies in exactly one diamond. As in [12], we require that a ring of diamonds contains at least 2 diamonds. Proposition 2.3 (Oum [12]). G is a connected claw-free simple bridgeless cubic graph, if and only if (1) G is isomorphic to K4, or (2) G is a ring of diamonds, or (3) there is a connected bridgeless cubic graph H, such that G can be obtained from H by replacing some edges of H with strings of diamonds, and by replacing any vertex of H with a triangle. The next auxiliary result allows us to relate coverings with even subgraphs to coverings with specific parity subgraphs. Like we stated in the introduction, some of the even subgraphs here might be empty. Theorem 2.4 ([7, Theorem 3.3]). For a graph G, the following two conditions are equivalent: (1) G contains five even subgraphs such that any edge of G belongs to exactly two of them; (2) G contains four parity subgraphs such that each edge belongs to either one or two of the parity subgraphs. Our final auxiliary result is a theorem proved by Giuseppe Mazzuoccolo which offers a new reformulation of Conjecture 1.3. Theorem 2.5. Conjecture 1.3 is equivalent to the statement that for all bridgeless claw-free cubic graphs we have k(G) < 4. Proof. Assume that for any claw-free bridgeless cubic graph G, we have k(G) < 4. Let us show that Conjecture 1.3 is also true. It is known that it suffices to prove Conjecture 1.3 for cubic graphs [18]. Let G be an arbitrary bridgeless cubic graph. Consider the cubic graph H obtained from G by replacing every vertex of G with a triangle. Observe that H is a claw-free bridgeless cubic graph. By our assumption, the edge-set of H can be covered with four perfect matchings. Observe that perfect matchings are parity subgraphs in cubic graphs, hence by Theorem 2.4, H admits a list of 5 even subgraphs covering each edge exactly twice. In order to complete the proof, let us observe that if a cubic graph K admits a list of 5 even subgraphs covering each edge exactly twice and it contains a triangle T, then the graph K/T also admits a list of 5 even subgraphs covering each edge exactly twice. In order to see this, let C = (Evi,..., Ev5) be the list of 5 even subgraphs covering the edges of K twice. Then it is easy to see that the edges of the 3-cut dK (T) are covered as follows: first edge belongs to Ev i and Ev 2, the second edge belongs to Ev i and Ev3, and finally A. Hakobyan and V Mkrtchyan: S12 and P12 -colorings of cubic graphs 437 the third edge belongs to Ev2 and Ev3. Moreover, Ev4 and Ev5 do not intersect the 3-cut. One can always achieve this by renaming the even subgraphs. Now, if we consider the restrictions of (Ev 1,..., Ev5) to K/T, we will have that they are forming a list of 5 even subgraphs covering each edge of K/T exactly twice. Applying this observation | V(G) | times to H, we will get the statement for the original cubic graph G. For the proof of the converse statement, let us assume that Conjecture 1.3 is true, and show that any claw-free bridgeless cubic graph G can be covered with four perfect match-ings. We prove the latter statement by induction on |V(G)|. If |V(G)| = 2, the the statement is trivially true. Assume that it is true for all claw-free bridgeless cubic graphs with less n vertices and let us consider a claw-free bridgeless cubic graph G with n > 4 vertices. Clearly, we can assume that G is connected. Let us show that we can assume that G is simple. On the opposite assumption, consider the vertices u and v that are joined with two edges. Let u' and v' be the the other neighbors of u and v, respectively. Consider a cubic graph G' obtained from G - {u, v} by adding a possibly parallel edge u'v'. Observe that G' is a bridgeless cubic graph with |V(G')| < n. Moreover, it is claw-free. Thus, by induction hypothesis, G' can be covered with four perfect matchings. Now, it is easy to see that using these list of four perfect matchings of G' we can construct a list of four perfect matchings of G covering E(G). Thus, without loss of generality, we can assume that G is simple. Hence, we can apply Proposition 2.3. If G meets the first or the second condition of the proposition, then it is easy to see that G is 3-edge-colorable, hence it can be covered with three perfect matchings. Thus, we can assume that there is a connected bridgeless cubic graph H such that G can be obtained from H by replacing some edges of H with strings of diamonds and every vertex of H with a triangle. Let us show that we can also assume that G has no string of diamonds. Assume it has one. Let it be S whose head and tails are u and v, respectively. Let u' and v' be the neighbors of u and v, respectively, that lie outside S. Consider a graph G' obtained from G - V(S) by adding a possibly parallel edge u'v'. Observe that G' is a bridgeless cubic graph with |V(G')| < n. Moreover, it is claw-free. Thus, by induction hypothesis, G' can be covered with four perfect matchings. Now, it is easy to see that using these list of four perfect matchings of G' we can construct a list of four perfect matchings of G covering E(G). Thus, without loss of generality, we can assume that G can be obtained from the connected bridgeless cubic graph H by replacing its every vertex with a triangle. By Conjecture 1.3, H has a list of five even subgraphs covering its edges exactly twice. By Theorem 2.4, we have that H admits a cover with four joins such that each edge of H is covered once or twice. Let v any vertex of H and let C = (T\, T2, T3, T4) be the cover of H with four joins. Since each edge of H is covered once or twice in C, we have that there is at most one join in C that contains all three edges incident to v. Thus, for any vertex v we have that either one of joins contains all three edges incident to v and the other three joins contain exactly one edge incident to v, or all joins contain exactly one edge incident to v. Now, it is not hard to see that these four joins covering H can be extended to four perfect matchings of G so that they cover G. The proof of the theorem is complete. □ 438 Ars Math. Contemp. 17 (2019) 349-368 3 Coloring-hereditary classes of cubic graphs In this section, we briefly discuss coloring-hereditary classes of cubic graphs. It is these classes that allowed us to come up with more conjectures related to Conjectures 1.1 and 1.4. If G and H are two cubic graphs with H -< G or G -< H, then we will say that G and H are comparable. A (not necessarily finite) set of cubic graphs is said to be an anti-chain, if any two cubic graphs from the set are not comparable. Let C be the class of all connected cubic graphs. If M CC is a class of connected cubic graphs, then we will say that M is coloring-hereditary, if for any cubic graphs G and H, if H G M and H -< G, then G G M. Assume that B C M is a subset of some coloring-hereditary class M of cubic graphs. We will say that B is a basis for M, if B is an anti-chain and for any connected cubic graph G we have that G G M if and only if there is a cubic graph H G B, such that H -< G. Our starting question is the following: does every coloring-hereditary class of cubic graphs admit a finite basis, that is, a basis with finitely many elements? It turns out that the answer to this question is negative. Let I be the infinite anti-chain of cubic graphs constructed in [14]. Consider the class M of connected cubic graphs G, such that for any G we have: G G M, if and only if there is a cubic graph H G I, such that H -< G. It is easy to see that M is a coloring-hereditary class of cubic graphs without a finite basis. Despite this, one may still look for interesting coloring-hereditary classes arising in graph theory, that admit a finite basis. Below, we discuss some such classes. The first class is C-the class of all connected cubic graphs. Clearly, it is coloring-hereditary. Observe that any connected cubic graph admitting an S10-coloring belongs to C. On the other hand, Conjecture 1.4 predicts that any cubic graph from C admits an Slo-coloring. Thus, we can view Conjecture 1.4 as a statement that S10 forms a basis for C. Let Cb be the class of all connected bridgeless cubic graphs. Statement (h) of Lemma 2.2 implies that Cb is a coloring-hereditary class of cubic graphs. Observe that any connected cubic graph admitting a P10-coloring belongs to Cb. On the other hand, Conjecture 1.1 predicts that any bridgeless cubic graph from Cb admits a P10-coloring. Thus, we can view Conjecture 1.1 as a statement that P10 forms a basis for Cb. Let C3 be the class of all connected 3-edge-colorable cubic graphs. Statement (b) of Lemma 2.2 implies that C3 is a coloring-hereditary class of cubic graphs. Let H be any connected 3-edge-colorable cubic graph. Observe that any cubic graph G is 3-edge-colorable if and only if H -< G. Thus, H forms a basis for C3. Figure 4: The graph S12. Figure 5: The graph P12. Let Cp be the class of all connected cubic graphs containing a perfect matching. Statement (c) of Lemma 2.2 implies that Cp is a coloring-hereditary class of cubic graphs. Ob- A. Hakobyan and V Mkrtchyan: S12 and P12 -colorings of cubic graphs 439 serve that any connected cubic graph admitting an S12-coloring (the graph from Figure 4) belongs to Cp. On the other hand, we suspect that Conjecture 3.1. Any cubic graph with a perfect matching admits an S12-coloring. Conjecture 3.1 predicts that all cubic graphs from Cp admit an S12-coloring. Thus, we can view Conjecture 3.1 as a statement that S12 forms a basis for Cp. Let us note that Conjecture 3.1 has been verified for claw-free cubic graphs in [5]. Let C(4) be the class of all connected cubic graphs G with k(G) < 4. Statement (d) of Lemma 2.2 implies that C(4) is a coloring-hereditary class of cubic graphs. Observe that any connected cubic graph admitting a P12 -coloring (the graph from Figure 4) belongs to C(4). On the other hand, we suspect that Conjecture 3.2. Any cubic graph G with k(G) < 4 admits a P12-coloring. Conjecture 3.2 predicts that all cubic graphs from C(4) admit a P12-coloring. Thus, we can view Conjecture 3.2 as a statement that P12 forms a basis for C(4). Also, note that (e) of Lemma 2.2 implies that Conjecture 4.9 from [7] is a consequence of Conjecture 3.2. 4 The main results In this section, we obtain our main results. First, we discuss the choice of graphs P12 and S12 in Conjectures 3.2 and 3.1, respectively. For this purpose, we recall the following two theorems that are proved in [11]. Theorem 4.1. If G is a connected bridgeless cubic graph with G -< P10, then G = P10. Theorem 4.2. If G is a connected cubic graph with G -< S10, then G = S10. The first theorem suggests that in Conjecture 1.1 the graph P10 cannot be replaced with any other connected bridgeless cubic graph. Similarly, the second theorem suggests that in Conjecture 1.4 the graph S10 cannot be replaced with other connected cubic graph. Now, we are going to obtain similar results for Conjectures 3.2 and 3.1. Theorem 4.3. Let G be a connected bridgeless cubic graph with G -< P12. Then either G = P10 or G = P12. Proof. Assume that f is a G-coloring of P12. Consider the triangle T in P12. Assume that the edges of T are e1, e2, e3. Since these three edges are pairwise adjacent in P12, we have that the edges f (e1), f (e2), f (e3) are pairwise adjacent in G. We have two cases to consider: Case 1: There is a vertex v of G, such that dG(v) = {f (e1), f (e2), f (e3)}. Observe that in this case the edges of the 3-edge-cut dPl2(V(T)) are colored by f (e1),f (e2),f (e3), respectively. Thus, if we contract T in P12, we will get a G-coloring of P10. Hence, by Theorem 4.1, G = P10. Case 2: The edges f (e1), f (e2), f (e3) form a triangle T0 in G. Observe that in this case the edges of the 3-edge-cut dPl2 (V(T)) are colored by the edges of the 3-edge-cut dG (V(T0)). Thus, f induces a G/T0-coloring of P12/T = P10. Hence, by Theorem 4.1, G/T0 = P10, which implies that G = P12. The proof of the theorem is complete. □ 440 Ars Math. Contemp. 17 (2019) 349-368 Corollary 4.4. Let G be a connected bridgeless cubic graph with k(G) < 4 and G -< P12. Then G = P12. Theorem 4.5. Let G be a connected cubic graph with G -< S12. Then either G = S10 or G = S\2. Proof. Assume that f is a G-coloring of Si2. Consider the central triangle T in Si2, that is, the unique triangle T such that all edges of dSl2 (V(T)) are bridges. Assume that the edges of T are e1, e2, e3. Since these three edges are pairwise adjacent in S12, we have that the edges f (e1), f (e2), f (e3) are pairwise adjacent in G. We have two cases to consider: Case 1: There is a vertex v of G, such that Oa(v) = {f (e1), f (e2), f (e3)}. Observe that in this case the edges of the 3-edge-cut dSl2 (V(T)) are colored by f (e1), f (e2), f (e3), respectively. Thus, if we contract T in S12, we will get a G-coloring of S10. Hence, by Theorem 4.2, G = S10. Case 2: The edges f (e1),f (e2), f (e3) form a triangle T0 in G. Observe that in this case the edges of the 3-edge-cut dSl2 (V(T)) are colored by the edges of the 3-edge-cut dG(V(T0)). Moreover, since all edges of dSl2 (V (T)) are bridges, by (g) ofLemma 2.2, we have that the three edges of dG(V(T0)) are bridges. This, in particular, means that each edge of T0 is of multiplicity one in G. Observe that f induces a G/T0-coloring of S12/T = S10. Hence, by Theorem 4.2, G/T0 = S10. Moreover, the new vertex vTo of G/T0 corresponding to T0, is incident to three bridges. Hence vTo is the unique cut-vertex of G/T0 = S10 that is incident to three bridges. This means that G = S12. The proof of the theorem is complete. □ Corollary 4.6. Let G be a connected cubic graph with a perfect matching such that G -< S12. Then G = S12. In the next statement, we discuss the following problem: assume that a bridgeless cubic G graph admits a P10-coloring such that one of the edges of P10 is not used. What can we say about G? We discuss the related problem for Conjecture 3.2 afterwards. Let us note that the following statement is proved by Eckhard Steffen. Proposition 4.7. Let G be a bridgeless cubic that admits a P10-coloring f, such that for an edge e of P10, we have: f-1(e) = 0. Then x'(G) = 3. Proof. ([16]) Assume that the edge e of P10 is not used in a P10-coloring f of G. We have that there exist two perfect matchings M1 and M2 of P10, such that M1 n M2 = {e}. By (c) of Lemma 2.2, we have that f -1(M1) and f-1 (M2) are perfect matchings in G. Since the edge e is not used in f, we have that the perfect matchings are edge-disjoint in G. Thus x'(G) = 3. The proof is complete. □ Next, we characterize the edges of P12, which can be fictive in a P12 -coloring of a graph with k(G) < 4. Proposition 4.8. Let G be a bridgeless cubic graph and let T be the unique triangle of P12. (a) If G admits a P12-coloring f, such that for an edge e 4. Let us show that we have equality here. Consider the three edges incident to v, and let it be our colors in a 3-edge-coloring of H. Now, color the remaining copies of P10 - v in G by edges of P10 - v, so that each edge is colored with its copy. As a result, we get a P12-coloring of G. Thus, by (d) of Lemma 2.2, we have k(G) < 4. Hence k(G) = 4. Moreover, in the P12 -coloring of G the edges of T are not used. The proof is complete. □ In the final part of the paper we establish some connections among the discussed conjectures. Theorem 4.9. Conjecture 3.1 implies Conjecture 1.4. Proof. Assume that Conjecture 3.1 is true. We claim that any cubic graph G admits an S10-coloring. In this proof, we will assume the following notation for the edges of S12: the three bridges of S12 are denoted by a, b, c, the edges of the unique contractible triangle of S12 are denoted by x, y, z, such that x and a, y and b, z and c are opposite edges. Finally, the edges of the end-block containing a vertex incident to a have the following labels: the edges incident to a are a1 and a2, and the parallel edges are a3 and a4. Similarly, we label other edges by b1, b2, b3, b4 and c1, c2, c3, c4. Let G be a cubic graph. If G contains a perfect matching, then it has an S12-coloring. Since S12 has an S^o-coloring, we have the statement in this case. Thus, without loss of generality, we can assume that G does not contain a perfect matching. Consider the graph Ga obtained from G by replacing all vertices of G with triangles. Observe that Ga contains a perfect matching. An example of such a matching would be E(G). Thus, there exists a smallest subset U C V(G), such that if we replace all vertices of U with triangles, we will get a cubic graph H containing a perfect matching. By Conjecture 3.1, H admits an S12-coloring f. Now, we claim that all triangles of H corresponding to vertices of U are colored with triangles of S12. Assume the opposite, that is, there is a triangle T corresponding to a vertex of H, such that f (E(T)) = dSl2 (v) for some vertex v of S12. Consider the graph H' obtained from H by contracting T. Observe that the resulting graph H' still has an S12-coloring, hence by (c) of Lemma 2.2 it contains a perfect matching. However, this violates the definition of the set U, since we found a smaller subset of vertices, whose replacement with triangles was leading to a cubic graph containing a perfect matching. Now, all triangles of H corresponding to vertices of U are colored with triangles of S12. Let us show that all these triangles corresponding to vertices of U are colored with the central triangle of S12, that is the only contractible triangle of S12. 442 Ars Math. Contemp. 17 (2019) 349-368 On the opposite assumption, assume that T, one of these triangles, is colored with other triangles of S12. Without loss of generality, we can assume that the edges of this triangle of S12 are a1, a2, a3. Thus the set of edges leaving T, are colored with a and a4. Two of them are colored with a4, and one is colored with a. Let M be a perfect matching of S12 containing the edges a and a3. By (c) of Lemma 2.2, we have that F = f-1(M) is a perfect matching in H. Now, observe that |F n dH(V(T))| = 1. Consider the cubic graph H" obtained from H by contracting T. Observe that F \ (F n E(T)) is a perfect matching of H". This violates the definition of the set U, since we found a smaller subset of vertices, whose replacement with triangles was leading to a cubic graph containing a perfect matching. Thus, all triangles of H corresponding to U are colored with the edges x, y, z of the central triangle of S12. Observe that G can be obtained from H by contracting all the triangles corresponding to U. Now, using the S12-coloring of H, we obtain an S10-coloring of G. Contract all triangles of H corresponding to U and the central triangle of S12 to obtain S10, and recolor the edges of H having color x with the color a, the edges of H with color y with color b and finally, the edges of H with color z with color c, respectively. Since x, y, z form an even subgraph in S12, by (f) of Lemma 2.2, the edges of f-1({x, y, z}) will form an even subgraph, that is vertex-disjoint union of cycles. Hence, after the re-coloring we obtain an S10-coloring of G. The proof of the theorem is complete. □ Theorem 4.10. Conjectures 1.3 and 3.2 imply Conjecture 1.1. Proof. Assume that Conjectures 1.3 and 3.2 are true, and let G be a bridgeless cubic graph. Let us show that G admits a P10--coloring. If k(G) < 4, then by Conjecture 3.2 it has a P12-coloring. Since P12 admits a P10-coloring, we have that G admits a P10--coloring. Thus, without loss of generality, we can assume that k(G) > 5. Consider the graph H obtained from G by replacing all vertices of G with triangles. Observe that H is a claw-free bridgeless cubic graph. Hence by Conjecture 1.3 and Theorem 2.5, k(H) < 4. Thus, by Conjecture 3.2, H admits a P12-coloring. Since P12 admits a P10-coloring, we have that H admits a P10-coloring f. Observe that since P10 is triangle-free, we have that for any triangle T of H there is a vertex v of P10, such that f (E(T)) = dPl0 (v). Thus, if we contract all the triangles of H that correspond to vertices of G, we will obtain a P10-coloring of G. The proof of the theorem is complete. □ The diagram from Figure 6 explains the relationship among the main four conjectures discussed in the paper. The first arrow shows that P12-conjecture implies P10-con-jecture if 5-CDC is true (Conjecture 1.3). The second arrow shows that the statement "P10-conjecture implies S12-conjecture" is the formulation of Conjecture 5.1. Finally, the third arrow shows that in Theorem 4.9 we showed that S12-conjecture implies the S10-conjecture. 5 Future work In this section, we discuss some open problems and conjectures that are interesting in our point of view. In the previous section, we established a connection between Conjectures 3.2 and 1.1, and Conjectures 3.1 and 1.4. We suspect that this relationship can be extended to a linear order among these four conjectures. Related with this, we would like to offer: A. Hakobyan and V Mkrtchyan: S12 and P12 -colorings of cubic graphs 443 Figure 6: The relationship among the main conjectures. Conjecture 5.1. Conjecture 1.1 implies Conjecture 3.1. All coloring-hereditary classes that we discussed up to now either had or are conjectured to have a basis with one element. One may wonder whether there is a coloring-hereditary class of cubic graphs arising from an interesting graph theoretic property, such that the basis of the class contains at least two graphs. For a positive integer k let Ck be the class of connected cubic graphs G with xN (G) < k. Statement (i) of Lemma 2.2 implies that Ck is a coloring-hereditary class of cubic graphs. Recently, it was shown that for any simple cubic graph xN (G) < 7 [10]. By using this result, a simple inductive proof can be obtained for the following extension of this result: Theorem 5.2. Let G be a cubic graph admitting a normal k-edge-coloringfor some integer k. Then x'N (G) < 7. Theorem 5.2 suggests that Ck is meaningful when k = 3,4, 5,6,7. Below, we discuss these classes for each of these values. When k = 3, Ck represents the class of connected 3-edge-colorable cubic graphs. Thus, our notation is consistent with that of Section 3. When k = 4, it can be easily seen that a cubic graph admits a normal 4-edge-coloring, if and only if it admits a 3-edge-coloring. Thus, C4 = C3. When k = 5, Jaeger has shown that a cubic graph admits a P10-coloring if and only if it admits a normal 5-edge-coloring. On the other hand, we have that any cubic graph admitting a P10-coloring, has to be bridgeless. Thus, if Conjecture 1.1 is true, then C5 = Cb. Finally, when k = 6 or k = 7, we suspect that the bases of the classes C6 and C7 contain infinitely many cubic graphs. We are able to show that the basis of C7 must contain at least two graphs. Let B be any basis of C7. It can be easily seen that we can assume that it does not contain a 3-edge-colorable graph. Moreover, by a simple inductive proof, one can show that all elements of B can be assumed to be simple graphs. Now, let S16 be the graph from Figure 7. The following two results are proved in [5]: 444 Ars Math. Contemp. 17 (2019) 349-368 Theorem 5.3. Let G be a simple graph with G — Si6. Then G = Si6. Theorem 5.4. Let G be a simple graph with G — Pi0. Then G = Pi0. Theorems 5.3 and 5.4 suggest that the only way to color the graphs Si6 and Pio with simple graphs is to take them in the basis B. Thus, B must contain at least two graphs. Finally, we would like to discuss some algorithmic problems. For a fixed connected cubic graph H consider a decision problem which we call the H-problem: Problem 5.5 (H-problem). Given a connected cubic graph G, the goal is to decide whether G admits an H-coloring. Observe that when H is 3-edge-colorable, we have that H-problem is equivalent to testing 3-edge-colorability of the input graph G, which is NP-complete [6]. When H = Si0, we have that all instances of H-problem have a trivial "yes" answer provided that Conjecture 1.4 is true. Thus, this problem is polynomial time solvable if Conjecture 1.4 is true. When H = Si2, Conjecture 3.1 implies that H-problem is equivalent to testing the existence of a perfect matching in the input graph G. This is known to be polynomial-time solvable. When H = Pi0, Conjecture 1.1 implies that H-problem is equivalent to testing bridgelessness of the input graph G. This problem is also polynomial time solvable. Finally, when H = Pi2, Conjecture 3.2 implies that H-problem is equivalent to testing whether the input graph G can be covered with four perfect matchings. The latter problem is proved to be NP-complete in [3]. Thus, depending on the choice of H, the H-problem may or may not be NP-complete. Let CNP be the class of all connected cubic graphs H, for which the H-problem is NP-complete. We suspect that: Conjecture 5.6. CNP is a coloring-hereditary class of cubic graphs. We also would like to offer the following conjecture, which presents a dichotomy for H-problems: Conjecture 5.7. Let H be a connected cubic graph. Then: • if H admits a Pi2 -coloring, then the H-problem is NP-complete; • if H does not admit a Pi2-coloring, then the H -problem is polynomial-time solvable. References [1] A. U. Celmins, On Cubic Graphs That Do Not Have an Edge-3-Colouring, Ph.D. thesis, Department of Combinatorics and Optimization, University of Waterloo, Waterloo, Canada, 1984. A. 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Zhang, Integer Flows and Cycle Covers of Graphs, volume 205 of Monographs and Textbooks in Pure and Applied Mathematics, Marcel Dekker, New York, 1997. /^creative ^commor ARS MATHEMATICA CONTEMPORANEA ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 17 (2019) 447-454 https://doi.org/10.26493/1855-3974.1917.ea5 (Also available at http://amc-journal.eu) The complement of a subspace in a classical polar space Krzysztof Petelczyc , Mariusz Zynel Institute of Mathematics, University of Bialystok, Ciolkowskiego 1 M, 15-245 Bialystok, Poland Received 22 January 2019, accepted 6 July 2019, published online 7 November 2019 Abstract In a polar space, embeddable into a projective space, we fix a subspace, that is contained in some hyperplane. The complement of that subspace resembles a slit space or a semiaffine space. We prove that under some assumptions the ambient polar space can be recovered in this complement. Keywords: Polar space, projective space, semiaffine space, slit space, complement. Math. Subj. Class.: 51A15, 51A45 1 Introduction Cohen and Shult coined the term affine polar space in [4] as a polar space with some hyperplane removed. They prove that from such an affine reduct the ambient polar space can be recovered. In [9] we prove something similar for the complement of a subset in a projective space. Looking at the results of these two papers it is seen that an interesting case has been set aside: the complement of a subspace in a polar space. We are trying to fill this this gap here, although under several specific assumptions: we consider classical polar spaces, i.e. embeddable into projective spaces (cf. [2]), and our subspace is contained in a hyperplane. A projective space with some subspace removed is called a slit space (cf. [5, 6, 8]) so, our complement can be seen as a generalized slit space. Singular subspaces in a polar space are projective spaces, in an affine polar space they are affine spaces (cf. [4]), while in our complement they are semiaffine or projective spaces. Adopting the terminology of [7], where the class of semiaffine spaces includes affine spaces, projective spaces and everything in between, we could say that singular subspaces of our complement are simply E-mail addresses: kryzpet@math.uwb.edu.pl (Krzysztof Petelczyc), mariusz@math.uwb.edu.pl (Mariusz Zynel) ©® This work is licensed under https://creativecommons.org/licenses/by/4.0/ 448 Ars Math. Contemp. 17(2019)447-454 semiaffine spaces. This let us call our complement a semiaffine polar space. Anyway, it is clear that the complement we examine is affine in spirit. A natural parallelism is there and the subspace we remove can be viewed as the horizon. As this paper is closely related to [4] and [9], it borrows some concepts, notations and reasonings from these two works. There are however new difficulties in this case. In contrast to [4], the horizon is not a hyperplane and thus, it induces a partial parallelism (cf. [8]). There are lines disjoint with the horizon in the ambient space and those lines, called non-affine, cannot be parallel to any line in the complement. If we had applied the definition of parallelism from [4] as it is, we would end up with non-affine lines in its equivalence classes. Therefore we use the Veblen condition to express parallelism in terms of incidence in the complement. This method unfortunately is viable only if we have at least 4 points per line in the polar space. Roughly speaking, the points of the horizon are identified with equivalence classes of parallelism, or, in other words, with directions of lines. On the horizon of an affine polar space a deep point emerges as the point which could be reached by no line of the complement. If the removed subspace is not a hyperplane then there is no deep point but a new problem arises. Some lines on the horizon are recoverable in a standard way, as directions of planes. For the others there are no planes in the complement that would reach them. An analogy to a deep point is clear, so we call them deep lines. To overcome the problem we introduce the following relation: a line K is anti-euclidean to a line L iff there is no line intersecting K that is parallel to L. Based on this relation is a ternary collinearity of points on deep lines. We do not know whether every subspace of a polar space is contained in a hyperplane. Any subspace can be extended to a maximal one, but does it have to be a hyperplane? If that is the case our assumptions could be weakened significantly. 2 Generalities A point-line structure M = (S, L}, where the elements of S are called points, the elements of L are called lines, and where L c 2S, is said to be a partial linear space, or a point-line space, if two distinct lines share at most one point and every line is of size (cardinality) at least 2 (cf. [3]). A line of size 3 or more will be called thick. If all lines in M are thick then M is thick. M is said to be nondegenerate if no point is collinear with all others, and it is called singular if any two of its points are collinear. It is called Veblenian iff for any two distinct lines Li, L2 through a point p and any two distinct lines Ki, K2 not through the point p whenever each of L1, L2 intersects both of K1, K2, then K1 intersects K2. A subspace of M is a subset X C S that contains every line, which meets X in at least two points. A proper subspace of M that shares a point with every line is said to be a hyperplane. If M satisfies exchange axiom, then a plane of M is a singular subspace of dimension 2. A partial linear space satisfying one-or-all axiom, that is for every line L and a point a G L, a is collinear with one or all points on L, will be called a polar space. The rank of a polar space is the maximal number n for which there is a chain of singular subspaces 0 = X1 c X2 c ■ ■ ■ C Xn (n = -1 if this chain is reduced to the empty set). For a G S by a^ we denote the set of all points collinear with a, and for X C S we put X^ = Q{a^ : a G X}, radX = X n X K. Petelczyc and M. Zynel: The complement of a subspace in a classical polar space 449 As an immediate consequence of one-or-all axiom we get (cf. [4]): Fact 2.1. For any point a G S the set is a hyperplane of p. Following [10], a subset X of S is called • spiky when every point a G X is collinear with some point b G X, • flappy when for every line L C X there is a point a G X such that L C aL. 2.1 Complement Let M = (S, L} be a thick partial linear space and let W be a proper subspace of M. By the complement of W in M we mean the structure Dm(W) := (Sw, £w}, where Sw := S \W and Lw := {k n Sw : kG LA k £ W}. The subspace W will be called the horizon of (W). Note that the complement (W) is a partial linear space. Following a standard convention we call the points and lines of the complement DM(W) proper, and points and lines of W are said to be improper. Given a subspace X of DM(W) its closure X is a subspace of M with X C X. We say that two lines K, L G Lw are parallel, and we write K ||W L iff K n L nW = 0. (2.1) This is always an equivalence relation. Its domain is Lw only in case W is a hyperplane, or in other words, a line L G Lw with L = L cannot be parallel to any line. A line L G Lw with the property that L ||w L will be called an affine line. The set of all affine lines, the domain of ||w, will be denoted by L*. For affine line L we write for the point of L in W, i.e. the point at infinity. A point a G W is said to be a deep point if there is no line L G Lw such that a = LTO. A plane of DM(W) containing an affine line is said to be a semiaffine plane. By we denote the set of points at infinity of a semiaffine plane n, i.e. n~ = {M~ : M G L* and M C n}. Note that n~ is a line iff n is an affine plane. A line L C W is said to be a deep line if there is no plane in Dp(W) with L = nTO. 3 Complement in a polar space Let p = (S, L} be a thick, nondegenerate polar space of rank at least 3. In the remainder of the paper we deal with Dp(W), where W is a proper subspace contained in some hyperplane of p. Let us emphasize, that we do not mean one particular hyperplane and it is not fixed in our reasonings in any way. If there was a unique hyperplane H containing W we would be able to recover the ambient space applying Proposition 2.7 from [4], which says that every automorphism of the complement Dp(H) can be uniquely extended to an automorphism of p. It is not however doable as there could be many hyperplanes containing W and none of them can be distinguished in terms of the complement. In polar spaces deep points appear only on hyperplanes and there could be at most one deep point on a hyperplane. 450 Ars Math. Contemp. 17(2019)447-454 Lemma 3.1. (i) If W is a hyperplane, then there is at most one deep point on W and it is in rad W. (ii) If W is not a hyperplane, then there are no deep points on W, that is W is spiky. Proof. (i): By Corollary 1.3 (ii) in [4]. (ii): Assume that a is a deep point in W. Then aL C W, and by Fact 2.1 we get that W contains a hyperplane. A contradiction, as a hyperplane in P is a maximal proper subspace (cf. [4, Lemma 1.1]). □ Lemma 3.2. Let P be an embeddable polar space and K,L e Lw be two distinct lines such that K || w L. The subspace W can be extended to a hyperplane of P not containing K and L. Proof. If W is a hyperplane of P then W itself is the required hyperplane. Assume that W is not a hyperplane. Let H be a hyperplane containing W, N be a projective space embracing p, and f be an embedding of P into N. Consider the projective subspace G spanned by f (H). By Proposition 5.2 from [4] G is a hyperplane of N. If f (K), f (L) £ G then the hyperplane H = f-1(G n f (S)) is the required one. Let H be the family of all hyperplanes in G containing f (W). _Now, assume that f (K) C G and f (L) £ G. Take aK e f (K) \ f (W) and aL e f (L) \ f (W) and choose a hyperplane G0 e H with aK e G0. Note that aK, aL meets G in aK. Take b e aK ,aL distinct from aK and aL. Assume that there is a line through b that intersects f (L) \ f (W) in some point c and meets G0 in a point d. Note that d e f (K) as otherwise we would have aK e G0. Lines aL, d and f (K) are on a plane spanned by the triangle aL, b, c. Therefore the line aL, d intersects f (K) in some point e distinct from d. Then d,e C G, and thus aL e G, a contradiction. Hence, G' = (G0,b) is a hyperplane of N such that f (L) £ G'. We have also aK, b n G0 = 0 since aK, b £ G. Thus f (K) £ G'. Finally, H' := f-1(G' n f (S)) is the hyperplane we are looking for. The case with f (K) £ G and f (L) C G goes the same way. Now, let f (K) C G and f (L) C G. As in the previous case we take aK e f (K) \ f (W), aL e f (L) \ f (W), but this time choose a hyperplane G0 e H with aK, aL e G0. Let b e G. Note that 0~b n G0 = 0 for i = K, L. So, if we set G' = (G0, b) then f (K) £ G' and f (L) £ G'. Moreover, G' is a hyperplane of N. Again, H' := f-1(G' n f (S)) is the required hyperplane. □ Lemma 3.3. Let K,L e Lw be two distinct lines such that K ||w L. There is a sequence ni,..., nn of planes in Dp(W) such that K= e n for i = 1,... ,n and K C ni, L C nn, and n, nj+1 share a line for j = 1,... ,n — 1. Proof. By Lemma 3.2 we can extend W to a hyperplane H of P such that K, L £ H. Take the point a = KBy (2.1) we have a = LTO. Now, take in P the bundle of all the lines together with all the planes through a. This structure is actually the quotient space a±/a, and it is, up to an isomorphism, a polar space (cf. [1, Lemma 2.1]), that we denote by P'. The set H', consisting of all the lines through a contained in H, is a hyperplane in P' induced by H. Then Dp/ (H') is an affine polar space, that in itself is connected (cf. [4]). So there is in Dp/ (H') a sequence of intersecting lines joining K and L as points of Dp/ (H'). However, lines of Dp/ (H') are planes of Dp(H). As W C H these planes are also planes of Dp (W). □ K. Petelczyc and M. Zynel: The complement of a subspace in a classical polar space 451 3.1 Parallelism Let K2 € Lw . Then K ||* K2 iff Ki n K2 = 0 and there are two distinct lines Li, L2 € Lw crossing both of Ki; K2, such that (3.1) Li n L2 = 0 and Li n L2 n K = 0 for i = 1, 2. In case there are are exactly 3 points per line in our polar space p, no two lines Ki, K2 on an affine plane in Dp(W) such that Ki || w K2 satisfy the right hand side of (3.1), as the required lines Li, L2 had to be of size 4. This is why from now on we assume that there are at least 4 points on every line of p. Let || be the transitive closure of ||*. It is clearly seen that || C L* x L*. Lemma 3.4. The relation || is reflexive on L*. Proof. Given a line Ki € Lw , considering that the rank of p is at least 3, take a plane n containing Ki in a maximal singular subspace through Ki. There are lines K2, Li, L2 on n such that Ki n K2 = 0 (that is K° = K2°°), Li = L2, Li n L2 = 0, and K, n Lj = 0 for i, j = 1, 2. Thus Ki ||* K2 by (3.1). This means that Ki || K2 and K2 || Ki, which by transitivity implies that Ki || Ki. □ Proposition 3.5. Let W be a subspace of p. The relation ||w defined in (2.1) and the relation || coincide on the set of lines of Dp(W). Proof. Let Ki, K2 € Lw. If Ki = K2, then Ki ||w K2 and Ki || K2. So, assume that Ki = K2. _ _ Consider the case where Ki ||W K2. By (2.1) it means that K n L n W = 0, and consequently K° = K°° = a for some a € W. This implies that Ki n K2 = 0. Assume that Ki and K2 are coplanar, and n is the plane of Dp(W) containing both of Ki, K2. The plane n is, up to an isomorphism, a projective plane, so it is Veblenian. Thus, by (3.1), Ki || * K2 .If Ki and K2 are not coplanar, then by Lemma 3.3 there is a sequence of planes ni,..., nn such that Ki C ni, K2 C nn, a € n, for i = 1,..., n, and nj, nj+i share a line for j = 1,..., n - 1. Let nj n nj+i = Mj. Note that a € Mi,..., Mn-i and Mj, Mj+i are coplanar. Therefore Mj ||* Mj+i. Moreover, Ki ||* Mi and M„_i ||* K2 by the same reasons. So finally we get Ki || K2. Now, assume that Ki ||* K2. Then Ki, K2 are disjoint and coplanar. Thus Ki, K2 meet in the closure of some plane, this means that they meet in W. By (2.1) it gives Ki ||w K2. If Ki || K2 then there is a sequence of proper lines Li,..., Ln such that Ki ||* Li ||* ••• ||* Ln ||* K2. So, from the previous reasoning we get Ki Li • • • || w Ln || w K2. As the relation ||w is transitive we have Ki || w K2. □ As an immediate consequence of Proposition 3.5 we get Corollary 3.6. Affine lines can be distinguished in the set Lw as those parallel to themselves. 452 Ars Math. Contemp. 17(2019)447-454 3.2 Recovering If W is a hyperplane it follows by [4, Proposition 2.7] that: Proposition 3.7. Let p be a thick nondegenerate polar space of rank at least 2 and let H be its hyperplane. The polar space p can be recovered in the complement Dp(H). So, from now on we additionally assume that W is not a hyperplane. By Proposition 3.5 the relation ||w can be expressed purely in terms of Dp(W). Recall that our parallelism is partial: it is defined only on affine lines. However it is not a problem in view of Corollary 3.6. From Lemma 3.1(ii) there is a bijection between the sets W = {LTO : L e C*} and {[L]|| : L e C*}. Thus we can recover W pointwise in a standard way: points of the horizon W are identified with equivalence classes of parallelism i.e. directions of affine lines of the complement Dp(W). Let us introduce a relation ~CC* x C* defined by the following condition: K1 - K2 iff for all M e C* if M n K = 0 then M ft K2. (3.2) In the sense of Euclid's Fifth Postulate it could be read as anti-euclidean parallelism. A lot more useful for us is its derivative = C C*/| x C*/| defined as follows: [Ki]y = [K2]| iff for all M e [Ki]| ,N e [K2]|: M - N and N - M. (3.3) Lemma 3.8. Let M, N be two nonparallel affine lines. The following conditions are equivalent: (i) [M]| = [N]|, (ii) there is a deep line L C W, such that MNe L. Proof. (i) ^ (ii): From one-or-all axiom, Mmust be collinear with at least one point of the line N. Moreover, Mcannot be collinear with a proper point of N, as [M] | = [N] |. Thus Mis collinear with the unique improper point of N, which is N Let L be the line through MNAssume, that n is a semiaffine plane with L = n~. Then, there are some affine lines M1, N1 C n with M~ = Mf° and N~ = Nf°. So, either M1 || N1 or M1 and N1 share a proper point. In view of (3.3), in both cases we get [M]| = [N]|. (ii) ^ (i): Assume that [M]| = [N]|. Due to (3.2) and (3.3) there is a proper point a e M and an affine line K such that a e K || N (or the symmetrical case holds). This means that a and Nare collinear in p. The one-or-all axiom implies, that either there are no other points on M that are collinear with Nor Nis collinear with all points on M. In the first case Nis not collinear with Min the latter (NM} ^ W is the plane containing the line MTO,N. □ One can note, that the relation = defined by (3.3) and the relation = introduced in [4] coincide, though their definitions are expressed differently. Besides, our relation is not transitive, but the reflexive closure of its analogue in [4] is an equivalence relation. This benefit is the result of some hyperplane properties (see Lemma 3.1(i)). Nevertheless, we can overcome this inconvenience and define ternary relation of collinearity on the horizon W. K. Petelczyc and M. Zynel: The complement of a subspace in a classical polar space 453 Lemma 3.9. If Ki, K2, K3 are pairwise nonparallel affine lines such that [Kj]|| = [K(i+i) mod 3]|| for i = 1, 2, 3, then points K°, K°°, K°° are on a line. Proof. Let a = K°, b = K°°, c = K°. By Lemma 3.8 there are improper lines L = a, b, M = b, c, N = cTa. Let H be a hyperplane containing W. If in Dp(H) there is a plane, which closure contains one of the lines L, M or N, then we also have such a plane in Dp(W), that contradicts Lemma 3.8. Thus, L, M, N C H are deep lines in relation to Dp(H). By Lemma 2.3 of [4] this means that each of L, M and N contains a point of rad H. Let d G rad H. Then, by Corollary 1.3 of [4], H = d^, {d} = rad H, and d is the unique deep point of H. As we have d G L, M, N, it must be L = M = N. □ Lemma 3.10. Let Ki, K2, K3 be pairwise nonparallel affine lines. Points K°, K°°, K°° are on a line iff one of the following holds: (i) there are affine lines Mi || Ki, M2 || K2, M3 || K3 such that M^ M2, M3form a triangle in Dp(W), (ii) [Ki]|| = [K2]|, [K2]| = [K3]|, and [K3]|| = [Ki]|. Proof. Assume that K~, K^, Kg° are on a line L. If (i) does not hold, then there is no plane n in Dp(W) with L = nTO. This means that L is a deep line and by Lemma 3.8 we get (ii). Now, assume that (i) is the case. Take a plane n spanned by the triangle M^ M2, M3. Then K b K2, K3 C n and Kf, K2°°, K^ are on a line n~. If (ii) is fulfilled then Kf, K2°, K°° are on a line directly by Lemma 3.9. □ The meaning of Lemma 3.10 is that we are able to recover improper lines regardless of whether W is flappy or not. Let [[K]|h [L]H]_ := {[M: [M= [K, [L] J. Then new lines can be grouped into two sets: L := | [[K]||, [L]|:[K]| = [L]| and K tf l}, C" := {n° : n is a semiaffine plane of Dp(W)}. All our efforts in this paper essentially amount to the following isomorphism P = (sw uc7||, cw u C uc", |). A new point [K] | is incident to a line L G cw iff K || L. It is incident to a line L G C iff there is M g cw such that [[K] |, [M] ^ = = L. Eventually, it is incident to a line L G C" iff K C n and L = n°. " Theorem 3.11. Let P be a nondegenerate, embeddable polar space of rank at least 3, with at least 4 points per line, and W be its subspace, that is contained in a hyperplane. The polar space P can be recovered in the complement Dp(W). References [1] J. Bamberg, J. De Beule and F. Ihringer, New non-existence proofs for ovoids of Her-mitian polar spaces and hyperbolic quadrics, Ann. Comb. 21 (2017), 25-42, doi:10.1007/ s00026-017-0346-0. 454 Ars Math. Contemp. 17(2019)447-454 [2] P. J. Cameron, Projective and Polar Spaces, volume 13 of QMWMaths Notes, Queen Mary and Westfield College, School of Mathematical Sciences, London, 1992, http://www.maths. qmul.ac.uk/~pjc/pps/. [3] A. M. Cohen, Point-line spaces related to buildings, in: F. Buekenhout (ed.), Handbook of Incidence Geometry: Buildings and Foundations, North-Holland, Amsterdam, pp. 647-737, 1995, doi:10.1016/b978-044488355-1/50014-1. [4] A. M. Cohen and E. E. Shult, Affine polar spaces, Geom. Dedicata 35 (1990), 43-76, doi: 10.1007/bf00147339. [5] H. Karzel and H. Meissner, Geschlitze Inzidenzgruppen und normale Fastmoduln, Abh. Math. Sem. Univ. Hamburg 31 (1967), 69-88, doi:10.1007/bf02992387. [6] H. Karzel and I. Pieper, Bericht uber geschlitzte Inzidenzgruppen, Jber. Deutsch. Math. Verein. 72 (1970), 70-114, http://eudml.org/doc/14 658 8. [7] A. Kreuzer, Semiaffine spaces, J. Comb. Theory Ser. A 64 (1993), 63-78, doi:10.1016/ 0097-3165(93)90088-p. [8] M. Marchi and S. Pianta, Partial parallelism spaces and slit spaces, in: A. Barlotti, P. V. Cec-cherini and G. Tallini (eds.), Combinatorics '81, North-Holland, Amsterdam-New York, volume 18 of Annals of Discrete Mathematics, 1983 pp. 591-600, doi:10.1016/s0304-0208(08) 73337-1, proceedings of the International Conference on Combinatorial Geometries and their Applications held in Rome, June 7-12, 1981. [9] K. Petelczyc and M. Zynel, The complement of a point subset in a projective space and a Grassmann space, J. Appl. Logic 13 (2015), 169-187, doi:10.1016/j.jal.2015.02.002. [10] K. Petelczyc and M. Zynel, Affinization of Segre products of partial linear spaces, Bull. Ira-nianMath. Soc. 43 (2017), 1101-1126, http://bims.iranjournals.ir/article_ 1010.html. /^creative ^commor ARS MATHEMATICA CONTEMPORANEA ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 17 (2019) 455-466 https://doi.org/10.26493/1855-3974.1925.57a (Also available at http://amc-journal.eu) Unicyclic graphs with the maximal value of Graovac-Pisanski index* Martin Knor Slovak University of Technology in Bratislava, Faculty of Civil Engineering, Department of Mathematics, Bratislava, Slovakia Jozef Komornik Faculty ofManagement, Comenius University, Odbojárov 10, P.O. Box 95, Bratislava, Slovakia Riste Skrekovski Faculty of Information Studies, Novo mesto and FMF, University of Ljubljana, Slovenia Aleksandra Tepeh Faculty of Information Studies, Novo mesto and Faculty of Electrical Engineering and Computer Science, University ofMaribor, Slovenia Received 31 January 2019, accepted 13 May 2019, published online 8 November 2019 Abstract Let G be a graph and let r be its group of automorphisms. Graovac-Pisanski index of G is GP(G) = ^(ffl ^ueV(g) Saer d(u, a(u)), where d(u, v) is the distance from u to v in G. One can observe that GP(G) = 0 if G has no nontrivial automorphisms, but it is not known which graphs attain the maximum value of Graovac-Pisanski index. In this paper we show that among unicyclic graphs on n vertices the n-cycle attains the maximum value of Graovac-Pisanski index. Keywords: Graovac-Pisanski index, modified Wiener index, unicyclic graphs. Math. Subj. Class.: 05C25 *The first author acknowledges partial support by Slovak research grants VEGA 1/0142/17, VEGA 1/0238/19, APVV-15-0220 and APVV-17-0428. The research was partially supported also by Slovenian research agency ARRS, program no. P1-00383, projects no. L1-4291 and J1-1692. E-mail addresses: knor@math.sk (Martin Knor), jozef.komornik@fm.uniba.sk (Jozef Komornik), skrekovski@gmail.com (Riste Skrekovski), aleksandra.tepeh@gmail.com (Aleksandra Tepeh) ©® This work is licensed under https://creativecommons.org/licenses/by/4.0/ 456 Ars Math. Contemp. 17(2019)447-454 1 Introduction Wiener index, the sum of distances in a graph, is an important molecular descriptor. It was introduced by Wiener in 1949, see [18], and since then many other molecular descriptor have appeared. One of them is the Graovac-Pisanski index [8], originally known as the modified Wiener index. With this index an algebraic approach for generalizing the Wiener index was presented. Namely, as the Wiener index also the Graovac-Pisanski index is based on distances but its advantage is in considering also the symmetries of a graph, and it is known that symmetries of a molecule have an influence on its properties [14]. In his pioneering paper, Wiener showed a correlation of the Wiener index of alkanes with their boiling points [18]. It turns out that the Graovac-Pisanski index combines the symmetry and topology of molecules to obtain a good correlation with some physico-chemical properties of molecules. Recently, Crepnjak et al. showed that the Graovac-Pisanski index of some hydrocarbon molecules is correlated with their melting points [6]. This index also drew attention from theoretical point of view. Researchers are interested in the difference between the Wiener and Graovac-Pisanski index. This difference was computed in [9] for some families of polyhedral graphs. The Graovac-Pisanski index of nanostructures was studied in [1, 2, 15, 16, 17] and for some classes of fullerenes and fullerene-like molecules in [3, 11, 12]. In [13] the symmetry groups and Graovac-Pisanski index of some linear polymers were computed. Upper and lower bounds for Graovac-Pisanski index were considered in [11]. In [7] and [16] Graovac-Pisanski index was further considered from computational point of view. Exact formulae for the Graovac-Pisanski index for some graph operations are presented in [4]. Recently it was proved that for any connected bipartite graph, as well as for any connected graph on even number of vertices, the Graovac-Pisanski index is an integer number [5]. Let G be a connected graph. The Graovac-Pisanski index of G is defined as where Aut(G) is the group of automorphisms of G and distG(u, v) denotes the distance from u to v in G. However, in the paper we will use a result from [5] to compute this index. To explain the method we need some additional definitions. Let G be a graph, u g V(G) and S C V(G). The distance of u in S, wS(u), is defined as The group of automorphisms of G partitions V(G) into orbits. We say that u, v g V(G) belong to the same orbit if there is an automorphism a g Aut(G) such that a(u) = v. Let V1, V2,..., Vt be all the orbits of Aut(G) in G. Moreover, for every i, 1 < i < t, let vi g Vj. That is, vj's are the representatives of Vj's. It was shown in [5] that By (1.1), if a graph has no nontrivial automorphisms, that is if all its orbits consist of single vertices, then its Graovac-Pisanski index is 0. Hence, all graphs with no nontrivial automorphisms achieve the minimum value of Graovac-Pisanski index. More interesting is the opposite problem. (1.1) i=1 M. Knor et al.: Unicyclic graphs with the maximal value of Graovac-Pisanski index 457 Problem 1.1. Find all graphs on n vertices with the maximum value of Graovac-Pisanski index. This problem was solved for trees in [10]. By a long H on n vertices we denote a tree obtained from a path on n - 4 vertices by attaching two pendent vertices to each endvertex of the path. Theorem 1.2. Let T be a tree on n > 8 vertices with the maximum value of Graovac-Pisanski index. Then T is either a path or a long H. Moreover, GP(T)= {if n isodd In" if n is even. For n < 7 there are also three other trees with the maximum value of Graovac-Pisanski index. However, they have the value of Graovac-Pisanski index as stated in Theorem 1.2. In this paper we prove the following statement. Theorem 1.3. Let G be a unicyclic graph on n vertices with the maximum value of Graovac-Pisanski index. Then G is the n-cycle and GP(C„) = ("r ifn isodd I IT lf n is even. Observe that Graovac-Pisanski index for extremal trees and for extremal unicyclic graphs has the same value. We believe the following holds. Conjecture 1.4. Let G be a graph on n vertices, n > 8, with the maximum value of Graovac-Pisanski index. Then G is either a path, or a long H, or a cycle. To support this conjecture we performed some computer experiments. They showed the validity of the conjecture for n = 8 and n = 9. We believe that the maximal degree of extremal graphs is small (at most 3), thus for the cases n = 10 and n =11 we limited our computer search to maximal degrees 5 and 4, respectively, and in these cases the conjecture was confirmed as well. The graphs from the conjecture are extremal also for n G {5, 6,7}, however when n equals 7 there exists an additional extremal graph. 2 Proof In this section we prove several claims which imply Theorem 1.3. Obviously, if we consider graphs of order n, we do not need to consider the multiplicative term " in (1.1). Therefore we define t GPa(G) = E (vi), (2.1) i=i where Vi, V2,..., Vt are all the orbits of Aut(G) in G and vi, v2,..., vt are their representatives, respectively. Then for given n, graphs on n vertices with the maximum value of GPa are the solutions of Problem 1.1. For a cycle on n vertices, GPa(Cn) = wV(v) where v is an arbitrary vertex of Cn and V = V (Cn). This implies the following statement. 458 Ars Math. Contemp. 17 (2019) 349-368 Proposition 2.1. We have GPa(C„) = TP" f n -odd, I — if n is even. In what follows we generalize the GPa-parameter. Let Z = {Z",..., Ztz } be a partition of V(G) and let z G Z,, 1 < i < tZ. Then tz GPaZ (G) = ^ wZi (zi). i=l In our proofs, sets Z, will usually be unions of orbits of Aut(G). Nevertheless, GPaZ(G) will depend on the choice of the representatives z,. To prove Theorem 1.3 we start with GPa(G), where G is an extremal unicyclic graph on n vertices different from the n-cycle. Then in a sequence of steps we modify either the graph or the partition and in each step we obtain a larger value of GPaZ. Since we terminate this process with Cn and GPa, we get the result. Hence, let G be a unicyclic graph on n vertices with the maximum value of Graovac-Pisanski index and such that G is not the n-cycle. Then G consists of a single cycle C and trees rooted at the vertices of the cycle. In what follows, orbits of vertices of C will be important. We start with modifying the partition by merging together some orbits of vertices which have the same distance from C. We denote by X the new partition of V(G), while the original partition into orbits is denoted by V. Let v be a vertex of C. If {v} is a trivial orbit of Aut(G), then orbits in the v-rooted tree form sets of the partition X. But if {v} is not a trivial orbit of Aut(G), we do the following. Let Ov be the orbit of Aut(G) containig v and let Ov (G) be the set of vertices of u-rooted trees where u G Ov. We partition the vertices of Ov(G) according to their distance from C. Hence, Ov alone is one set of X, another set of X contains those vertices of Ov (G) which are adjacent to a vertex of C, etc. We have the following statement. Lemma 2.2. For arbitrary choice of the representatives of sets in X we have GPa(G) < GPaX(G). Proof. Let X, be a set from X and let x, be an abitrary vertex of X,. Observe that X, is a union of several orbits of Aut(G). Let V0 be an orbit of Aut(G) such that V0 C X,. Then wVo (u) is the same for every u g V0. So let v0 be a vertex of V0 at the shortest distance from xj. Then both x, and v0 are in the same tree rooted at a vertex of C. Assume that they are in a v-rooted tree T. Let u be a vertex of V0. If u is not in T then distG(xj,u) = distG(v0,u) since distG(xj, v) = distG(v0, v). So let u be a vertex in T. Let z be a vertex on the (unique) (v0, u)-path at the shortest distance from v. Since v0 is a vertex of V0 at the shortest distance from x,, the shortest (x,, u)-path must contain z. Thus distG(v0, u) < distG(x,, u) and so wVo (v0) < wVo (x,). Consequently, GPa(G) < GPaX(G) as required. □ Now we modify the graph G, and we consider a partition Y of the vertex set of the modified graph inherited from the partition X of G. So let v be a vertex of C. If {v} is a M. Knor et al.: Unicyclic graphs with the maximal value of Graovac-Pisanski index 459 trivial orbit of Aut(G) then we do not change the v-rooted tree, and its orbits form sets of the partition Y. Hence, in this case the sets of Y coincide with the sets of X (and also with the orbits of V). But if {v} is not a trivial orbit of Aut(G) then we change the v-rooted tree. If the v-rooted tree has p vertices in G then we replace it by a path on p vertices rooted at the endvertex, which we again denote by v. Denote by F the graph which results when all these replacements are made. Since we did not change the cycle, we denote the cycle of F again by C. Let Ov be the orbit of Aut(G) containing v. By our assumption |Ov | > 2. Analogously as above, let Ov (F) be the set of vertices of u-rooted trees where u e Ov. Partition Ov (F) into p disjoint sets of Y according to their distance from C. Observe that for every Y e Y and for every two vertices y1, y? e Y we have wYi(y1) = wYi (y2). Hence when computing GPaY(F), we can choose the representatives yi in Y arbitrarily. However, orbits of F may be strictly larger than the sets Y. This is caused by the fact that two non-isomorphic rooted trees may have the same numbers of vertices. Our next statement follows. Lemma 2.3. For arbitrary choice of representatives of sets in Y we have GPaX (G) < GPaY (F). Proof. Let H be a graph. A ray in H is a subgraph of H which is isomorphic to a path, its first vertex has degree at least 3 in H, its last vertex has degree 1 in H and all the other vertices have degree 2 in H. We do not prove the inequality directly. Instead, we construct a sequence of graphs G = G0, G1,..., Gq = F, each Gi with a partition Xi, such that GPaX (Gi) < GPaX+ (Gi+1) for a special choice of representatives in Xi+1, where 0 < i < q—1, X0 = X and Xq = Y. We remark that for every i, Gi will be a unicyclic graph with the cycle C such that if O = {v1,..., vt} is an orbit of vertices of C in G, then all vj -rooted trees in Gi are mutually isomorphic, 1 < j < t. If t =1 then the v1-rooted trees in G, G1,..., F are mutually isomorphic and all X, X1,..., Y coincide on the vertex sets of these trees. However if t > 2, then the vertex set Ovi (Gi) of the vj -rooted trees, 1 < j < t, is partitioned in Xi according to the distance from C, and we assume that all the representatives of these sets are in the v1 -rooted tree. This assupmtion is possible since O is an orbit in G, and although O does not need to be an orbit of Gi, the vertices of O are nicely distributed along the cycle C in Gi. So consider i, 0 < i < q. We assume that Gi is already known and we construct Gi+1. For this, let O = {v1,..., vt} be an orbit of vertices of C in G, where t > 2. If the v1-rooted tree (and so also vj -rooted trees for 2 < j < t) is a path rooted at the endvertex, then we are done with this orbit of G. So suppose that the v1-rooted tree has at least two endvertices different from v1, and consequently, at least two distinct rays starting at a common vertex. Let R1 and R2 be two rays starting at a vertex c such that distG (v1, c) is maximum possible. We assume that R1 is not shorter than R2. If there is a representative xj of Xj which is in R2, then replace it by a vertex of Xj in R1. Observe that this replacement does not change GPaX (Gi). Now delete R2 from the v1-rooted tree and attach it to the second vertex of R1. Moreover, repeat the same procedure in all the other vj -rooted trees, 2 < j < t, and denote by Gi+1 the resulting graph. Denote by Ti and Ti+1 the v1-rooted 460 Ars Math. Contemp. 17 (2019) 349-368 tree in G® and G®+1, respectively. If R and R2 have the same length, then this operation may create a new set in X®+1, because T® may have smaller depth than T®+1. (As is the custom, by depth we denote the largest distance from the root.) In such a case choose a representative of this new set in R2. This is the unique case when a representative will be in R2 in T®+1. Let d = distGi (v1, c) and let £ be the length of R2. Assume that the indices of sets in X® and X®+1 are chosen so that XJ+1 in G®+1 was obtained from XJ in G® and the representatives of X® and XJ+1 coincide whenever possible. Then wXi (xj) in G® may j j j j differ from wXi+1 (xj+1) in G®+1 only if XJ+1 contains vertices of vk-rooted trees, 1 < j j j k < t, which are at distance d +1, d + 2,..., d +£ +1 from C. We distinguish three cases. Case 1: XJ+1 contains vertices at distance d +1 from C. Then in the v1-rooted tree, XJ+1 is smaller than XJ by exactly one vertex. Consequently |XJ | — |XJ+11 = t. Comparing to yi yi+i ' 1 - GPaX (G®), GPaX (Gi+1) is decreased by 2 due to a missing vertex in Ti+1 and it is decreased by (t — 1)2(d +1) + c due to missing vertices in vk-rooted trees 2 < k < t. Here c represents the distances using the edges of C, that is c = J2k=2 distG(v1, vk). Hence, wXi+i (xj+1) — Wxi (xj) = —2 — (t — 1)2(d + 1) — c. (2.2) Xj j Case 2: XJ+1 contains vertices at distance d + a from C, where 2 < a < £ Then |XJ+1| = |XJ| and comparing to GPaXi(G®), GPaXi+1 (Gi+1), is decreased by 2 due to a shorter distance to a vertex of XJ+1 in R2. Hence, Case 3: X®+1 contains vertices at distance d + I + 1 from C. Then in the v1-rooted tree, -i+i (xf1) — wxi (x® ) = —2. (2.3) j 1 contains vertices XJ+1 is larger than XJ by exactly one vertex. Consequently |XJ+11 — |XJ | = t. Comparing to GPaXi (G®), GPaXi+1 (G®+1) is increased by (t — 1)2(d +1 +1) + c due to new vertices in vk -rooted trees, 2 < k < t. Here c is the very same constant as in Case 1, that is c = J2k=2 distG(v1, vk). In some cases, namely if XJ is not empty, GPaX is increased by at least 2 due to a new vertex in T®+1, but we do not need to consider this contribution in our calculations. Hence, wxi+i (x!+1) — wxi (x®) > (t — 1)2(d + £ +1) + c. (2.4) Xj j Since wxi+i (x®+1) = wXi (x®) when X^ % OVi(G®), summing the expressions (2.2), (2.3) and (2.4j) we get j GPaXi+i (G®+1) — GPaXi(G®) > (—2 — (t — 1)2(d + 1) — c) — (£ — 1)2 + ((t — 1)2(d + £ +1)+ c) = (t — 2)2£ > 0 since t > 2. □ Let Y® € Y. Observe that if Y® n V(C) = 0, then Y® C V(C). Let Y' be those sets of Y which contain vertices of V(C). We define a new partition Z of F as follows: Z = Y\Y' u V(C). M. Knor et al.: Unicyclic graphs with the maximal value of Graovac-Pisanski index 461 That is, we merge together all sets Y containing vertices of V(C). All the other sets of Z coincide with the sets of Y. We have the following statement. Lemma 2.4. For arbitrary choice of representatives of sets in Z we have GPaY (F) < GPaZ (F). Proof. Observe that there are three types of sets in Z. First, if vi G V(C) is a trivial orbit in G, then orbits of vertices of the vi-rooted tree are sets of Z. Second, if jvi,..., vt} C V(C) is a non-trivial orbit in G, that is if t > 2, then the vj-rooted trees are paths with endvertices vj, and sets of vertices of OVl (F) in Z contain vertices of these t rooted trees which are at the same distance from C. Finally, Z contains V(C). If Zj is a set of Z of the first type or of the second type and u, v G Zj, then wZi (u) = wZi (v). Hence, to prove the statement it suffices to show that Y wY(yj) < wv(C)(z) = wY(z) where z is an arbitrary vertex of V(C) and Y' is defined before Lemma 2.4. (Recall that yj is a representative of Y in Y.) Thus, let z G V(C) and let Y G Y'. In what follows we show that wYi (yj) < wYi (z). We distinguish four cases. Case 1: |Yj| = 1. Since wYi(yj) = 0 < distF(z, yj) = wYi(z), we have wYi(yj) < WYi (z). Case 2: |Yj| = 2. let Yj = {yj, y}. Then by triangle inequality wy;(yj) = distF(yj, y) < distF(z,yj) + distF(z,y) = w7i (z). Hence, in the sequel we assume that | Y | > 3. Since Yj is an orbit of vertices of C in G, there is a nontrivial rotational automorphism a in Aut(G) such that {ak(yj) | k G N} C Yj. Let r be the biggest order of a rotational automorphism of this type and let a be the corresponding automorphism. Observe that r > 2. Since wYi (u) = wYi (v) for u, v G Yj, we assume that yj is chosen so that distF (z, yj) is smallest possible. Case 3: r is even. Let Y/ = {ak(yj) | 0 < k < r}. We rename vertices of Y/ as {y0, y1,..., yr-1} so that distF(yj,yk) < distF(yj,yfc+1) whenever 0 < k < r — 1. Observe that y0 = yj, the vertices y2£-i and y2£ have the same distance from yj if 1 < I < r/2 and yr-i is the unique vertex of Y/ with the largest distance from yj. Since yj is the vertex of Yi with the smallest distance from z, we have distF (y2£-i, yj) + distF (yj, y2£) = distF (y2£-i, z) + distF (z, y2£) for 1 < I < r/2 and also distF (yj, yr-i) = distF (yj, z) + distF (z, yr-i) = 2 |V (C )|. Hence, wY/(yj) = wY/(z). If Yj = Y/, we are done. Therefore, in the sequel assume that there is also a reflexion p such that P(Y') C Y andP(Y')n Yj = 0. Then Y = Y'uP(Y/) and |Y| = 2r. Observe that all the vertices of P(Y') are obtained from arbitrary one of 462 Ars Math. Contemp. 17 (2019) 349-368 them using a. Thus, let yf be a vertex of ft(Y/) with the smallest distance from yi. Then using the same arguments as above we get Wf(Y!)(Vf) = wf(Yf)(Vi). Since Wf(Y!) (yf) = wY! (yi), we get WYi (yi) = 2wyi (yi). Analogously, if yfz is a vertex of ft(Y/) at the smallest distance from z, we get wf(Y')(yfz) = wf(Yi)(z). Since Wf(Yf) (yfz) = wY!(yi), we obtain wy,(yi) = »y,(z). Case 4: r is odd. Let Y/ = {ak(yi) | 0 < k < r}. Then proceeding analogously as in Case 3 one gets wY! (z) = wy; (yi) +distF (yi ,z), and so wYi (yi) < wYi (z) if Yi = Y/. Hence, assume that there is a reflexion ft such that ft(Y/) C Yi and ft(Y/) n Y/ = 0. Again, Yi = Y/ U ft(Y/) and |Yi| = 2r, and all the vertices of ft(Y/) are obtained from arbitrary one of them using a. Thus, let yf be a vertex of ft (Y/) with the shortest distance from yi. Then analogously as in Case 3 one gets WY'(yi) = wf(Y')(yf) = wf{YD(yi) - distF(yf,yi) and so »Yi (yi) = 2wy,' (yi) + distF (yf ,yi). Also, let yfz be a vertex of ft(Y/) with the shortest distance from z. Then analogously as above we get WY' (yi) = Wf(Y!)(yfZ ) = Wf(Y')(z) - distF (yfz, z) and so WY, (z) = 2»y' (yi) + distF(yfz, z) + distF(yi,z). Since distF(yf ,yi) < distF(yfz,yi) < distF(yfz,z) + distF(yi,z), we get »y(yi) < »y, (z). □ Finally, we are in a position to prove the last lemma which implies Theorem 1.3. Observe that there is a strict inequality in Lemma 2.5. Lemma 2.5. We have GPaZ(F) < GPa(Cn). Proof. Analogously as in Lemma 2.3, we prove the statement by a sequence of steps. Let O1,..., Oq be all orbits of vertices of C in G, such that for every v G Oi the v -rooted tree is nontrivial (i.e., it has more than one vertex). Observe that if the v-rooted tree is nontrivial in G, then the v-rooted tree in F is also nontrivial. Assume that IO11 > |O2 | > • • • > |Oq|. M. Knor et al.: Unicyclic graphs with the maximal value of Graovac-Pisanski index 463 We consecutively create unicyclic graphs F = F0, F1,..., Fq = Cn with partitions Z = Z0, Z1,..., Zq, respectively, and for every i, 0 < i < q, we show that GPaZ (F®) < GPaZ (F®+1). The graph F®+1 is obtained from F® by moving the vertices of v -rooted trees, where v G O®+1, into the unique cycle C® of F®. Now we describe the process in detail. Choose i, 0 < i < q. For v G O®+1, let p + 1 be the number of vertices of v-rooted tree in F® (or in G, since the numbers of vertices of v-rooted trees in G and F are the same). By our assumption p > 1. Orient the cycle C® of F® and for every vertex u G V(C®) let uf be the vertex following u on C®. Let v G O®+1. Delete the p non-root vertices of the v-rooted tree from F® and subdivide the edge vvf exactly p times. Repeat this procedure for all vertices of O®+1 and denote by F®+1 the resulting unicyclic graph. The partition Z®+1 is exactly the same as Z®, the only exception is that instead of the set V(C®) and various sets partitioning Ov (F®) for v G O®+1 we have just the set V(C®+1) in Z®+1. We assume that if a set of Z® is identical with a set of Z®+1, then they have the same representatives. Let Z ' g Z ® and Z * g Z ®+1 such that Z ' = Z *. Then Z ' is a collection of vertices of Ou(F®), i.e., of u-rooted trees for u g Oj, where j > i + 1. Since the distances between these vertices cannot be shorter in F®+1 than in F® (they can be only larger due to the extension of C® to C®+1), we have (z') < (z*) where z' is a representative of Z' in F® and z* is a representative of Z* in F®+1. Hence, it suffices to check the contribution of V(C®+1) in GPaZ+1 (F®+1) and in GPaZ (F®) the contribution of V(C®) and of the sets of non-root vertices of v-rooted trees for v g O®+1. Let t = |O®+1|. Analogously as in the proof of Lemma 2.4 we distinguish four cases. In these cases, we set c = | V(C®) |. Moreover, by ¿a we denote the parity of a. That is ¿a = 1 if a is odd and = 0 if a is even. Case 1: t = 1. By Proposition 2.1, V(C®) contributes 4(c2 - ¿c) to GPaZ(F®) and V(C®+1) contributes 1 ((c + p)2 - ¿c+p) to GPaZ'+1 (F®+1). Let O®+1 = {v1}. Denote by T the v1 -rooted tree in F®. Since T is a tree on p + 1 vertices, the orbits of T contribute to GPaZ (F®) at most 1 ((p + 1)2 - ¿^+1) by Theorem 1.2. So 4(GPaZi+1 (F®+1) - GPaZ(F®)) > (c + p)2 - ¿c+p - c2 + ¿c - (p + 1)2 + ¿p+1 > (c + p)2 - c2 - (p +1)2 - 1 = 2p(c - 1) - 2 > 0 since c > 3 and p > 1. Case 2: t = 2. In this case the v-rooted trees are paths whenever v g O®+1. Since the contribution to GPaZ (F®) of j-th vertices of these paths (i.e., of vertices at distance j from the roots) is at most j + c/2 + j, the total contribution of non-root vertices of v-rooted trees, v G O®+1, is at most 2(p+1) + 1 cp. Since the contribution of V(C®) is 4(c2 - ¿c) and the contribution of V(C®+1) is 4 ((c + 2p)2 - ¿c+2p), where ¿c+2p = ¿c, we get 4(GPaZ'+1 (F®+1) - GPaZ(F®)) > (c + 2p)2 - ¿c - c2 + ¿c - 8(p+1) - 2cp 2 2 2 > (c + 2p)2 - c2 - 4p2 - 4p - 2cp = 2p(c - 2) > 0 since c > 3 and p > 1 . 464 Ars Math. Contemp. 17 (2019) 349-368 In the remaining cases we may assume that there is a nontrivial rotational automorphism a of F® such that when v G Oi+1 then also a(v) G Oi+1. Let r be the biggest order of such a rotational automorphism a, and moreover, let a be such that the distance s between v and a(v) is the smallest possible. Then c = r • s. Case 3: r is even. Then r > 2. Let v0 G Oi+1. Denote O' = {ak(v0) | 0 < k < r}. First assume that |Oi+1| = 2r. Hence there is also a reflexion P such that P(O') Ç Oi+1 and P(O') n O' = 0. Let v1 g Oi+1 such that the distance t between v0 and v1 is the smallest possible. Then t < s/2 and v1 G P(O'). Observe that now t > 1 and s > 2. Since c = rs is even, the contribution of V (C®) to GPaZ (F®) is 4r2s2. The contribution of V(Ci+1) to GPaZ + ( Fi+1 ) is 1 r2 ( s + 2p)2. Now we calculate the contribution of non-root vertices of v-rooted trees when v g Oi+1. These trees are paths and the contribution of j-th vertices is 2(s + 2j) + 2(2s + 2j ) + • • • + 2(( § - 1)s + 2j) + (§ s + 2j) + (t + 2j ) + (s + t + 2j) + ••• + (( 2 - 1)s+1 + 2j) +(s - t + 2j) + (2s - t + 2j) + • • • + ((2 - 1)s - t + 2j) + (2s - t + 2j) = 4(s + 2j) + 4(2s + 2j) + • • • + 4((f - 1)s + 2j) + 2(2s + 2j) + 2j = 2r2s + 2j(2r - 1). So the contribution of non-root vertices of v-rooted trees, v G Oi+1, is r2s + 2j(2r - 1)) = 2r2sp + (p2 + p)(2r - 1). j=1 Hence GPaZ (Fi+1) - GPaZ(F®) > 1 r2s2 + r2sp + r2p2 - 1 r2s2 - 2r2sp - 2rp2 - 2rp + p2 + p = p2(r-1)2 + p(r(2rs-2) + 1) > 0 since r > 2, s > 2 and p > 1. In the case when |Oi+1| = r, the contribution of V(Ci+1) is 4r2(s + p)2 and the contribution of j-th vertices in v-rooted trees, v G Oi+1, is 2(s + 2j) + 2(2s + 2j) + • • • + 2((2 - 1)s + 2j) + (2s + 2j) = 1 r2s + 2j(r - 1). So the contribution of non-root vertices of v-rooted trees, v G Oi+1, is r2s + 2j(r - 1)) = 1 r2sp + (p2 + p)(r - 1). j=1 Hence GPaZi+1 (Fi+1) - GPaZ(F®) > 4r2s2 + 1 r2sp + 4r2p2 - 4r2s2 12 2 i 2 i - 4 r sp - rp - rp + p + p ^ r , = p2(2 - 1)2 + p(r(4rs - 1) + 1) > 0 M. Knor et al.: Unicyclic graphs with the maximal value of Graovac-Pisanski index 465 since in this case r > 4, s > 1 and p > 1. Case 4: r is odd. Then r > 3. Let v0 G Oi+1 and O' = {ak(v0) | 0 < k < r}. First assume that |Oi+1| = 2r. Hence there is also a reflexion ft such that ft(O') Ç Oi+1 and ft(O') n O' = 0. Let v1 g Oi+1 such that the distance t between v0 and v1 is the smallest possible. Then t < s/2 and v1 G ft(O'). The contribution of V(C4) to GPaZ(F®) is 4(r2s2 - Jrs). The contribution of V(Ci+1) to GPaZ(Fi+1) is 4(r2(s + 2p)2 -¿r(s+2p)). Now we calculate the contribution of non-root vertices of v-rooted trees when v g Oi+1. These trees are paths and the contribution of j-th vertices is 2(s + 2j) + 2(2s + 2j) + • • • + 2( ^ s + 2j ) + (t + 2j) + (s + t + 2j) + • • • +2 (^s + t + 2j) + (s - t + 2j) + (2s - t + 2j) + • • • + (^s - t + 2j) = 4(s + 2j) + 4(2s + 2j) + • • • + 4( ^ s + 2j) + (t + 2j) = 1 (r2 - 1)s + 2j(2r - 1) + t. So the contribution of non-root vertices of v-rooted trees, v G Oi+1, is E (2(r2 - 1)s + 2j(2r - 1) +t) = 1 (r2 -3=1 < 1 r2sp +(p2 + p)(2r - 1) (r2 - 1)s + 2j(2r - 1) + t) = ±(r2 - 1)sp + (p2 + p)(2r - 1) + pt since -1 sp + pt < 0. Hence GPaZi+1 (Fi+1) - GPaZ (F®) > 1 r2s2 + r2sp + r2p2 - 1 Jrs - 4r2s2 + 1 J, - 2r2sp - 2rp2 - 2rp + p2 + p = p2(r - 1)2 + p(r(2rs - 2) + 1) > 0 since r > 3, s > 2 and p > 1. In the case when |Oi+1| = r, the contribution of V(Ci+1 ) to GPaZ'+1 (Fi+1) is 4 (r2 (s + p)2 - ¿r(s+p)) and the contribution of j-th vertices in v-rooted trees, v G Oi+1, is 2(s + 2j) + 2(2s + 2j) + • • • + 2(^s + 2j) = 4(r2 - 1)s + 2j(r - 1). So the contribution of non-root vertices of v-rooted trees, v G Oi+1, is p E (4(r2 - 1)s + 2j(r - 1)) = 1 (r2 - 1)sp + (p2 + p)(r - 1). j=1 Hence GPaZi+1 (Fi+1) - GPaZ(F®) > 4r2s2 + 2r2sp + 4r2p2 - 4 - 4r2s2 - 4 (r2 - 1)sp - rp2 - rp + p2 + p = p2(§ - 1)2 + p(r( 1 rs - 1) + 4s + 1) - 1 > 0 since in this case r > 3, s > 1 and p > 1. (Observe that the second bracket is at least | if r = 3 and s = 1.) □ 466 Ars Math. Contemp. 17(2019)447-454 References [1] A. R. Ashrafi and M. V. Diudea (eds.), Distance, Symmetry, and Topology in Carbon Nano-materials, volume 9 of Carbon Materials: Chemistry and Physics, Springer, Cham, 2016, doi:10.1007/978-3-319-31584-3. [2] A. R. Ashrafi, F. Koorepazan-Moftakhar and M. V. Diudea, Topological symmetry of nanos-tructures, Fuller. Nanotub. Car. N. 23 (2015), 989-1000, doi:10.1080/1536383x.2015.1057818. [3] A. R. Ashrafi, F. Koorepazan-Moftakhar, M. V. Diudea and O. Ori, Graovac-Pisanski index of fullerenes and fullerene-like molecules, Fuller. Nanotub. 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Chem. 2 (2004), 2692-2699, doi:10.1039/b407105k. [15] H. Shabani and A. R. Ashrafi, Symmetry-moderated Wiener index, MATCH Commun. Math. Comput. Chem. 76 (2016), 3-18, http://match.pmf.kg.ac.rs/electronic_ versions/Match7 6/n1/match7 6n1_3-18.pdf. [16] N. Tratnik, The Graovac-Pisanski index of zig-zag tubulenes and the generalized cut method, J.Math. Chem. 55 (2017), 1622-1637, doi:10.1007/s10910-017-0749-5. [17] N. Tratnik and P. Zigert Pletersek, The Graovac-Pisanski index of armchair tubulenes, J. Math. Chem. 56 (2018), 1103-1116, doi:10.1007/s10910-017-0846-5. [18] H. Wiener, Structural determination of paraffin boiling points, J. Am. Chem. Soc. 69 (1947), 17-20, doi:10.1021/ja01193a005. ars mathematica contemporanea ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 17 (2019) 467-479 https://doi.org/10.26493/1855-3974.1892.78f (Also available at http://amc-journal.eu) On separable abelian p-groups Grigory Ryabov * Novosibirsk State University, 1 Pirogova st., 630090, Novosibirsk, Russia, and Sobolev Institute of Mathematics, 4 Acad. Koptyug avenue, 630090, Novosibirsk, Russia Received 29 December 2018, accepted 3 April 2019, published online 13 November 2019 An S-ring (a Schur ring) is said to be separable with respect to a class of groups K if every algebraic isomorphism from the S-ring in question to an S-ring over a group from K is induced by a combinatorial isomorphism. A finite group is said to be separable with respect to K if every S-ring over this group is separable with respect to K. We provide a complete classification of abelian p-groups separable with respect to the class of abelian groups. Keywords: Isomorphisms, Schur rings, p-groups. Math. Subj. Class.: 05E30, 05C60, 20B35 1 Introduction Let G be a finite group. A subring of the group ring ZG is called an S-ring (a Schur ring) over G if it is determined in a natural way by a special partition of G (the exact definition is given in Section 2). The classes of the partition are called the basic sets of the S-ring. The concept of the S-ring goes back to Schur and Wielandt. They used S-rings to study a permutation group containing a regular subgroup [19, 20]. For more details on S-rings and their applications we refer the reader to [13]. Let A and A1 be S-rings over groups G and G' respectively. An algebraic isomorphism from A to A' is a ring isomorphism inducing a bijection between the basic sets of A and the basic sets of A'. Another type of an isomorphism of S-rings comes from graph theory. A combinatorial isomorphism from A to A' is defined to be an isomorphism of the corresponding Cayley schemes (see Subsection 2.2). Every combinatorial isomorphism induces the algebraic one. However, the converse statement is not true (the corresponding examples can be found in [6]). *The work is supported by the Russian Foundation for Basic Research (project 17-51-53007). E-mail address: gric2ryabov@gmail.com (Grigory Ryabov) Abstract ©® This work is licensed under https://creativecommons.Org/licenses/by/4.0/ 468 Ars Math. Contemp. 17(2019)447-454 Let K be a class of groups. Following [3], we say that an S-ring A is separable with respect to K if every algebraic isomorphism from A to an S-ring over a group from K is induced by a combinatorial one. We call a finite group separable with respect to K if every S-ring over G is separable with respect to K (see [18]). The importance of separable S-rings comes from the following observation. Suppose that an S-ring A is separable with respect to K. Then A is determined up to isomorphism in the class of S-rings over groups from K only by the tensor of its structure constants (with respect to the basis of A corresponding to the partition of the underlying group). Given a group G denote the class of groups isomorphic to G by KG. If G is separable with respect to KG then the isomorphism of two Cayley graphs over G can be verified efficiently by using the Weisfeiler-Leman algorithm [12]. In the sense of [10] this means that the Weisfeiler-Leman dimension of the class of Cayley graphs over G is at most 3. More information concerned with separability and the graph isomorphism problem is presented in [3, 17]. Denote the classes of cyclic and abelian groups by KC and KA respectively. The cyclic group of order n is denoted by Cn. In the present paper we are interested in abelian groups and especially in abelian p-groups which are separable with respect to KA. The problem of determining of all groups separable with respect to a given class K seems quite complicated even for K = KC. Examples of cyclic groups which are non-separable with respect to KC were found in [6]. In [5] it was proved that cyclic p-groups are separable with respect to KC. We prove that a similar statement is also true for KA. Theorem 1.1. For every prime p a cyclic p-group is separable with respect to KA. The result obtained in [18] implies that an abelian group of order 4p is separable with respect to KA for every prime p. From [9] it follows that for every group G of order at least 4 the group G x G is non-separable with respect to KGxG. One can check that a normal subgroup of a group separable with respect to KA is separable with respect to Ka (see also Lemma 2.5). The above discussion shows that a non-cyclic abelian p-group separable with respect to KA is isomorphic to Cp x Cpk or Cp x Cp x Cpk, wherep e {2,3} and k > 1. The separability of the groups from the first family was proved in [17]. In the present paper we study the question on the separability of the groups from the second family. Theorem 1.2. The group Cp x Cp x Cpk, where p e {2, 3} and k > 1, is separable with respect to KA if and only if k = 1. As an immediate consequence of Theorem 1.1, Theorem 1.2, and the above mentioned results, we obtain a complete classification of abelian p-groups separable with respect to Ka. Theorem 1.3. An abelian p-group is separable with respect to KA if and only if it is cyclic or isomorphic to one of the following groups: C2 x C2k , C3 x C^k , C23, C3 , where k 1. Throughout the paper we write for short "separable" instead of "separable with respect to Ka". The text is organized in the following way. Section 2 contains a background of G. Ryabov: On separable abelian p-groups 469 S-rings. Section 3 is devoted to S-rings over cyclic p-groups. We finish Section 3 with the proof of Theorem 1.1. In Section 4 we prove Theorem 1.2. The author would like to thank Prof. I. Ponomarenko for the fruitful discussions on the subject matters and Dr. Sven Reichard for the help with computer calculations. Notation. • The ring of rational integers is denoted by Z. • Let X C G. The element J2xex x of the group ring ZG is denoted by X. • The order of g G G is denoted by |g|. • The set {x-1 : x G X} is denoted by X-1. • The subgroup of G generated by X is denoted by (X}; we also set rad(X) = {g G G : gX = Xg = X}. • If m G Z then the set {xm : x G X} is denoted by X(m). • Given a set X C G the set {(g, xg) : x G X, g G G} of edges of the Cayley graph Cay(G, X) is denoted by R(X). • The group of all permutations of a set Q is denoted by Sym(Q). • The subgroup of Sym(G) induced by right multiplications of G is denoted by Gright. • For a set A C Sym(G) and a section S = U/L of G we set AS = {fS : f G A, Sf = S}, where Sf = S means that f permutes the L-cosets in U and fS denotes the bijection of S induced by f. • If a group K acts on a set Q then the set of all orbtis of K on Q is denoted by Orb(K, Q). • If H < G then the normalizer of H in G is denoted by NG(H). • If K < Sym(Q) and a G Q then the stabilizer of a in K is denoted by Ka. • The cyclic group of order n is denoted by Cn. 2 S-rings In this section we give a background of S-rings. The most of definitions and statements presented here are taken from [13, 17]. 2.1 Definitions and basic facts Let G be a finite group and ZG the group ring over the integers. The identity element of G is denoted by e. A subring A C ZG is called an S-ring over G if there exists a partition S = S(A) of G such that: (1) {e} G S, (2) if X G S then X-1 G S, (3) A = SpanZ{X : X G S}. The elements of S are called the basic sets of A and the number |S| is called the rank of A. Given X, Y, Z G S the number of distinct representations of z G Z in the form z = xy with x G X and y G Y is denoted by cX Y. If X and Y are basic sets of A then X Y = J2zes(A) cX yZ. So the integers cX Y are structure constants of A with respect 470 Ars Math. Contemp. 17(2019)447-454 to the basis {X : X G S}. It is easy to verify that given basic sets X and Y the set XY is also basic whenever |X| = 1 or |Y| = 1. A set X C G is said to be an A-set if X G A. A subgroup H < G is said to be an A-subgroup if H is an A-set. One can check that for every A-set X the groups (X} and rad(X) are A-subgroups. A section U/L is said to be an A-section if U and L are A-subgroups. If S = U/L is an A-section then the module AS = SpanZ {Xn : X G S(A), X C U} , where n: U ^ U/L is the canonical epimorphism, is an S-ring over S. If K < Aut(G) then the set Orb(K, G) forms a partition of G that defines an S-ring A over G. In this case A is called cyclotomic and denoted by Cyc(K, G). Let G be abelian. Then from Schur's result [19] it follows that X(m) G S(A) for every X G S(A) and every m coprime to |G|. We say that X, Y G S(A) are rationally conjugate if Y = X(m) for some m coprime to |G|. 2.2 Isomorphisms and schurity Throughout this and the next two subsections A and A' are S-rings over groups G and G' respectively. A bijection f: G ^ G' is called a (combinatorial) isomorphism from A over to A' if {R(X)f : X G S(A)} = {R(X') : X' G S(A')}, where R(X )f = {(gf, hf) : (g, h) G R(X)}. If there exists an isomorphism from A to A' we write A = A'. The group Iso(A) of all isomorphisms from A onto itself has a normal subgroup Aut(A) = {f G Iso(A) : R(X)f = R(X) for every X G S(A)}. This subgroup is called the automorphism group of A. Note that Aut(A) > Gright. If S is an A-section then Aut(A)S < Aut(AS). An S-ring A over G is said to be normal if Gright < Aut(A). One can check that NAut(A) (Gright)e = Aut(A) O Aut(G). (2.1) Now let K be a subgroup of Sym(G) containing Gright. As Schur proved in [19], the Z-submodule V(K, G) = SpanZ{X : X G Orb(Ke, G)}, is an S-ring over G. An S-ring A over G is called schurian if A = V(K, G) for some K such that Gright < K < Sym(G). Not every S-ring is schurian. The first example of a non-schurian S-ring was found by Wielandt in [20, Theorem 25.7]. It is easy to see that A is schurian if and only if S(A) = Orb(Aut(A)e, G). (2.2) Every cyclotomic S-ring is schurian. More precisely, if A = Cyc(K, G) for some K < Aut(G) then A = V(Gright x K, G). G. Ryabov: On separable abelian p-groups 471 2.3 Algebraic isomorphisms and separability A bijection y: S(A) ^ S(A') is called an algebraic isomorphism from A to A' if cx,y = cXv,Y v for all X, Y, Z G S(A). The mapping X ^ Xv is extended by linearity to the ring isomorphism of A and A'. This ring isomorphism we denote also by y. If there exists an algebraic isomorphism from A to A' then we write A =Alg A'. An algebraic isomorphism from A to itself is called an algebraic automorphism of A. The group of all algebraic automorphisms of A is denoted by AutAlg (A). Every isomorphism f of S-rings preserves the structure constants and hence f induces the algebraic isomorphism y f. However, not every algebraic isomorphism is induced by a combinatorial one (see [6]). Let K be a class of groups. An S-ring A is defined to be separable with respect to K if every algebraic isomorphism from A to an S-ring over a group from K is induced by a combinatorial isomorphism. Put AutAlg(A)0 = {y G AutAlg(A) : y = yf for some f G Iso(A)}. It is easy to see that yf = yg for f, g G Iso(A) if and only if gf-1 G Aut(A). Therefore | AutAlg(A)o| = | Iso(A)|/| Aut(A)|. (2.3) One can verify that for every group G the S-ring of rank 2 over G and ZG are separable with respect to the class of all finite groups. In the former case there exists the unique algebraic isomorphism from the S-ring of rank 2 over G to the S-ring of rank 2 over a given group of order |G| and this algebraic isomorphism is induced by every bijection. In the latter case every basic set is singleton and hence every algebraic isomorphism is induced by an isomorphism in a natural way. Let y: A ^ A' be an algebraic isomorphism. One can check that y is extended to a bijection between A- and A'-sets and hence between A- and A'-sections. The images of an A-set X and an A-section S under these extensions are denoted by Xv and Sv respectively. If S is an A-section then y induces the algebraic isomorphism yS: AS ^ A'S,, where S' = Sv. The above bijection between the A- and A'-sets is, in fact, an isomorphism of the corresponding lattices. One can check that (X= (X and rad(Xv) = rad(X )v for every A-set X (see [4, Equation (10)]). Since cX?}Y = SY,X -i |X|, where X, Y G S(A) and ¿Y,X-i is the Kronecker delta, we conclude that |X| = cXe}X-i, (X-1)v = (Xv)-1, and |X| = |Xfor every A-set X. In particular, |G| = |G'|. , 2.4 Cayley isomorphisms A group isomorphism f: G ^ G' is called a Cayley isomorphism from A to A' if S(A)f = S(A'). If there exists a Cayley isomorphism from A to A' we write A =Cay A'. Every Cayley isomorphism is a (combinatorial) isomorphism, however the converse statement is not true. 472 Ars Math. Contemp. 17(2019)447-454 2.5 Algebraic fusions Let A be an S-ring over G and $ < AutAlg(A). Given X G S(A) put X$ = |JX The partition {X$ : X G S(A)} defines an S-ring over G called the algebraic fusion of A with respect to $ and denoted by A$. Suppose that $ = {yf : f G K} for some K < Iso(A) and A is schurian. Then one can verify that A$ = V(Aut(A)K, G). In particular, the following statement holds. Lemma 2.1. Let A be a schurian S-ring over G and K < Iso(A). Then A$, where $ = {yf : f G K}, is also schurian. 2.6 Wreath and tensor products Let A be an S-ring over a group G and S = U/L an A-section. The S-ring A is called the S-wreath product if L < G and L < rad(X) for all basic sets X outside U. In this case we write A = Au Is AG/L. The S-wreath product is called non-trivial or proper if e = L and U = G. If U = L we say that A is the wreath product of AL and AG/L and write A = AL IAG/L. Let Ai and A2 be S-rings over groups Gi and G2 respectively. Then the set S = S(A1) x S(A2) = {X1 x X2 : X1 G S(A1),X2 G S(A2)} forms a partition of G = G1 x G2 that defines an S-ring over G. This S-ring is called the tensor product of A1 and A2 and denoted by A1 ( A2. Lemma 2.2. The tensor product of two separable S-rings is separable. Proof. As noted in [18, Lemma 2.6], the statement of the lemma follows from [1, Theorem 1.20]. □ Lemma 2.3 ([17, Lemma 4.4]). Let A be the S-wreath product over an abelian group G for some A-section S = U/L. Suppose that Au and AG/L are separable and Aut(AU )S = Aut(AS). Then A is separable. In particular, the wreath product of two separable S-rings is separable. Let Q be a finite set. Permutation groups K, K' < Sym(Q) are called 2-equivalent if Orb(K, Q2) = Orb(K', Q2). A permutation group K < Sym(Q) is called 2-isolated if it is the only group which is 2-equivalent to K. Lemma 2.4. Let A be the S-wreath product over an abelian group G for some A-section S = U/L. Suppose that Au and AG/L are separable, Au is schurian, and the group Aut(AS) is 2-isolated. Then A is separable. Proof. Since Au is schurian, the groups Aut(AU)S and Aut(AS) are 2-equivalent. Indeed, Orb(Aut(AU)S, S2) = Orb(Aut(AS),S2) = {R(X) : X G S(AS)}. G. Ryabov: On separable abelian p-groups 473 This implies that Aut(A^)S = Aut(AS) because Aut(AS) is 2-isolated. Therefore the conditions of Lemma 2.3 hold and A is separable. □ Lemma 2.5. Let H be a normal subgroup of a group G, B an S-ring over H, ^ G AutAig(B) \ AutAig(B)0. Then there exists ^ G AutAig(A) \ AutAig(A)o, where A = B l Z(G/H), such that ^H = y. Proof. Define ^ as follows: X^ = X^ for X G S(AH) and X^ = X for X G S(A) \ S(Ah). Let us prove that ^ G AutAig (A). To do this it suffices to check that cXv yv — c|,Y for all X, Y, Z G S(A). Suppose that X, Y G S(Ah). If Z G S(Ah) then yv = y v = cXy .If Z G S(Ah ) then Z^ G S(AH) and hence cXV y v = cX y = 0. Now suppose that exactly one of the sets X, Y, say X, lies inside H. Then Y^ = Y and X u X^ C H < rad(Y). So XY = X^Y = |X|Y. This implies that , yv = cX , y = |X| whenever Z = Y and cXV y v = cX , y = 0 otherwise. Finally, suppose that X, Y G S(Ah ). In this case X ^ = X and Y ^ = Y .If Z G S(Ah) then Z^ = Z and hence cfV y v = cX,y. If Z G S(AH) then Z and Z^ enter the element XY with the same coefficients because H = rad(X) n rad(Y). Therefore cX;y v = cX,y. Thus, ^ G AutAig (A). If ^ is induced by an isomorphism then [4, Lemma 3.4] implies that = ^ is also induced by an isomorphism. We obtain a contradiction with the assumption of the lemma and the lemma is proved. □ 3 S-rings over cyclic p-groups In this section we prove Theorem 1.1. Before the proof we recall some results on S-rings over cyclic p-groups. The most of them can be found in [7, 8]. Throughout the section p is an odd prime, G is a cyclic p-group and A is an S-ring over G. We say that X G S(A) is highest if X contains a generator of G. Put rad(A) = rad(X), where X is highest. Note that rad(A) does not depend on the choice of X because every two basic sets are rationally conjugate and hence have the same radicals. Lemma 3.1. The S-ring A is schurian and one of the following statements holds for A: (1) | rad(A)| = 1 and rk(A) = 2; (2) | rad(A)| = 1, A is normal, and A = Cyc(K, G) for some K < K0, where K0 is the subgroup of Aut(G) of order p — 1; (3) | rad(A) | > 1 and A is the proper generalized wreath product. Proof. The S-ring A is schurian by the main result of [16]. The other statements of the lemma follow from [8, Theorem 4.1, Theorem 4.2 (1)] and [7, Lemma 5.1, Equation (1)]. □ Lemma 3.2. Let S be an A-section with |S| > p2. The following statements hold: (!) If Statement (2) of Lemma 3.1 holds for A then Statement (2) of Lemma 3.1 holds for As; 474 Ars Math. Contemp. 17(2019)447-454 (2) If rk(AS) = 2 then Aut(A)S = Sym(S). Proof. Statement (1) of the lemma follows from [8, Corollary 4.4] and Statement (2) of the lemma follows from [8, Theorem 4.6 (1)]. □ Lemma 3.3. Suppose that Statement (2) of Lemma 3.1 holds for A. Then Aut(A) is 2-isolated. Proof. By [15, Lemma 8.2], it suffices to prove that Aut(A)e has a faithful regular orbit. The S-ring A is normal. So (2.1) implies that Aut(A)e < Aut(G). Let X G S(A) be highest. Since A is cyclotomic, each element of X is a generator of G. If f g Aut(A)e fixes some x G X then f is trivial because f G Aut(G) and x is a generator of G. Besides, A is schurian and hence X G Orb(Aut(A)e, G) by (2.2). Therefore X is a regular orbit of Aut(A)e. The group Aut(G) is cyclic because p is odd. So both of the groups Aut(A)e and Aut(A)X are cyclic groups of order |X |. Thus, X is a faithful regular orbit of Aut(A)e and the lemma is proved. □ Lemma 3.4. Suppose that Statement (2) of Lemma 3.1 holds for A and ^ is an algebraic isomorphism from A to an S-ring A' over an abelian group G'. Then G' is cyclic. Proof. By the hypothesis, A = Cyc(K, G) for some K < Aut(G) with |K| < p - 1. The group E = {g G G : |g| = p} is an A-subgroup of order p because A is cyclotomic. The group E' = Ev is an A'-subgroup of order p by the properties of an algebraic isomorphism. Assume that G' is non-cyclic. Then there exists X' G S(A') containing an element of order p outside E'. Let X g S(A) such that Xv = X'. The set X consists of elements of order greater than p because G is cyclic and all elements of order p from G lie inside E. The identity element e of G enters the element Xp with a coefficient dividing by p because xp = e for each x G X. The identity element e' of G' enters the element (X')p with a coefficient which is not divided by p because (x')p = e' for some x' G X' and |X'| < p - 1. Since ^ is an algebraic isomorphism, we have (XT = (X')p and {e}v = {e'}. This implies that e and e' must enter Xp and (X')p respectively with the same coefficients, a contradiction. Therefore G' is cyclic and the lemma is proved. □ Lemma 3.5. Suppose that | rad(A)| > 1. Then there exists an A-section S = U/L such that A is the proper S-wreath product, | rad(Ay) | = 1, and |L| = p. Proof. From [17, Lemma 5.2] it follows that there exists an A-section U/L1 such that A is the proper U/L1-wreath product and | rad(Ay )| = 1. Let L be a subgroup of L1 of order p. Then the lemma holds for S = U/L. □ Lemma 3.6 ([5, Theorem 1.3]). Every S-ring over a cyclic p-group is separable with respect to . G. Ryabov: On separable abelian p-groups 475 Proof of the Theorem 1.1. The statement of the theorem for p e {2, 3} was proved in [17, Lemma 5.5]. Further we assume thatp > 5. Let A be an S-ring over a cyclicp-group G of order pk, where k > 1. Prove that A is separable. We proceed by induction on k. If k = 1 then G is the unique up to isomorphism group of order p and the statement of the theorem follows from Lemma 3.6. Let k > 2. One of the statements of Lemma 3.1 holds for A. If Statement (1) of Lemma 3.1 holds for A then rk(A) = 2 and hence A is separable. Suppose that Statement (2) of Lemma 3.1 holds for A. Let p be an algebraic isomorphism from A to an S-ring A' over an abelian group G'. Due to Lemma 3.4, the group G' is cyclic. So p is induced by an isomorphism by Lemma 3.6. Therefore A is separable. Now suppose that Statement (3) of Lemma 3.1 holds for A. Then A = Av lS AG/L for some A-section S = U/L with L > e and U < G. The S-rings Av and AG/L are separable by the induction hypothesis. Due to Lemma 3.5 we may assume that rad(A^) = e and |L| = p. In this case rk(A^) =2 or Statement (2) of Lemma 3.1 holds for Av. If rk(Ay) =2 or |S| = 1 then U = L and A is separable by Lemma 2.3. Assume that Statement (2) of Lemma 3.1 holds for Av. If |S| > p2 then Statement (2) of Lemma 3.1 holds for AS by Statement (1) of Lemma 3.2. Lemma 3.3 implies that Aut(AS) is 2-isolated. The S-ring Av is cyclotomic and hence it is schurian. Therefore A is separable by Lemma 2.4. It remains to consider the case when |S| = p. In this case |U| = p2. If rad(X) > L for every X e S(A) outside U then rad(X) > U for every X e S(A) outside U because G is cyclic. This yields that A = Av l AG/V and hence A is separable by Lemma 2.3. Suppose that there exists X e S(A) outside U with rad(X) = L. The remaining part of the proof is divided into two cases. Case 1: (X} < G. In this case put Si = (X}/L. The S-ring A is the Si-wreath product and |S1| > p2. Note that | rad(ASl) | = 1 because rad(X) = L. So Statement (1) or Statement (2) of Lemma 3.1 holds for ASl. In the former case Aut(A^X) )Sl = Aut(ASl) = Sym(S1) by Statement (2) of Lemma 3.2 and A is separable by Lemma 2.3. In the latter case Aut(ASl) is 2-isolated by Lemma 3.3. Since A^ is schurian, the conditions of Lemma 2.4 hold for S1 and A is separable by Lemma 2.4. Case 2: (X} = G. In this case | rad(AG/L) | = 1 because rad(X) = L. Let n: G ^ G/L be the canonical epimorphism. Clearly, n(U) is an AG/L-subgroup and n(X) lies outside n(U). So rk(AG/L) > 2 and hence Statement (2) of Lemma 3.1 holds for AG/L. Let p be an algebraic isomorphism from A to an S-ring A' over an abelian group G'. Put U' = U^ and L' = Lv. Clearly, L' < U'. (3.1) The algebraic isomorphism p induces the algebraic isomorphism pv from Av to Av>, where U' = UFrom Lemma 3.4 it follows that U' = Cp2. (3.2) Also p induces the algebraic isomorphism pG/L from AG/L to AG//L. Lemma 3.4 implies that G'/L' is cyclic. Since |L'| = |L| = p, we conclude that G = Cpk or G = Cp x Cpk-l. 476 Ars Math. Contemp. 17(2019)447-454 However, in the latter case L' is not contained in a cyclic group of order p2 because G'/L' is cyclic. This contradicts to (3.1) and (3.2). So G' is cyclic and p is induced by an isomorphism by Lemma 3.6. Therefore A is separable and the theorem is proved. □ 4 Proof of Theorem 1.2 Proposition 4.1. The group Cp is separable for p e {2, 3}. Before we prove Propostion 4.1 we give the lemma providing a description of S-rings over these groups. Lemma 4.2. Let A be an S-ring over C:^, where p e {2,3}. Then A is schurian and one of the following statements holds: (1) rk(A) = 2; (2) A is the tensor product of smaller S-rings; (3) A is the proper S-wreath product of two S-rings with |S| < p; (4) p = 3 and A =Cay Aj, where Ai is one of the 14 exceptional S-rings whose parameters are listed in Table 1. Remark 4.3. In Table 1 the notation km means that an S-ring have exactly m basic sets of size k. Table 1: Parameters of the 14 exceptional S-rings Ai, A2,..., A14. S-ring rank sizes of basic sets Ai 3 1, 132 A2 4 1, 6, 8, 12 A3 4 1, 2, 122 a4 5 1, 42, 6, 12 A5 5 1, 2, 83 Ae 6 1, 2, 64 a7 7 1, 2, 44, 8 Ag 7 1, 2, 32, 63 A9 8 1, 2, 4e A10 9 1, 23, 45 Aii 10 1, 25, 44 A12 10 13, 3e, 6 A13 11 13, 38 A14 14 1, 213 Proof. The statement of the lemma can be checked with the help of the GAP package COCO2P [11]. □ G. Ryabov: On separable abelian p-groups 477 Proof of the Proposition 4.1. From [17, Theorem 1, Lemma 5.5] it follows that the group Cp is separable for p e {2,3} and k < 2. Let A be an S-ring over G = Cp, where p e {2,3}. Then one of the statements of Lemma 4.2 holds for A. If Statement (1) of Lemma 4.2 holds for A then, obviously, A is separable. If Statement (2) of Lemma 4.2 holds for A then A is separable by Lemma 2.2. Suppose that Statement (3) of Lemma 4.2 holds for A. Then A is the proper schurian S-wreath product for some A-section S = U/L with |S| < 3. Since A is schurian, Av is also schurian. Note that Aut(AS) is 2-isolated becasue |S| < 3. Therefore A is separable by Lemma 2.4. Suppose that Statement (4) of Lemma 4.2 holds for A and y is an algebraic isomorphism from A to an S-ring A' over an abelian group G'. Clearly, if A' is separable then y-1 is induced by an isomorphism and hence y is also induced by an isomorphism. If G' = Cp 3 then A' is separable by Theorem 1.1; if G' = Cp x Cp2 then A' is separable by [17, Theorem 1]; if G' = Cp and one of the Statements (1)-(3) of Lemma 4.2 holds for A' then A' is separable by the previous paragraph. So in the above cases y is induced by an isomorphism. Thus, we may assume that G' = Cp and Statement (4) of Lemma 4.2 holds for A'. Two algebraically isomorphic S-rings have the same rank and sizes of basic sets. So information from Table 1 implies that A¿ ^Alg Aj whenever i = j. Therefore we may assume that == == 2. Proof. In view of Lemma 2.5 to prove that the group Cp x Cp x Cpk is non-separable for p e {2,3} and k > 2 it is sufficient to construct an S-ring A over Cp x Cp x Cp2, p e {2, 3}, and an algebraic isomorphism y from A to itself which is not induced by an isomorphism. Let G = (a) x (6) x (c), where |a| = |6| = p and |c| = p2. Put A = (a), B = (6), C = (c), c1 = cp, and C1 = (c1). Firstly consider the case p = 2. Let f e Aut(G) such that f: (a, 6, c) ^ (a, 6ac1, ca) and A = Cyc((f), G). It easy to see that |f | =2 and the basic sets of A are the following To = {e}, Ti = {a}, T2 = {cj, T3 = {acj, X1 = cA, X2 = c3A, Y1 = 6(ac1), Y2 = 6a(ac1), Z1 = 6cC1, Z2 = 6caC1. Define a permutation y on the set S(A) as follows: TO = T0, Tf = T1, Tf = T3, Tf = T2, X? = X1, Jiff = X2, 478 Ars Math. Contemp. 17(2019)447-454 Y1 = Zi, Y2f = Z2, Zf = Yi, Z2f = Y2. It easy to see that |p| = 2. The straightforward check implies that p is an algebraic t tv isomorphism from A to itself. Let us check, for example, that cy2 = cy2^ yv. We have Y1Y2 = 2a + 2c1 and YfYf = Z1Z2 = 2a + 2ac1. So cT, Y2 = c^ Y ^ = 2. Note that A corresponds to a Kleinian quasi-thin scheme of index 4 in the sense of [14]. The S-ring A is cyclotomic and hence it is schurian. Assume that p is induced by an isomorphism. Then the algebraic fusion A^ is schurian by Lemma 2.1. However, computer calculations made by using the package COCO2P [11] (see also [21]) imply that Al 6, then min s +t subject to: 0 < s < n 0 < t < m (1.1) s, t G Z. 8n - 7 is not a perfect square, 8n - 7 is a perfect square. See [12, Theorem 2.1]. V. Irsic and M. Konvalinka: Strong geodetic problem on complete multipartite graphs 483 In the following section, we generalize the above result to all complete bipartite graphs. To conclude the introduction, we state the following interesting and surprisingly important fact. Lemma 1.1 (Shifting Lemma). Let Kni..,nr be a complete multipartite graph with the multipartition Xi,..., Xr, |Xj| = nj for i G [r]. Let S = Si U • • • U Sr be an optimal strong geodetic set, with Si C Xj and |Sj| = sj for i G [r]. If si < s2, s3 and s2 < n2, s3 < n3, then there exist x G Si and y G S2 U S3, such that S U {y} — {x} is also an optimal strong geodetic set. Proof. Let G = Kn1}...,nr, |Xj| = nj, |Sj| = sj for i G [r]. Suppose Sj = 0,Xj for i G {1, 2, 3}. Without loss of generality, si = min{si, s2, s3}, and let geodesics between vertices of Si cover fewer vertices in X2 — S2 than in X3 — S3. Select vertices x g Si and y G X2 — S2. Geodesics between vertices from Si can be fixed in such a way that no vertex in X2 is covered with a geodesic containing x. This is trivial for si G {1, 2,3}, and follows from m 2 < (si-i) for si > 4. Now consider T = S U {y} — {x}, T| = |S|. We will prove that T is a strong geodetic set of G. Fix geodesics between vertices in T in the same way as in S, except those containing x or y. As x G T, at most si — 1 vertices U in V(G) — Xi — X2 are uncovered. But geodesics containing y can cover the vertex x, as well as s2 — 1 other vertices in V(G) — X2. As we have s2 — 1 > si — 1, those geodesics can be fixed in such a way that U is covered. □ Proposition 1.2. For every complete multipartite graph there exist an optimal strong geodetic set such that its intersection with all but two parts of the multipartition is either empty or the whole part. Proof. Let G = Kni..,nr, |Xj| = nj, be a multipartite graph. The Shifting Lemma states that every strong geodetic set with three or more parts of size not equal to 0 or nj can be transformed into a strong geodetic set of the same size, where one of these parts becomes smaller and one larger. After repeating this procedure on other such triples, at most two parts can have size different from 0 or nj. □ The rest of the paper is organized as follows. In the next section, some further results about the strong geodetic number of complete bipartite graphs are obtained. In Section 3 we discuss the strong geodetic problem on complete multipartite graphs. Finally, in Section 4 the complexity of the strong geodetic problem on multipartite and complete multipartite graphs is discussed. 2 On complete bipartite graphs In this section, we give a complete description of the strong geodetic number of a complete bipartite graph. Instead of giving an explicit formula for sg(Kn m), we classify the triples (n, m, k) for which sg(Kn,m) = k. Define f ( O) 1 + /max{ft — 1, 2^ f (a,O ) = a — 1+1 2 484 Ars Math. Contemp. 17 (2019) 493-514 Theorem 2.1. For positive integers n, m and k, (n,m) 3 and max{n,m} > 3. The statement follows from the following (note that the sum s1 + s2 equals k for every (s1, s2) that appears below). Note that all different optimal solutions are described here, hence some of the conditions overlap. 1. If n < 3 and m = f (k, n) = k, or m = f (k, i — 1) and f (k, k — i — 1) < n < f (k, k — i) for i < 4, or m = f (k, i — 1) and n = f (k, k — i — 1) for i < 4, then (0, k) is an optimal solution. Symmetrically, if m < 3 and n = f (k, m) = k, or f (k, i — 1) < m < f (k, i) and n = f (k, k — i — 1) for i > k — 4, then (k, 0) is an optimal solution. V. Irsic and M. Konvalinka: Strong geodetic problem on complete multipartite graphs 485 Figure 1: All pairs (n, m) for which sg(Kn,m) = 12. 2. If 3 < n < k and m = f (k, n), then (n, k—n) is an optimal solution. Symmetrically, if 3 < m < k and n = f (k, m), then (k — m, m) is an optimal solution. 3. If f (k, i — 1) < m < f (k, i) and f (k, k — i — 1) < n < f (k, k — i) for 4 < i < k — 4, or m = f (k, i — 1) and f (k, k — i — 1) < n < f (k, k — i) for i > 3, or f (k, i — 1) < m < f (k, i) and n = f (k, k — i — 1) for i < k — 3, or m = f (k, i — 1) and n = f (k, k — i — 1) for 3 < i < k — 3, then (i, k — i) is an optimal solution. 4. If f (k, i — 1) < m < f (k, i) and n = f (k, k — i — 1) for i < k — 4, or m = f (k, i — 1) and n = f (k, k — i — 1) for 2 < i < k — 4, then (i + 1, k — i — 1) is an optimal solution. 5. If m = f (k, i — 1) and f (k, k — i — 1) < n < f (k, k — i) for i > 4, or m = f (k, i — 1) and n = f (k, k — i — 1) for 4 < i < k — 2, then (i — 1, k — i + 1) is an optimal solution. It is easy to see that the above solutions give rise to the strong geodetic sets of size k. For example, in the first case, the part of the bipartition of size m is a strong geodetic set with parameters (0, k). What remains to be proved is sg(Kn,m) > k for each case. This can be shown by a simple case analysis. As the reasoning is similar in all cases, we demonstrate only two of them. Let X be the part of the bipartition of size n and Y the part of size m. Also, let S = Si U S2, where Sx Ç X, S2 Ç Y, be some strong geodetic set. • The case k>n > 3 and m = k — 1 + (n22i) = k — n + Q): If Si = X, then geodesics between these vertices cover at most 2 vertices in Y, so at least k — n vertices in Y must also lie in a strong geodetic set. Hence, |S| > n — (k — n) = k. If Si = X, geodesics between these vertices cover at most (n2"i) vertices in Y, so at least k — 1 vertices from Y must lie in a strong geodetic set. Hence, |S| > |Si| + (k — 1). If Si = 0 or |S21 > k, we have |S| > k. Otherwise, S = S2 and contains exactly k — 1 vertices. But then the remaining vertices in Y are not covered. • The case where f (k, i — 1) < m < f (k, i) and f (k, k — i — 1) < n < f (k, k — i) 486 Ars Math. Contemp. 17 (2019) 493-514 for 4 < i < k — 4: We can write fk - i - 2\ n = k — 1+i j + l, l € {1, ...,k — i — 2}, m = k — 1+ ^ — ^ + j, j €{1,...,i — 2}. Suppose |S| < k — 1. If |S1| < i — 2, these vertices cover at most (i-2) vertices in X, thus at least k vertices remain uncovered and |S | > k. Hence, |S1| > i — 1. Similarly, |S2| > k — i — 1. If |S1| = i — 1, then (i-1) vertices in Y are covered. As k + j — i + 1 are left uncovered, it holds that |S2| > k — i + 2 and thus |S| > k +1. If |S2| = k — i — 1, then (fc-2-1) vertices in X are covered. As l + i + 1 are left uncovered, it holds that |S1| > i + 2 and thus |S | > k + 1. Hence |S11 > i and S | > k — i and thus |S| > k. □ The first condition from Theorem 2.1 can be simplified as follows. Corollary 2.3. If n > 3 and m > ("), then sg(K„jm) = m +1 — (n-1). f n < 3 and m > n, then sg(Kn,m) = m. When m < , Theorem 2.1 is harder to apply. Note, however, that the theorem suggests that m is approximately equal to k — 1 + , and n is approximately equal to k — 1 + (fc-2-1). Furthermore, note that we can rewrite the system of equations (with known m,n and variables k, i) m = k —1+(i"21), n = k — 1+ fc-2-1) as a polynomial equation of degree 4 for k (say by subtracting the two equations, solving for i, and plugging the result into one of the equations), and solve it explicitly. It seems that one of the four solutions is always very close to sg(Km,n). Denote the minimal distance between sg(Km,n) and a solution k of m = k — 1 + (i-1), n = k — 1 + (fc-2-1) by e(m, n). Then our data is indicated in Table 2. Table 2: The difference between the exact and estimated values of sg(Km,n) for different values of n. n 10 100 1000 10000 100000 max{e(m,n) : n < m < } 1.094 1.774 1.941 1.983 1.995 We conjecture the following. Conjecture 2.4. If n < m < , then e(m,n) < 2. If the conjecture is true, sg(Km,n) is among the (at most 16) positive integers that are at distance < 2 from one of the four solutions of the system m = k - 1 + (i-1), n = k - 1 + (fc-2-1). For each of these (at most) 16 candidates, there are at most three (consecutive) i's for which f (k, i -1) < m < f (k, i), found easily by solving the quadratic equation m = k — 1 + (i-1). For each such i, check if f (k, k — i — 1) < n < f (k, k — i). This allows for computation of sg(Km,n ) with a constant number of operations. V. Irsic and M. Konvalinka: Strong geodetic problem on complete multipartite graphs 487 3 On complete multipartite graphs The optimization problem (1.1) can be generalized to complete multipartite graphs. However, solving such a program seems rather difficult. Hence, we present an approximate program which gives a nice lower bound for the strong geodetic number of a complete multipartite graph. If i vertices from one part are in a strong geodetic set, geodesics between them cover at most (2) other vertices. In the following, we do not take into account the condition that they can only cover vertices in other parts, and that the number of selected vertices must be an integer. Recall the notation (1mi,..., kmk} which describes a partition with m2 parts of size i, 1 < i < k. Let G be a complete multipartite graph corresponding to the partition n = (1mi,..., kmk} and let a2j denote the number of parts of size j with exactly i vertices in the strong geodetic set. Thus we must have J]j =0 aij — mj and XjLiXjLo (2)aij > Ej^Ej^Ci - i)a2j. The second condition simplifies to Ejk=i Ej=i C+1)aij > Ejk=i Ej=o jaij = Ejk=i jmj = n. As aoj's do not appearin it anymore, we also simplify the first condition to J2j=i a2j < mj and get k j " "ij min ^^ ¿a. j=1i=1 j subject to: ^^ aj < mj i=1 (3.1) ib b C +1) aij >n j=i ¿=^2 / 0 < ajj < mj. As the sequence (2) - k is increasing for k > 3, it is better to select more vertices in a bigger part. Hence, the optimal solution is ak,k = mk ak-1,k- 1 = mk-1 ai+i,i+i = mi+1 ai,i = Imi +-----+ 1m1 - (2)mk - • • • - C+1)mi+1 C+1) where l is the smallest positive integer such that (++) mk + ••• + (+2 ) ml+i < kmk + + 1m1 = |V (K^m, + (i121)mi+1,and ,kmk > |, which is equivalent to Zmi + • • • + 1m1 > (2)mk + sg(K( 1mi, ,kmfc >) > kmk +----+ (Z + 1)mi+1 + Zmi +-----+ m1 - (2) mk----- ('+1) mi+1 i+1 2 488 Ars Math. Contemp. 17 (2019) 493-514 The result is particularly interesting in the case when n = (km), i.e. when we observe a multipartite graph with m parts of size k, as we get l = k and Sg(K(fcm) ) > 2km k + 1 On the other hand, considering a strong geodetic set consisting only of the whole parts of the bipartition yields an upper bound. At least l G Z, where 1(k + (2) ) > mk, parts must be in a strong geodetic set. Hence, 2m k + 1 • k. Sg(K f m)) < This implies the following result. Proposition 3.1. If k, n G N and (k + 1} | 2m, then sg(Kkm)} = f+f • 4 Complexity results for multipartite graphs The strong geodetic problem can be naturally formed as a decision problem. Problem 4.1 (Strong geodetic set). Input: a graph G, an integer k. Question: does a graph G have a strong geodetic set of size at most k? The strong geodetic problem on general graphs is known to be NP-complete [1]. In the following we prove that it is also NP-complete on multipartite graphs. The reduction uses the dominating set problem. Recall that a set D C V(G) is a dominating set in the graph G if every vertex in V(G) — D has a neighbor in D. Problem 4.2 (DOMINATING SET). Input: a graph G, an integer k. Question: does a graph G have a dominating set of size at most k? The dominating set problem is known to be NP-complete on bipartite graphs [15]. The idea of the following proof is similar to the proof that the ordinary geodetic problem restricted to chordal bipartite graphs is NP-complete [6]. Theorem 4.3. Strong geodetic set restricted to bipartite graphs is NP-complete. Proof. To prove NP-completeness, we describe a polynomial reduction of DOMINATING set on bipartite graphs to Strong geodetic set on bipartite graphs. Let (G, k) be an input for DOMINATING SET, and (X, Y} a bipartition of the graph G. Define a graph G', V(G') = V(G) U jwi, u2} U {x' : x G X} U {y' : y G Y}, with the edges E(G), u1 ~ u2, and x ~ u2 ~ x' for all x G X, y ~ u1 ~ y' for all y G Y. Define the sets X' = X U {mi} U jx' : x G X}, Y' = Y U{w2}U{y' : y G Y}, V. Irsic and M. Konvalinka: Strong geodetic problem on complete multipartite graphs 489 and observe that (X', Y') is a bipartition of the graph G'. Define the parameter k' = k + |V (G)|. Suppose D is a dominating set of the graph G of size at most k. Define D' = D U {x' : x € X} U {y' : y G Y}. Notice that |D'| < k'. For each x G X n D, fix geodesics x ~ y ~ u1 ~ y', y G NG(x). Similarly, for each y G Y n D, fix y ~ x ~ u2 ~ x', x G NG(y). As D is a dominating set, these geodesics cover all vertices in V(G). Additionally, fix geodesics x ~ u2 ~ x' for some x G X, and y ~ ui ~ y' for some y G Y, to cover the vertices ui, u2. Hence, D' is a strong geodetic set of the graph G'. Conversely, suppose D' is a strong geodetic set of G' of size at most k'. Vertices {x' : x G X} U {y' : y G Y} are all simplicial, hence they all belong to D'. Geodesics between them cannot cover any vertices in V (G), thus V (G) n D' = 0. Let D = D' n V(G). Clearly, |D| < k. Consider x G V(G) — D. Thus x is an inner point of some y, z-geodesic. At most one of y, z does not belong to D. The structure of the graph ensures that at least one of y, z is a neighbor of x. Hence, D is a dominating set of the graph G. □ Corollary 4.4. STRONG GEODETIC SET restricted to multipartite graphs is NP-complete. In the following we consider the complexity of Strong geodetic set on complete multipartite graphs. Proposition 1.2 gives rise to the following algorithm. Let G be a graph and (X1,..., Xr) its multipartition. Denote ni = |Xj|, i G [r]. For all {i,j} C (H), for all subsets R of [r] — {i, j}, for all si G {0,... ,n4}, for all Sj G {0,..., nj}, set Si C Xi of size Sj, and Sj C Xj of size Sj. Check if Si U Sj U |JfceR Xk is a strong geodetic set for G. The answer is the size of the smallest strong geodetic set. The time complexity of this algorithm is O(n2r22r). This confirms the already known result that Strong geodetic set restricted to complete bipartite graphs is in P, which is an easy consequence of Theorem 2.1. Moreover, it is now clear that the problem is solvable in quadratic time. The same holds for complete r-partite graphs (when r is fixed). But for a general complete multipartite graph (when the size of the multipartition is part of the input), the algorithm tells us nothing about complexity. But we also observe an analogy between the Strong geodetic set problem on complete multipartite graphs and the Knapsack problem, which is known to be NP-complete [17]. Recall that in this problem, we are given a set of items with their weights and values, and we need to determine which items to put in a backpack, so that a total weight is smaller that a given bound and a total value is as large as possible. The approximate reduction from the Strong geodetic set on complete multipartite graphs to the Knapsack problem is the following. Let (X1,..., Xr) be the parts of the complete multipartite graph. The items x1,..., xr represent those parts, a value if xi is (^X^1) and the weight is |Xi|. Thus selecting the items such that their total value is as large as possible and the total weight as small as possible, is almost the same as finding the smallest strong geodetic set of the complete multipartite graph (as Proposition 1.2 states that at most two parts in the strong geodetic set are selected only partially). We were not able to find a reduction from the Knapsack problem to the Strong geodetic set on complete multipartite graphs. But due to the connection with the Knapsack problem, it seems that the problem is not polynomial. Hence we pose Conjecture 4.5. STRONG GEODETIC set restricted to complete multipartite graphs is NP-complete. 490 Ars Math. Contemp. 17 (2019) 493-514 However, as already mentioned, determining the strong geodetic number of complete r-partite graphs for fixed r is polynomial. Using a computer program (implemented in Mathematica [18]) we derive the results shown in Table 3. Table 3: The strong geodetic numbers for some small complete multipartite graphs. n sg(K) n sg(K) n Sg(Kn ) (1) 1 (6) 6 (1, 32) 4 (2) 2 (1, 5) 5 (22, 3) 4 (12) 2 (2, 4) 4 (12,2, 3) 4 (3) 3 (12,4) 4 (14, 3) 4 (1, 2) 2 (32) 3 (1, 23) 5 (13) 3 (1, 2, 3) 3 (13,22) 5 (4) 4 (13, 3) 3 (15, 2) 6 (1, 3) 3 (23) 4 (17) 7 (22) 3 (12, 22) 4 (12, 2) 3 (14, 2) 5 (14) 4 (16) 6 (5) 5 (7) 7 (1,4) 4 (1, 6) 6 (2, 3) 3 (2, 5) 5 (12, 3) 3 (12, 5) 5 (1, 22) 4 (3, 4) 4 (13, 2) 4 (1, 2,4) 4 (15) 5 (13,4) 4 References [1] A. Arokiaraj, S. Klavzar, P. Manuel, E. Thomas and A. Xavier, Strong geodetic problems in networks, Discuss. Math. Graph. Theory, in press, doi:10.7151/dmgt.2139. [2] M. Atici, Computational complexity of geodetic set, Int. J. Comput. Math. 79 (2002), 587-591, doi:10.1080/00207160210954. [3] B. Bresar, M. Kovse and A. Tepeh, Geodetic sets in graphs, in: M. Dehmer (ed.), Structural Analysis of Complex Networks, Birkhauser/Springer, New York, pp. 197-218, 2011, doi:10. 1007/978-0-8176-4789-6_8. [4] L. R. Bueno, L. D. Penso, F. Protti, V. R. Ramos, D. Rautenbach and U. S. Souza, On the hardness of finding the geodetic number of a subcubic graph, Inform. Process. Lett. 135 (2018), 22-27, doi:10.1016/j.ipl.2018.02.012. [5] G. Chartrand, F. Harary and P. Zhang, Geodetic sets in graphs, Discuss. Math. Graph Theory 20 (2000), 129-138, doi:10.7151/dmgt.1112. [6] M. C. Dourado, F. Protti, D. Rautenbach and J. L. Szwarcfiter, Some remarks on the geodetic number of a graph, Discrete Math. 310 (2010), 832-837, doi:10.1016/j.disc.2009.09.018. [7] T. Ekim and A. Erey, Block decomposition approach to compute a minimum geodetic set, RAIRO Oper. Res. 48 (2014), 497-507, doi:10.1051/ro/2014019. V. Irsic and M. Konvalinka: Strong geodetic problem on complete multipartite graphs 491 [8] T. Ekim, A. Erey, P. Heggernes, P. van't Hof and D. Meister, Computing minimum geodetic sets of proper interval graphs, in: D. Fernández-Baca (ed.), LATIN 2012: Theoretical Informatics, Springer, Berlin, Heidelberg, volume 7256 of Lecture Notes in Computer Science, 2012 pp. 279-290, doi:10.1007/978-3-642-29344-3_24, proceedings of the 10th Latin American Symposium held at the Universidad Católica San Pablo (UCSP), Arequipa, April 16-20, 2012. [9] V. Gledel, V. Irsic and S. Klavzar, Strong geodetic cores and Cartesian product graphs, Appl. Math. Comput. 363 (2019), 124609 (10 pages), doi:10.1016/j.amc.2019.124609. [10] P. Hansen and N. van Omme, On pitfalls in computing the geodetic number of a graph, Optim. Lett. 1 (2007), 299-307, doi:10.1007/s11590-006-0032-3. [11] F. Harary, E. Loukakis and C. Tsouros, The geodetic number of a graph, Math. Comput. Modelling 17 (1993), 89-95, doi:10.1016/0895-7177(93)90259-2. [12] V. Irsic, Strong geodetic number of complete bipartite graphs and of graphs with specified diameter, Graphs Combin. 34 (2018), 443-456, doi:10.1007/s00373-018-1885-9. [13] V. Irsic and S. Klavzar, Strong geodetic problem on Cartesian products of graphs, RAIRO Oper. Res. 52 (2018), 205-216, doi:10.1051/ro/2018003. [14] S. Klavzar and P. Manuel, Strong geodetic problem in grid-like architectures, Bull. Malays. Math. Sci. Soc. 41 (2018), 1671-1680, doi:10.1007/s40840-018-0609-x. [15] M. Liedloff, Finding a dominating set on bipartite graphs, Inform. Process. Lett. 107 (2008), 154-157, doi:10.1016/j.ipl.2008.02.009. [16] P. Manuel, S. Klavzar, A. Xavier, A. Arokiaraj and E. Thomas, Strong edge geodetic problem in networks, Open Math. 15 (2017), 1225-1235, doi:10.1515/math-2017-0101. [17] S. Martello and P. Toth, Knapsack Problems: Algorithms and Computer Implementations, Wiley-Interscience Series in Discrete Mathematics and Optimization, John Wiley & Sons, New York, NY, 1990. [18] Wolfram Research, Inc., Mathematica, Champaign, Illinois, https://www.wolfram. com/mathematica. ARS MATHEMATICA CONTEMPORANEA ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 17 (2019) 493-514 https://doi.org/10.26493/1855-3974.1825.64c (Also available at http://amc-journal.eu) Archimedean toroidal maps and their minimal almost regular covers* Kostiantyn Drach t Jacobs University Bremen, Bremen, Germany Yurii Haidamaka V. N. Karazin Kharkiv National University, Kharkiv, Ukraine Mark Mixer Wentworth Institute of Technology, Boston, United States of America Maksym Skoryk V. N. Karazin Kharkiv National University, Kharkiv, Ukraine Received 20 October 2018, accepted 26 June 2019, published online 20 November 2019 The automorphism group of a map acts naturally on its flags (triples of incident vertices, edges, and faces). An Archimedean map on the torus is called almost regular if it has as few flag orbits as possible for its type; for example, a map of type (4.82) is called almost regular if it has exactly three flag orbits. Given a map of a certain type, we will consider other more symmetric maps that cover it. In this paper, we prove that each Archimedean toroidal map has a unique minimal almost regular cover. By using the Gaussian and Eisenstein integers, along with previous results regarding equivelar maps on the torus, we construct these minimal almost regular covers explicitly. Keywords: Maps, polytopes, groups, covers, Gaussian and Eisenstein integers. Math. Subj. Class.: 52B15, 51M20, 52C22 *This project began during the Fields-MITACS 2011 Summer Undergraduate Research Program. We would like to thank the Fields Institute for its hospitality during that program. Additionally we thank Daniel Pellicer, Isabel Hubard, and Asia Weiss for their constant support, and for providing the question which has been solved in this paper. Also, we thank the referees for suggesting several improvements. ^The research of the first author was partially supported by the advanced grant 695 621 "HOLOGRAM" of the European Research Council (ERC), which is gratefully acknowledged. E-mail addresses: k.drach@jacobs-university.de, kostya.drach@gmail.com (Kostiantyn Drach), ga1damakay@gmail.com (Yurii Haidamaka), mixerm@wit.edu (Mark Mixer), skoryk.maksym@gmail.com (Maksym Skoryk) ©® This work is licensed under https://creativecommons.Org/licenses/by/4.0/ Abstract 494 Ars Math. Contemp. 17 (2019) 493-514 1 Introduction Throughout the last few decades there have been many results about polytopes and maps that are highly symmetric, but that are not necessarily regular. In particular, there has been recent interest in the study of discrete objects using combinatorial, geometric, and algebraic approaches, with the topic of symmetries of maps receiving a lot of interest. There is a great history of work surrounding maps on the Euclidean plane or on the 2-dimensional torus. When working with discrete symmetric structures on a torus, many of the ideas follow the concepts introduced by Coxeter and Moser in [5, 6], where they present a classification of regular (reflexible and irreflexible) maps on the torus. Such "toroidal" maps can be seen as quotients of regular tessellations of the Euclidean plane. More recently, Brehm and Kuhnel [1] classified the equivelar maps on the two dimensional torus. Several more results have appeared about highly symmetric maps (see [4, 24] for example), and about highly symmetric tessellations of tori in larger dimensions (see [15, 16]). There is also much interest in finding minimal regular covers of different families of maps and polytopes (see for example [10, 17, 22]). In a previous paper [7], two of the authors constructed the minimal rotary cover of any equivelar toroidal map. Here we extend this idea to toroidal maps that are no longer equivelar, and construct minimal toroidal covers of the Archimedean toroidal maps with maximal symmetry. We call these covers almost regular; they will no longer be regular (or chiral), but instead will have the same number of flag orbits as their associated tessellation of the Euclidean plane (see the definition in Section 2). Our main results can be summarized by the following theorem. Theorem 1.1. Each Archimedean map on the torus has a minimal almost regular cover on the torus, this cover is unique and can be constructed explicitly. The paper is organized as follows. Section 2 contains the necessary background on maps and their symmetries, including the definition of an almost regular Archimedean map. In Section 3, almost regular Archimedean toroidal maps are characterized in terms of their lifts to the planar tessellations and the translation subgroups generating respective quotients. Section 4 contains our main results (Theorems 4.2-4.4) regarding the relationship between maps and their minimal almost regular covers; these theorems together constitute the Main Theorem. 2 Preliminares In this section we provide definitions and results necessary for our main theorems; many of these ideas, as well as further details, can be found in [2, 16, 25]. A finite graph X embedded in a compact 2-dimensional manifold S such that every connected component of S \ X (which is called a face) is homeomorphic to an open disc is called a map (on the surface S). In this paper, we consider Archimedean maps on a flat 2-dimensional torus, which we call Archimedean toroidal maps. A map M on the torus is Archimedean if the faces of M are regular polygons (in the canonical flat metric on the torus) and every pair of vertices of M can be mapped into each other by an isometry of the torus (here an isometry of the torus is a distance preserving diffeomorphism of the torus, again with respect to the canonical flat metric). A map M is equivelar of (Schlafli) type {p, q} if all of its vertices are q-valent, and all of its faces are regular p-gons. If the map is Archimedean, it can be described, as in [9], K. Drach et al.: Archimedean toroidal maps and their minimal almost regular covers 495 by the arrangement of polygons around a vertex, where a map of type (p1vp2 ■ ■-pk) has k polygons (ap1-gon, p2-gon,..., and apk-gon) in the given order incident to each vertex. A particular vertex structure of a map is called a type. We note here that there is some debate over how the word Archimedean should be used. In some settings it means just that all of the faces are regular polygons and all of the vertex figures are the same. When you add the requirement that, for any pair of vertices, there exists a symmetry mapping one to the other, these maps are sometimes called uniform (for example, see [8, 20, 21, 23]). As we will see below, Archimedean toroidal maps arise naturally as quotients of tessellations of the Euclidean plane with regular polygons; these tessellations are called Archimedean, and they are described in the same way as Archimedean maps. The following classical theorem gives a complete classification of planar Archimedean tessellations. Theorem 2.1 (Classification of Archimedean tessellations on the plane [9]). There are only 11 tessellations on the plane by regular polygons so that any vertex can be mapped to every other vertex by the symmetry of the tessellation. These are the following tessellations: {3, 6}, {4,4}, {6, 3}, (4.82), (3.122), (3.6.3.6), (3.4.6.4), (4.6.12), (32.4.3.4), (34.6), (33.42) (see Figure 1). The equivelar Archimedean tessellations of type {3, 6}, {4,4}, and {6,3} and the corresponding toroidal maps were considered in [7]. In this paper we will mainly work with the non-equivelar Archimedean tessellations of the Euclidean plane; denote A to be the set of all non-equivelar tessellations. Furthermore, for reference, on each such tessellation we can place a Cartesian coordinate system with the origin at a vertex of the tessellation. For the tessellations {4, 4}, (32.4.3.4), and (4.82), the coordinate system is further specified by assuming that the vectors ei := (1,0) and e2 := (0,1) represent the shortest possible translational symmetries of the tessellation. Similarly, for the remaining Archimedean tessellations, other than (33.42), we assume that the vectors e1 := (1,0) and e2 := (1/2, %/3/2) represent the shortest possible translational symmetries. This choice of coordinate system will allow us to utilize the geometry of the Gaussian and Eisenstein integers in our results (see Subsection 4.1). Given an Archimedean tessellation t, we call (e1, e2) the basis for t. For the tessellation (33.42) the basis will be specified separately in the later sections. Let (e1, e2) be the basis for t g A\ {(33.42)}. The set {Ae1 + pe2: A, p G Z} forms the vertex set of a regular tessellation which we call the tessellation associated with t, and which we denote by t*. By construction, (e1, e2) is also the basis for t*. To clarify this notation, we note here that for example if t is of type {6, 3}, then t* is of type {3,6}. Given an Archimedean tessellation t, denote by Tt the maximal group of translations that preserve t . As we already mentioned, Archimedean toroidal maps can be seen as quotients of planar Archimedean tessellations. These quotients can be written explicitly in terms of possible subgroups of Tt , due to the following theorem. Theorem 2.2 (Archimedean toroidal maps are quotients [19]). Let M be an Archimedean map on the torus. Then there exists an Archimedean tessellation t of the Euclidean plane and a subgroup G < Tt so that M = t/G. 496 Ars Math. Contemp. 17 (2019) 493-514 (a) {3, 6} (d)(4.82) (b) {4, 4} (e) (3.122) (c) {6, 3} (f) (3.6.3.6) (g) (3.4.6.4) (h) (4.6.12) (i) (32.4.3.4) (j) (34.6) (k) (33.42) Figure 1: Archimedean tessellations. This theorem shows that, for each type, there is a one-to-one correspondence between Archimedean toroidal maps and translation subgroups of Tt ; clearly, the pair of generators of every such subgroup should be comprised of non-collinear vectors. We point out here that the converse of Theorem 2.2 is also true, as clearly any map on the torus that is obtained as a quotient of an Archimedean tessellation by a translation subgroup is an Archimedean toroidal map. Let M be an Archimedean toroidal map; by Theorem 2.2, it can be written as t / (a, b), where t is the planar Archimedean tessellation (of the same type as M) and (a, b) < Tt is the translation subgroup with generators a, b e Tt . We use the standard notation Ta,b := t/ (a, b) for the map M. Note that the pair a, b is not uniquely defined by M, but the quotient is independent of possible choices. A flag of a planar tessellation t is a triple of an incident vertex, edge, and face of the tessellation. We can then define a flag of a toroidal map Ta,b as the orbit of a flag under the group (a, b). We note that when the map is combinatorially equivalent to an abstract polytope (see [16]), this is equivalent to a flag equaling a triple of an incident vertex, edge, and face of the map itself. Two flags of a map on the torus are said to be adjacent if they lift to flags in the plane that differ in exactly one element. K. Drach et al.: Archimedean toroidal maps and their minimal almost regular covers 497 Let N = t/H and M = t/G be Archimedean maps on the torus, where H is a subgroup of G. Then there is a surjective function n: N ^ M that preserves adjacency and sends vertices of N to vertices of M (and edges to edges, and faces to faces). The function n is called a covering of the map M by the map N. This is denoted by N \ M, and we say that N is a cover of M. We can use the notion of covering to create a partial order < on any non-empty set S of toroidal maps, where M < N if and only if N is a cover of M. A minimal cover of a map M G S is minimal with respect to this partial order in S. We note here that this notion of covering can also be generalized to maps on different surfaces, and to abstract polytopes of higher ranks. If S is the set of all regular maps that cover a given map M, then the minimal elements of the partial order are the minimal regular covers of M, as studied in [10, 18] for example. For a map M = t/G, where G = (a, b), we call the parallelogram spanned by the vectors a, b a fundamental region of M. Then a fundamental region for the covering map N = t/H, H < G, can be viewed as k fundamental regions of M 'glued together' (see Figure 2). It is easy to show that the number k is equal to the index [G : H] of the subgroup H in G. / / / / 3 \ / / 1 / / \ \ / I / 2 : 5 5 \ : / \ : : \ / / \ 5 -J b / 3 u / 7 1 t / \ / / a Figure 2: {4,4}u,v \ {4,4}a,b is a 5-sheeted covering, and the covering map {4,4}u,v is obtained by gluing together 5 fundamental regions of {4,4}a b. 2.1 Symmetries and automorphisms of tessellations and maps In this section we follow [3, 11, 12] and [13] in our notation and definitions. Let t be a tessellation of the Euclidean plane, and let Aut(T) be its symmetry group (the collection of isometries of the Euclidean plane that preserve the tessellation). Let G and H be two subgroups of Aut(T) generated by two linearly independent translations. The maps t/G and t/H are isomorphic if G and H are conjugate in Aut(T). A symmetry 7 G Aut(T) induces an automorphism of a toroidal map t/G if and only if it normalizes G, that is 7G7-1 = G; denote NormAut(T)(G) for the group of elements in Aut(T) that normalize G. Geometrically, such 7 maps fundamental regions of t/G to fundamental regions of t/G. Finally, the we define the automorphism group Aut(T/G) as the group induced by the normalizer NormAut(T)(G); in other words Aut(T/G) = NormAut(T) (G)/G. We will also denote the collection of symmetries NormAut(T)(G) as simply Sym(T/G). We note here that an automorphism of a map can equivalently be defined as an automorphism of the underlying graph that can be extended to a homeomorphism of the surface. 498 Ars Math. Contemp. 17 (2019) 493-514 A map M is called regular if its automorphism group acts transitively on the set of flags. A map M is called chiral if its automorphism group has two orbits on flags with adjacent flags lying in different orbits. A map M is called rotary if it is either regular or chiral. For the toroidal maps of type {4,4}, {3, 6}, and {6,3} the minimum possible number of flag orbits is one, given by regular maps of those types (which have been previously classified, see also Subsection 2.2). For other types of maps on the torus, the minimum number of flag orbits will not be one. However, we still would like to understand the maps of each type that achieve the fewest possible number of flag orbits. An Archimedean map on the torus is called almost regular if it has the same number of flag orbits under the action of its automorphism group as the Archimedean tessellation on the plane of the same type has under the action of its symmetry group. 2.2 Regular and chiral toroidal maps The classification in the next sections depends heavily on the classification of regular and chiral toroidal maps. Here we summarize the relevant details about toroidal maps of type {4,4} and {3, 6} that are needed in our results. The results in this subsection, and much more can all be found in [13]. Let t be a tessellation of the Euclidean plane of type {4,4} or {3,6}. Then Aut(r) is of the form Tt x S, where S is the stabilizer of a vertex of t, which we can assume to be the origin without loss; S is called a point stabilizer. Then let M = t/G be a toroidal map. Notice that every translation in Tt induces an automorphism of M (where the elements of G induce the trivial automorphism). Define x as the central inversion of the Euclidean plane, that is the symmetry that sends any vector u to -u. Then, as seen in Lemma 6 of [13], Aut(M) is induced by a group K so that Tt x (x) < K < Aut(T). Furthermore, there is a bijection between such groups K and subgroups K' of S containing x, and thus one needs to determine which symmetries in the point stabilizer S normalize G. Finally, the number of flag orbits of the toroidal map M is the index of NormAut(T) (G) in Aut(T), which is the same as the index of K' in S. First let us consider toroidal maps of type {4,4}; let t be the regular tessellation of the Euclidean plane of this type, and (ei, e2) be the basis for t. The point stabilizer S is generated by two reflections R1 and R2, where R1 is reflection across the line spanned by e1 + e2, sending vectors (x, y) to (y, x), and R2 is reflection across the line spanned by e1, sending vectors (x, y) to (x, -y). There are exactly three conjugacy classes of proper subgroups K' of S containing x but not equal to (x). In other words there are exactly five possible point stabilizers: all of S, only (x), and finally the three groups described next. The three subgroups are K' are (x, Ri), (x, R2), and (x, R1R2), and each has index 2 in S, where (x) has index 4 in S. We note that it is important for our classification to notice that a toroidal map of type {4,4} is regular if and only if K' contains both R1 and R2, as well as R1R2 which is the rotation by n/2 around the origin. This occurs only in the two well known families of regular toroidal maps, {4,4}(a 0),(o,a) and {4,4}(a a) (a _a), both of which have squares as fundamental regions. The chiral toroidal maps of type {4,4} also have squares as their fundamental regions, but have no reflections in their automorphism groups. Finally, the remaining classes of toroidal maps have fundamental regions that are not squares. Next, let us consider toroidal maps of type {3, 6}; let t be the regular tessellation of K. Drach et al.: Archimedean toroidal maps and their minimal almost regular covers 499 the Euclidean plane of this type. We use the previously described basis of ei = (1,0) and e2 = (1/2, %/3/2) to describe the symmetries of these maps. The point stabilizer S is again generated by two reflections R1 and R2, where R1 is reflection across the line spanned by e1 + e2, sending vectors (x, y) to (y, x), and R2 is reflection across the line spanned by e1, sending vectors (x, y) to (x + y, -y). Now there are exactly two conjugacy classes of proper subgroups K' of S containing X but not equal to (x). These subgroups K' are (x, RiR2}, with index 2 in S, and (x, R2) with index 3 in S, where (x) has index 6 in S. We note that it is important for our classification to notice that a toroidal map of type {3, 6} is regular if and only if K' contains both Ri and R2, as well as RiR2 which is the rotation by n/3 around the origin. This occurs only in the two well known families of regular toroidal maps, {3, 6}(a 0)i(o,a) and {3, 6}(aa)(2a_a). For those two families the fundamental regions are parallelograms composed of two regular triangles. The chiral toroidal maps of type {3, 6} also have parallelograms composed of two regular triangles as their fundamental regions, but have no reflections in their automorphism groups; this is similar to the type {4,4}. 3 Almost regular maps In this section we consider Archimedean tessellations of the torus with as much symmetry as possible. As we already mentioned, one natural way to understand the symmetry of a map is to consider the action of its automorphism group on its flags. Here we want to understand the maps on the torus with as few flag orbits as possible. Theorem 3.1 (Regular to almost regular maps). For Areg := {(4.82), (3.6.3.6), (3.122), (4.6.12), (3.4.6.4), (32.4.3.4)} , let t G Areg be an Archimedean tessellation of one of these types. Then rUjV is an almost regular Archimedean map if and only if (t*)u,v is a regular map on the torus, with t* being the regular tessellation associated with t. The proof of this theorem will follow from the following six propositions, each separately dealing with a type of map. In each case we will use the translational symmetries to simplify the problem by considering a fundamental region of t/Tt . Proposition 3.2 (Almost regular maps of type (4.82)). A map M = t/G on the torus of type (4.82) is almost regular (with three flag orbits) if and only if t */G is regular. Proof. Notice first that t* is of type {4,4}, and let (e1, e2) be the basis for t (and hence for t *). Assume that a map M on the torus of type (4.82) has exactly three flag orbits. For this to be the case, there must be the following symmetries in Sym(M), as shown in Figure 3: • reflection across a line in the direction e1 + e2 through the center of a square of the map; • reflection across a line in the direction e1 through the center of a square and the edge of an octagon. 500 Ars Math. Contemp. 17 (2019) 493-514 It was summarized in Subsection 2.2 that the existence of these symmetries is enough to show that t*/G is regular. Conversely, if t*/G is regular, then the fundamental region of M is a square, and each of the previous listed symmetries are elements of Sym(M). Furthermore, every trans-lational symmetry in Sym(T/TT) is also in Sym(T/G), and thus M has only three flag orbits. Note that the translations in Sym(T/TT) act on the flags in 24 flag orbits, and then the listed symmetries force there to only be three orbits. Figure 3: Minimum number of flag orbits for t of type (4.82), and a fundamental region of t/Tt , where the vectors ei and e2 form the boundary of the fundamental region. On the left of Figure 3, and other figures to follow, all the faces incident to single vertex are shown, and the flags in these faces, which can be seen as triangles in the barycentric subdivision of the tessellation, are colored based on their orbit. On the right of Figure 3, the fundamental region of t/Tt is shown in blue, with the underlying tessellation shown in black. One can see, for example, that this fundamental region contains 24 flags of t. □ The proofs of the following five propositions is similar to the proof of Proposition 3.2. Proposition 3.3 (Almost regular maps of type (3.6.3.6)). A map M = t/G on the torus of type (3.6.3.6) is almost regular (with two flag orbits) if and only if t */G is regular. Proof. Notice first that t* is of type {3, 6}, and let (ei, e2) be the basis for t. Assume that a map M on the torus of type (3.6.3.6) has exactly two flag orbits. For this to be the case, there must be the following symmetries in Sym(M), as shown in Figure 4: • reflection across a line in the direction e1 + e2 going through the centers of a hexagon and an adjacent triangle; • reflection across a line in the direction e1 going through the centers of two hexagons incident to the same vertex. As summarized in Subsection 2.2, the existence of these symmetries is enough to conclude that t* /G is regular. Conversely, if t*/G is regular, then each of the previous listed symmetries are elements of Sym(M). Furthermore, every translational symmetry in Sym(T/TT) is also in Sym(T/G), and thus M has only two flag orbits. Note that the translations in Sym(T/TT) act on the flags in 24 flag orbits, and then the listed symmetries force there to only be two orbits. □ K. Drach et al.: Archimedean toroidal maps and their minimal almost regular covers 501 Figure 4: Minimum number of flag orbits for t of type (3.6.3.6), and a fundamental region of t/Tt (as above, the vectors ei and e2 form the boundary of the fundamental region). Proposition 3.4 (Almost regular maps of type (3.122)). A map M = t/G on the torus of type (3.122) is almost regular (with three flag orbits) if and only if t */G is regular. Proof. Notice again that t* is of type {3, 6}, and let (e1, e2) be the basis for t. Assume that amap M on the torus of type (3.122) has exactly three flag orbits. For this to be the case, there must be the following symmetries in Sym(M), as shown in Figure 5: • reflection across a line in the direction e1 + e2 going through the centers of a 12-gon and an adjacent triangle; • reflection across a line in the direction e1 going through the centers of two adjacent 12-gons. As in the previous proposition, the existence of these symmetries again forces t*/G to be regular. Figure 5: Minimum number of flag orbits for t of type (3.122), and a fundamental region of t/H (as above, the vectors e1 and e2 form the boundary of the fundamental region). Conversely, if t*/G is regular, then each of the previous three listed symmetries are elements of Sym(M). Furthermore, every translational symmetry in Sym(T/TT) is also in Sym(T/G), and thus M has only three flag orbits. Note that the translations in Sym(T/TT) act on the flags in 36 flag orbits, and then the listed symmetries force there to only be three orbits. □ Proposition 3.5 (Almost regular maps of type (4.6.12)). A map M = t/G on the torus of type (4.6.12) is almost regular (with six flag orbits) if and only if t */G is regular. 502 Ars Math. Contemp. 17 (2019) 493-514 Proof. Notice again that t* is of type {3, 6}, and let (ei, e2) be the basis for t. Assume that a map M on the torus of type (4.6.12) has exactly six flag orbits. For this to be the case, there must be the following symmetries in Sym(M): • reflection across a line in the direction e1 + e2 going through the centers of a 12-gon and an adjacent hexagon; • reflection across a line in the direction e1 going through the centers of a 12-gon and an adjacent square. As in the previous proposition, the existence of these symmetries again forces t*/G to be regular. Figure 6: Minimum number of flag orbits for t of type (4.6.12), and a fundamental region of t/Tt (as above, the vectors e1 and e2 form the boundary of the fundamental region). Conversely, if t*/G is regular, then each of the previous three listed symmetries are elements of Sym(M). Furthermore, every translational symmetry in Sym(T/TT) is also in Sym(T/G), and thus M has only six flag orbits. Note that the translations in Sym(T/TT) act on the flags in 72 flag orbits, and then the listed symmetries force there to only be six orbits. □ Proposition 3.6 (Almost regular maps of type (3.4.6.4)). A map M = t/G on the torus of type (3.4.6.4) is almost regular (with four flag orbits) if and only if t */G is regular. Proof. Notice again that t* is of type {3,6}. Let (e1, e2) be the basis for t. Assume that a map M on the torus of type (3.4.6.4) has exactly four flag orbits. For this to be the case, there must be the following symmetries in Sym(M): • reflection across a line in the direction e1 + e2 going through the centers of a hexagon and a triangle sharing an incident vertex; • reflection across a line in the direction e1 going through the centers of a 12-gon and an adjacent square. Again the existence of these symmetries again forces t*/G to be regular. Conversely, if t*/G is regular, then each of the previous three listed symmetries are elements of Sym(M). Furthermore, every translational symmetry in Sym(T/TT) is also in Sym(T/G), and thus M has only four flag orbits. Note that the translations in Sym(T/TT) act on the flags in 48 flag orbits, and then the listed symmetries force there to only be four orbits. □ K. Drach et al.: Archimedean toroidal maps and their minimal almost regular covers 503 Figure 7: Minimum number of flag orbits for t of type (3.4.6.4), and a fundamental region of t/Tt (as above, the vectors ei and e2 form the boundary of the fundamental region). Proposition 3.7 (Almost regular maps of type (32.4.3.4)). A map M = t/G on the torus of type (32.4.3.4) is almost regular (with five flag orbits) if and only if t */G is regular. Proof. Notice again that t* is of type {4,4}. Let (ei, e2) be the basis for t. Assume that a map M on the torus of type (32.4.3.4) has exactly five flag orbits. For this to be the case, there must be a rotation by n/2 around the center of a square in Sym(M). This symmetry can be represented by R\R2 as described in Subsection 2.2, and thus t*/G is either regular or chiral. However, Sym(M) must also contain a reflection across the edge of adjacent triangles in the direction of e1 + e2. This means that the 8 flags in the fundamental region of t*/Tt are all in the same orbit, and thus t*/G is regular. Figure 8: Minimum number of flag orbits for t of type (32.4.3.4), and a fundamental region of t/Tt drawn in two equivalent ways so that to show existence of mirror symmetries (second picture) and rotational symmetry (third picture); as above, the vectors ei and e2 form the boundary of the fundamental region. Conversely, if t*/G is regular, then there is a rotation by n/2 around the center of a square, as well as a reflection across the edge of adjacent triangles in Sym(M). Furthermore, every translational symmetry in Sym(T/TT) is also in Sym(T/G), and thus M has only five flag orbits. □ Theorem 3.8 (Rotary to almost regular toroidal map). Let t be the Archimedean tessellation of type (34.6). Then tu,v is an almost regular Archimedean map (with ten flag orbits) if and only if tu* v is a rotary map on the torus. 504 Ars Math. Contemp. 17 (2019) 493-514 Proof. Suppose that the map ru v has exactly ten flag orbits. For this to be the case, there must be a rotation by n/3 around the center of a hexagon in Sym(rUjV). The existence of this symmetry forces t* / (u, v) to be rotary. Note, these are the only reflexive symmetries in Sym(Tu,v). Figure 9: Minimum number of flag orbits for t of type (34.6), and a fundamental region of t/H (as above, the vectors ei and e2 form the boundary of the fundamental region). Conversely, if tu*,v is rotary, then there is a rotation by n/3 around the center of a hexagon in Sym(TU,V). Furthermore, every translational symmetry in Sym(T/TT) is also in Sym(TU V), and thus tu v has only ten flag orbits. Note that the translations in Sym(T/TT) act on the flags in 60 flag orbits, and then the listed symmetry forces there to only be ten orbits. □ Theorems 3.1 and 3.8 provide us with a fair understanding of how almost regular maps of type Areg and (34.6) look: for each of them the associated map on the torus must be regular, respectfully rotary. The only remaining Archimedean tessellation not covered by the previous two results is (33.42). Since the translation subgroup of the symmetry group of (33.42) does not coincide with the symmetry group of one of the regular planar tessellations (as it is for all tessellations in Aeg), and does not contain the rotation subgroup of a regular planar tessellation (as it is for the tessellation (34.6)), we have to deal with (33.42) separately and with different techniques. In order to state a complete characterization of almost regular maps of type (33.42), we introduce the following notation: write (e1, e2) for the positively-oriented basis of the plane E2 represented by the shortest non-parallel translations that are in the symmetry group of (33.42). Recall also that Tt stands for the maximal translation subgroup of the symmetry group Sym(T) of a given Archimedean tessellation t. Theorem 3.9 (Almost regular maps of type (33.42)). Let t = (33.42). Then M = t/G, with G < Tt, is an almost regular Archimedean map (with five flag orbits) if and only if the translation subgroup G is of the form (ce1, — de1 + 2de2), or (ce1, — de1 + (c + 2d) e2) (3.1) for non-zero integers c and d. K. Drach et al.: Archimedean toroidal maps and their minimal almost regular covers 505 Remark 3.10. Observe that the statement above, in fact, does not depend on the choice of the basis for E2. The explicit coordinate form (3.1) was used only to simplify later search for minimal almost regular covers in the proof of Theorem 4.4. Remark 3.11. The groups listed in (3.1) are never isomorphic. Proof. The beginning of the proof is similar to the proofs given for the tessellations in A \ {(33.42)}. Assume that a map M has exactly five flag orbits. For this to be the case, there must be the following symmetries in Sym(M): • reflections across a horizontal line through the center of a square (see Figure 10(a)); • reflections across a vertical line through the center of a square (see Figure 10(b)). We will write h, respectively v, for a reflection across a horizontal, resp. vertical, line through the center of a square, where a horizontal line is in the direction of ei. (a) Reflection across the horizontal line; 10 flag orbits in the fundamental region. (b) Reflection across the vertical line; 10 flag orbits in the fundamental region. (c) The minimal number of flag orbits in the fundamental region. Figure 10: Flag orbits in the fundamental region of (33.42) /T(33.42). Let us find all possible subgroups G < Tt such that the listed symmetries preserve G by conjugation. Suppose G is generated by a pair of non-parallel vectors a, b e Tt, and assume that h o u o h-1 e G and v o u o v-1 e G for every u e G. Because the basis (e1, e2) of E2 was chosen in such a way that both e1 and e2 are the symmetries of t which generate the group Tt (i.e. (e1, e2) = Tt), there exist two pairs of integers a1, a2 and b1, b2 such that a = a1 ei + a2e2 = (a1, a2), b = 61 ei + 62e2 = (61,62). Note that if 2a1 + a2 = 261 + b2 = 0, then a and b are parallel, which is impossible. Hence, without loss of generality we can assume 2a 1 + a2 = 0; this technical assumption will be used later in the proof. In order to understand the structure of the group G, observe that Rv(e1) := v o e1 o v 1 = —e1, Rv(e2)= e2 — e1, Rh(e1) := h o e1 o h-1 = e1, Rh(e2) = e1 — e2. (3.2) For an element u e Tt, the action of Rv and Rh on u is defined using (3.2) by linearity. Since the reflection across a vertical line through the center of a square preserves G by conjugation, the group G must contain the vector Rv(a) = (—a1 — a2, a2), as it is easy to compute from (3.2). Hence G contains the vector a — Rv(a) = (2a1 + a2, 0), which is 506 Ars Math. Contemp. 17 (2019) 493-514 not zero since 2a1 + a2 = 0 by our assumption. Therefore, G contains a proper non-trivial subgroup G' of vectors with vanishing second coordinate. Pick c = (c1,0) with c1 > 0 to be a generator of G'. Observe that, in fact, c is the shortest vector among all vectors in G with positive first coordinate and vanishing second coordinate. Let d € G be a vector such that G = (c, d). Moreover, since (c, d) = (c, d + k c) for every k € Z, by picking an appropriate k we may assume that d is chosen in such a way that in coordinates d = (d1, d2) we have Now we use the second symmetry from our list: since the conjugation by a reflection across a horizontal line through the center of a square preserves G, the vectors (2d1 + d2, 0) and (2d1 + d2 - c1,0) must be in G'. Indeed, (3.2) and linearity of Rh gives us Rh (d) = (d1 +d2, —d2), and this vector must be in G. Hence, Rh (d)+d = (2d1+d2,0) € G' < G. Similarly, Rh (d) + d — c = (2d1 + d2 — c1,0) € G' < G. Recall that c = (c1,0) was the shortest vector in G' among all vectors with positive first coordinate. Therefore, as (2d1 + d2,0) € G' we must have either 2d1 + d2 = 0, or 2d1 + d2 > c1 (note that 2d1 + d2 is necessarily non-negative by assumption (3.3)). In the latter case it then follows from (3.3) that 0 < 2d1 + d2 — c1 < c1, which is possible, again by minimality of c1, only if 2d1 + d2 — c1 =0. Therefore, either d2 = — 2d1, or d2 = —2d1 + c1, and hence setting c := c1, d := —d we obtain that the group G must be of one of the types in (3.1). One implication in Theorem 3.9 is proven. Conversely, it is straightforward to see that a group of one of the types in (3.1) is preserved by conjugation with both h and v, and hence the symmetry group Sym(r/G) contains both h and v, which implies that t/G has only five flag orbits, and thus is an almost regular Archimedean map. □ 4 Minimal covers of Archimedean toroidal maps In this section we will prove the Main Theorem. This will be done by combination of three statements — Theorem 4.2, 4.3 and 4.4. Prior the proofs, in the following subsection we recall some of the facts about the Gaussian and the Eisenstein integers — a number-theoretic tool that provides a natural 'language' for our main results. 4.1 Gaussian and Eisenstein integers The Gaussian and Eisenstein integers provide an essential ingredient for understanding the Archimedean toroidal maps. The plots of these domains in the complex plane are the vertex sets of regular tessellations: a tessellation of type {4,4} for the Gaussian integers and {3,6} for the Eisenstein integers. We will follow [7] and [14] in our notation. The Gaussian integers Z[i] are defined as the set {a + bi : a, b € Z} c C, where i is the imaginary unit, with the standard addition and multiplication of complex numbers. Similarly, the Eisenstein integers Z[w] are defined as {a + bw : a, b € Z} c C, where w := (1 + i\/3)/2. We adopt a unifying notation Z[a] with a € {i, w} to denote either of these two sets. Since we are dealing with different types of integers, to avoid confusion we will call rational integers the elements of Z. (3.3) K. Drach et al.: Archimedean toroidal maps and their minimal almost regular covers 507 As in [7], in this paper we use the following notation: given a = a + 6a e Z[a], we call its conjugate the number a := a + 6a, where a is the conjugate complex number to a e C. Also we call Re a := a and Im a := 6 the real and imaginary parts, respectively. Note here that if a = i, then this is the traditional notion of 'real part' and 'imaginary part' of a complex number. However, if a = w, then the 'traditional' real and imaginary parts of a + 6w are a + 6/2 and 6%/3/2, respectively. For every a = a + 6a e Z[a], we assign the norm N(a) := aa. An integer a e Z[a] \ {0} divides ft e Z[a] if and only if there is y e Z[a] such that ft = aY. Recall that, in the ring of Gaussian integers, the units are only ±1, ±i, while in the ring of Eisenstein integers the units are only ±1, ±w, ±w. Two integers a, ft e Z[a] are called associated if a = fte for some unit e. Let us recall the concept of a greatest common divisor for rings of integers. An integer Y e Z[a] is a greatest common divisor (GCD) of a, ft e Z[a] with N(a) + N(ft) = 0, if Y divides both a and ft and for every y' with the same property it follows that y' divides Y. It is well-known that both Z[i] and Z[w] are Unique Factorization Domains (see [14]), that is the rings with unique (up to associates) factorization into primes. Hence, a greatest common divisor is well-defined, again up to associates. Because of that, we write y = GCD(a, ft) implying that y is defined up to multiplication by an associate. We also agree that if there is a rational integer n among associates to GCD(a, ft), then we specifically take GCD(a, ft) := |n|. For example, GCD(3, 6i) e {3, -3,3i, -3i}, and thus by our convention GCD(3,6i) = 3. The power of Gaussian and Eisenstein integers is coming from the natural identification of these sets with the vertex set of a regular tessellation {4,4} or {3, 6}. In particular, we can identify the basis (ei, e2) (see Section 2) with the ordered pair (1, a) from Z[a]. This identification leads to the group homomorphism of T{4,4} (resp. T{3,6}) with Z[i] (resp. Z[w]) — where the latter groups are regarded as Abelian groups with respect to addition. From this point of view, we will identify every two vectors a = (ai, a2), b = (61, 62) with the complex numbers a = a1 + a2a and ft = 61 + 62a. Finally, write := Ta,b, and, for further brevity, Tn := rn,CTn. Note here that the pair of vectors n, an span a square if a = i, or a rhombus with angle n/3 if a = w. Therefore, if n and n' are equal up to associates, then Tn = rv>. 4.2 Proof of the Main Theorem In Theorems 4.2-4.4 we will prove that each Archimedean map on the torus has a unique minimal almost regular cover on the torus, which we will construct explicitly. To accomplish these proofs, we will use some known results for equivelar toroidal maps (see [7]). Proposition 4.1 (Covering correspondence of maps and their associates). Let t be an Archimedean tessellation of the plane, not of type (33.42), and G and H two subgroups of Tt . Then t/H covers t/G if and only if t */H covers t */G. Proof. Suppose t/H \ t/G. Then H is a subgroup of G, and both of them are subgroups of Tt . By construction, the groups Tt and Tt* are equal, and so the same subgroup structure holds in Tt*, which means that t* /H covers t* /G. Conversely, if t* /H covers t* /G, then H is a subgroup of G, and G is a subgroup of Tt*. The latter group is again equal to Tt by the very definition of t*. Hence, t/H \ t/G. □ Observe that for every t e A \ {(33.42)} we have a well-defined basis (e1, e2) such 508 Ars Math. Contemp. 17 (2019) 493-514 that (ei, e2) = Tt = Tt*, where t* = {p, q} is the regular tessellation associated to t. Therefore, every u G Tt with the coordinates (ui, u2) in the basis (e1, e2) can be identified by the homomorphism discussed in Subsection 4.1 with the integer u1 + u2a G Z[a] (here a depends on t*). We will use this equivalent language instead of the vector language in order to state the following two theorems which are the first two main results of the paper. Theorem 4.2 (Minimal almost regular covers for toroidal maps of type Areg). Let Ta,p be an Archimedean map given as a quotient of an Archimedean tessellation t g Areg by a translation subgroup (a, ft) < Tt generated by two integers a, ft G Z[a] \ {0} with a/ft G Z. Then the map Tn with Ngogc (1 + a), if N(1 + a) divides ^ - ^ and ^ - ^, Im(aß), otherwise, where c = GCD (Re a, Im a, Re ß, Im ß) is a unique minimal almost regular cover of Ta,ß. Moreover, the number kmin of fundamentals regions of Ta,ß glued together in order to the fundamental region of Tn is equal to N(l+ff) c km'n ^ IIm(aß)l Im(aß)l if N(1 + a) divides ^ - ^ and ^^ - ^, otherwise. Proof. This theorem is a direct consequence of [7, Theorem 3.6]. Indeed, let t* be the regular tessellation associated to the Archimedean tessellation t. By Theorem 3.1, there is one-to-one correspondence between the translation subgroups of Tt* that generate regular maps on the torus and translation subgroups of Tt that generate almost regular maps on the torus. By Proposition 4.1, any such correspondence preserves the covering order and, in particular, sends minimal elements to minimal elements. Therefore, t*/H is the minimal regular cover of t*/G, where G := (a, ft), if and only if t/H is the minimal almost regular cover of t/G. By [7, Theorem 3.6], every map t*/G has a unique minimal regular cover t*/H, where H < Tt* can be given explicitly in terms of number-theoretical properties of a, ft G Z[a]. Hence, the same holds for t/G, which yields existence and uniqueness of a minimal almost regular cover. The explicit form of H from [7, Theorem 3.6] translates verbatim into the explicit form given in Theorem 4.2; this finishes the proof. □ Theorem 4.3 (Minimal almost regular covers for toroidal maps of type (34.6)). Let Ta^ be an Archimedean map given as a quotient of an Archimedean tessellation t = (34.6) by a translation subgroup (a, ft) < Tt generated by two integers a, ft G Z[w] \ {0} with a/ft G Z. Then the map Tn with Im(aft) n = 1 where y = GCD(a,ft) is a unique minimal almost regular cover of Ta,p. Moreover, the number kmin of fundamentals regions of Ta,p glued together in order to obtain the fundamental region of Tn is equal to _ |Im(aft)| km ^min N (y) n 2 c K. Drach et al.: Archimedean toroidal maps and their minimal almost regular covers 509 Proof. The proof is similar to the proof of Theorem 4.2, where instead of [7, Theorem 3.6] we use [7, Theorem 3.5]. □ Recall that we associate to the tessellation of type (33.42) the basis (ei, e2) comprised of a pair of translations that are in the symmetry group of (33.42) such that e1 connects the centers of two adjacent squares and e2 is the shortest translation vector forming an acute angle to e1. Everywhere below we assume that the coordinate representation of a translation from T(3342) is given with respect to the basis (e1, e2). Theorem 4.4 (Minimal almost regular covers for toroidal maps of type (33.42)). Let Ta,b be an Archimedean toroidal map given as a quotient of the tessellation t = (33.42) by a translation subgroup (a, b) < Tt generated by two vectors a = (ai, a2) and b = (bi, b2) with A := a1b2 — a2b1 = 0. Then for Ta,b there exists and is unique a minimal almost regular cover tu,v generated by the subgroup (u, v) where '((c, 0), (—di,c + 2di)) <(c, 0), (—d2, 2d2)), provided 02 — and ^ — are even, r 9i 92 gi g2 ' otherwise, gi = GCD(a2, 62), 92 = GCD(2ai + a2, 2bi + 62), A j A f 11 \ , A c = —, di = — —--1--, «2 =--. gi 2 \gi 92 J 92 Moreover, the number kmin of fundamentals domains of Ta,b glued together in order to obtain the fundamental region of tu,v is equal to if o, — 201+^2 and 99l — are even, gi g2 gi g2 otherwise. Proof. Our strategy in proving Theorem 4.4 will be the following: we explicitly describe all almost regular covers for Ta b and then determine the one which is the smallest (under the covering relation). Suppose that an Archimedean map tu v is a cover of Ta b. This is equivalent of saying that the group (u, v) is a proper subgroup of (a, b), which in algebraic terms is equivalent to existence of a linear integer relation between the generators of both groups: I nia + mib = u, I n2a + m2b = v, (4.1) for some integers ni, n2, mi, m2 with nim2 = n2mi. (The last condition guarantees that u and v are, in fact, non-parallel.) If, on top, tu v is almost regular, then the generators u, v might be chosen to be of one of the types in (3.1) (see Theorem 3.9). We now consider these two cases one by one. Case 1: suppose u = c e1 and v = —d e1 + 2d e2 for some non-zero integers c and d. Then, in order to find the generators of such type, we must solve the following system of k 510 Ars Math. Contemp. 17 (2019) 493-514 vector Diophantine equations {ni a + mi b = c e1, 1 + 1 1 (4.2) n2a + m2b = — d e1 + 2d e2, for the variables n1, n2, m1, m2 treating c and d as parameters. The first equation in (4.2) in coordinates is equivalent to the system of linear Diophan-tine equations iniai + mibi = c, (4 3) |nia2 + mi&2 = 0. By the standard methods the full family of solutions for (4.3) is n1 = — k, m1 = — — k, c =— k, k € Z*, (4.4) gi gi gi where we recall that — = a1 62 — a261 and g1 = GCD(a2, 62) (here Z* stands for the set of all non-zero integers). Similarly, the second equation in (4.2) in coordinates reads |n2ai + m-2&i = —d, (4 5) |n2a2 + m-2&2 = 2d; Multiplying the first equation by 2 and adding the second we get n2(2ai + 0,2) + m2(2bi + 62) = 0, from which we conclude, after straightforward cancellations, that the full family of solutions for (4.5) is 2bi + 62 2oi + 0^ — n2 = -s, m2 =--s, d =--s, s € Z , (4.6) g2 g2 g2 where g2 = GCD(2o1 + o2,261 + 62). Concluding Case 1 from obtained solution (4.4) and (4.6), we see that Tu v is an almost regular Archimedean cover of Ta b and is of the first type in (3.1) if and only if (u, v) = ^ k, ^ s, —2—^ =: Gm. Finally, observe that for every given pair of non-zero integers k and s the almost regular map T/Gfc,s covers t/G1,1. Hence, by definition of a minimal cover, t/G1,1 is the minimal almost regular cover (for Ta,b) of the first type in (3.1). Note that the full family of toroidal maps T/Gfc,s does not form a totally ordered set with respect to covering; however, this poset has a unique minimal element. We will see a similar type of covering behavior later. Finishing Case 1, we compute the number of fundamental regions of (a, b) one should glue together in order to obtain the fundamental region of G11. This is done by comparing areas of those regions. In the standard basis in E2 the area of the fundamental region of (a, b) is equal to Ao := |a x b| = |—| • |e1 x e2| . K. Drach et al.: Archimedean toroidal maps and their minimal almost regular covers 511 Similarly, by using (4.4) and (4.6) we compute the area of the fundamental region of G11: 2A2 A1 := |c ei x (-d ei + 2d e2)| = 1-r •lei x e2|. |g1g2| Therefore, the number we are looking for is equal to A 41 = 2 Ao g1g2 (4.7) Observe that 2(0462 - «2^1) = (2«4 + 02)61 - (261 + 62)0,2, (4.8) and hence the right hand side in (4.7) is an integer. Of course, the same conclusion likewise follows from the geometric meaning of 41/40. Case 2: suppose u = c e1 and v = -d e1 + (c + 2d)e2 for some non-zero integer c and an integer d; this is the second type in (3.1). We proceed similarly to Case 1, with a bit more involved computation. Again, in order to find all almost regular covers of the second type we have to find all solutions of the system {«4a + m1b = c ei, n2a + m2b = -d ei + (c + 2d)e2, for integers n1, n2, m1, m2 treating c and d as parameters. Similarly to Case 1, the solutions to the first equation in this system have the following form: n1 = —k, m1 = -02k, c = —k, k € Z*. (4.9) g1 g1 g1 The second equation from the system above in coordinates reads: {n2o1 + m261 = — d, 21 + 2 1 , (4.10) «202 + m.262 = c + 2d. Again, multiplying the first equation by 2 and adding the second one we obtain, by using (4.9), «2(201 + 02) + m2(261 + 62) = c = — k. (4.11) g1 This is a linear nonhomogeneous Diophantine equation in n2 and m2. The standard theory of linear Diophantine equations tells us that the any solution of (4.11) is the sum of a partial solution to the given equation and of the general solution of the corresponding homogeneous equation n2 (201 + 02) + m2 (261 + 62) = 0. Moreover, a necessary condition for (4.11) to have a solution is that A g2 divides — k. (4.12) g1 512 Ars Math. Contemp. 17 (2019) 493-514 Using (4.8) it is straightforward to check that A/(gig2) G Z* if and only if «2 2ai + a2 62 2bi + 62 ,,, and---are even. (4.13) gi g2 gi g2 Therefore, if condition (4.13) is met, then the necessary condition (4.12) is satisfied for every integer k. On the other hand, if (4.13) is violated, then (4.12) can be only satisfied if k is even (as it follows from (4.8)). Let us consider these two sub-cases. Sub-case 2.1: Suppose that condition (4.13) is violated; hence k is necessarily even. Then it is straightforward to check that the pair of integers / 62 , , «2 , n2 = -k, m2 =--k 2 2gi , 2 2gi provides a partial solution of equation (4.11). Therefore, in Sub-case 2.1 the full family of solutions of (4.11), and thus of (4.10), has the form 62 , , 26i + 62 «2 = -k +--s, 2gi g2 «2 , 2«i + 02 „ „ m2 =---k--——2s, k G 2Z, s G Z. (4.14) 2gi g2 d A , A d =--k--s, 2gi g2 Concluding Sub-case 2.1 of Case 2 by combining (4.9) and (4.14), we see that, provided condition (4.13) is not met, tu,v is an almost regular Archimedean cover of Tab and is of the second type in (3.1) if and only if (u, v) = / f 2 Al, o) ,(Al + As, —2As) \ =: HiiS, W JW \gi g2 g2 // where l, s G Z*. Now let us check the covering relations. Note that for a given pair of non-zero integers l and s, the almost regular map t/H; ,s non-trivially covers t/Gi, i. Therefore, in the poset of coverings of almost regular maps of the form T/Gfc,s (see Case 1) and t/HIjS the quotient of t by Gi i is the unique minimal element. Sub-case 2.2: assume condition (4.13) is satisfied. This is equivalent of saying that both pairs ai/gi, (2ai + «2)/g2 and 62/gi, (26i + 62)/g2 consist of integers with the same parity. As we showed above, this is also equivalent to A/(gig2) G Z*. If k is even, then we can run verbatim the same arguments as in Sub-case 2.1, with the same conclusion. Hence we can assume that k is odd. But if k is odd and condition (4.13) is met, then the pair of numbers , 62 , 26i + 62 , «2 , 2«i + 02 «2 = -k +--, m-2 =--k-- 2gi 2g2 2gi 2g2 are necessarily full integers, and moreover provide a partial solution to (4.11). Therefore, we can write down a complete solution to (4.11): 62 . , 26i + 62 26i + 62 60 26i + 62 n2 = -k +---1--s = -k +--(2s + 1), 2gi 2g2 g2 2gi 2g2 «^ 2«i + «2 2«i + «2 «2 , 2«i + «2 1. m2 =--k----s =--k--(2s + 1), 2gi 2g2 2g2 2gi 2g2 K. Drach et al.: Archimedean toroidal maps and their minimal almost regular covers 513 where s G Z. Substituting this solution in either of the equations in (4.10), we obtain d=- £k - 4(2s+1>- where both k and 2s + 1 are some odd integers. Similarly to as we did before, concluding Sub-case 2.2 of Case 2, we see that, provided condition (4.13) is satisfied and k is odd (the case k was already discussed), tu,v is an almost regular Archimedean cover of Ta b and is of the second type in (3.1) if and only if (u, v) = ((A(21 +1), , (A(21 +1) + A(2s +1), - A(2s +) =: where l, s G Z. Observe that t/F1jS covers t/F0j0 for any pair l, s G Z*. Finally, comparing the groups Gi,i = ((A, °),(A, -2A)) and Fo,o = ((A,°),(A + -A \\01 / \g2 92)1 \\91 J V2gl 2g2 92 we conclude that the almost regular map t/G1;1 non-trivially covers the almost regular map t/Fo,o. Therefore, summing up the results of Case 1 and Case 2, we obtain that if condition (4.13) is satisfied, then t/F00 is a minimal almost regular Archimedean map that covers Ta,b. Otherwise, t/G1;1 is a minimal almost regular cover. In both cases these minimal covers are unique elements in the corresponding posets of possible almost regular covers. This almost finishes the proof of Theorem 4.4. The only thing that is left to check is that in Sub-case 2.2 of Case 2, provided k is odd, the number of fundamental regions of (a, b) one should glue together to obtain the fundamental region of F0 0 is equal to A 9192 (note that this is an integer). This computation is similar to the one found at the end of Case 1, and is thus omitted; this completes the proof of Theorem 4.4. □ References [1] U. Brehm and W. Ktihnel, Equivelar maps on the torus, European J. Combin. 29 (2008), 18431861, doi:10.1016/j.ejc.2008.01.010. [2] U. Brehm and E. Schulte, Polyhedral maps, in: J. E. Goodman and J. O'Rourke (eds.), Handbook of Discrete and Computational Geometry, CRC Press, Boca Raton, Florida, CRC Press Series on Discrete Mathematics and its Applications, pp. 345-358, 1997. [3] J. Collins and A. Montero, Equivelar toroids with few flag-orbits, arXiv:1802.07837 [math.CO] . [4] M. Conder and P. 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ARS MATHEMATICA CONTEMPORANEA 17 (2019) 515-524 https://doi.org/10.26493/1855-3974.1763.6cb (Also available at http://amc-journal.eu) Symplectic semifield spreads of PG(5, qf), q even Valentina Pepe * Sapienza University of Rome, Italy * Received 23 July 2018, accepted 14 July 2019, published online 24 November 2019 Abstract Let q > 2-34t be even. We prove that the only symplectic semifield spread of PG(5, qf'), whose associate semifield has center containing Fq, is the Desarguesian spread. Equiva-lently, a commutative semifield of order q3t, with middle nucleus containing Fqt and center containing Fq, is a field. We do that by proving that the only possible Fq -linear set of rank 3t in PG(5, qt) disjoint from the secant variety of the Veronese surface is a plane of PG(5,qt). Keywords: Semifields, spreads, symplectic polarity, linear sets, Veronese variety. Math. Subj. Class.: 05B25, 51E15, 51E23, 14M12 1 Introduction Let PG(r - 1, q) be the projective space of dimension r - 1 over the finite field Fq of order q. An (n — 1)-spread S of PG(2n — 1, q), which we will call simply spread from now on, is a partition of the point-set in (n — 1)-dimensional subspaces. With any spread S it is associated a translation plane A(S) of order qn via the Andre-Bruck-Bose construction (see e.g. [7, Section 5.1]). Translation planes associated with different spreads of PG(2n — 1, q) are isomorphic if and only if there is a collineation of PG(2n — 1, q) mapping one spread to the other (see [1] or [16, Chapter 1]). A spread S is said to be Desarguesian if A(S) is isomorphic to AG(2, qn) and hence a plane coordinatized by the field of order qn. The spread S is said to be a semifield spread if A(S) is a plane of Lenz-Barlotti class V and this is equivalent to saying that A(S) is coordinatized by a semifield. *The author acknowledges the support of the project "Polinomi ortogonali, strutture algebriche e geometriche inerenti a grafi e campi finiti" of the SBAI Department of Sapienza University of Rome. E-mail address: valepepe@sbai.uniroma1.it (Valentina Pepe) ©® This work is licensed under https://creativecommons.Org/licenses/by/4.0/ 516 Ars Math. Contemp. 17 (2019) 493-514 A finite semifield S = (S, +, *) is a finite algebra satisfying all the axioms for a skew-field except (possibly) associativity of multiplication. The subsets N; = {a e S : (a* b) *c = a* (b*c), Vb,c e S}, Nm = {b e S : (a* b) *c = a* (b*c), Va,c e S}, Nr = {c e S : (a* b) *c = a* (b*c), Va,b e S} and K = {a e N; n Nm n Nr : a*b = b*a, Vb e S} are fields and are known, respectively, as the left nucleus, the middle nucleus, the right nucleus and the center of the semifield. A finite semifield is a vector space over its nuclei and its center. If A(S) is coordinatized by the semifield S, then S has order qn and its left nucleus contains Fq. Semifields are studied up to an equivalence relation called isotopy, which corresponds to the study of semifield planes up to isomorphisms (for more details on semifields see, e.g., [7]). The spread S is said to be symplectic if the elements of S are totally isotropic with respect to a symplectic polarity of PG(2n -1, q). If A(S) is coordinatized by the semifield S, then S is called symplectic semifield and if its center contains Fs < Fq, then from S we get by the cubical array (see [13]) a semifield isotopic to a commutative semifield with middle nucleus containing Fq and center containing Fs ([11]). Let q be even. For n = 2, there is the following remarkable theorem due to Cohen and Ganley. Theorem 1.1 ([6]). A commutative semifield of order q2 with middle nucleus containing Fq is afield. For n > 2, the only known commutative semifields, that are not a field, are the KantorWilliams symplectic pre-semifields of order qn and n > 1 odd ([12]) and their commutative Knuth derivatives ([11]). Symplectic semifield spreads in characteristic 2 with odd dimension over F2 give arise to Z4-linear codes and extremal line sets in Euclidean spaces ([4]). Most of the above mentioned results are obtained with an algebraic approach, whereas ours is mainly geometric. For small n, the study of semifield spreads has shown to be a good way to classify semifields. Let M (n, Fq) be the set of all n x n matrices over Fq. Without loss of generality, we may always assume that S(to) := {(0, y) : y e Fj;} and S(0) := {(x, 0) : x e FJ^} belong to S, hence we may write S = {S(A) : A e C} U S(to), with S(A) := {(x, xA) : x e F;}, with C c M(n, Fq) such that |C| = qn and C contains the zero matrix. The set C is called the spread set associated with S. In order to have a semi-field spread, the non-zero elements of C must be invertible and C must be a subgroup of the additive group of M(n, Fq) ([7, Section 5.1]), hence C is a vector space over some subfield of Fq. If we choose the symplectic polarity induced by the alternating bilinear form ^((xi, yi), (x2y2)) = xiyif - yixj, x4, yi e F;, then the subspace S(A) e S is totally isotropic if and only if A is symmetric. The symmetric matrices form an n(n2+1) -dimensional subspace of M (n, Fq) that then induces a PG ^ n(n2+1) - 1, qj. The rank-1 symmetric matrices form the Veronese variety V of degree 2 of PG ^n(n2+1) - 1, qj (this V Pepe: Symplectic semifield spreads of PG(5 ,qt), q even 517 is the so called determinant representation of the Veronese variety of degree 2, see [8, Example 2.6]). Hence the singular symmetric matrices form the (n - 2)-th secant variety, say Vn_2, of the Veronese variety. If C is an Fs-vector space, q = s4, then dimFs C = nt and it defines a subset L of PG(""+11 - 1 , q called Fs-linear set of rank nt (for a complete overview on linear sets see [18]). So to a symplectic semifield spread of PG(2n - 1, q) there corresponds an Fs-linear set L, q = s4, of PG ^"("+1) - 1, qj of rank tn such that L n V"_2 = 0 (see also [15]). We recall the associated semifield has left nucleus containing Fq and if Fs is the maximum subfield with respect to L is linear, then the center of the semifield is isomorphic to Fs. So the isotopic commutative semifield we get has middle nucleus containing Fq and center isomorphic to Fs. In this article, we are focused on the case n = 3, i.e., on symplectic semifield spreads of PG(5, q), when q is even. In such a case, only two non-sporadic examples are known: the Desarguesian spread and one of its cousin (see [10]), so they are both obtained by slicing the so called Desarguesian spread of Q+(7, q). In the former case, the associated translation plane is the Desarguesian plane, hence it is coordinatized by the finite field of order q3 and the relevant linear set is actually linear on Fq. In the latter case, the semifield spread is associated to a spread set C that is an F2-linear set L of PG(5, q), where F2 is the maximum subfield of Fq for which L is linear, and the associate semifield has order q3 and center F2 . In [5], it is proven that the only symplectic semifield spread of PG(5, q2), q > 214, whose associate semifield has center containing Fq, is the Desarguesian spread, meaning that a commutative semifield of order q6, with middle nucleus containing Fq2 and center containing Fq is a field, provided q is not too small. That was done by studying the intersection of the five non-equivalent Fq-linear sets of PG(5, q2) with the secant variety V1 of the Veronese variety and the only one that can have empty intersection with V1 is a plane. A classification of the Fq-linear sets of PG(5, q4) of rank 3t is not feasible, as the number of non-equivalent ones quickly grows with t. In fact, the present paper, we had a slightly different approach which allowed us to generalize the result of [5] in PG(5, q4) for any t: by field reduction, a PG(5, q4) can be seen as PG(6t - 1, q), a linear set of rank 3t as a subspace = PG(3t - 1, q) and V1 an algebraic variety, say Vf, of codimension t in PG(6t - 1, q). Hence, we have studied when a subspace of dimension 3t - 1 can have empty intersection with V (over Fq), regardless the geometric feature of the linear set in PG(5, q4). 2 Preliminary results 2.1 Fq-linear sets and the Fq-linear representation of PG(r — 1 ,qt) The set L c PG(V, Fqt) = PG(r - 1,q4), with V an r-dimensional vector space over Fqt, is said to be an Fq-linear set of rank m if it is defined by the non-zero vectors of an Fq-vector subspace U of V of dimension m, i.e. L = Lu = j(u)v : u e U \{0}}. If r = m and (Ly) = PG(r - 1, q4), then Ly = PG(r - 1, q). In this case, Ly is said to be a subgeometry (of order q) of PG(r - 1, q4). Throughout this paper, we shall extensively use the following result: a subset E of PG(r - 1, q4) is a subgeometry of order q if and only if there exists an Fq-linear collineation a of PG(r - 1, q4) of order t such 518 Ars Math. Contemp. 17 (2019) 493-514 that E = Fix a, where Fix a is the set of points fixed by a. This is a straightforward consequence of the fact that there is just one conjugacy class of Fq-linear collineations of order t in PrL(r, q4), namely that of ^. (xi^ x^ ..., xr_1) i y (x^ Xl,..., i). In particular, all subgeometries = PG(r -1, q) of PG(r -1, q4) are projectively equivalent to the subgeometry induced by {(x0, xi,..., xr-1) : xj G Fq}. A subspace n of PG(r — 1, q4) defines a subspace of Fix a = PG(r — 1,q) of the same dimension if and only if n = n (see [14, Lemma 1]). It will be more convenient for us to explicitly state the following equivalent result. Notation. Let F be any field containing Fq. Throughout the paper we will denote by n(F) the unique subspace of PG(r — 1, F) containing n. Lemma 2.1. If we consider PG(r — 1, q) embedded as a subgeometry of PG(r — 1, q4) and n is a subspace of PG(r — 1, q) of dimension s — 1, then the subspace n(Fqt) of PG(r — 1, q4) containing n has dimension s — 1 as well. Analogously, if W is an algebraic variety of PG(r — 1, q4), then W n Fix a c W n Wn • • • n Wand hence W n Fix a has the same dimension and degree of W if and only if W = W Remark 2.2. An algebraic variety W is said to be a variety of PG(r — 1, q) if it consists of the set of zeros of polynomials /1, /2,..., fk G Fq [xo , x* 1, . . . , x* r — 1 1, and we will write W = V(f1, f2,..., fk). By dimension and degree of W we will mean the dimension and degree of the variety when considered as variety of PG(r — 1, Fq), with Fq the algebraic closure of Fq. In the remaining part of this section, we will describe the setting we adopt to study the Fq -linear sets of PG(V, Fqt) = PG(r — 1,q4). When we regard V as an Fq-vector space, dimFq V = rt and hence PG(V, q) = PG(rt — 1, q). Furthermore, a point (v)F t G PG(r — 1, q4) corresponds to the (t — 1)-dimensional subspace of PG(rt — 1,q) given by {Av : A G Fqt}. This is the so-called Fq-linear representation of (v)F t and the set S, consisting of the (t — 1)-subspaces of PG(rt — 1, q) that are the linear representation of the points of PG(r — 1, q4), is a partition of the point set of PG(rt — 1, q). Such a partition S is called Desarguesian spread of PG(rt — 1, q). In this setting, a linear set Lv is the subset of the Desarguesian spread S with non-empty intersection with the projective subspace ny of PG(rt — 1, q) induced by U. We shall adopt the following cyclic representation of PG(rt — 1, q) in PG(rt — 1, q4). Let PG(rt — 1, q4) = PG(V', q4), with V' the standard rt-dimensional vector space over Fqt and let ej the ¿-th element of the canonical base of V'. Consider the semi-linear collineation a with accompanying automorphism x i xq and such that ej i ei+r, where the subscript are taken mod rt. Then a is an Fq-linear collineation of order t and Fix a = {(x, xq,..., xq ) : x = (x0, x1,..., xr-1), xj G Fqt, x = 0} is isomorphic to PG(rt — 1, q). The elements of S are the subspaces nP := (P, PCT,..., Pfft 1) n Fix a, with P G n0 = PG(r — 1, q4) and n0 defined by xj =0 Vi > r — 1 (see [14]). Let nj be njf. In the following, we shall identify a point P of n0 = PG(r — 1, q4) with the spread V Pepe: Symplectic semifield spreads of PG(5 ,qt), q even 519 element nP. We observe that P is just the projection of nP from (ni, n2,..., nt_i} on n0. If Lu is a linear set of rank m, then it is induced by an (m - 1)-dimensional subspace ny c PG(rt - 1, q) = Fix a and it can be viewed both as the subset of n0 that is the projection of nu from (n1, n2,..., nt_1} on n0 as well as the subset of S consisting of the elements nP such that nP n nu = 0. We stress out that we have defined the subspaces nu and nP as subspaces of Fix a = PG(rt - 1, q). Let F be any field containing Fqt, then the projection of nu(F) on n0 from (n1, n2,..., nt_1} is (Lu}F. Let H be a hypersurface of PG(r - 1, q4) and let f G Fqt [x0, x1,..., xr-1] a polynomial defining H, i.e., H = V(f). In the linear representation of PG(r - 1, q4) = n0, the points of H correspond to the spread elements nP such that P G H, hence it is the intersection of the variety V (f, f,..., f) of PG(rt - 1, q4) with Fix a, where, by abuse of notation, we extend the action of a also to polynomials. We observe that the variety V(f, fa,..., fa ) is the join of the varieties H, Ha,..., Ha (see [8, Chapter 8]) and hence it has dimension t(dim H +1) - 1 = t(r - 1) - 1 = tr - t - 1 and degree deg(H)4. We observe that V(f, fa,..., fa ) it is defined by t polynomials and dim V (f,f,...,f) = tr - t - 1 = dimPG(rt - 1, q4) - t, hence V (f,f,..., f) is a complete intersection (see [8, Example 11.8]). We will denote the join of the varieties W1, W2,..., Wk by Join(W1, W2,..., Wk). Let TP (W) be the tangent space to the algebraic variety W at the point P G W. Proposition 2.3 (Terracini's Lemma [20]). Let W = Join(Y1, Y2) and let P = (P1, P2} G W with Pi G Yi. Then (Tp (Y1), Tp2 (Y2)} C Tp(W). The variety V(f, fa,..., f1) is the join of the varieties , i = 0,1,..., t - 1. We recall that is a hypersurface of ni, hence TPi ) is a hypersurface of ni for a non-singular point Pi G . By ni n (n¿, j = i} = 0, we get dim(Tp0 (H), Tpi (H),..., Tpt_i (Hff'-1)} = rt - 1 -1 for non-singular points P0, P1,..., Pt_ 1. Since for a non-singular point P G V(f, fa,..., ffft-1), dim TP (V (f,f, ...,ffft-1)) = rt - 1 - t, we have (Tpo(H),TP1 (H),... ,Tpt-1 (Hfft-1)} = Tp(V(f,f,..., f"t-1)) for a non- singularP G V(f,f,...,f). Let Sing(W) be the set of the singular points of a variety W; we recall that Sing(W) is a subvariety of W. From the discussion above, it is clear that t_1 Sing(V(f,f,...,f))= (J Si, i=0 with Si = Join(Sing(Hffi), H^, j = i). 2.2 The Veronese surface and its secant variety In this section we denote by Pn-1 the (n - 1)-dimensional projective space over a generic field F. 520 Ars Math. Contemp. 17(2019)447-454 The Veronese map of degree 2 v2 : (x0,xbx2) € P2 i—> (. .., x1,. ..) € P5 is such that x1 ranges over all monomials of degree 2 in x0, x1, x2. The image V := v2(P2) is the quadric Veronese surface, a variety of dimension 2 and degree 4. A section H n V, where H is a hyperplane of P5, consists of the points of v2 (C), where C is a conic of P2. If we use the so-called determinantal representation of V (see [8, Example 2.6]), then we can take P5 as induced by the subspace of M(3, F) consisting of symmetric matrices and v2(x0,x1,x2) = A such that aij = XjXj, i.e., V consists of the rank 1 matrices of M (3, F). Hence, the secant variety of V, say V1, consists of the symmetric matrices of rank at most 2, i.e., V1 consists of the singular symmetric 3 x 3 matrices. So V1 is a hypersurface of P5 of degree 3. It is well known that the singular points of V1 are the points of V. The automorphism group G of V is the lifting of G = PGL(3, F) acting in the obvious way: v2(p)^ = v2(pg) V# € PGL(3, F). The group G obviously fixes V1. The maximal subspaces contained in V1 are planes and they are of three types: the span of v2(l), with l a line of P2, the tangent planes TP(V) for P € V, and, when the characteristic of F is even, the nucleus plane nN. Let the characteristic of F be even. The plane nN of P5 consists of the symmetric matrices with zero diagonal, hence nN is contained in V1. By the Jacobi's formula, det A = tr(adj(A)), where tr(M) is the trace of a matrix M and adj(M) is the adjoint matrix of M. Let Eij be the 3 x 3 matrix with 1 in the ij-position and 0 elsewhere, so we have det A = tr(adj(A) ) = tr(adj(A)(Eij + E^)) = 0 Vi = j. It follows that a hyperplane is tangent to V1 if and only if it contains nN. Also, each point of nN is the nucleus of a point of a unique conic v2 (l). If P € V1, then the tangent hyperplane H to V1 at P is such that H n V = v2(l2), where l = (p1,p2) if P € nN and hence P € (v2(p1), v2(p2)), or l is such that P is the nucleus of v2(l) if P € nN. The tangent plane at v2 (p) to V is the intersection of three hyperplanes K1, K2, K3 such that Ki n V = v2(l U li), where li, li are lines through p. If F is an algebraically closed field, then any subspace of P5 of dimension at least 1 has non-empty intersection with V1. If F = Fq, then there are subspaces of larger dimension disjoint from V1 and, by the Chevalley-Warning Theorem, we know that they can have dimension at most 2. For q even we have the following result. Theorem 2.4 ([5]). Let q > 4 be even, then there exists one orbit of planes under the action of G disjoint from V1. 3 Proof of the main result Through this section, we assume q to be even. Let Fq be the algebraic closure of Fq. We adopt the Fq-linear representation of PG(5, qf), i.e., we regard the points of PG(5, q4) as elements of a Desarguesian spread of PG(6t - 1, q) and Lu as the subset of the spread with non-empty intersection with a (3t - 1)-dimensional subspace nu of PG(6t - 1, q); also, we consider PG(6t - 1, q) as subgeometry of PG(6t - 1, q4) (cf. Section 2). Let f be the polynomial with coefficients in F2 such that V1 = V(f), hence the Fq-linear representation of V1 is V(f, f,..., f) n Fix a. Let Vf be V(f, f,..., f). We have that V1 n Lu = 0 ^ Vf n nu = 0 ^ Vf n Fix an nu (Fqt) = 0. Let W be nu (Fq) n V f. We observe that W = Wa, hence dim W = dim W n Fix a. We stress V Pepe: Symplectic semifield spreads of PG(5 ,qt), q even 521 out that W is defined by polynomials in Fqt [x0,x 1,..., x6t-1] but it might not contain any Fqt -rational point. The linear representation of nN is the (3t - 1)-dimensional subspace nN of Fix a that is partitioned by the spread elements {np : P G nN}. As Ly n nN = 0, we must have ny n nN = 0 and hence, by Lemma 2.1, ny (Fqt) n nN (Fqt) = 0 and nu (Fq) n nN (Fq) = 0. Theorem 3.1. Let P G W, then dim Tp (Vt) n (Fq) = dim Tp (V t) - 3t. Proof. The subspace ny (Fq) has codimension 3t, hence dimTP(Vt) n ny(Fq) > dimTP(Vt) - 3t. Let P G (Po, Pi,..., Pt-i) with Pi g n¿(Fq). We have that Tp (Vt) = (Tp0 (Vi ),Tp (Vf),...,Tp- (Vf'-)) and nN cTp. (Vf) Vi, hence nN (Fq) c Tp (Vt )._Since ny (Fq) n nN (Fq) = 0 and dimnN(Fq) = 31 - 1, we have dimTp(Vt) n n v(Fq) < dimTp(Vt) - 3t, hence the statement follows. □ Corollary 3.2. We have dim W = 2t - 1, hence W is a complete intersection. Proof. If P is non-singular for Vt, then dim Tp(Vt) = dim(Vt) = 5t - 1, whereas dim Tp (Vt) > 5t - 1 for P G Sing(Vt). As W = Vt n (Fqi , Tp (W) = Tp (Vt) n ny (Fqt). By Theorem 3.1, dimTp(Vt) n ny(Fq) = dimTp(Vt) - 3t > 2t - 1, and dim Tp (Vt) n ny (Fq) > 2t - 1 only if P G Sing(Vt). Hence dim W = 2t - 1. We observe that 2t -1 = dimny(Fq) -t, hence W is a complete intersection. □ Corollary 3.3. Sing(W) = Sing(Vt) n (Fq). Proof. By Theorem 3.1, dimTp(W) = dimTp(Vt) - 3t, hence dim Tp(W) > dim W = 2t - 1 if and only if dim Tp (Vt) > 5t - 1 = dim( Vt), i.e., P G Sing(Vt). □ If a variety Y is a complete intersection and dim Y - dimSing(Y) > 2, then Y is normal (see [19, Chapter 2, Section 5.1] for the general definition of normal varieties). An important tool for our proof is the following reformulation of the Hartshorne connectedness theorem ([9]). Theorem 3.4 ([3, Theorem 2.1]). If Y is a normal complete intersection, then Y is absolutely irreducible. Theorem 3.5. If W is reducible and Ly n V 1 = 0, then Ly is a plane which is isomorphic to PG(2, q4) disjoint from V1. 522 Ars Math. Contemp. 17(2019)447-454 Proof. If W is reducible, then W is not normal and hence dim Sing(W) = dim W — 1 = 2t — 2. A point P G W is singular if and only if P G Sing(V4 ) n ny (Fq). We have Sing(Vt) = Ut-o1 Si, with Si = Join(Sing(Vf ), Vf*, j = i) = Join(V, Vf* , j = i) (see Section 2), so Sf * = Si and hence dim Sing(Vt) n ny (Fq) = dimS0 n ny(Fq) = 2t — 2. Let P G So n nu (Fq) with P = (P0, PO,... ,Pt-o ), Po G V, P G Vf, i = 1, 2,..., t — 1, then the tangent space TP(S0 n ny(Fq)) is (Tp0 (V), Tp (Vf ),..., Tp- (Vft-1 )) n nu(Fq) = Kl n K n K3 n Hi n • • • n h4*_O n ny(Fq), with Ki, Hj hyperplanes of PG(6t — 1, q4) such that Ki projects on the hyperplane Ki of n0 for i = 1,2,3, H1 projects on the hyperplane Hj of n Vj = 1, 2,..., t — 1, Ko n K2 n K3 = Tp0(V) and Hj = TPj (Vfj). We can take Ko,K2,Ks such that KO n V = v2(^), K2 n V = v2(^2) and K3 n V = v2(^ O U ¿2). Hence, K\ n K2l n H n • • • n Htl_ O contains nN and so dim K2l n K| n Hi n • • • n H4- O n ny (Fq) is the smallest possible, i.e., 2t — 2. Hence, Ki d K2l n K3 n Hi n • • • n H_O n ny(Fq) and the projection of ny (Fq) on n0 is a subspace n0 such that the tangent space of P0 at V n n0 has codimension 2 in n0. So either the codimension of n0 n V in n0 is 2 or n0 n V has codimension 3 in n0 but it has singular points. Suppose we are in the latter case. The Veronese variety V is smooth, hence n0 can by a 3 or 4-dimensional subspace of n0. If n0 is a hyperplane of n0 and n0 n V (Fq ) has singular points, then n0 n V is is either v2(^2) or v2(^ O U ¿2). In the first case, n0 contains . A plane = PG(2, q4) is a Fq-linear set of rank 3t, so n0(Fqt ) = PG(4, q4) contains two linear sets of rank 3t that must intersect by Grassmann, i.e., n VO = 0. If n0 n V = v2(^ O U ¿2), then n0 contains the tangent space at V of the point P = v2 (^ O n ) and it is the unique tangent space at V contained in n0. Let t be the collineation induced by the field automorphism x ^ xq , then both n0 and V(Fq) are fixed by t, hence ï> (V)T = Tp(V) and, by Lemma 2.1, Tp (V) contains a PG(2, q4). Again, by Grassmann, n VO = 0. Suppose that n0 is a 3-dimensional space, so it contains 4 points counted with their multiplicity and at least one of them is multiple. If P is a multiple point and it is Fqt -rational, i.e., P = PT, then n0 contains a line tangent to V at P that it is fixed by t and hence contains a PG(1,q4), so, by Grassmann, nVO = 0. So a multiple point P must be Fqst -rational, but also PT G n0 n V would be, hence s = 2 and we have n0 n V = {P, PT }, with P g n0(q24). The line joining P and PT is set-wise fixed by t and so it contains a PG(1, q4), yielding again n VO = 0. So suppose that the codimension of n0 n V(Fq) in n0 is 2. Hence n0 is either a 3-dimensional space or a plane. Suppose that n0 is a 3-dimensional space and so dimn0 n V (Fq ) = 3 — 2 = 1. Since n0 n V(Fq) is the Veronese embedding of the intersection of two distinct conics, n0 contains the Veronese embedding of a line I and it cannot contain the embedding of any other line. Hence v2(^)T c n0 implies v2(^)T = v2(^) and so (v2(^)) contains a V Pepe: Symplectic semifield spreads of PG(5 ,qt), q even 523 plane = PG(2, qt). By Grassmann, LU n Vi = 0. Hence n0(qt) is a plane and so Lu = n0(qt). □ Theorem 3.6. If W is absolutely irreducible and q > 2 ■ 34t, then W n Fix a has at least one point. Proof. By [2, Corollary 7.4], an absolutely irreducible algebraic variety of PG(n — 1, q) with dimension r and degree S for q > max{2(r + 1)S2,2S4} has at least one Fq-rational point. By r = 2t — 1 and S < 3t = deg Vf, we have the statement. □ We conclude the section with our main result. Theorem 3.7. Let q > 2 ■ 34t be even. The only symplectic semifield spread of PG(5, qt) whose associate semifield has center containing Fq, is the Desarguesian spread. Proof. By Theorems 3.6 and 3.5, we have that the only Fq-linear set of rank 3t disjoint from V1 is a plane. The planes disjoint from V1 form a unique orbit under the action of G (see Theorem 2.4). In this case, the linear set is Fqt -linear as well, hence the semifield associated to the spread is 3-dimensional over its center. By [17], in even characteristic this implies that the semifield is a field, hence the spread is Desarguesian. □ Corollary 3.8. Let q > 2 ■ 34t be even. Then a commutative semifield of order q3t, with middle nucleus containing Fqt and center containing Fq, is afield. Remark 3.9. We emphasize that the hypothesis of even characteristic is crucial for all our arguments: only for even q the variety V1 contains the plane nN, and using L n nN = 0 we can prove that W is a complete intersection, i.e. W has codimension t, and the singular points of W are just the ones coming from V. References [1] J. 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Shafarevich, Basic Algebraic Geometry 1: Varieties in Projective Space, Springer, Heidelberg, 3rd edition, 2013, doi:10.1007/978-3-642-37956-7. [20] A. Terracini, Sulle Vk per cui la varieta degli Sh(h + 1)-seganti ha dimensione minore dell'ordinario, Rend. Circ. Mat. Palermo 31 (1911), 392-396, doi:10.1007/bf03018812. /^creative ^commor ARS MATHEMATICA CONTEMPORANEA ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 17 (2019) 525-533 https://doi.org/10.26493/1855-3974.1610.03d (Also available at http://amc-journal.eu) On the Hamilton-Waterloo problem: the case of two cycles sizes of different parity* Melissa S. Keranen Michigan Technological University, Department of Mathematical Sciences, Houghton, MI 49931, USA Adrian Pastine Instituto de Matemática Aplicada San Luis (IMASL), Universidad Nacional de San Luis, CONICET, Ejército de los Andes 950 (D5700HHW), San Luis, Argentina Received 19 February 2018, accepted 3 October 2019, published online 25 November 2019 Abstract The Hamilton-Waterloo problem asks for a decomposition of the complete graph of order v into r copies of a 2-factor F and s copies of a 2-factor F2 such that r + s = \_v-r\. If F1 consists of m-cycles and F2 consists of n cycles, we say that a solution to (m, n)-HWP(v; r, s) exists. The goal is to find a decomposition for every possible pair (r, s). In this paper, we show that for odd x and y, there is a solution to (2kx, y)-HWP(vm; r, s) if gcd(x, y) > 3, m > 3, and both x and y divide v, except possibly when 1 G {r, s}. Keywords: 2-factorizations, Hamilton-Waterloo problem, Oberwolfach problem, cycle decomposition, resolvable decompositions. Math. Subj. Class.: 05C51, 05C70 1 Introduction The Oberwolfach problem asks for a decomposition of the complete graph Kv into copies of a 2-factor F. To achieve this decomposition, v needs to be odd, because the vertices must have even degree. The problem with v even asks for a decomposition of Kv into ^jj2 copies of a 2-factor F, and one copy of a 1-factor. The uniform Oberwolfach problem (all cycles of the 2-factor have the same size) has been completely solved by *The authors would like to thank Stefaan De Winter and Martín De Borbon for wonderful discussions about rings of polynomials that led to the original constructions used in previous versions of the paper, and the anonymous referees for helpful comments which greatly improved the readability of the final version. This work was partially supported by the Universidad Nacional de San Luis, grant PROICO 03-0918. E-mail ¡addresses: msjukuri@mtu.edu (Melissa S. Keranen), adrian.pastine.tag@gmail.com (Adrian Pastine) ©® This work is licensed under https://creativecommons.org/licenses/by/4.0/ 526 Ars Math. Contemp. 17 (2019) 493-514 Alspach and Haggkvist [1] and Alspach, Schellenberg, Stinson and Wagner [2]. The nonuniform Oberwolfach problem has been studied as well, and a survey of results up to 2006 can be found in [8]. Furthermore, one can refer to [6, 7, 9, 23, 24] for more recent results. In [19] Liu first worked on the generalization of the Oberwolfach problem to equipartite graphs. He was seeking to decompose the complete equipartite graph K(m:n) with n partite sets of size m each into (n-21)m copies of a 2-factor F. For such a decomposition to exist (n - 1)m has to be even. In [14] Hoffman and Holliday worked on the equipartite generalization of the Oberwolfach problem when (n - 1)m is odd, decomposing into copies of a 2-factor F, and one copy of a 1-factor. The uniform Oberwolfach problem over equipartite graphs has since been completely solved by Liu [20] and Hoffman and Holliday [14]. For the non-uniform case, Bryant, Danziger and Pettersson [7] completely solved the case when the 2-factor is bipartite. In particular, Liu showed the following. Theorem 1.1 ([20]). For m > 3 and u > 2, K(h:u) has a resolvable Cm-factorization if and only if hu is divisible by m, h(u — 1) is even, m is even if u = 2, and (h, u, m) ^ {(2, 3, 3), (6, 3, 3), (2, 6, 3), (6, 2, 6)}. The Hamilton-Waterloo problem is a variation of the Oberwolfach problem, in which we consider two 2-factors, F1 and F2. It asks for a factorization of Kv when v is odd or Kv — I (I is a 1-factor) when v is even into r copies of F1 and s copies of F2 such that r + s = L^J, where F1 and F2 are two 2-regular graphs on v vertices. Most of the results for the Hamilton-Waterloo problem are uniform, meaning F1 consists of cycles of size m (Cm-factors), and F2 consists of cycles of size n (Cn-factors). If there is a decomposition of Kv into r Cm-factors and s Cn-factors we say that a solution to (m, n)-HWP(v; r, s) exists. The case where both m and n are odd positive integers and v is odd is almost completely solved by [11, 12]; and if m and n are both even, then the problem again is almost completely solved (see [5, 6]). However, if m and n are of differing parities, then we only have partial results. Most of the work has been done in the case where one of the cycle sizes is constant. The case of (m, n) = (3,4) is solved in [4, 13, 21, 25]. Other cases which have been studied include (m, n) = (3, v) [18], (m, n) = (3,3x) [3], and (m,n) = (4, n) [16,21] . In this paper, we consider the case of m and n being of different parity. This case has gained attention recently, where it has been shown that the necessary conditions are sufficient for a solution to (m, n)-HWP(v; r, s) to exist whenever m | n, v > 6n > 36m, and s > 3 [10]. We provide a complementary result to this in our main theorem, which covers cases in which m \ n and solves a major portion of the problem. Theorem 1.2. Let x, y, v, k and m be positive integers such that: (i) v, m > 3, (ii) x, y are odd, (iii) gcd(x,y) > 3, (iv) x and y divide v, (v) 4k divides v. Then there exists a solution to (2kx, y)-HWP(vm; r, s) for every pair r, s with r + s = |_(vm — 1)/2_|, r, s = 1. M. S. Keranen and A. Pastine: On the Hamilton-Waterloo problem: the case of two cycles ... 527 2 Preliminaries The complete cyclic multipartite graph C(x:n) is the graph with n partite sets of size x, where two vertices (g, i) and (h, j) are neighbors if and only if i - j = ±1 (mod n), with subtraction being done modulo n. The directed complete cyclic multipartite graph ~C(x:n) is the graph with n parts of size x, with arcs of the form e(g, i), (h, i +1)) for every 0 < g, h < x - 1, 0 < i < n - 1. One of the main tools in [17] is a Lemma that combines decompositions of C(x:k) to obtain decompositions of K(v:m). We present a version of the Lemma for uniform decompositions, as those are the focus of this manuscript. Lemma 2.1 ([17]). Let m, x, y, and v be positive integers. Let si,..., sm-i be non- 2 negative integers. Suppose the following conditions are satisfied: • There exists a decomposition of Km into Cn-factors. • For every 1 < t < m-1 there exists a decomposition of C(v:n) into st Cxn-factors and rt Cyn-factors. Let (m-1) (m-1) 2 2 s = ^^ st and r = ^^ rt. t=i t=i Then there exists a decomposition of K(v:m) into s Cxn-factors and r Cyn-factors. In order to decompose (x:n), x and n odd, into Cn-factors and Cxn-factors, the authors of [17] labeled the vertices by Zx x {0,..., n - 1}. They build a 2-factor F by providing n permutations of G. The ith permutation is used to connect vertices in column i - 1 to vertices in column i, in particular the n-th permutation is used to connect vertices in column n - 1 to vertices in column 0. It must be said that these permutations were used implicitly in [17], as no permutation language was used for this part of the construction. Notice that in general, if the columns are labeled by an abelian group G, f is the ith permutation and g G G, in the 2-factor F, vertex (g, i - 1) is connected to vertex (f (g), i). Let F be the composition of all n permutations of the 2-factor F, such that (F(g), 0) is the vertex at which we finish if we start at vertex (g, 0) and move through F until we reach column 0 again. In the constructions in [17], G is abelian, and g - F(g) depends only on F and not on g. If this is the case, the length of the cycles of F is n times the order of the element g - F(g). Lemma 2.2. Assume F is a 2-factor built with the permutation F, and g - F(g) depends only on F. If q is the order of g - F(g), then F is a qn-factor of ~C(xy4k:n). As we will need to use the permutations of Zx, we will introduce them. For a G Zx, let fa be the permutation that adds a to every element of Zx, i.e. fa (g) = g + a. Let H(a, ,0) be the 2-factor made with the following permutations: • fa from column i - 1 to column i if 1 < i < n - 3/2; • f-a from column i - 1 to column i if n - 3/2+1 < i < n - 3; • fa from column n - 3 to column n - 2; 528 Ars Math. Contemp. 17 (2019) 493-514 • /-2a from column n - 2 to column n - 1 (this is a permutation because x is odd); • /p from column n — 1 to column 0. In [17] the first n — 3 permutations were different, but the end result was the same. Notice that F(g) = g — a + 3. For every r e {0,1,2,..., x — 3, x — 2, x}, the authors of [17] gave permutations 0 of Zx that satisfied: (a) 0(a) = a for r elements of Zx; (b) gcd(a — 0(a), x) = 1 for the remaining x — r elements of Zx. Then, the decomposition of cC(x:n) was given by the 2-factors H(a, 0(a)), a e Zx. In order for such a decomposition to work, for every a, 3 e Zx the permutations /a, /p needed to satisfy /a = /p if and only if a = 3, as otherwise some arcs would be repeated in the factor H(a, 0(a)) and the factor H(3,0(3)). Then, in [17], decompositions of it(x:n), cC(y:n), and (4:n) were combined using a graph product and permutations of Zx x Zy x Z4 to decompose G (4xy:n). Instead of doing so, we will use group products to label the vertices of (4kxy:n), although we will make use of permutations of the group product. In Section 3, we give permutations of Z2k x Z2k, and show that they satisfy the necessary conditions to be used for decompositions. In Section 4, we use multivariate bijections to give decompositions of (4fcxy:n) into 2kxk-factors and ityk-factors. Finally, in Section 5, we use these decompositions to prove our main results. 3 The permutation fa(a, b) of Z2k x Z2k Consider the group G = Z2k x Z2k, an element a = (a^ a2) and the function /a(a, b) = (—b + ai, a — b + a2). Lemma 3.1. /a is a permutation of G. Proof. As |G| is finite, it is enough to prove that /a is an injective function. Assume /a(a, b) = /a(c, d). Then (—b + ai, a — b + a2) = (—d + ai, c — d + a2). The equality —b + a1 = —d + a1 implies b = d. Using b = d, the equality a — b + a2 = c — d + a2 implies a = c. Therefore, /a is a permutation of G. □ Lemma 3.2. /p(/2(a, b)) = (a, b) — a + 3. Proof. We will prove this lemma by computing /p(/2 (a, b)). /a(a, b) = (—b + ai, a — b + a2) /2(a, b) = /(—b + ai, a — b + a2) = (—a + b — a2 + ai, —b + ai — a + b — a2 + a2) = (—a + b — a2 + ai, ai — a) M. S. Keranen and A. Pastine: On the Hamilton-Waterloo problem: the case of two cycles ... 529 fg(fa(a, b)) = fg(-a + b - + ai, ai - a) = (a — ai + fti, —a + b — a2 + ai — ai + a + ft2) = (a - ai + fti, b - a2 + ^2) = (a, b) - a + ft. □ Letting ft = a in Lemma 3.2 yields f(a, b) = (a, b). Corollary 3.3. f(a, b) = (a, b). As it was mentioned in Section 2, we need to show that if a = ft, then fa(a, b) = fg (a, b) for every (a, b) G Z2k x Z2k; so that each arc is used exactly once. The statement of the following lemma is an equivalent claim. Lemma 3.4. fa(a, b) = fg (a, b) for some (a, b) G Z2k x Z2k if and only if a = ft. Proof. Assume fa(a, b) = fg(a, b). Then (-b + ai, a - b + a2) = (-b + fti, a - b + ft2). Hence, ai = fti and a2 = ft2. Therefore a = ft. □ 4 Decomposing (C(4kxy:n) into ((yn-factors and itx2fcn-factors Let G = Z2k x Z2k and label each column of " (4fcKy:n) with the elements of the group Gj X Zx X Zy . Let R = G x Zx x Zy. For every A G R, let A = (a, ft, 7), with a G G, ft G Zx and 7 G Zy. For a G G, let fa be defined as in Section 3. For ft G Zx let fg be the permutation of Zx defined by fg (a) = a + ft. Similarly, for 7 G Zy let fY be the permutation of Zy defined by fY(a) = a + 7. Finally, for A = (a, ft, 7) G R let f be the permutation of R defined by fA(a, b,c) = (fa(a),fg(b),f7(c)). Let ^ be a permutation of R, and for each A g G let H4kxy (A, y(A)) be the 2-factor formed with the following permutations: 1. f from column i to i + 1 if 1 < i < n - 3/2; 2. f—i from column i to i +1 if n - 3/2 + 1 < i < n - 3; 3. f from column n - 2 to column n - 1; 4. f (a, -2ft, -27) from column n - 1 to column n; 5. fv(a) from column n to column 1. Notice that if you start in column 1 at vertex (a, b, c) the first time you reach column 1 again you reach vertex (a, b, c) - (a, ft, 7) + y>(a, ft, 7) = (a, b, c) - A + <^(A). Hence, we can apply Lemma 2.2 to obtain the length of the cycles in the 2-factor. Let A G R. If A - <^(A) = (a,b, 0) with a G ±{(1,0), (0,1), (1,1)} and gcd(b,x) = 1, then by Lemma 2.2 the 2-factor H4k (A, y(A)) is a "t2kxn-factor. If A - y(A) = (0,0, c) with gcd(c, y) = 1, Lemma 2.2 implies that H4k (A, y(A)) is a "tyn-factor. Therefore, to obtain a decomposition of "t(4kxy:n) into r "t2kxn-factors and s "ty„-factors, we need a permutation ^ satisfying 530 Ars Math. Contemp. 17 (2019) 493-514 (A) A - p(A) = (a, b, 0) with a G ±{(1,0), (0,1), (1,1)} and gcd(b,x) = 1 for r elements A g R; (B) A — p(A) = (0,0, c) with gcd(c, y) = 1 for s = 4kxy — r elements A G R. In order to obtain the permutation p, consider the subgroup 2G of G of index 4, and let K = {(0,0), (1, 0), (0,1), (1,1)}. Notice that K is a set of representatives of the cosets of 2G in G. Let e G 2G, and let ^ be a permutation of G. If g, ^(g) G e + K, then either g = ^(g) or |g — ^(g)| = 2k because g — ^(g) G ±K. Hence, we can obtain p by providing 4k-1 permutations pe of K x Zx x Zy satisfying (A') A — pe(A) = (a, b, 0) with a G ±{(1, 0), (0,1), (1,1)} and gcd(b,x) = 1 for re elements A g K x Zx x Zy; (B') A—pe(A) = (0,0, c) withgcd(c, y) = 1 for se = 4xy —re elements A G KxZx xZy; having r = J2ee2G r^, and having p act in each (e + K) x Zx x Zy as pe, i.e. if g = (e, c, d) + (p, 0, 0), with p G K, p(g) = (e, c, d) + pe(p, c, d). Notice that if a G K, a G ±{(1,0), (0,1), (1,1)} if and only if a = (0,0). In [17], for every r G {0, 2,3,..., 4xy — 3,4xy — 2,4xy}, permutations ^ of Z4 x Zx x Zy were given satisfying: (A'') A — ^(A) = (a, b, 0), with a = 0 and gcd(b, x) = 1 for r elements A G Z4 x Zx x Zy; (B'') A — ^(A) = (0,0, c), with gcd(c, y) = 1 for the remaining 4xy — r elements A G Z4 x Zx x Zy . Let n: K ^ Z4 be a bijection such that n(0,0) = 0, and let ^: K x Zx x Zy ^ Z4 x Zx x Zy be the bijection that fixes the coordinates of Zx and Zy, and that behaves like n in the coordinate of K. Then if is a permutation of Z4 x Zx x Zy, satisfying Conditions (A'') and (B'') with re and se, pe = is a permutation of K x Zx x Zy satisfying Conditions (A') and (B') with re and se. If we wanted either x = 1 or y = 1, we would need to change Conditions (A) and (B), but it is easy to see that the necessary permutations to decompose (C(4kxy:n) exist. Therefore we have the following. Lemma 4.1. Let r G {1,4kxy — 1}, then there is a decomposition of ~C(4kxy:n) into r ~C2kxn-factors and s = 4kxy — r Cyn-factors. 5 Main results The complete solution to the uniform case of the Oberwolfach problem will be vital to the proof of our main result. Theorem 5.1 ([1, 2, 15, 22]). Kv can be decomposed into Cm-factors (and a 1-factor if v is even) if and only if v = 0 (mod m), (v, m) = (6, 3) and (v, m) = (12, 3). We now apply the results from Section 4 to produce the following important result for the uniform equipartite version of the Hamilton-Waterloo problem where the two factor types consist of cycle sizes of distinct parities. M. S. Keranen and A. Pastine: On the Hamilton-Waterloo problem: the case of two cycles ... 531 Theorem 5.2. Let x, y, z, v, m, k be positive integers v,m,k > 3 satisfying the following: (i) v,m > 3, (ii) k > 2, (iii) x, y, z odd, (iv) z > 3, (v) gcd(x,y) = 1, (vi) vm = 0 (mod 4kxyz), v = 0 (mod 4kxy), / --1 v(m — 1) ■ (vii) (4kxy ) is even, (viii) ,m, z) £ {(2, 3, 3), (6, 3, 3), (2, 6, 3), (6, 2, 6)}, then there is a decomposition of K(v:m) into r C2kxz-factors and s Cyz-factors, for any s,r =1. Proof. Let v1 = v/4kxy. Consider K(vi:m). Item (vi) ensures that z divides v1m; and items (vii), (i), and (viii) give us v1(m - 1) is even, m = 2, and m, z ) £ {(2, 3, 3), (6, 3, 3), (2, 6, 3), (6, 2, 6)}. 4k xy Thus by Theorem 1.1 there is a decomposition of K(vi:m) into Cz-factors. Replace each vertex in a £ K(vi:m) by 4kxy vertices (a, b), with 0 < b < 4kxy — 1, having an edge between (a1, b1) and (a2, b2) if and only if there was an edge between a1 and a2. This yields K(v:m). Even more, each Cz-factor becomes a copy of C(4kxy:z). By Lemma 4.1, we have that each C(4kxy:z) can be decomposed into rp C2kxz-factors and sp Cyz-factors as long as rp, sp = 1. Choosing sp such that J2p sp = s and sp, rp = 1, provides a decomposition of K(v:m) into r C2kxz-factors and s Cyz-factors by Lemma 2.1. □ The next lemma, given in [17] shows how to find solutions to the Hamilton-Waterloo problems by combining solutions for the problem on complete graphs and solutions for the problem on equipartite graphs. Lemma 5.3 ([17]). Let m and v be positive integers. Let F1 and F2 be two 2-factors on vm vertices. Suppose the following conditions are satisfied: • There exists a decomposition of K(v:m) into sa copies of F1 and ra copies of F2. • There exists a decomposition of mKv into sp copies of F1 and rp copies of F2. Then there exists a decomposition of Kvm into s = sa + sp copies of F1 and r = ra + rp copies of F2. We are now in a position to provide a proof of the main theorem. Theorem 5.4. Let x, y, v, k and m be positive integers such that: (i) v, m > 3, (ii) x, y are odd, 532 Ars Math. Contemp. 17 (2019) 493-514 (iii) gcd(x,y) > 3, (iv) x and y divide v, (v) 4k divides v. Then there exists a solution to (2kx, y)-HWP(vm; r, s) for every pair r, s with r + s = |_(vm - 1)/2_|, r, s = 1. Proof. Let r and s be positive integers with r + s = |_(vm - 1)/2J and r, s = 1. Write r = ra + rp and s = sa + sp, where ra,rp, sa, sp are positive integers that satisfy ra,sa = 1, ra + sa = v(m-1)/2, rp + sp = |(v - 1)/2j,andrp, sp G {0, |(v - 1)/2j}. Start by decomposing Kvm into K(v:m) © mKv. Let z = gcd(x, y), xi = x/z, yi = y/z. By Theorem 5.2 there is a decomposition of K(v:m) into ra C2kxiz-factors and sa Cyiz-factors. This is a decomposition of K(v:m) into ra C2kx-factors and sa Cy-factors. By Theorem 5.1 there is a decomposition of mKv into rp C2kx-factors and sp Cy-factors. Lemma 5.3 shows that all of this together yields a decomposition of Kvm into r Cx-factors and s Cy-factors. □ References [1] B. Alspach and R. Haggkvist, Some observations on the Oberwolfach problem, J. Graph Theory 9 (1985), 177-187, doi:10.1002/jgt.3190090114. [2] B. Alspach, P. J. Schellenberg, D. R. Stinson and D. Wagner, The Oberwolfach problem and factors of uniform odd length cycles, J. Comb. Theory Ser. A 52 (1989), 20-43, doi:10.1016/ 0097-3165(89)90059-9. [3] J. Asplund, D. 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ARS MATHEMATICA CONTEMPORANEA 17 (2019) 535-554 https://doi.org/10.26493/1855-3974.1884.5cb (Also available at http://amc-journal.eu) The finite embeddability property for IP loops and local embeddability of groups into finite IP loops Martin Vodicka Max-Planck-Institut für Mathematik in den Naturwissenschaften, Inselstrasse 22, 04103 Leipzig, Germany Pavol Zlatos * Faculty of Mathematics, Physics and Informatics, Comenius University, Mlynska dolina, 842 48 Bratislava, Slovakia Received 19 December 2018, accepted 11 September 2019, published online 29 November 2019 We prove that the class of all loops with the inverse property (IP loops) has the Finite Embeddability Property (FEP). As a consequence, every group is locally embeddable into finite IP loops. The first one of these results is obtained as a consequence of a more general embeddability theorem, contributing to a list of problems posed by T. Evans in 1978, namely, that every finite partial IP loop can be embedded into a finite IP loop. Keywords: Group, IP loop, finite embeddability property, local embeddability. Math. Subj. Class.: 20E25, 20N05, 05B07, 05B15, 05C25, 05C45 The Finite Embeddability Property (briefly FEP), was introduced by Henkin [17] for general algebraic systems already in 1956. For groupoids (i.e., algebraic structures (G, ■) with a single binary operation), which is sufficient for our purpose, it reads as follows: A class K of groupoids has the FEP if for every algebra (G, ■) G K and each nonempty finite subset X C G there is a finite algebra (H, *) G K extending (X, ■), i.e., X C H and x ■ y = x * y for all x, y G X, such that x ■ y G X. Using this notion an earlier result of Henkin [16] can be stated as follows: The class of all abelian groups has the FEP (see also Gratzer [14]). A more general notion of local embeddability can be traced back to even earlier papers by Mal'tsev [21, 22] (see also the posthumous monograph [23]). It was explicitly *The second author acknowledges with thanks the support by the grant no. 1/0333/17 of the Slovak grant agency VEGA. E-mail addresses: vodicka@mis.mpg.de (Martin Vodicka), zlatos@fmph.uniba.sk (Pavol Zlatos) Abstract ©® This work is licensed under https://creativecommons.Org/licenses/by/4.0/ 536 Ars Math. Contemp. 17 (2019) 493-514 (re)introduced and studied in detail mainly for groups by Gordon and Vershik [28]: A groupoid (G, ■) is locally embeddable into a class of groupoids M if for every nonempty finite set X C G there is (H, *) G M such that X C H and x ■ y = x * y for all x,y G X satisfying x ■ y G X. Informally this means that every finite cut-out from the multiplication table of (G, ■) can be embedded into an algebra from M. A standard model-theoretic argument shows that this condition is equivalent to the embeddability of (G, ■) into an ultraproduct of algebras from M (for the ultraproduct construction see, e.g., Chang, Keisler [3]). Thus a class K has the FEP if and only if every (G, ■) G K is locally embeddable into the class Kfin of all finite members in K. As proved by Evans [9], for a variety (equational class) K this is equivalent to the condition that every finitely presented algebra in K is residually finite, i.e., embeddable into a direct product of finite algebras from K. The groups locally embeddable into (the class of all) finite groups were called LEF groups in [28]. The authors also noticed that, unlike the abelian ones, not all groups are LEF, in other words, the class of all groups doesn't have the FEP. As examples of finitely presented groups which are not residaully finite, hence not locally embeddable into finite groups, can serve the Baumslag-Solitar groups BS(m,n) = (a,b | a-1bma = bn) for |m|, |n| > 1, |m| = |n| (see Meskin [24]). A complete list of minimal partial Latin squares embeddable into a closely related infinite group but not embeddable into any finite group, even under a weaker concept of embedding, was recently described by Dietrich and Wanless [6]. This immediately raises the question of finding some classes of finite grupoids into which all the groups were locally embeddable and which, at the same time, would be "as close to groups as possible". The question is of interest for various reasons: The class of all LEF groups properly extends the class of all locally residually finite groups and plays an important role in dynamical systems, cellular automata, etc. (see, e.g., Ceccherini-Silberstein, Coornaert [2], Gordon, Vershik [28]). Glebsky and Gordon [13] have shown that a group is locally embeddable into finite semigroups if and only if it is an LEF group. It follows that looking for a class of finite groupoids into which one could locally embed all the groups one has to sacrifice the associativity condition. They also noticed that the results about extendability of partial Latin squares to (complete) Latin squares imply that every group is locally embeddable into finite quasigroups. Refining slightly the original argument they have shown that every group can even be locally embedded into finite loops (see also their survey article [12]). A further decisive step in this direction was done by Ziman [30]. Building upon the methods of extension of partial Latin squares preserving some symmetry conditions (see Cruse [4] and Lindner [20]), he has shown that the class of all loops with antiautomor-phic inverses, i.e., loops with two-sided inverses satisfying the identity (xy)-1 = y-1x-1 (briefly AAIP loops), has the FEP (though he didn't use this notion explicitly). As a consequence, every group is locally embeddable into finite AAIP loops. Quasigroups and loops experts consider the class of all AAIP loops still as a "rather far going extension" of the class of all groups. On the other hand, they find the class of all loops with the inverse property, i.e., loops with two-sided inverses satisfying the identities x-1(xy) = y = (yx)x-1 (briefly IP loops), which is a proper subclass of the class of all AAIP loops, a much more moderate extension of the class of all groups (Dripal [8]). In the present paper we are going to show that Ziman's result can indeed be strengthened in this sense. Using mainly graph-theoretical methods and Steiner triple systems, we will prove that the class of all IP loops still has the FEP. As a consequence, every group is locally M. Vodicka and P. Zlatos: The finite embeddability property for IP loops and local. 537 embeddable into finite IP loops. When the original version of this paper was already submitted, A. Dripal turned our attention towards the point that the problem we are solving was implicitly formulated in Evans' paper [10] in the last line of the table on page 798. At the same time he remarked that we have proved even more, namely that every finite partial IP loop can be embedded into a finite IP loop. Later on, the same point was made by the anonymous referee. We will discuss these issues in the next section, after introducing the respective notions and formulating our results more precisely. For basic definitions and facts about quasigroups and loops the reader is referred to the monographs Belousov and Belyavskaya [1] and Pflugfelder [26]. 1 Formulation of the main results, discussion and plan of the proof In order to guarantee that the class of all IP loops forms a variety, we define an IP loop as an algebra (L, •, 1, -1) with a binary operation of multiplication •, a distinguished element 1 denoting the unit, and a unary operation -1 of taking inverses, satisfying the identities 1x = x = x1, and x-1(xy) = y = (yx)x-1. Then the identities x-1x = 1 = xx-1 and (x-1) 1 = x easily follow. Also it is clear that, for any a, b e L, the equations ax = b, ya = b have unique solutions x = a-1b, y = ba-1 in L. Since the unit element 1 and the inverse map x ^ x-1 in every IP loop are uniquely determined by the multiplication •, referring to an IP loop (L, •, 1, -1) as just (L, •) is unambiguous. However, usually it will be denoted simply by L. A partial IP loop (P, •) is a set P endowed with a partial binary operation • defined on a subset D(P, •) C P x P, called the domain of the operation •, satisfying the following three conditions: (1) there is an element 1 e P, called the unit of P, such that (1, x), (x, 1) e D(P, •) and 1x = x1 = x for all x e P; (2) for each x e P there is a unique y e P, called the inverse of x and denoted by y = x-1, such that (x, y), (y, x) e D(P, •) and xy = yx = 1; (3) for any x,y e P such that (x,y) e D(P, •), we have (x-1,xy), (xy,y-1) e D(P, •) and x-1(xy) = y, (xy)y-1 = x. In most cases we will denote a partial IP loop (P, •) as P and its domain as D(P), only; the more unambiguous notation (P, •) and D(P, •) will be used mainly in case we need to distinguish the operations on two or more (partial) IP loops. A partial IP loop (Q, *) is called an extension of a partial IP loop (P, •) if P C Q, D(P, •) C D(Q, *) and x • y = x * y for each pair (x, y) e D(P). Suppressing the names of the operations, we write P < Q or Q > P. Alternatively we say that the partial IP loop P is embedded in the partial IP loop Q. Obviously, the relation < between partial IP loops is reflexive, antisymmetric and transitive. Our main results are the following three theorems. Theorem 1.1. Every finite partial IP loop P can be embedded into some finite IP loop L. Given an IP loop L and a finite set X C L, we can form the finite partial IP loop p = X u{1}u X-1, 538 Ars Math. Contemp. 17 (2019) 493-514 where X 1 = {x 1 : x G X}, by restricting the original loop operation on L to the set D(P) = {(x, y) G P x P : xy G P}. Then, obviously, P < L. Thus Theorem 1.1 readily implies our next, already announced result. Theorem 1.2. The class of all IP loops has the Finite Embeddability Property. Equiva-lently, every finitely presented IP loop is residually finite. As a special case of Theorem 1.2 we obtain Theorem 1.3. Every group can be locally embedded into the class of all finite IP loops. Equivalently, every group can be embedded into some ultraproduct of finite IP loops. The second, equivalent formulation of Theorem 1.2 answers in affirmative the question posed by Evans [10] in the IP loop row and R. F. column of the table on page 798. In order to discuss the relation of our results to Evans' table we recall the following three abbreviations used in [10]. Unlike the author, who used them for various classes of algebras, we apply them just to the class of all IP loops. E1: Every finite partial IP loop can be embedded into some finite IP loop. E2: Every finite partial IP loop can be embedded into some (finite or infinite) IP loop. E3: Every finite partial IP loop which can be embedded into some IP loop, can be embedded into some finite IP loop. Obviously, condition E3 is equivalent to the FEP for IP loops, and condition E1 is equivalent to the conjunction E2 A E3, depicted (among other relations) in the chart on the top of page 798 in [10] (the equivalence E1 A E2 ^ E3 on page 797 is clearly a typo). Theorem 1.1 seems to contradict the sign "X" (meaning "No") in the IP loop row and E1 column of the table in the middle of page 798 in [10]. Evans, however, considered a seemingly weaker notion of a partial IP loop (hence a stronger concept of embeddability) there. Paraphrasing and slightly adapting his definition to apply to our situation, we obtain a rather vague formulation: "A partial IP loop is a set P in which the operations of multiplication and taking inverses are defined on some subsets D(P) C P x P and D'(P) C P, respectively, which satisfies the defining IP loop identities, insofar as they can be applied to the partial operations on P" (cf. [10, §3, page 796]). In particular it is not clear (though not crucial) whether P has to contain the unit element 1 or not. Thus, at least at a glance, it seems possible that there could exist some finite partial IP loop in his sense, which is not embeddable into any finite IP loop. However, the responses "X" (i.e. "No") to E1 and " (meaning "Yes") to E2 in Evans' table, together with our Theorem 1.2 responding E3 affirmatively, still contradict the equivalence E1 ^ E2 A E3, regardless of the details of the definition of a partial IP loop. Unfortunately, Evans provided neither any counterexample to E1 nor any proof or reference in favor of E2 in [10]. A possible clue to resolving this problem lies in the PhD thesis [27] by Evans' student C. Treash from 1969. Her definition of the concept of an incomplete IP loop on page 27 is namely equivalent to that of our partial IP loop (cf. our Lemma 2.1). According to her Theorem 1, stated and proved on page 28: Every (finite or infinite) incomplete IP loop can be embedded into some IP loop, which implies E2 as a special case. Thus Evans' negative response to E1 seems to be indeed a shortcoming or just another typo. M. Vodicka and P. Zlatos: The finite embeddability property for IP loops and local... 539 After this digression we are returning to the main theme of our paper. From the course of our arguments it is clear that it suffices to prove just Theorem 1.1. We divide its proof into three steps consisting of the three propositions below. Their formulation requires some additional notions and notation. In the absence of associativity there is no obvious way how to define the order of an element. Nonetheless, the sets of elements of order 2 and 3, respectively, can still be defined for any partial IP loop P : O2(P) = {x G P : x =1, (x,x) G D(P) andxx = 1}, O3(P) = {x G P : x =1, (x, x), (x, xx), (xx, x) G D(P) and x(xx) = (xx)x = 1}. In other words, for x = 1 in P we have x G O2(P) if and only if x-1 = x, and x G O3(P) if and only if (x, x) G D(P) and x-1 = xx. The number of elements of the sets O2(P), O3(P) in a finite partial IP loop P will be denoted by o2(P), o3(P), respectively. The number of elements of any finite set A is denoted by #A. Proposition 1.4. Let (P, •) be a finite partial IP loop. Then there exists a finite partial IP loop (Q, *) such that P < Q and 3 | o3(Q). A pair (x, y) in a partial IP loop P will be called a gap if (x, y) G D(P). The set of all gaps in P will be denoted by r(P, •) = (P x P) \ D(P, •) = {(x, y) G P x P : (x, y) G D(P, •)}, or just briefly by r (P ). Obviously, both D(P ), r (P ) are binary relation on the set P, and a partial IP loop P is an IP loop if and only if it contains no gaps, i.e., r(P) = 0. Proposition 1.5. Let P be a finite partial IP loop such that 3 | o3 (P). Then there exists a finite partial IP loop Q > P satisfying the following four conditions: (4) 3 | o3(Q), #Q > 10, #Q = 4 (mod 6) and P (Q) Ç O2(Q) x O2(Q). Proposition 1.6. Let P be a finite partial IP loop satisfying the above conditions (4), such that r(P) = 0. Then there exists a finite partial IP loop Q > P satisfying the conditions (4), as well, such that #P(Q) < #P(P). Theorem 1.1 follows from Propositions 1.4, 1.5 and 1.6. Indeed, if P is a finite partial IP loop (such that P(P) = 0, because otherwise there is nothing to prove) then, using Proposition 1.4, we can find a finite partial IP loop Q > P such that 3 | 03 (Q). If r(Q) = 0 then L = Q is already a finite IP loop extending P, and we are done. Otherwise, applying Proposition 1.5, we obtain a finite partial IP loop Q1 > Q satisfying conditions (4) from Proposition 1.5. If P(Q1) = 0 then we are done, again. Otherwise, we can apply Proposition 1.6 and get a finite partial IP loop Q2 > Q1 satisfying conditions (4), as well, such that #P(Q2) < #P(Q1). Iterating this step finitely many times we finally arrive at some finite partial IP loop Qn extending P such that r(Qn) = 0. Then L = Qn > P is a finite IP loop we have been looking for. Thus it is enough to prove Propositions 1.4, 1.5 and 1.6. This will take place in the next four sections. 540 Ars Math. Contemp. 17 (2019) 493-514 2 Some preliminary results In this section we list the auxiliary results we will use in the proofs of Propositions 1.4, 1.5 and 1.6. Lemma 2.1. Let P be a partial IP loop and x, y, z G P. Then the following six conditions are equivalent: (i) (x,y) G D(P) and xy = z; (ii) (z,y-1) G D(P) and zy-1 = x; (iii) (x-1, z) G D(P) and x-1z = y; (iv) (y,z-1) G D(P) and yz-1 = x-1; (v) (z-1,x) G D(P) and z-1x = y-1; (vi) (y-1, x-1) G D(P) and y-1x-1 = z-1. Proof. Using property (3) from the definition of partial IP loops and (if necessary) the fact that (a-1) = a for any a G P, we can get the following cycle of implications: (i) (ii) (v) (vi) (iv) (iii) (i). We show just the first implication, leaving the remaining ones to the reader. If (x, y) G D(P) and xy = z then, according to (3), (z, y-1) = (xy, y-1) G D(P) and zy-1 = x. □ In particular, we will frequently use the fact that, as a consequence of Lemma 2.1, the conditions (x, y) G D(P) and (y-1, x-1) G D(P) are equivalent for any x, y G P. In the generic case all the six equivalent conditions in Lemma 2.1 are different. There are just two kinds of exceptions: first the trivial ones, when at least one of the elements x, y, z equals the unit 1 (which never produce gaps), and second, if x = y G O3(P), when the six conditions reduce to just two: (i3) (x, x) G D(P) and xx = x-1, (ii3) (x-1,x-1) G D(P) and x-1x-1 = x. Since the first condition is satisfied by the definition of order 3 elements, so is the second one, hence this situation produces any gap, neither. From now on we will preferably use a more relaxed language: when writing xy = z for elements x, y, z of some partial IP loop P we will automatically assume that (x, y) G D(P), without mentioning it explicitly. The number of gaps in any finite partial IP loop P is related to the size of P and that of the set O3(P) of order three elements through a congruence modulo 6. Lemma 2.2. Let P be a finite partial IP loop. Then #r(P) = (#P - 1)(#P - 2) - 03(P) (mod 6). M. Vodicka and P. Zlatos: The finite embeddability property for IP loops and local... 541 Proof. We know that (x, 1), (1, x), (x, x G D(P) for any x G P .At the same time, (a, a) G D(P) for all a G O3(P), as well. Except for these pairs, there are other (#P - 1)(#P - 2) - o3(P) pairs which can be either in D(P) or in r(P). Those which are in D(P) can be split into sixtuples according to Lemma 2.1, hence their number is divisible by 6, proving the above congruence. □ We will also use the Dirac's theorem from [7], giving a sufficient condition for the existence of a Hamiltonian cycle in a graph. For our purpose, the term graph always means an undirected graph without loops and multiple edges. For the basic graph-theoretical concepts the reader is referred to Diestel [5]. Lemma 2.3. Let G be a graph with n > 3 vertices in which every vertex has degree at least n/2. Then G has a Hamiltonian cycle. 3 Extensions of partial IP loops and the proof of Proposition 1.4 All the three Propositions 1.4, 1.5 and 1.6 deal with extensions (Q, *) of a partial IP loop (P, •), which can be combined using two more specific types of this construction: first, extensions preserving (the domain of) the binary operation • on the original partial IP loop P and extending the base set of P, and, second, extensions preserving the base set P and extending (the domain of) the binary operation on P. In symbols, the extending partial IP loop Q > P satisfies D(P) = D(Q) n (P x P) in the first case, while P = Q and D(P, •) C D(Q, *) in the second. We start with the first type of extensions. Let P, Q be two partial IP loops such that P n Q = {1}, i.e., their base sets have just the unit element 1 in common. Then, obviously, the set P U Q can be turned into a partial IP loop, which we denote by P U Q, extending both P and Q, with domain D(P U Q) = D(P) U D(Q), i.e., preserving the original operations on both P and Q, and leaving undefined all the products xy, yx, for x G P \ {1}, y G Q \ {1}. The partial IP loop P U Q is called the direct sum of the partial IP loops P and Q. Let us fix the notation for some particular cases of this construction, considered as extensions of the IP loop P fixed in advance. In all the particular cases below A denotes a nonempty set disjoint from P. Then the set A U {1} will be turned into a partial IP loop (A U {1}, •), depending on some map a: A ^ A. Let a: A ^ A be an involution, i.e., a(a(a)) = a for any a G A. Then the minimal partial IP loop [A, a] has the base set A U {1} and the partial binary operation given by 1 • 1 = 1, and 1a = a1 = a, aa(a) = a(a)a = 1, for any a G A, leaving the operation result ab undefined for any other pair of elements a, b G A. The reader is asked to realize that [A, a] is indeed a partial IP loop, and that it is minimal (concerning its domain) among all partial IP loops with the base set A U {1}, which satisfy a -1 = a(a) for any a G A. Then, obviously, O2[A, a] = {a G A : a(a) = a}, 542 Ars Math. Contemp. 17 (2019) 493-514 i.e., order two elements in [A, a] coincide with the fixpoints of the map a. The direct sum of the partial IP loops P and [A, a] is denoted by P[A, a] = P U [A, a]. Order two elements in P[A, a] split into two disjoint easily recognizable parts O2(P[A, a]) = O2(P) U O2 [A, a]. If a = idA: A ^ A is the identity on A, we write P [A, idA] = P [A], in which case O2(P[A]) = O2(P) U A. If A = jai,..., an} is finite, we write P [A] = P [ai,...,a„]. In particular, if A = {a} is a singleton (and a = idA is the unique map A ^ A), then P [ja}]= P [a]. If A = ja, a'} where a = a', and a is the transposition exchanging a and a', we denote P[A, a] = P[a ^ a']. From among the second type of extensions of a partial IP loop P, preserving its base set P and extending just (the domain of) its operation the simplest ones attempt at filling in just a single gap in P. This type of extension will be called a simple extension through the relation xy = z. More precisely, having x, y, z G P such that (x, y) G r(P), we want to put xy = z. From Lemma 2.1 it follows that then we have to satisfy the remaining five relations, too. This is possible only if all the pairs (x, y), (z, y-1), (x-1, z), etc., occurring there are gaps in P. This is a sufficient condition, as well, since in that case we can define all the products as required by Lemma 2.1. Thus filling in the gap (x, y) enforces to fill in some other related gaps, too. In that case we automatically assume that the remaining five relations are defined in accord with Lemma 2.1. Iterating simple extensions through particular relations we have to check in each step whether any new relation uv = w (and its equivalent forms) does not interfere not only with the pairs in D(P) but also with the gaps already filled in by previous simple extensions. In other words, we are interested in situations when we can fill in a whole set of gaps at once. If (P, •) is a partial IP loop and * is a partial operation on the set P with domain T C P x P, such that T C r(P) then, since D(P) n T = 0, we can extend the original operation • to the set D(P) U T by putting xy = x * y for (x,y) G T. The resulting structure will be called the extension of the IP loop P through the operation * and denoted by P*. The next lemma tells us when such an extension gives us a partial IP loop, again. In its formulation x-1 denotes the inverse of the element x G P with respect to the original operation • in P. M. Vodicka and P. Zlatos: The finite embeddability property for IP loops and local... 543 Lemma 3.1. Let (P, ■) be a partial IP loop and * be a partial binary operation on the set P with domain T C r (P). Then the extension of the operation ■ through the operation * to the set D(P) U T yields a partial IP loop extending P if and only if T and * satisfy the following condition: (5) for any x,y,z G P, if (x, y) G T and x * y = z then also all the pairs (z,y-1), (x-1 ,z), (y, z-1), (z-1,x), (y-1,x-1) belong to T and satisfy all the relations z * y-1 = x, x-1 * z = y, y * z-1 = x-1, z-1 * x = y-1, y-1 * x-1 = z-1. Proof. In view of Lemma 2.1, condition (5) obviously is necessary. By the same reason, condition (5) implies that each of the particular relations xy = x * y, for (x, y) G T, can be separately added to P. Since T C r(P), no particular relation xy = x * y can interfere with the remaining added relations uv = u * v. □ The following simple combination of both the types of extensions will be used in the proof of Proposition 1.4. Let A be a set (disjoint from P) and a: A ^ A be a fixpointfree involution (i.e., a(a) = a for every a G A). Then [A, a]3 denotes the extension of the minimal partial IP loop [A, a] through (just) the additional relations aa = a(a) for any a G A. Formally, [A, a}3 is the extension of [A, a] through the operation * defined on the set T = {(a, a) : a G A} by a * a = a (a) for any a G A. It is clear that each pair (a, a) is indeed a gap in [A, a] and that the condition (5) from Lemma 3.1 is satisfied. Hence [A, a}3 is a partial IP loop extending [A, a] in which a-1 = a(a) = aa for each a G A, i.e., every element a G A has order three. For the direct sum P[A,a]3 = P U [A,a]3 we have O3(P [A,a]3)= O3(P) U A. If A = {a, a'}, where a = a', then the denotations [A, a ^ a']3 and P[a ^ a']3 = P[A, a ^ a']3 are already self-explanatory, and similarly for [A, a ^ a',b ^ b']3 and P[a ^ a',b ^ b']3 = P[A, a ^ a',b ^ b']3 where the set A consists of four distinct elements a, a', b, b'. Proof of Proposition 1.4. Let a, a', b, b' be four distinct elements not belonging to P. Let us form the extensions Q = P[a ^ a']3 and R = P[a ^ a',b ^ b']3. Obviously, 03(Q) = 03(P ) + 2 and o3(R) = 03 (P) + 4. Since one of the numbers o3(P), o3(P) + 2, o3(P) + 4 is divisible by 3, one of the partial IP loops P, Q, R has the desired property. □ 544 Ars Math. Contemp. 17 (2019) 493-514 4 The proof of Proposition 1.5 A more subtle combination of the two types of extensions introduced in Section 3 will be required in the proof of Proposition 1.5. Proof of Proposition 1.5. Let P be a finite partial IP loop such that 3 | o3 (P), and A be a finite set disjoint from P with the number of its elements satisfying #A > max{5(#P) - 1, #r(P)/2} and 10 < #P + #A = 4 (mod 6). First we construct the minimal extension P[A], in which every element a of A has order two, while O3(P [A]) = O3(P). Hence the partial IP loop P [A] has the base set P U A with the required number of elements and the same number of elements of order three as P. Next, we construct an extension of P [A] in which all the original gaps in r(P) will be filled. We take T = r(P) C r(P[A]) and introduce a binary operation * on T, assigning to each pair of gaps (x, y), (x-1, y-1) G T a (self-inverse) element x * y = y-1 * x-1 = (x * y)-1 from A. At the same time we arrange that (with the above exception) x * y = u * v whenever (x, y) and (u, v) are different gaps in P. This is possible, as #A > #r(P)/2. Since all the pairs (x, y) G P[A] such that x G A or y G A, except for (1, a), (a, 1) and (a, a) where a G A, are gaps in P[A], condition (5) of Lemma 3.1 is obviously satisfied. Thus we can construct the partial IP loop P [A]*, extending P [A] through the operation *. It still has the base set P U A, while r(p[A]*) n (P x P) = 0. At the same time, D(P[A]*) n (A x A) = {(a, a) : a G A}, so that ab =1 G P for any (a, b) G D(P[A]*) n (A x A). Finally, we construct an extension Q of P[A]* with the same base set P U A, such that r(Q) c o2(q) x o2(q). As all the elements of A are of order two, and P[A] * has no gap (x, y) G P x P, it suffices to manage that (x, a), (a, x) G D(Q) for all a G A, x G P \ O2(P), x =1. We will proceed by an induction argument. To this end we represent the set P \ (O2 (P) U{1}) = {x1,x-1,...,xn,x-1}, in such a way that each pair of mutually inverse elements x, x-1 G P \ (O2(P) U {1}) occurs in this list exactly once. To start with we put Q0 = P[A] *. Now we assume that, for some 0 < k < n, we already have an IP loop Qk > P[A]* with the same base set P U A, satisfying the following three conditions: (6) au, va G A for any a G A, u, v G P \ {1} such that (a, u), (v, a) G D(Qk), (7) ab G P for any (a, b) G D(Qfc) n (A x A), and (8) (x;, a), (a, x;) G D(Qk) for all 0 < l < k, a G A. M. Vodicka and P. Zlatos: The finite embeddability property for IP loops and local... 545 Observe that Qo trivially satisfies all these conditions (with k = 0), and condition (8) jointly with Lemma 2.1 imply that (x-1, a), (a, x-1) G D(Qk) for all 0 < l < k, a G A, too. For x = xk+1, we have to fill in all the gaps in Qk in which x occurs, preserving all the conditions (6), (7), (8) with k replaced by k +1. That way all the gaps in Qk containing x-1 will be filled in, as well. Let us introduce the sets Lx = {a G A : (a, x) G r(Qk)} and Rx = {a G A : (x, a) G r(Qk)}. Then Lemma 2.1 implies that (a, x) G r(Qk) if and only if (x-1, a) G r(Qk) for each a G A, hence Lx = Rx-i and Rx = Lx-i. Claim 1. We have #Lx = #Rx. Proof. Since xu = v implies v-1x = u-1 for any u, v G P U A, we have a bijection between the sets (P U A) \ Lx = {u G P U A : (x, u) G D(Qfc)}, (P U A) \ Rx = {v G P U A : Cv-1,x) G D(Qfc)}, which implies that the sets Lx and Rx have the same number of elements. □ Thus there exists a bijective map n : Lx ^ Rx (with inverse map n-1 : Rx ^ Lx); latter on we will specify some additional requirements concerning it. We intend to use n in defining the extending operation * on the set Tx = CLx x {x}) U C{x} x Rx) U CRx x {x-1}) U C{x-1} x Lx) U {(a, n(a)) : a G Lx} U {(n(a), a) : a G Lx} by putting a * x = n(a) for any a G Lx. Then we have to satisfy the remaining five conditions of Lemma 3.1, i.e. (remembering that the elements of A are self-inverse), a * n(a) = x, x-1 * a = n(a), n(a) * a = x-1, n(a) * x-1 = x * n(a) = a. The substitution b = n(a) into the last two relations yields b * x-1 = x * b = n-1 (b) for any b G Rx. Thus we have to guarantee that each pair (a, n(a)), where a G Lx, will be a gap in Qk. Since (a, a) G D(Qk ) for all a G A, this will imply n(a) = a for a G Lx n Rx (if any). Additionally, n should avoid any "crossing", i.e., the situation that n(a) = b and n(b) = a for some distinct a, b G Lx n Rx. This namely, according to Lemma 2.1, would imply that a * b = x = b * a, and, since (a * b)-1 = b * a, produce a contradiction x = x-1. Now it is clear that once we succeed to satisfy all the above requirements, the partial IP loop Qk+1, to be obtained as the extension of Qk through the operation * constructed from the 546 Ars Math. Contemp. 17 (2019) 493-514 bijection n as described, will satisfy all the conditions (6), (7), (8) (with k +1 in place of k). Thus it is enough to show that there is indeed a "crossing avoiding" bijection n: Lx ^ Rx such that (a, n(a)) G r(Qfc) for each a G Lx. To this end we denote the common value of #Lx = #Rx by m, enumerate the sets Lx = {ai, .. ., am}, Rx = {bi, .. ., bm} in such a way that i = j whenever aj = bj G Lx n Rx, and introduce the graph Gx on the vertex set V = {1,..., m}, joining two vertices i, j by an edge if and only if i = j and both (aj, bj), (aj, bj) G r(Qfc). Claim 2. The graph Gx has a Hamiltonian cycle. Proof. According to Lemma 2.3, it suffices to show that m > 3 and that the minimal degree of vertices in Gx is at least m/2. We keep in mind that both the right side and the left side multiplication in Qk by a fixed element are injective maps. Since ax G P \ {1} for every a G A such that (a, x) G D(Qk), there are at most #P - 1 pairs (a, x) in D(Qk). Hence m = #Lx > #A - #P +1 > 4(#P) > 3. Let i be any vertex in Gx. Then i is not adjacent to a vertex j if and only if at least one of the pairs (aj, bj), (aj, bj) belongs to D(Qk). However, for fixed aj or bj, all such products ajbj or ajbj belong to P and, in both cases, every element of P occurs as a result at most once. Thus there are at most 2(#P) vertices in Gx not adjacent to i. Therefore, mm deg(i) > m - 2(#P) > m - — = —. □ Let n be a cyclic permutation of the set V such that (1,n(1),... ,nm-1(1)) isaHamil-tonian cycle in Gx. We define n: Lx ^ Rx by n(aj) = bn(j) for any i G V. Obviously, n is bijective, (aj, n(aj)) G r(Qk) for each i G V, and, since m > 3, it avoids any crossing. It follows that in the extension Qk+1 of the partial IP loop Qk through the operation * all the gaps from the set Tx are filled in, and the conditions (6), (7), (8) are preserved. The last partial IP loop Q = Qn satisfies already all the requirements of Proposition 1.5. □ 5 Steiner triples and the proof of Proposition 1.6 In the proof of Proposition 1.6 we will make use of Steiner loops and Steiner triple systems. A Steiner loop is an IP loop satisfying the identity xx = 1, i.e., an IP loop in which every element x = 1 has order two. Steiner loops are closely related to Steiner triple systems, which are systems S of three element subsets of a given base set X such that each two element subset {x, y} of X is contained in exactly one set {x, y, z} G S. Namely, if L is a Steiner loop L then X = L \ {1} becomes a base set of the Steiner triple system Sl = {{x, y, xy} : x, y G X}. M. Vodicka and P. Zlatos: The finite embeddability property for IP loops and local... 547 Conversely, if S is a Steiner triple system with the base set X then, adjoining to X a new element 1 G X, we obtain a Steiner loop with the base set X + = X U {1}, the unit 1 and the operation given by the casework {1, if x = y, z, where {x, y, z} G S, if x = y, for x, y G X. Based on this definition, we will call a Steiner triple any three-element set {x, y, z} C O2 (P) in any partial IP loop P, such that the product of any two of its elements equals the third one. It is well known that there exists a Steiner triple system S on an n-element set X if and only if n = 1 or n = 3 (mod 6) (see, e.g., Hwang [18]). The construction reducing eventually the number of gaps in a given partial IP loop P, satisfying certain conditions which will be emerging gradually, is composed of several simpler steps, we are going to describe, now. At the same time, it depends on a six term progression a = (a0, ai, a2, a3, a4, a5) of pairwise distinct order two elements of P chosen in advance; the criteria for its choice will be clarified later on. The first step is the triplication construction, which uses Steiner loops heavily. Given an arbitrary finite partial IP loop P such that #P = 2 or #P = 4 (mod 6) we denote n = #P - 1. Then Steiner triple systems on n-element sets, as well as Steiner loops on (n + 1)-element sets exist; assume that Y, Z are two n-element sets, such P, Y, Z are pairwise disjoint, and that both the sets Y + = Y U {1}, Z + = Z U {1} are equipped with binary operations turning them into Steiner loops. In rather an ambiguous way, we denote by 3P = P U Y + U Z+ the direct sum of the partial IP loop P with the Steiner loops Y + and Z + (see Section 4). It is a partial IP loop with the base set P U Y U Z, consisting of 3n +1 elements, and the domain D(3P) = D(P) U (Y x Y) U (Z x Z) U ({1} x (Y U Z)) U ((Y U Z) x {1}). We will extend the partial operation on 3P by filling in all the gaps consisting of pairs of elements of different sets P, Y, Z. That way we'll obtain an extension 3P* of 3P with the same base set P U Y U Z, such that r(3P*) = r(P). The extending operation * is defined on the set T = (P0 x (Y U Z)) U ((Y U Z) x P0) U (Y x Z) U (Z x Y) C r(P U Y+ U Z+), where P0 = P \ {1}. It depends on some arbitrary fixed enumerations Po = {xo, ...,xn_i}, Y = {yo,... ,y„_i}, Z = {zo,..., z„_i} of the sets P0, Y, Z, respectively. Once having them we put yi * zi+fc = xi+2fc for 0 < i, k < n, with the addition of subscripts modulo n. Then, in order to satisfy the conditions of Lemma 2.1, we define xi+2k * zi+k = yi, yi * xi+2k = zi+k, x_i2fc * yi = zi+fc, zi+k * x_+2k = yi, zi+k * yi = x_+2k, 548 Ars Math. Contemp. 17 (2019) 493-514 using the fact that all the elements of Y and Z are self-inverse. As all the pairs (x, y), (y, x), (x, z), (z, x), (y, z), (z, y), where x G P0, y € Y, z G Z, are gaps in 3P, Lemma 3.1 guarantees that the extension 3P * of the partial IP loop 3P through the operation * is a partial IP loop, again. For lack of better terminology we will call it a Steiner triplication of the partial IP loop P and suppress the Steiner loops Y +, Z + and the particular enumerations in its notation. The Steiner triplication 3P* of P satisfies r(3P*) = r(P), hence it still has the same number of gaps as P. However, Proposition 1.6 requires us to decrease this number. This will be achieved in a roundabout way. First we cancel some pairs in the domain D(3P*), creating that way the potential to fill in more gaps than we have added. In order to allow for this next step, P has to satisfy some additional conditions, namely, #P > 10 (i.e., n > 9) and o2(P) > 6. Though the enumerations of the sets P0, Y, Z, used in the definition of the extending operation *, could have been arbitrary, we now assume that these sets were enumerated in such a way that the six term progression a = (a0, a^ a2, a3, a4, a5) chosen in advance coincides with the sixtuple (x0, x2, xi, x5, x3, xn-3) and that {y0, yi, y3} is a Steiner triple in Y +. This artificial trick will facilitate us the description of the next step of our construction. As special cases of the above defining relations for the operation * in 3P* we get zo = yoxo = y3x„_3, zi = yox2 = yixi and z3 = yix5 = y3x3. In other words, we have the following seven Steiner triples in 3P*: {xo,yo,zo}, {x2,yo,zi}, {xi,yi,zi}, {x5 ,yi,z3}, {x3,y3,z3}, {x„_3,y3,zo}, {yo,yi,y3}. We delete these triples from the domain of 3P *. More precisely, for any one of these three-element sets we delete from D(3P*) all the six pairs consisting of its distinct elements. That way we obtain a partial IP loop 3P- < 3P *, called the reduction of 3P*, which still is an extension of P, however, it has 42 more gaps than 3P* (6 for each Steiner triple), and, since r(P) = r(3P*), than P, as well. Instead we introduce some new triples consisting of the same elements, namely which are intended to become Steiner triples, after we define a partial operation o on the set {x0, x2, xi, x5, x3, xn-3, y0, yi, y3, z0, zi, z3} by putting the product of any pair of distinct elements of a given three-element set from this list equal to the third one. That way we obtain an extending operation of the partial IP loop 3P- if and only if all the pairs entering this new operation are gaps in 3PThis is obviously true for the 18 pairs arising from the three triples in the last row above. However, this need not be the case for the pairs arising from the six triples in the first two rows. The problem can be reduced to the question which of the pairs (x0,x2), (x2,xi), (xi,x5), (x5,x3), (x3,xn-3), (xn-3,x0) belong to r(P). If, e.g., (x0,x2) G r(P) then we cannot put x0 o x2 = y0, so that {x0,x2,y0} cannot become a Steiner triple. Therefore, we include just those triples {xj, xj, yk} or {x4, xj, zk} for which the pair (xj, xj) is a gap in P. Every such "good" triple results in filling in six gaps. We already have 18 gaps filled in thanks to the last row. Thus we need at least five "good" triples in the {xo,x2,yo}, {x5,x3,z3}, {yo ^i^iL {x2, xi, zi}, {x3,x„_3,y3}, {yi,y3, z3}, {xi,x5,yi}, {x„_3,xo, zo}, {y3,yo, zo}, M. Vodicka and P. Zlatos: The finite embeddability property for IP loops and local... 549 first two rows in order to fill in additional 30 gaps; this would give 18 + 30 = 48 > 42 gaps, while having just four "good" triples results in refilling back 42 gaps, only. In general, we can fill in 6(3 + g) gaps, where 0 < g < 6 is the number of gaps (xj, Xj) in the list. We refer to this last step of the construction as to "filling in the gaps along the path" a and denote the final resulting extension of the reduction 3P- by 3P(a). Obviously, 3P(a) is an extension of the original IP loop P, as well, having 6(3 + g) - 42 = 6(g - 4) less gaps than P. This number can be negative, 0 or positive, depending on whether g < 4, g = 4, or g > 4. That's why we are interested just in the case when g > 4. After all these preparatory accounts we can finally approach the proof of Proposition 1.6. Proof of Proposition 1.6. Let P be a finite partial IP loop satisfying the conditions (4), i.e., 3 | o3(P), #P > 10, #P = 4 (mod 6) and r(P) C O2(P) x O2(P), such that r(P) = 0. We are to show that there is a finite partial IP loop Q > P satisfying these conditions, as well, with less gaps than P. Since r(P) C O2 (P) x O2 (P), it is an antireflexive and symmetric relation on the set O2(P). Thus we can form the gap graph G(P) = (V, E) with the set of vertices V = {x € O2(P) : (x, y) G r(P) for some y G O2(P)} and the set of edges E = {{x, y} : (x, y) G P(P)}. From the definition of the set of vertices V it follows there are no isolated vertices in G(P). Let's record some less obvious useful facts about this graph. Claim 3. (a) The degree of each vertex in G(P) is even. (b) The number of edges in G(P) is divisible by three. Proof. (a): Let x G O2(P). Then the conditions xy = z and xz = y are equivalent for any y, z G P. Additionally, as x = 1, from xy = z it follows that y = z. Thus the elements y G P such that (x, y) G D(P) can be grouped into pairs, hence their number is even. As #P is even, too, so is the degree deg(x) = #{y G O2(P) : (x,y) G P(P)} = #P - #{y G P : (x,y) G D(P)}. (b): By Lemma 2.2 we have #r(P) = (#P - 1)(#P - 2) - 03(P) = 0 (mod 6). On the other hand, #P = 4 (mod 6) and 3 | 03 (P), yielding 3 | #r(P). Obviously, the number of edges in G(P) is half of the number of gaps #r(P), hence the number of edges in G(P) must be divisible by three. □ The structure of connected components in G(P) obeys the following alternative. 550 Ars Math. Contemp. 17 (2019) 493-514 Claim 4. Let C be a connected component of the graph G(P). Then either C contains a triangle or a path of length five, or, otherwise, C is isomorphic to one of the following graphs: the cycle C4 of length four, the cycle C5 of length five or the complete bipartite graph K2,m where m > 4 is even. Proof. Let C be any connected component in G(P). As G(P) has no isolated vertices and the degree of every vertex in C is even (and therefore at least two), there is a cycle in C. Assume that C contains no triangle and no path of length five. Then the length of this cycle must be bigger than three and less than six. Thus there are just the following two options: (a) There is a cycle of length five in C. Then there cannot be any other edge coming out from its vertices since then there would be a path of length five contained in C. Thus C coincides with this cycle. (b) There is a cycle of length four in C; let us denote it by (v0, v1, v2, v3). Then, as G(P) contains no triangle, neither {v0, v2} nor {v1, v3} is an edge in G(P). If there are no more vertices in C then C is a cycle of length four. Otherwise, we can assume, without loss of generality, that there is a fifth vertex u0 g C adjacent to v0. As u0 has an even degree, it must be adjacent to some other vertex, too. If it were adjacent to some vertex u1, distinct from all the vertices v0, v1, v2, v3, there would be apath (u1, u0, v0, v1, v2, v3) of length five in C. If u0 were adjacent to v1 or to v3, there would be a triangle (u0, v0, v1) or (u0, v0, v3) in C. That means that {u0, v2} must be an edge in G(P) and deg(u0) = 2. It follows that every other vertex in C must have degree two and it must be adjacent either to v0 and v2 or to v1 and v3. However, the second option is impossible, since in that case (u0, v0, v1, u1, v3, v2) would be a path of length five. This means that C is isomorphic to the complete bipartite graph K2,m, where one term of this partition is formed by the set {v0, v2} and the second one by the rest of the vertices in C. Since every vertex has an even degree, m must be even. At the same time, m > 4, as K2 2 has just four vertices (and it is isomorphic to the cycle C4). □ Thus the proof of Proposition 1.6 will be complete once we show how to construct the extension Q in each of the cases listed in Claim 4. (a) G(P) contains a triangle, i.e., a three-element set of vertices {x, y, z} such that all its two-element subsets are edges. Then we can extend P through the operation * turning {x, y, z} into a Steiner triple. The corresponding extension Q of P has all the properties required and six less gaps than P. (b) G(P) contains a path a = (a0, a1, a2, a3, a4, a5) of length five. Then we can form the Steiner triplication 3P* of P and, filling in the gaps along the path a in its reduction 3Pwe obtain the final extension Q = 3P (a) satisfying the condition (4), again. If (a5, a0) is a gap in P (i.e., if a is a cycle of length five in G(P)) then Q has twelve gaps less than P, otherwise it still has six gaps less than P. We still have to prove Proposition 1.6 in the case there is neither any triangle nor any path of length five in G(P). To this end it is enough to construct, in each of the remaining cases listed in Claim 4, an extension Q of P such that the graph G(Q) has the same number of edges as G(P), however, there is a path b of length five in G(Q). From such a Q we can construct another extension 3Q (b) > Q > P with a smaller number of gaps and still M. Vodicka and P. Zlatos: The finite embeddability property for IP loops and local... 551 satisfying the condition (4), similarly as we did in the case (b). So let us have a closer look at the remaining cases. (c) G(P) contains a connected component isomorphic to K2,m, where m > 4. Let {w0, ui} be the two-element partition set and {v0, vi, v2, v3} be any four-element subset of the m-element partition set in that component of G(P). We denote by a the six-term progression (v0, w0, v1, v2, u1, v3) and construct the extension Q = 3P(a) of the partial IP loop P with the gap graph G(Q). Then {v0, w0}, {w0, v1}, {v2, u1}, and {u1, v3} are edges in G(P), while {v1, v2} and {v3, v0} are not. Hence the new graph G(Q) has the same number of edges as G(P) and Q has the same number of gaps as P. At the same time, there are two distinct new vertices y1, z3 in G(Q), occurring in the enumerations of the sets Y, Z, respectively. Now, one can easily verify that b = (v0, u1, v1, y1, v2, w0) is a path of length five in G(Q). (d) There are two distinct connected components C and D in G(P), each of them isomorphic to the cycle C4 or C5. Let m and l denote any of the numbers 4 or 5. We assume that (w0, u1,..., wm-1) and (v0, v1,..., v;-1) are the cycles forming the components C = Cm and D = C;, respectively. Now we take the six term progression a = (w0, w1, w2, v0, v1, v2) and form the extension Q = 3P(a). Once again, {w0, w1}, {u1,u2}, {v0,v1} and {v1,v2} are edges in G(P), while {w2,v0} and {v2, w0} are not. Hence G(Q) has the same number of edges as G(P) and Q has the same number of gaps as P. Now, picking the new distinct vertices y1 G Y, z3 G Z, we obtain the path b = (w3, w2, y1, v0, v;-1, v;-2) of length five in G(Q). (e) G(P) consists of a single connected component isomorphic either to C4 or to C5. However, this is impossible, since the number of edges in G(P) is divisible by three. This concludes the proof of Proposition 1.6, as well as of Theorems 1.1, 1.2 and 1.3. 6 Final remarks The discussion from the introduction together with Theorem 1.3 naturally lead to the following question. Problem 6.1. Is there some minimal (ore even the least) axiomatic class K of IP loops such that every group is locally embeddable into Kfin? Does this class (if it exists) satisfy the Finite Embeddability Property? The first candidate which should be examined in this connection seems to be the class of all Moufang loops introduced in [25]: A Moufang loop is a loop satisfying one (hence all) of the following four equivalent identities cf. Pflugfelder [26], Kunnen [19]. It is well known that every Moufang loop is an IP loop. The following is not the usual definition of the concept of a sofic group (see Gromov [15], Weiss [29], Ceccherini-Silberstein, Coornaert [2]), however, as proved by Gordon and Glebsky [12], it is equivalent to it. A group (G, •, 1) is sofic if for every nonempty □ x(y(xz)) = ((xy)x)z, x(y(zy)) = ((^y»^ (xy)(zx) = (x(yz))x, (xy)(zx) = x((yz)x), 552 Ars Math. Contemp. 17 (2019) 493-514 finite set X C G and every e > 0 there exists a finite quasigroup (Q, *) such that X C Q, for all x,y e X satisfying xy e X we have xy = x * y, as well as No example of a non-sofic group is known up today, however, there is a general belief that not every group is sofic. Theorem 1.3 together with the above description of sofic groups indicate that the sofic groups could perhaps be characterized as groups locally em-beddable into some "nice" subclass of the class of finite IP loops, fulfilling some "reasonable amount of associativity". A natural candidate is the class of all finite Moufang loops, once again. For some additional reasons in favor of this choice see [11]. As already indicated, one should start with trying to clarify the following question. Problem 6.2. Does the class of all Moufang loops have the FEP? 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ARS MATHEMATICA CONTEMPORANEA 17 (2019) 555-579 https://doi.org/10.26493/1855-3974.1581.b47 (Also available at http://amc-journal.eu) On graphs with the smallest eigenvalue at least -1 - V2, part III Tetsuji Taniguchi t Hiroshima Institute of Technology, Department of Electronics and Computer Engineering, 2-1-1 Miyake, Saeki-ku, Hiroshima, Japan Received 15 January 2018, accepted 13 September 2019, published online 3 December 2019 There are many results on graphs with the smallest eigenvalue at least -2. In order to study graphs with the eigenvalues at least -1 - %/2, R. Woo and A. Neumaier introduced Hoffman graphs and H-line graphs. They proved that a graph with the sufficiently large minimum degree and the smallest eigenvalue at least -1 - a/2 is a slim {[h2], [hs], [hr], [h9]}-line graph. After that, T. Taniguchi researched on slim {[h2], [hs]}-line graphs. As an analogue, we reveal the condition under which a strict {[hi], [h4], [h7]}-cover of a slim {[hr]}-line graph is unique, and completely determine the minimal forbidden graphs for the slim {[hr]}-line graphs. Keywords: Hoffman graph, line graph, smallest eigenvalue. Math. Subj. Class.: 05C50, 05C75 *S. K. was supported by JSPS KAKENHI; grant number: 18J10656. tT. T. was supported by JSPS KAKENHI; grant number: 16K05263. E-mail addresses: kubota@ims.is.tohoku.ac.jp (Sho Kubota), t.taniguchi.t3@cc.it-hiroshima.ac.jp (Tetsuji Taniguchi), kiyoto.yosino.r2@dc.tohoku.ac.jp (Kiyoto Yoshino) Sho Kubota * Tohoku University, Graduate School of Information Sciences, 6-3-09 Aoba, Aramaki-aza Aoba-ku, Sendai, Miyagi, Japan * Kiyoto Yoshino Tohoku University, Graduate School of Information Sciences, 6-3-09 Aoba, Aramamaki-aza, Aoba-ku, Sendai, Miyagi, Japan Abstract ©® This work is licensed under https://creativecommons.org/licenses/by/4.0/ 556 Ars Math. Contemp. 17 (2019) 493-514 1 Introduction Throughout this paper, we will consider only undirected graphs without loops or multiple edges, and denote by Amin(T) and ¿(r) the minimum eigenvalue and the minimum degree of a graph r, respectively. P. J. Cameron, J. M. Goethals, J. J. Seidel and E. E. Shult have characterized generalized line graphs as the graphs with the smallest eigenvalue at least —2 except for finitely many graphs with at most 36 vertices in [3]. After that, A. Hoffman proved the following theorem in [6]. Theorem 1.1. There exists an integer valued function f defined on the intersection of the half-open interval (-1 — a/2, -2] and the set of the smallest eigenvalues of graphs, such that if r is a connected graph with ¿(r) > f (Amin(r)) then (i) if — 1 > Amin(r) > —2 then r is a complete graph and Amin(r) = —1. (ii) if—2 > Amin(r) > —1 — a/2 then r is a generalized line graph and Amin(r) = —2. In [12], R. Woo and A. Neumaier introduced Hoffman graphs and H-line graphs, where H is a family of isomorphism classes of Hoffman graphs, to extend the result of A. Hoffman, and proved Theorem 1.2. Moreover, they raised the problem [12, Open problem 3] to reveal the list of minimal forbidden graphs for the slim {[h2], [hs], [hr], [h9]}-line graphs. These Hoffman graphs and some ones that appear in this paper and [12] are listed in Figure 1 (here, the names hi, h2,... depend on [12]). hi h2 A -V h4 = r= • • hs = • h6 = •• = Xlj Figure 1: Hoffman graphs with slim (resp. fat) vertices depicted as small (resp. large) black dots. Theorem 1.2. Let a4 (« —2.4812) be the smallest root of the polynomial x3 + 2x2 — 2x — 2. There exists an integer valued function f defined on the intersection of the half-open interval (a4, —1 — a/2] and the set of the smallest eigenvalues of graphs, such that if r is a graph with Amin(r) G (a4, —1 — a/2] and ¿(r) > f (Amin(r)), then r is an {[h2], [hs], [hr], [ho]}-line graph. S. Kubota et al.: On graphs with the smallest eigenvalue at least — 1 — \/2, part III 557 Since it is difficult to solve the open problem, T. Taniguchi considered a partial problem. In [11], he completely determined the 38 minimal forbidden graphs for the slim {[h2], [hs]}-line graphs by using Theorem 1.3 [10]. Theorem 1.3. A slim {[h2], [hs]}-line graph with at least 8 vertices has a unique strict {[h2], [h3], [hs]}-cover up to equivalence. As an analogue of his result, we reveal the minimal forbidden graphs for the slim {[hr] }-line graph. In Section 2, we introduce a part of the basic theory of Hoffman graphs summarized in detail in [7]. In Section 3, we introduce minimal forbidden graphs. In Section 4, we aim to prove Theorem 4.11 which reveals the necessary and sufficient condition that a strict {[hi], [h4], [hr]}-cover of a slim {[hr]}-line graph becomes unique up to equivalence. Furthermore, when the condition is not satisfied, the theorem shows the shape of the slim {[hr]}-line graph and indicates its strict {[h1], [h4], h]}-covers are exactly two up to equivalence. This helps us to examine the minimal forbidden graphs for the slim {[hr]}-line graphs. In order to prove our main result Theorem 5.1, in which we determine the minimal forbidden graphs for the slim {[hr]}-line graphs, we computed the minimal forbidden graphs with at most 9 vertices by the software Mamga [2]. In Section 5, we determine the minimal forbidden graphs apart from those with at most 9 vertices. 2 Hoffman graphs We introduce definitions related to Hoffman graphs. Details are in [7]. Definition 2.1. A Hoffman graph h is a pair (H, p) of a graph H = (V, E) and a labelling map p: V ^ {f, s}, satisfying the following conditions: (i) every vertex with label f is adjacent to at least one vertex with label s; (ii) vertices with label f are pairwise non-adjacent. We call a vertex with label s a slim vertex, and one with label f a fat vertex. We denote by Vs(h) (resp. Vf (h)) the set of slim (resp. fat) vertices of h. For a vertex x of a Hoffman graph h, we denote by Nf (x) (resp. Nj®(x)) the set of neighbors labelled f (resp. s) of x, and set N (x) = Nf (x) U N^x). For a set X of vertices of h, we let Nf (X) := |JxeX Nf (x) and N^X) := |JN^x). We regard an ordinary graph H without labelling as a Hoffman graph (H, p) without fat vertices, that is, p(x) = s for any vertex x of H. Such a graph is called a slim graph. Definition 2.2. A Hoffman graph h' = (H', p') is called an induced Hoffman subgraph of a Hoffman graph h = (H, p), if H' is an induced subgraph of H and p|V(H') = p'. For a subset S of Vs(h), we denote by ((S}}(, the induced Hoffman subgraph of h by SU Nf (S). We denote by (S}r the ordinary induced subgraph by S of a graph r for a subset S of V(r). For a Hoffman graph h, (Vs(h)}h is called the slim subgraph of h. The diameter of a graph is the maximum distance between two distinct vertices. Let r be a graph and C be a subset of V(r). Then, C is a clique in r if the induced subgraph (C}r is a complete graph. The size of the largest clique in r is called the clique number. A partition n = {C1, C2,..., Ct} of V(r) is called a clique partition if all cells Cj are cliques. Focusing on cliques is useful for discovering the structure of line graphs. Also in this paper, we may focus on clique numbers and clique partitions. 558 Ars Math. Contemp. 17 (2019) 493-514 Definition 2.3. Let h be a Hoffman graph, and let h1 and h2 be two induced Hoffman subgraphs of h. The Hoffman graph h is said to be the sum of h1 and h2, written as h = h1 ® h2, if the following conditions are satisfied: (i) V(h) = V(h1) u V(h2); (ii) Vs(h) = Vs(h1) u Vs(h2) and Vs(h1) n Vs(h2) = 0; (iii) if x G Vs(hi), y G Vf (h) for i = 1 or 2, and x ~ y, then y G Vf (h®); (iv) if x G Vs (h1) and y G Vs (h2), then x and y have at most one common fat neighbor, and they have one if and only if they are adjacent. If h is the sum of some two nonempty Hoffman graphs, then it is said to be decomposable. Otherwise, h is said to be indecomposable. Remark that the sum of Hoffman graphs satisfies commutative and associative laws. Definition 2.4. Let h and m be Hoffman graphs, and let ^ be a graph morphism from the underlying graph of h to that of m. Mapping ^: h ^ m is called a morphism if it preserves the labelling, that is, ^(Vs(h)) C Vs(m) and ^(Vf (h)) C Vf (m). If ^ is a morphism and a graph isomorphism, then it is called an isomorphism, and h and m are said to be isomorphic, written as h — m. Let [h] denote the isomorphism class of h. Definition 2.5. Let H be a family of isomorphism classes of Hoffman graphs. A Hoffman graph m is called a H-line graph if it is an induced subgraph of some Hoffman graph h = 0"=1 h®, where [h®] G H for every i. In this case, m is called a slim H-line graph if m is a slim graph, and h is called a strict H-cover of a graph r if Vs(h) = V(r). Two strict H-covers h and h' of a graph r are said to be equivalent, if there exists an isomorphism ^: h ^ h' such that ^>|r is the identity automorphism of r. Lemma 2.6. Let H be a family of isomorphism classes of Hoffman graphs, and let H' be the family of the isomorphism classes of indecomposable induced Hoffman subgraphs by a nonempty set of slim vertices in a member of H. Then, every slim H-line graph has a strict H'-cover. Proof. Let h = 0"=1 h®, where [h®] G H for every i. Then, it holds that n «S»h = 0 ((S n Vs(h®)»h ®=1 forasubset S of Vs(h). Therefore ((S}}^ isastrict H'-cover of the induced subgraph by S since every addend is the sum of some indecomposable induced Hoffman graphs of h. □ 3 Minimal forbidden graphs In graph theory, various important families of graphs can be described by a set of graphs that do not belong to that family. This is the concept of so-called minimal forbidden graphs. First, we give the definition. Suppose that a family G of graphs is closed under the operation to take induced subgraphs, that is, G satisfies the condition that for a graph G in G, any induced subgraph of G is also in G. Then, we say that a graph F is a minimal forbidden graph for G if both of the following are satisfied: S. Kubota et al.: On graphs with the smallest eigenvalue at least — 1 — \/2, part III 559 (i) F is not in G; (ii) Every proper induced subgraph of F is in G. On the family of ordinary line graphs [1] and the family of slim {[h2], [hs]}-line graphs [11], their minimal forbidden graphs are revealed. Besides this, characterizations of forests, perfect graphs [4] and Threshold graphs [5] by minimal forbidden graphs are also known. In addition, Sumner [9] claimed that if r is a connected K1j3-free graph of even order, then r has a 1-factor. As such, there are also known results that properties of a family of graphs in the case that "forbidden graphs" are specified in advance. As we can see from these results, revealing the minimal forbidden graphs is one way to understand families of graphs. Unfortunately not being finite, but we are able to reveal the minimal forbidden graphs for the slim {[hr]}-line graphs. 4 The condition that an {[hi], [h4], [h7]}-strict cover of a slim {[h7]}-line graph is unique up to equivalence In this section, set H = {[h1], [h4], [hr]}. Note that every slim {[hr]}-line graph has a strict H-cover by Lemma 2.6. For exmaple, the graph r in Figure 2 is a slim {[hr]}-line graph. Indeed, considering the sum h = h7 ® h1 ® h4 of Hoffman graphs in Figure 9, we see that r h = hr e hi e h4 = Figure 2: A slim {[hr]}-line graph and its strict {[h1], [h4], [h7]}-cover. the slim subgraph of h is the graph r. (In Figure 2, the dotted lines are used for convenience to show what kind of small Hoffman graphs the graph r is decomposed by. In addition, for two vertices x and y which belong to distinct addends, we omit the edge between x and y if they have a common fat neighbor since the existence of edge between x and y depends only on that of their common fat neighbor by Definition 2.3 (iv).) In addition, h 1 and h4 are induced subgraphs of hr, so the graph r is certainly a slim {[hr]}-line graph. On the other hand, since Vs(h) = V(T) holds, the Hoffman graph h is a strict {[h1], [h4], [h7]}-cover of r. Let h = 0n=1 h® where [h®] € H for every i. Then, we can regard Nf as a mapping from Vs (h) to Vf (h) since every slim vertex is adjacent to exactly one fat vertex. For a slim vertex x of h, let h(x) denote the addend h® containing x, and let C\(x) = N^(Nf (x)) 560 Ars Math. Contemp. 17 (2019) 493-514 and cov h := IN» | u e Vf (h)} = {Ch(x) | x e Vs(h)}. Let x be a slim vertex of h. We show that C, (x) is a clique. First, we take u e Vf (h) such that Nf ( x) = {u}. We arbitrarily take two slim vertices y and z in NS(u) (= C, (x)). It suffices to show that y and z are adjacent. If y and z are contained in the same indecomposable addend of h, then they are adjacent. Otherwise, so are they by Definition 2.3 (iv). Hence, the desired result follows. Note that cov h is a clique partition of Vs (h). Moreover, it holds clearly that Nf | a = for any subset A c Vs (h). Lemma 4.1. Let h = 0"=1 hl> where [h®] e H for every i, and let C be a clique of the slim subgraph of h. Then, the following hold: (i) two distinct slim vertices x and y are adjacent if and only if h(x) = h(y) or Nf (x) = Nf (y); (ii) C c C, (x) for any x e C, or C c Vs(h(y)) for any y e C; (iii) If C c Ch (x) n Vs(h(y)) for some x,y e C, then |C| < 2. Proof. Statements (i) and (iii) hold clearly. Assume that C C C, (x) for some x e C. There exists y e C such that Nf (x) = Nf (y). Thus, h(x) = h(y) holds by (i). Statement (ii) follows since Nf (x) = Nf (z) or Nf (x) = Nf (z) for each z e C. □ We introduce some definitions to determine the strict H-covers of a graph. Definition 4.2. Let r be a graph, and let {Cj}ie/ be a partition of the vertex set of r. Then, define n(x) = Nr(x) — C® for x e C®. In addition, define n0(x) = {x} and nk (x) = nk-1(x) U J n(y) yGnk—1 (x) for a positive integer k. A vertex x of r is said to be good for the given partition {Cj}ie/ if x satisfies one of the following conditions: (i) n(x) = 0; (ii) n(x) = {y} for some y, and n(y) = {x}; (iii) n(x) = {y, z} for some y and z, n(y) = {x}, n(z) = {x}, and y ~ z; (iv) n(x) = {y} for some y, n(y) = {x, z} for some z, n(z) = {y}, and x ~ z. Furthermore, a set of vertices is said to be good if every element is good. Let Or be the set of clique partitions for which every vertex is good. We can regard cov as a mapping from the set of equivalent classes of strict H-covers of r to Or, and Proposition 4.3 holds. It is clear that if n(u) has a good vertex then u is good, and if u is good then n(u) is good. Proposition 4.3. The mapping cov for a graph is bijective. S. Kubota et al.: On graphs with the smallest eigenvalue at least — 1 — \/2, part III 561 Proof. We construct the inverse mapping of cov. Let {Ci}iei G Or. A Hoffman graph m is defined as Vs(m) := V(r), Vf (m) := {Ci}iej and E(m) := E(r) U {{x, C} | x G V(r), and C G {CJie/ and x G C}. For x G V(r), define the induced Hoffman graph mx := ((n2(x)))m. It holds that m = 0{mx | x G V(r)}, and [mx] = [hi], [h4] or [hr] for each vertex x. Hence, m is a strict H-cover of r. The mapping ^: Or 3 {Ci}ie/ ^ m G the set of strict H-covers of r is the inverse mapping of the mapping cov. □ We have the following lemma: Lemma 4.4. Let r be a graph with a partition {Ci}ie1 of the vertex set. Then, a vertex x is good for {Ci}ie/ if and only if x is good for {n3(u) n Ci}ie/ in (n3(u))r. Let r be a connected graph, and let K be a nonempty set of vertices. Then, let dKr(x) = dK (x) = d(x) := min d(x, k) ' keK for x G V(r), where d(x, y) is the distance between x and y. Define dmax = max dK,r(y), yev(r) ' and let (K) denote the family {{y G{x}U N(x) | dK,r(y) > 0K,r(x)} | x G V(r) and dK,r(x) G 2N + 1} U {K} of sets of vertices. If K g cov h then ^r(K) = cov h for every strict H-cover h of r. This means that we can restore the clique partition if we find a member of a partition in Or. We have the following lemmas: Lemma 4.5. Let r be a graph with a clique K. If (K) is a partition of V(r) and r has no induced subgraph isomorphic to K1,3, then ^r(K) is a clique partition. Lemma 4.6. Let r be a connected slim {[hr] }-line graph with a clique C of size c. Let h be a strict H-cover of r. If the following (i) or(ii) holds, then C (x) is the maximal clique containing C for any x G C, and a strict H-cover of r is unique up to equivalence. (i) c > 4, (ii) c = 3, and |Nm (C)| = 1 for any strict H-cover m of r. Proof. In the case of (i), for each clique D which contains C and any x G C, D C C (x) holds by Lemma 4.1 (ii) and | D | > 4. Hence, C^ (x) is a unique maximal clique containing C for any x G C .In the case of (ii), we can prove as well by Lemma 4.1 (ii) and (iii). Next, we show the uniqueness of a strict H-cover of r. The maximal clique D containing C is defined independently of a choice of a strict H-cover. Hence, ^r(D) is also defined independently of one. By Proposition 4.3, a strict H-cover of r is unique. □ 562 Ars Math. Contemp. 17 (2019) 493-514 We define Si and S2 Lemma 4.7. Let h be a strict H-cover of a graph r. If r has an induced subgraph isomorphic to Si or S2, then the vertices of the triangle of the induced subgraph are adjacent to the same fat vertex in h. Proof. Let A be the triangle in the induced subgraph S ~ Si or S2. Let m be a strict H-cover of r. We suppose that |N*m(A)| > 2 to prove INÎ(A)| = 1 by contradiction. Then, we have A is not contained in Cm(x) for every x G V(r) since every slim vertex in Cm (x) are adjacent to the same fat vertex. This together with Lemma 4.1 (ii) implies that A c Vs(m(y)) for any y G A. We take a vertex y G V(A). Then, A c Vs(m(y)), and hence [m(y)] ((V(S)}}m is a strict H-cover of S. Hence, we have [hr]. Moreover, ((V(S)»m = «A»m(tf) ©((V(S) \ A))m/ ^ hr ©((V(S) \ A))m', where m' denotes the Hoffman graph so that m = m(y) © m'. It is easy to verify that the slim subgraph of hr © ((V(S)))m/ is distinct from Si and S2. This is a contradiction to S ~ S1 or S2. Therefore the desired result follows. □ The Lemma 4.6 gives conditions that a strict H-cover is unique, and Lemma 4.7 gives a concrete situation satisfying one of the conditions. Lemma 4.8. If the slim subgraph of a Hoffman graph h = 0N=1 h® with [h®] G H for every i is connected, then that of 0 i=1 ®=k h® is connected for some k. Proof. Note that an {[hr]}-line graph is connected if and only if the slim subgraph is connected. Let r be the graph with the vertices {1,..., N} whose two distinct vertices x and y are adjacent if and only if V (h®) n V (hj) = 0. Since r is connected, there exists integer k such that r - k is also connected. Hence, 0N=1 ®=k h® is connected, and the slim subgraph is connected. □ Let t = (t®)"=1 be a finite sequence of positive integers. Then, define the graphs Pt and Ct by V (Pt) = V (Ct) E (Pt) E (Ct) {(i,j) | 1 < i < n, 1 < j < ti}, {{(i, j), (i', j')} I i - i' = 1, or i = i' and j = j'}, {{(i, j), (i', j')} I i - i' = 1 (mod n), or i = i' and j = j'}, respectively (see Example 4.10). Let K,...,afc ] := {(a,,j ) G V (r) | 1 < i < k, 1 < j < ta.} S. Kubota et al.: On graphs with the smallest eigenvalue at least — 1 — \/2, part III 563 for jai,..., ak} C {1,..., n}, where r = Pt or Ct. In addition, let TP := {(ti)"=i G{1, 2}n | n e Z>2,ti + ti+i < 3(1 < Vi < n - 1)} and (4.1) TC := {(ti)n=i e {1, 2}n | n e (2Z)>4, t» +t(i+1 mod n) < 3(1 < Vi < n)}. (4.2) Furthermore, a vertex u of a graph is said to be end if the graph is isomorphic to Pt for some t e TP with the length n, and u e [1] or [n]. In the following lemma, we see that Pt and Ct are slim {[hr] }-line graphs, and reveal their strict H-covers. Lemma 4.9. For t e TP of length n, we have OPt is the set of {[1], [2, 3], [4, 5], [6, 7],...} and {[1, 2], [3,4], [5, 6],...}. (4.3) For t e CP of length n, we have OCt is the set of {[1, 2], [3,4],..., [n - 1, n]} and {[n, 1], [2, 3],..., [n - 2, n - 1]}. (4.4) Namely, for t e TP (resp. TC), the graph Pt (resp. Ct) has precisely two strict H-covers up to equivalence. Proof. Recall that ^r(C) = n holds for every C e n where r is a slim {[hr]}-line graph and n e Or. In order to reveal Or, it suffices to verify whether (K) is in Or for every clique K of r. We fix a sequence t e TP of length n, and determine OPt. Since if n = 2 then desired result holds, we may assume that n > 3. On the other hand, every clique of Pt is contained in [i, i + 1] for an integer i e {1,..., n - 1}. The clique partitions in (4.3) are obtained from cliques [1], [n] and [i, i + 1] for i e {1,..., n - 1}. Moreover we can verify that (K) is not in OPt for one clique K of the other following cliques: (i) non-empty subsets of [i] for i e {2,..., n - 1}; (ii) {(i, 1), (i + 1,1)} and {(i, 2), (i + 1,1)} for i e {1,..., n - 1} with t» = 2; (iii) {(i, 1), (i + 1,1)} and {(i, 1), (i + 1, 2)} for i e {1, ...,n - 1} with t»+i = 2. Similarly, we can determine OCt for every t e TC. Finally, by Proposition 4.3, which claims that cov is a bijection from the set of strict H-cover of a slim {[hr] }-line graph r to Or, Pt and Ct have precisely two strict H-covers up to equivalence. □ Example 4.10. We give examples of strict H-covers of Pt and Ct. In the case of t = 564 Ars Math. Contemp. 17 (2019) 493-514 By Proposition 4.3, these clique partitions give strict H-covers. They are and ■ respectively. The similar consideration is applied to Ct for t g TC. For example, we consider t = (1,1, 2,1,2,1). Then the graph Ct is and By Proposition 4.3, we have the following two strict H-covers: and Theorem 4.11. If a connected slim {[h7]}-line graph r with the clique number c satisfies one of the following conditions: S. Kubota et al.: On graphs with the smallest eigenvalue at least — 1 — \/2, part III 565 (a) c =1 or c > 4; (b) r has an induced subgraph isomorphic to S1 or S2 then it has a unique strict {[hi], [h4], [h7]}-cover up to equivalence. Otherwise, r is isomorphic to Pt for some t e TP or Ct for some t e TC, and it has precisely two strict {[hi], [h4], [hr]}-covers up to equivalence. Proof. If (a) or (b) holds then a strict H-cover is unique by Lemma 4.6 and Lemma 4.7 (see Example 4.12). Otherwise, it is proved that r is isomorphic to either Pt for some t G TP or Ct for some t e TC by induction on the number of addends of a strict H-covers of r. Fix a strict H-cover h = 0 N=1 h®, where [h®] G H for every ¿.If N =1 then r ~ P{i,i} or P{1,2}. Otherwise, we can take an integer k such that the subgraph r' induced by the slim vertices of h' = 0N=1 ®=k h® is connected by Lemma 4.8. Each of S1 and S2 is not isomorphic to any induced subgraph in r'. Note that the clique number c' of r' is at most 3. Suppose c' = 2 or 3 since the result follows if c' = 1. h = h' ® hk = mj ® hk The slim subgraph of h k h Figure 3: An example of the case that the slim subgraph of h' is isomorphic to Ct for t e TC. If r' ~ Ct/ for some t' e TC, then h = h' ® hk must have an induced subgraph isomorphic to either S1 or S2 (see Figure 3). Otherwise, r' ~ Pt> for some t' e TP. Let mi := cov-1({[1], [2, 3], [4, 5], [6, 7],...}), m2 := cov-1({[1, 2], [3, 4], [5, 6],...}), and let n denote the length of t. By the induction hypothesis, we can take j e {1, 2} so that h' and mj are equivalent. Then, we show that the following two conditions hold: (A) |Nm (u)| + |NSk (u)| < 3 holds for every u e Vf (mj) n Vf (hk); 566 Ars Math. Contemp. 17 (2019) 493-514 (B) N^.(u) = [1], [1, 2], [n] or [n - 1,n] holds for every u e Vf (mj) n Vf (hk). First, if |N£. (u) | + IN^k (u) | > 4 holds for u e Vf (mj) n Vf (hk), then N£. (u) U N^ (u) is a clique of size greater than 4 in r, a contradiction to the assumption that r does not satisfy the condition (a). Second, we suppose that the condition (B) does not hold. Then we can take a fat vertex u e Vf (mj ) n Vf ( h k ) so that NS j (u) = [i,i +1] for i e {2,..., n - 2}. Thus, r has an induced subgraph isomorphic to Si (see Figure 4), a contradiction to the assumption that r does not satisfy the condition (b). Therefore the two conditions (A) and (B) are proved. h = h' ® hk [1] [2, 3] [4] Figure 4: An example of the case that r' ~ Pt> and the condition (B) does not hold. In the case of [hk] = [hi], by the condition (B), the fat vertex u in hk equals Nfj ([1]), Nfj ([1, 2]), Nfj ([n]) or Nf ([n - 1,n]) for some j = 1 or 2. If u = Nf ([1]) or Nf ([n]) then r is isomorphic to Py for some y G TP. Otherwise, without loss of generality we can assume that u = Nf([1, 2]). t1 = = 1 by the condition (A). Hence, r is isomorphic to Py for some y e TP. Then ti = ¿2 We consider the case of [hk] = [h4] or [hr]. If n = 2 then the desired result holds. Thus, we may assume that n > 3. Then u = Nf. ([i,i + 1]) for every fat vertex u e Vf (hk) and 1 < i < n - 1 (4.5) since if (4.5) does not hold then h has an induced subgraph isomorphic to either S1 or S2 (see Figure 5), a contradiction. Let u and v are distinct fat vertices of hk. Then one of the following holds: (i) u = Nf j ([1]) and v £ Vf (h'), or u = Nf. ([n]) and v £ Vf (h'); (ii) u = Nf j ([1]),v = Nf j ([n]) by exchanging u and v if necessary. Hence, h = h' ® hk is isomorphic to either Py for some y e TP or Cy for some y e TC. □ S. Kubota et al.: On graphs with the smallest eigenvalue at least — 1 — \/2, part III 567 [1,2] [3,4] [5] Figure 5: An example of the case that r' = Pt>, hk — h4 and u = Nf j ([1,2]). Example 4.12. In Theorem 4.11, there are the two conditions that a slim j[hr]}-line graph has a unique strict H-cover up to equivalence. For each condition, we give an example. We let G and h denote the slim j[hr]}-line graph and its strict H-cover in Figure 6, respectively. Then, the clique number c of G is equal to 4, and the set K of small circles of G is a maximal clique. Namely, G satisfies the condition (a) in Theorem 4.11. Take a vertex x in K. As shown in Lemma 4.6, K = C§ (x) holds. Since K = C§ (x) G cov h, we have ^G(K) = ^G(Ch(x)) = cov h. As with the proof of Proposition 4.3, we derive the Hoffman graph h by adding fat vertices according to ^g(K). Figure 6: A slim {[h7]}-line graph whose clique number c is 4 and its strict H-cover corresponding to ^g(K). Next, we let H and m denote the slim {[hr]}-line graph and its strict H-cover in Figure 7, respectively. Let H' be the subgraph induced by the small circles in H. Then, the clique number c of H is equal to 3, and H' is isomorphic to Si. Namely, H satisfies the condition (b) in Theorem 4.11. Let K be the triangle of H'. Take a vertex x in K. As shown in Lemma 4.7, the vertices in K are adjacent to the same fat vertex of m. In 568 Ars Math. Contemp. 17 (2019) 493-514 particular, K = Cm(x) holds. Since K = Cm(x) G cov m, we have ^h (K) = ^h (Cm(x)) = cov m. As with the proof of Proposition 4.3, we derive the Hoffman graph h by adding fat vertices according to (K). Similarly, for a slim j[hr]}-line graph in Figure 8, we can construct its strict H-cover. Figure 7: A slim {[hr]}-line graph containing Si induced by the small circles and its strict H-cover corresponding to ^G(K). (Hgho-oàAf Figure 8: A {[h7]}-line graph containing S2 induced by the small circles and its strict H-cover corresponding to (K). Remark 4.13. Let r be a connected slim j[hr]}-line graph with the clique number at least 2. Let K be a maximal clique of r. We suppose that |K| > 4 when r satisfies the condition (a) in Theorem 4.11, and that K contains the triangle of Si or that of S2 when r satisfies the condition (b). Then, as shown in Lemma 4.6, the strict {[h1], [h4], [hr]}-cover is cov-1(^r(K)). S. Kubota et al.: On graphs with the smallest eigenvalue at least — 1 — \/2, part III 569 5 The minimal forbidden graphs for the slim {[h 7] }-line graphs The following theorem is the main result in this paper. Theorem 5.1. A graph is a minimal forbidden graph for the slim {[hr]}-line graphs if and only if it is one of the following graphs: (i) Mi (i = 1, 2,3,4,6, 7,11,12,19) in Figure 9; (ii) odd cycles with at least 5 vertices; (iii) graphs in Figures 11 and 13. We explain the reason that the graphs in Figure 9 are minimal forbidden graphs for the slim {hr}-line graphs. They are obtained by enumeration by Magma. The following briefly describes the program. The MAGMA program is available at [8]. It is also available at https://doi.org/ 10.26493/1855-397 4.1581.b4 7. Hoffman graphs can construct large new graphs little by little from small graphs by using the concept of sum. With this method, all possible {[hr]}-line graphs with a small number of slim vertices can be obtained by considering all cases where fat vertices can be stuck together. Therefore, we can obtain all slim {[hr]}-line graphs with a small number of vertices. On the other hand, the graphs up to 10 vertices have databases in Magma [2]. Using this, the list F of graphs with at most 10 vertices that are not slim {[hr] }-line graphs is completely revealed. After that, the set of minimal elements of F can be calculated. We will prove Theorem 5.1 separately. (C1) r has an induced subgraph isomorphic to Si, S2 or the complete graph K4; (C2) For any maximal clique K containing the largest clique of some induced subgraph isomorphic to S1, S2 or K4, ^r(K) is a partition of V(T). Proposition 5.2. Let r be a minimal forbidden graph for the slim {[hr]}-line graphs with at least 10 vertices. Then, r does not satisfy the condition (C1) if and only if r is an odd cycle. Proof. It is easy to verify that every odd cycle with at least 5 vertices is a minimal forbidden graph. Thus, the necessity is proved. Next, we prove the sufficiency. Pick two vertices x and y to determine the diameter of r. Then, r - x and r - y are connected and slim {[hr]}-line graphs. By Theorem 4.11, r — x is isomorphic to either Pt for some t G TP or Ct for some t G TC. We have |Nr(x)|< 4 (5.1) since r — y is also isomorphic to either Pt for some t G TP or Ct for some t G TC. By (4.1) and (4.2), which are the definition of TP and TC, we have the length l of t is at least 6 since i E ti = ir| — 1 > 9. i=i In the rest of this proof, we will consider the decision on whether the vertex is end or non-end in r — x. 570 Ars Math. Contemp. 17 (2019) 493-514 Figure 9: The minimal forbidden graphs for the slim {[hr]}-line graphs, with at most 9 vertices. S. Kubota et al.: On graphs with the smallest eigenvalue at least — 1 — \/2, part III 571 Step 1: Show that Nr(x) n is non-empty if z is a non-end vertex in Nr(x). Suppose that Nr(x) n Nr-x(z) is empty. Then, there exists two distinct vertices u and v of r - x such that u ~ z ~ v and u v. It is a contradiction that Mi ~ (jx, z, u, v})r. Step 2: Show that |Np(x) nNp-x(z)| < 1 for z G Np(x). Suppose |Np(x) nNp_K(z)| > 2, and let zi and z2 be two distinct vertices in Nr(x) n Nr-x(z). We have zi z2 since K4 ~ (jx, z, z1, z2})r if z1 ~ z2. Thus, z is non-end. Let i be the integer such that z G [i]. Then, the following hold: (i) if (ti_i, ti, ti+i) = (1,1,2), then M3 ~ (jx} u [i - 1, i, i + 1])r; (ii) if (ti-i, ti, ti+i) = (1,2,1), then M2 ~ (jx} u [i - 1, i, i + 1])r; (iii) if (ti-i, ti, ti+i) = (2,1,2), then M3 is isomorphic to an induced subgraph in jx} U [i - 1, i, i + 1]. Then, r is isomorphic to either M2 or M3 by the minimality of r. This is a contradiction to |V(r)| > 10. Consider the case of (ti-i,ti,ti+i) = (1,1,1). If |Np(x)| = 3 then r ~ Pv or Cf where t' = (ti,...,ti-i, 2,ti+i,... ,t). Otherwise, |Nr(x)| = 4 holds by (5.1), and hence we let jz, zi, z2, z3} = Nr(x). Then, the following hold: (i) if z3 G [i - 2, i + 2] then M3 ~ (jx} U Nr(x))r; (ii) if z3 G [i - 2, i + 2] then Mi ~ (jx, zi, z2, z3})r (see Figure 10). These are contradictions to |V(r)| > 10. Figure 10: Examples of the case that (tj— 1, tj, tj+1 ) — (1,1,1) and |Np (x) | — 4 in Step 2. Step 3: Show that the vertices in Nr(x) are end. Suppose that a vertex z is non-end in Nr(x). By Step 1 and 2, |Nr(x) n Nr—x(z)| — 1 holds. Thus, we can take a vertex z1 so that {zi} — Nr(x) n Nr—x(z). There are i and j such that z G [i] and z1 G [j]. Let I — [i, j, i ± 1, j ± 1]. It follows that Nr(x) n I — {z, z1} by Step 2. If z1 is non-end, then some induced subgraph of r is isomorphic to M1 if i — j, S1 otherwise, a contradiction. Otherwise, we may assume that i — 2 and j — 1 without loss of generality. Then, the following hold: (i) if (t1, t2) — (2,1), then M1 is isomorphic to some induced subgraph of r; (ii) if (t1, t2) — (1,2), then M3 is isomorphic to some induced subgraph of r; 572 Ars Math. Contemp. 17 (2019) 493-514 (iii) if (ii, t2) = (1,1) and |Nr(x)| > 3, then r has an induced subgraph isomorphic to Si or S2; (iv) if (ti,t2) = (1,1) and |Nr(x)| = 2, then r ~ P(t1 + i,t2,...,t,). The result follows. Moreover, r - x is isomorphic to Pt. Step 4: For i = 1 or l, if Nr(x) n [i] = 0 then Nr(x) n [i] = [i] since Nr(x) n [i] = [i] implies that r has an induced subgraph isomorphic to Si by Step 3. Step 5: If Nr(x) = [1] then r ~ P(i,t1,...,tt) is an j[hr]}-line graph, a contradiction. Hence, r ~ C(i,il,...,ii). If l is odd then C(i,tl,...,tl) is an j[hr]}-line graph, a contradiction. Thus, r is an odd cycle. □ Proposition 5.3. Let r be a minimal forbidden graph for the slim {[h7]}-line graphs, with at least 10 vertices and the condition (C1). Then, r is isomorphic to one of the graphs in Figure 11 if and only if r does not satisfy the condition (C2). Axo xi x2l-i x2l / \ >-•—-•-d--*-• . A ■■ ■ A .. Figure 11: Minimal forbidden graphs for the slim j[hr]}-line graphs, with at least 10 vertices and the condition (C1), without (C2). Proof. It is easy to verify that ^r(K) is not a partition, where r is one of the graphs of Figure 11, and K is the rightmost clique of size at least 3. Next, we prove the necessity. By the condition (C1), r has an induced subgraph isomorphic to Si, S2 or K4. Hence, we can take a maximal clique K containing the largest clique of some induced subgraph S isomorphic to S1, S2 or K4. Since the condition (C2) is not satisfied, we may suppose that ^r(K) is not a partition of V(r). If |K| > 3 then we replace S by (K')r, where K' is a set of 4 vertices in K. Let l = |_(dmax - 1)/2j and V' = {x g V(r) | 3k(x) < 2l}. Step 1: There is a vertex g not in V(S) U K with 3k(g) = dmax. Then, {C g ^r(K) | 3k(c) < 2l for any c g C} = {C g (K) | 3k(c) < 2l for any c g C} is a clique partition of V' since r - g is a slim {[h7]}-line graph satisfying the condition (a) or (b) in S. Kubota et al.: On graphs with the smallest eigenvalue at least — 1 — \/2, part III 573 Theorem 4.11. Therefore, (K) is not a partition if and only if there exists vertices x, y, p and q such that dK (x) = dK (y) = 2/ +1, and q e ((N(x) U{x}) n (N(y) U{y})) - V' p e ((N(x) U {x}) - (N(y) U {y})) - V'. If z e V(r) - ({x, y,p, q} U S) with Ok(z) > 2/ + 1, then r - z is an {[hr]}-line graph and p — y by Remark 4.13. This is a contradiction to p — y. It follows that {u e V(r) | dK(u) > 2/ + 1} - V(S) = {x, y,p, q}. (5.2) Step 2: Assume that / = 0. In the case of |K| > 4 (i.e., S ~ K4), {K - {k}, {x, y,p, q}} C ^r-fc(K - {k}) is a clique partition for some k e K by Remark 4.13 and (5.2), i.e., |K| > 6. This is a contradiction to p — y. In the case of |K| = 3 (i.e., S ~ S1 or S2), we obtain |V(r)| < 9 and a contradiction. Hence, / is a positive integer. Step 3: Show that x — y. In addition, it holds that x = p and x = q. Suppose that x and y are not adjacent. Then, p = x and q e {x, y} by Remark 4.13. If x and y are adjacent to a vertex r with dK (r) = 2/, then r has an induced subgraph isomorphic to K1j3, a contradiction. When dK (q) = 2/ + 1, there is a neighbor q' of q with dK (q') = 2/ such that q' e N(x) n N(y). Without loss of generality we assume that q' and y are not adjacent. Then, the induced subgraph r' obtained by deleting the neighbors of y expect q has a clique partition (K) which contains a set {x, y, q}. This is a contradiction, and dK (q) = 2/ + 2 follows. Then, r has an induced subgraph isomorphic to an odd cycle with at least 5 vertices. This is a contradiction since (K) is a clique partition in Or-q. Step 4: Let P denote a path (x0,..., x2i+1) such that x0 e K and x = x2I+1. If S ~ K4, then replace S by (K'}r, where K' is a set of 4 vertices in K containing x0 and a vertex not adjacent to x1. The graph r' by deleting vertices other than V(P) U V(S) U {x, y,p} is a slim {[hr]}-line graph with a clique partition (K). Thus, it holds that V(r) = V(P) U V(S) U {x, y,p} and y - x2i. Step 5: If dK(p) = 2/ + 1, then p — x2l. Thus, ({y,p, x2l, x2i-1}}r ~ K13 holds, a contradiction. Thus, dK (p) =2/ + 2. Step 6: In the case of |K| > 4 (i.e., S ~ K4), deg(x1) < 3 since r - p is an {[hr]}-line graph. Obtain the second and third graphs in Figure 11. Step 7: In the case of |K| = 3 (i.e., S ~ S1 or S2), if x1 £ S then |N(x1) n K| = 1 since M1 and M3 are minimal forbidden graphs for the slim {[hr]}-line graphs. Then, we can replace S by the induced subgraph by K U {x1, w}, where w is a vertex of S not adjacent to x0. Hence, x1 e S holds. We can draw r as Figure 12. The edge e exists if and only if the edge e' does in Figure 12. If the edges e and e' exist, then r - x0 is not an {[hr]}-line graph. Otherwise, we obtain the first graphs in Figure 11. □ Let r be a connected graph, and let K and D be nonempty subsets of V(r). D is said to be deletable for K if K - D = 0, r - D is connected, and (K - D) = {C - D | 574 Ars Math. Contemp. 17 (2019) 493-514 xo xi x P Figure 12: In the proof of Proposition 5.3. C € ^r(K )}-{0}. In addition, a vertex v is said to be deletable for K if {v} is deletable for K. Lemma 5.4. Let r be a connected graph, and let K and D be nonempty subsets of V(r). If ^r(K) is a partition of V (r) and dK,r |r-D = dK-D,r-D, then D is deletable for K. Proof. Write d = dK,r for short. Then, it holds that by dK,r |r-D = dK-D,r-D, ^r-D(K - D) -{K - D} = {{y € ({x} U N(x)) - D | d(y) > d(x)} | x € V(r - D), d(x) € 2N + 1}, {C - D | C € ^r(K)} - {K - D} = {{y € ({x} U N(x)) - D | d(y) > d(x)} | x € V(r), d(x) € 2N + 1}. Let x € D with odd d(x), and take Cx € (K) containing x. Assuming that Cx -D = 0, there exists z € C such that d(z) = d(x) by the assumptions. It holds that Cx - D = {y € ({x} U N(x)) - D | d(y) > d(x)} Lemma 5.5. Let r be a minimal forbidden graph for the slim {[h7] }-line graphs, with the condition (C1) and(C2). Let S be an induced subgraph isomorphic to Si, S2 or K4. Let K be a maximal clique of r contains the largest clique of S. Then, the following hold: (i) ^r(K) is a clique partition; (ii) if u is a non good vertex for ^r(K), and v € V(S) U K is a deletable vertex for K, then v € n3(u) and v is non good for (K), where nk (■) is defined for ^r(K) and a non negative integer k. Proof. By the condition (C2), ^r(K) is a partition of V(r). Moreover, it is a clique partition of V(r) by Lemma 4.5 since r has no induced subgraph isomorphic to M1 ~ K1,3. Next, suppose that the vertex v is not in n3(u). Then, {C n n3(u) | C € ^r(K)} = {C n n3(u) - {v} | C € (K)} = {C n n3(u) | C € ^r-v(K)} holds. Thus, the vertex u is good for ^r-v (K) if and only if it is good for ^r-v (K) in r - v by Lemma 4.4. On other hand, v € V(S) U K and r - v is connected. Hence, ^r-v(K) € Or-v, that is, every vertex of r - v is good for ^r(K) by Remark 4.13 since r is a minimal forbidden graph for the slim {[hr]}-line graphs. Therefore, u is good for ^r-v (K) and (K). This is a contradiction that u is non good for ^r(K). □ and {y G ({z} u N (z)) - D | d (y) > d(z)} G (K - D). □ S. Kubota et al.: On graphs with the smallest eigenvalue at least — 1 — \/2, part III 575 Proposition 5.6. Let r be a minimal forbidden graph for the slim {[hr ]}-line graphs, with at least 10 vertices and the condition (C1). Then, r is one of the graphs in Figure 13 if and only if r satisfies the condition (C2). " \** the number of vertices is odd ^ ' ------® KT------13 Figure 13: The minimal forbidden graphs for the slim j[hr]}-line graphs, with at least 10 vertices and the conditions (C1) and (C2). Proof. The sufficiency obviously holds. Prove the necessity. Fix an induced subgraph S isomorphic to S1, S2 or K4, and let K be a maximal clique containing the largest clique of S. By Lemma 5.5 (i), ^r(K) is a clique partition. Then, nk ( ) is defined for ^r(K) and a non negative integer k. If r has an induced subgraph isomorphic to K4, then replace S by it. Let l = |_(dmax -1)/2J. By the definition of ^r(K), we can pick the subset {{y G N(x) U {x} | 3k(y) > Ok(x)} | x G V(r), 3k(x) = 2l + 1} of ^r(K). We denote by {Cj}n=1 the subset. Note that C® are pairwise distinct. If l = 0 then let D1 = K and m =1. Otherwise we pick the subset {{y G N(x) U {x} | 3k(y) > 3k(x)} | x G V(r), 3k(x) = 2l - 1} of ^r(K). We denote by {A}n= 1 the subset. Note that D® are pairwise distinct. Without loss of generality, we can take an integer m such that D® - V(S) is empty if and only if i > m. In the case of dmax = 2l + 1, we show that r is isomorphic to one of the graphs in Figure 13. 576 Ars Math. Contemp. 17 (2019) 493-514 Step 1: Show that l = 0. Suppose that l = 0 to prove by contradiction. Set B = V(r) — (V(S) U K). Then, every vertex in B is deletable by Lemma 5.4. Moreover, the deletable vertex is non good for ^r(K) by applying Lemma 5.5 (ii) to a non good vertex for (K) and the deletable vertex. We obtain a contradiction by checking the following: (i) |B| < 3; (ii) if 7 < |V(r) — K|, then 5 < |V(r) — K| — 2 < |B|; (iii) if 4 < |V(r) — K| < 6, then S ~ K4 and 4 < |B|; (iv) if |V(r) — K| < 3, then r is an {[hr]}-line graph. Assume that |B| > 4. If we find a vertex k e K with |n(k) | > 3, then we can take a vertex b e B such that |n(k) — {b}| > 3. This is a contradiction since the vertex b is deletable and ^r-b(K) e Or-b by Remark 4.13. Thus, |n(k)|< 2 (5.3) holds for every k e K. Fix a vertex b e B. Then, |n(b)| > 3 holds by (5.3) and applying Lemma 5.5 (ii) to the non good vertex b and each vertex in B — {b}. We obtain a contradiction as well and |B| < 3. Next, In the case of |V(r) — K| < 3, we have |B| < 3, |K| > 7 and hence S ~ K4. If 2 < |B| < 3, then |n(b)| < 2 for every vertex b e B. Hence, we can pick a vertex k in K — J n(b). beB It is clear that k is deletable and (K — {k}) e Or-k. Thus every vertex of r is good for (K), a contradiction to r being a non {[hr]}-line graph. Step 2: Every vertex x with dK(x) = dmax is not in V(S) U K and deletable by l > 1 and Lemma 5.4. Moreover, such a vertex x is non good for ^r(K) by applying Lemma 5.5 (ii) to a non good vertex for ^r(K) and the deletable vertex x. Fix a vertex u with dK (u) = dmax. By applying Lemma 5.5 (ii) to u and each vertex x with dK (x) = dmax, we have {x e V(r) | dK(x)= dmaxK n2(u) since r has no induced subgraph isomorphic to Mi ~ Kij3. For a vertex x with dK (x) = dmax — 1 = 2l, it holds that x e n3(u) since if n(x) = 0 then x is deletable by Lemma 5.4 and x e n3(u) holds by applying Lemma 5.5 (ii) to the vertices u and x. Thus, n3(u) = {x e V(r) | Bk(x) > 2l}. (5.4) Furthermore, Dj contains a vertex v with dK (vj) = dmax — 1 for every 1 < i < m, since every deletable vertex not in V(S) U K is contained in n3(u) by Lemma 5.5 (ii). Step 3: Show that n = m =1. If n > 2 then r has an induced subgraph isomorphic to M1 by (5.4), a contradiction. Thus, n =1. Suppose that m > 2 to prove by contradiction. Without loss of generality, we can assume that vi ^ u ^ V2 by (5.4). In the case of m > 3, S. Kubota et al.: On graphs with the smallest eigenvalue at least — 1 — \/2, part III 577 the vertex v3 is deletable clearly, and the vertex u is non good for ^r-V3 (K) in r - v3. This is a contradiction to ^r-d3 (K) G Or-d3 by Remark 4.13. In the case of m = 2, we have C1 = {u} since if we find an vertex u' in C1, then the vertex u' is deletable by Lemma 5.4 and u is non good for (K) in r - u', a contradiction to (K) G Or_«'. Fix a vertex v' G Dj with dK (v') = 21 -1 for i = 1 and 2, respectively. Then, the set Di U D2 - (V(S) U{vi,v2,vi,v2}) is deletable and u is good. Hence, the set is empty. If v1 and v2 are not adjacent in r, then r has an induced subgraph isomorphic to an odd cycle with at least 5 vertices since u is deletable and ^r(K) - {C1} = ^r_„(K) g Or-u. This is a contradiction to the minimality of r. Hence, v1 and v2 are adjacent in r. First, consider the case of 1 > 2. Let d be the vertex adjacent to v2 with dK (d) = 21 - 2. Note that d is not in S. Then, r - d is not an j[hr]}-line graph since ^r—) G Or—d (cf. Figure 14). Second, consider the case of l = 1. Suppose that S ~ K4. Note that n' = m = 2 holds. If |K| > 5, then we can take a deletable vertex k in K. By Remark 4.13, ^r-k(K - {k}) G Or—k holds since K - {k} is a maximal clique with at least 4 vertices. However, u is non good for k(K - {k}), a contradiction to the minimality of r. We have |V(r)| = |K| + |Di | + |D21 + |Ci | = 4 + 2 + 2 +1 = 9, a contradiction to |V(r) | > 10. Thus, r has no induced subgraph isomorphic to K4. Suppose that S ~ S1 or S2. We define the vertices wi of S as Figure 15. Note that the vertex v' is not in V(S) for i = 1,2 since | V(r) | > 10. If both v1 and v2 are adjacent only to w1, then r has an induced subgraph isomorphic to M1, a contradiction. Hence, it is not so. Without loss of generality, we can assume that v2 and w2 are adjacent. Since r has no induced subgraph isomorphic to K4, v2 is not adjacent to some vertex w G {w1, w3}. The vertices v'2 and w5 are adjacent since (v2, w, w2, w5)r is not isomorphic to M1. Furthermore, v2 is also adjacent to w5 since ^r(K) is a clique partition. Hence, v2 is deletable for K, a contradiction to We have n = m =1. Step 4: Let p = |C11 and q = |{x G V(r) | Ok(x) = 2l}|. The induced subgraph (C1 U D1)r is an {[h7]}-line graph by the minimality of r. Hence, ^(CiuDi)r(D1) = 578 Ars Math. Contemp. 17 (2019) 493-514 {Ci, Di j holds by |Di| > 4 and Remark 4.13. This is a contradiction since non good vertices for ^r(K) in Ci U Di are also non good for ^(ClUDl)r (Di). Thus, q < 2. When q = 1, p = 3 holds and we obtain the second and third graph in Figure 13 in the same way as the Step 4 in the proof of Proposition 5.3. Consider the case of q = 2. If p =1 then r is an {[h7]j-line graph. When p = 2, we obtain the first and second graph in Figure 13 since r has no induced subgraph isomorphic to M3. If p > 3 then we can assume that n(u) = {x G V(r) | dK(x) = 21} by (5.4). Then, r - w is not an {[hr] j-line graph for some w G Ci — {v j, a contradiction. Suppose that dmax = 21 + 2. We have 1 > 0 and every vertex u with dK (u) = 21 + 2 is deletable for K. By Lemma 5.5 (ii), the vertex u is non good for ^r(K). Moreover, every vertex v with dK (v) < 21 + 1 is good for ^r(K) since (n3(v))r is the induced subgraph of r — u, where dK (u) =21 + 2. Hence, a vertex u is good if and only if dK (u) < 21 + 1. We have n =1 since a vertex v with dK(v) = 21 + 2 is non good. Let v be a vertex with dK(v) = 21 + 1. If v is not a vertex of S, then we have a contradiction to Lemma 5.5 (ii) that v is deletable for K. Hence, v is a vertex of S. Thus, S ~ Si, 1 = 0 and n = 2. Then, | V(r) | < 9 since |K| = 3, |Ci | < 3 and |C2| < 3, a contradiction. □ Proof of Theorem 5.1. The minimal forbidden graphs with at most 9 vertices are revealed in Figure 9. This theorem follows by Proposition 5.2, 5.3 and 5.6. □ References [1] L. W. Beineke, Characterizations of derived graphs, J. Comb. Theory 9 (1970), 129-135, doi: 10.1016/s0021-9800(70)80019-9. [2] W. Bosma, J. Cannon and C. Playoust, The Magma algebra system I: The user language, J. Symbolic Comput. 24 (1997), 235-265, doi:10.1006/jsco.1996.0125. [3] P. J. Cameron, J.-M. Goethals, J. J. Seidel and E. E. Shult, Line graphs, root systems, and elliptic geometry, J. Algebra 43 (1976), 305-327, doi:10.1016/0021-8693(76)90162-9. S. Kubota et al.: On graphs with the smallest eigenvalue at least — 1 — \/2, part III 579 [4] M. Chudnovsky, N. Robertson, P. Seymour and R. Thomas, The strong perfect graph theorem, Ann. of Math. 164 (2006), 51-229, doi:10.4007/annals.2006.164.51. [5] M. C. Golumbic, Trivially perfect graphs, Discrete Math. 24 (1978), 105-107, doi:10.1016/ 0012-365x(78)90178-4. [6] A.J. Hoffman, On graphs whose least eigenvalue exceeds — 1 — \[2, Linear Algebra Appl. 16 (1977), 153-165, doi:10.1016/0024-3795(77)90027-1. [7] H. J. Jang, J. Koolen, A. Munemasa and T. Taniguchi, On fat Hoffman graphs with smallest eigenvalue at least —3, Ars Math. Contemp. 7 (2014), 105-121, doi:10.26493/1855-3974.262. a9d. [8] S. Kubota, On (h7}-line graphs, https://sites.google.com/view/s-kubota/ home/magma-programs/on-h7-line-graphs, accessed on April 8, 2019. [9] D. P. Sumner, 1-factors and antifactor sets, J. London Math. Soc. 13 (1976), 351-359, doi: 10.1112/jlms/s2-13.2.351. [10] T. Taniguchi, On graphs with the smallest eigenvalue at least —1 — -^/2, part I, Ars Math. Contemp. 1 (2008), 81-98, doi:10.26493/1855-3974.35.fe9. [11] T. Taniguchi, On graphs with the smallest eigenvalue at least —1 — -^/2, part II, Ars Math. Contemp. 5 (2012), 239-254, doi:10.26493/1855-3974.182.139. [12] R. Woo and A. Neumaier, On graphs whose smallest eigenvalue is at least —1 — \[2, Linear Algebra Appl. 226/228 (1995), 577-591, doi:10.1016/0024-3795(95)00245-m. ARS MATHEMATICA CONTEMPORANEA ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 17 (2019) 581-589 https://doi.org/10.26493/1855-3974.1866.fd9 (Also available at http://amc-journal.eu) Lobe, edge, and arc transitivity of graphs of connectivity 1 Jack E. Graver, Mark E. Watkins Department of Mathematics, Syracuse University, Syracuse, NY Received 27 November 2018, accepted 3 October 2019, published online 9 December 2019 We give necessary and sufficient conditions for lobe-transitivity of locally finite and locally countable graphs whose connectivity equals 1. We show further that, given any biconnected graph A and a "code" assigned to each orbit of Aut(A), there exists a unique lobe-transitive graph r of connectivity 1 whose lobes are copies of A and is consistent with the given code at every vertex of r. These results lead to necessary and sufficient conditions for a graph of connectivity 1 to be edge-transitive and to be arc-transitive. Countable graphs of connectivity 1 the action of whose automorphism groups is, respectively, vertex-transitive, primitive, regular, Cayley, and Frobenius had been previously characterized in the literature. Keywords: Lobe, lobe-transitive, edge-transitive, orbit, connectivity. Math. Subj. Class.: 05C25, 05C63, 05C38, 20B27 1 Introduction Throughout this article, r denotes a connected simple graph. Consider the equivalence relation = on the edge-set £T of r whereby e1 = e2 whenever the edges e1 and e2 lie on a common cycle of r. A lobe is a subgraph of r induced by an equivalence class with respect to =. Equivalently, a lobe is a subgraph that either consists of a cut-edge with its two incident vertices or is a maximal biconnected subgraph1. A vertex is a cut-vertex if it belongs to at least two different lobes. Connected graphs other than K2 have connectivity 1 if and only if they have a cut-vertex. Clearly no finite vertex-transitive graph admits a cut-vertex. E-mail addresses: jegraver@syr.edu (JackE. Graver), mewatkin@syr.edu (MarkE. Watkins) 1The term "lobe" is due to O. Ore [6]. We eschew the term "block" for this purpose, as it leads to ambiguity when discussing imprimitivity. Abstract ©® This work is licensed under https://creativecommons.org/licenses/by/4.0/ 582 Ars Math. Contemp. 17 (2019) 493-514 Graphs of connectivity 1 whose automorphism groups have certain given properties have been characterized. Those whose automorphism groups are, respectively, vertex-transitive, primitive, and regular were characterized in [5]. In particular, primitive planar graphs of connectivity 1 were characterized in [11]2. Cayley graphs of connectivity 1 were characterized in [9]. Graphs of connectivity 1 with Frobenius automorphism groups were characterized in [10]. In the present work, we complete this investigation; we characterize graphs of connectivity 1 whose automorphism groups act transitively on their set of lobes. As a consequence, we obtain characterizations of edge-transitive graphs and arc-transitive graphs of connectivity 1. The conditions for a graph of connectivity 1 to be lobe-transitive or to be vertex-transitive are independent; such a graph may have either property or neither one or both. Such is not the case for edge- and arc-transitivity. In Section 3 we give necessary and sufficient conditions for a graph to be lobe-transitive. We further show that, given any biconnected graph A and a "list" of orbit-multiplicities of copies of Aut(A), one can construct a lobe-transitive graph of connectivity 1 all of whose lobes are isomorphic to A and locally respects the given list. We give necessary and sufficient conditions for a countable graph of connectivity 1 to be edge-transitive in Section 4 and to be arc-transitive in Section 5. As the sets of conditions for these latter two properties are more intertwined with lobe-transitivity than the characterization of vertex-transitivity (for graphs of connectivity 1), scattered throughout are examples that illustrate some algebraic distinctions among these various properties. 2 Preliminaries Throughout this article, the symbol N denotes the set of positive integers. The symbols I, J, and K, often subscripted, denote subsets of N of the form {1,2,... ,n} or the set N itself; they appear as sets of indices. All graphs (and their valences) in this article are finite or countably infinite. The symbol Si,j (the so-called "Kronecker delta") assumes the value 1 if i = j and 0 if i = j. For a graph A and any subgroup H < Aut(A), the set of orbits of H acting on VA is denoted by O(H). The set of lobes of a graph r is denoted by L(r). We let {Lk : k e K} denote the partition of L(r) into isomorphism classes of lobes. For given k e K and a lobe A e Lk, we let O (Aut(A)) = {(VAj : j e Jk}, and we understand that if a: A ^ © is an isomorphism between lobes in Lk, then a((VA)j) = (V©)j for all j e Jk. Finally, for each k e K and j e Jk, we define the function r(k): Vr ^ N by Tj(k)(v) = |{A e Lk : v e (VAj}|. (2.1) For A0 e L(r) and n e N, we recursively define the subgraphs ro(Ao) = Ao, r„+i(Ao) = U{A e L(r) : VA n Vr„(Ao) = 0}. Lemma 2.1 ([5, Lemma 3.1]). Let A, © G L (T) and let n G N. If for each k G K and j G Jk, the function rjfc) is constant on Vr, then any isomorphism an: rn (A) ^ rn(©) admits an extension to an automorphism a G Aut(r). 2For a short algebraic proof that all 1-ended planar graphs with primitive automorphism group are biconnected, see [8]. J. E. Graver and M. E. Watkins: Lobe, edge, and arc transitivity of graphs of connectivity 1 583 This lemma was used in [5] to prove the following characterization of vertex-transitive graphs of connectivity 1. Theorem 2.2 ([5, Theorem 3.2]). Let r be a graph of connectivity 1. A necessary and suffic V r. (k) sufficient condition for r to be vertex-transitive is that all the functions t( ) be constant on Notation. When all the lobes of the graph r are pairwise isomorphic, that is, the index set K has but one element, then in Equation (2.1) the index k is suppressed; we simply replace Jfc by J and j by Tj. 3 Lobe-transitivity Let r be a graph of connectivity 1. It is immediate from the above definitions that the edge-sets of the lobes of r are blocks of imprimitivity of the group Aut(r) acting on Er. Hence any automorphism of r must map lobes onto lobes, and therefore, if Aut(r) is to act transitively on L(r), then all the lobes of r must be pairwise isomorphic. However, pairwise-isomorphism of the lobes alone is not sufficient for lobe-transitivity, even when every vertex of r lies in the same number of lobes. Let us first dispense with trees; the proof is elementary and hence omitted. Proposition 3.1. A finite or countable tree is lobe-transitive (and simultaneously, edge-transitive) if and only if there exist nl,n2 G N U{K0} such that every edge has one incident vertex of valence n and the other of valence n2. If n = n2, the tree is also arc-transitive. For graphs of connectivity 1 other than trees, we have the following characterization of lobe-transitivity. Theorem 3.2. Let r be a graph of connectivity 1, and let A0 be an arbitrary lobe of r. Let {Pi : i G I} be the set of orbits of Aut(r), and let Q = {Qj : j G J} be the set of those orbits of the stabilizer in Aut(r) of A0 that are contained in A0. Then necessary and sufficient conditions for the graph r to be lobe-transitive are: (1) For each lobe A G L (r), there exists an isomorphism aA : A0 ^ A. (2) For each j G J, there exists a function Tj : Vr ^ N U {0, K0} such that (a) for all v G Vr, Tj(v) = |{A G L(r): v G aA(Qj)}| (3.1) and (b) for each i G I, Tj is constant on Pi and is nonzero if and only if Qj C Pi. Proof. (Necessity) Suppose that r is lobe-transitive. For each lobe A G L(r), there is an automorphism aA G Aut(r) that maps the fixed lobe A0 onto A. The restriction to A0 of aA is an isomorphism aA: A0 ^ A that satisfies condition (1). For any lobe A, an automorphism a G Aut(r) is in the stabilizer of A if and only if a—1aaA is in the stabilizer of A0. It follows that the partition {aA(Qj) : j G J} of VA is the set of orbits of the stabilizer of A that are contained in A. Furthermore, since the stabilizer of A0 is a subgroup of Aut(r), the partition {aA(Qj) : j G J} of VA 584 Ars Math. Contemp. 17 (2019) 493-514 refines the partition {Pj n VA : i e I}. If for some indices i and j, the vertex v satisfies v e aA(Qj) C Pj, then for any lobe ©, the vertex a0a-1(v) lies in Pj n ae(Qj). This implies that, for all j e J, the function Tj as given in Equation (3.1) is well-defined and constant on Pj. Suppose that for an arbitrary index i e I, the vertex v lies in Pj. Since by Equation (3.1), Tj (v) counts for each j e J the number of lobes A such that v lies in aA(Qj ), it follows that Tj (v) is positive exactly when a-1(v) e Qj C Pj holds, concluding the proof of condition (2.b). (Sufficiency) Assume conditions (1) and (2). To prove that r is lobe-transitive, it suffices to prove that every isomorphism ae : A0 ^ © is extendable to an automorphism of r. (Note that in this direction of the proof, ae is not presumed to be the restriction to A0 of an automorphism ae e Aut(r) but, in fact, it is.) Fix a lobe ©0 and a vertex v e VA0 and let w = a0o (v). For some j e J, the vertex v lies in Qj, and so w e a0o (Qj). Since both Tj(v) and Tj(w) are therefore positive, both v and w lie in the same orbit Pj of Aut(r) by condition (2.b). Furthermore, since Tj is constant on Pj for each j e J, there exists a bijection Pj from the set of lobes A such that v e aA(Qj) onto the set of lobes © such that w e ae(Qj). Let A1 be a lobe in the former set, and let ©1 = Pj(A1). Although v and w lie in the images of the same orbit Qj in lobes A1 and ©1, respectively, the vertices a-^v) and ct—1(w) need not be the same vertex of A0. However, since both vertices lie in the same orbit Qj of A0, there exists an automorphism a e Aut(A0) such that aa-1(v) = a-^(w). Then ct01 aa-1 is an isomorphism from A1 onto ©1 that maps v onto w and therefore agrees with aeo at the vertex v common to A0 and A1. The amalgamation of 2. By Theorem 2.2, r is vertex-transitive. By Theorem 3.2, r is lobe-transitive, with O(Aut(r)) being the trivial partition (with just one big cell P1). Also |J| = 1 and t1(v) = m for all v e vr. Remark 3.4. There exists a "degenerate" family of lobe-transitive graphs r of connectivity 1 that have but a single cut-vertex. For some cardinal K > 2, consider a collection of K copies of a biconnected graph A0, and let v0 e VA0. Let aA : A0 ^ A be an isomorphism as in Theorem 3.2, and let aA(v0) = vA for each copy A of A0 in the collection. We obtain r by identifying v0 and all the vertices vA and naming the new amalgamated vertex w, which forms a singleton orbit {w} of Aut(r). Clearly r is lobe-transitive and w is its unique cut-vertex. If A0 has finite diameter, then r has zero ends (see [3]) when A0 is finite and has K ends when A0 is infinite; if A0 has infinite diameter, then r has at least K ends. Other than the graphs just described, all countable lobe-transitive graphs of connectivity 1 are "tree-like" with H0 cut vertices and either 2 or 2N° ends. Theorem 3.5. Let A0 be any biconnected graph. Let H < Aut(A0), let Q = O(H) = {Qj : j e J}, and let R = {Rk : k e K} be a partition of VA0 refined by Q. For each k e K, let the function : J ^ N U {0, H0} satisfy (j) > 0 if and only if Qj C and (to avoid the triviality of a single lobe) jej pk(j) > 2 for at least one k e K. Then there exists (up to isomorphism) a unique lobe-transitive graph r of connectivity 1 such that J. E. Graver and M. E. Watkins: Lobe, edge, and arc transitivity of graphs of connectivity 1 585 (1) for each lobe A G L (T), there exists an isomorphism aA : Ao ^ A; (2) for each vertex v G Vr and each j G J, we have Pk(j) = |{A G L(r) : a_^v) G Qj C Rk}|. Proof. Let A0, H, Q, and pk be as postulated. Let r0 = A0 from which we construct r as follows. Let v be any vertex of A0. For some j, k, it must hold that v G Qj C Rk, and so pk(j) > 0. For each i such that Qe C Rk, we postulate the existence of pk(i) copies A of A0 (including A0 itself when i = j) such that, if aA: A0 ^ A is an isomorphism, then some vertex in aA(Qe) is identified with the vertex v. The graph r is produced by repeating this process for each vertex of A0. We repeat this process starting at each vertex w G Vr \ Vr0, the only notational change being that, if specifically w G aA(Qj) for some j' G J, then we consider the subset aA(Qj) of VA (instead of Qj in A0) to which w belongs. Thus we construct r2. Inductively, suppose that rn has been constructed for some n > 2. Let w G Vrn \ Vrn_i, and so w G aA(Qm) holds for some m G J and a unique lobe A G L(rn) \ L(rn-1). Supposing that Qm C Rk, we postulate the existence of pk (m) new copies of A0 that share only the vertex w with rn according to the above identification. In this way we construct rn+1. Finally, let r = |Jrn. It remains only to prove that r so-constructed is lobe-transitive. Let © be any lobe of r. By the above construction, all lobes of r are pairwise isomorphic, and so there exists an isomorphism a0 : A0 ^ ©. Starting with T0 = © and by using the technique in the proof of Sufficiency in Theorem 3.2, one constructs a sequence r0, ri,... so that for all n G N, we have r^ = rn, and a0 is extendable to an isomorphism from rn to r^. Thus r — Urn, and a0 can be extended to an automorphism of r. □ Example 3.6. In the notation of Theorem 3.5, let A0 be the 5-cycle with one chord as shown in Figure 1(a), and let H = Aut(A), yielding the orbit partition {Q1, Q2, Q3} as indicated. Let Ri = Qi U Q3 and R2 = Q2, giving J = {1,2, 3} and K = {1,2}. Define p1(1) = 3, p1(3) = 1, and p2(2) = 2. Note that all other values of p1 and p2 must equal 0. Then r1 is as seen in Figure 1(b). Q1 Q2 Q3 (a) Aq (b) ri Figure 1: A0 and r1 from Example 3.6. The pairs of conditions in Theorems 3.2 and 3.5 may appear alike, but there is a notable difference between them. This occurs when the arbitrarily chosen subgroup H < Aut(A0) 586 Ars Math. Contemp. 17 (2019) 493-514 of Theorem 3.5 is a proper subgroup of the stabilizer of A0 in Aut(T), where r is the graph constructed from A0 and the functions of Theorem 3.5. We illustrate this distinction with following example. Example 3.7. Our initial lobe A0 is a copy of K4, with vertices labeled as in Figure 2(a), and so Aut(A0) = Sym(4) of order 24. We use A0 to "build" the lobe-transitive graph r shown four times in Figure 2(b). The action on A0 by the stabilizer of A0 in Aut(r) is the 4-element group (gi) x (g2) whose generators have cycle representation g1 = (vi, v2) and g2 = (v3, v4). The shadings of the vertices in the four depictions of r in Figure 2(b) correspond respectively to the four different subgroups of (g1) x (g2) described below. For the sake of simplicity, we assume R = Q. (a) (b) Figure 2: The clothesline graph. (i) H is the trivial group {i}. Thus H induces four orbits Qj = {vj} for j e {1, 2, 3,4}. The functions are then given by (j) = 2j for k e {1,2} and (j) = Sj,k for k e {3,4}. (ii) H = (gi}. There are three orbits of H: Qi = {vi, v2}, Q2 = {v3}, and Q3 = {v4}, which give ^(1) = 2, ^2(2) = m3(3) = 1. All other functional values are zero. (iii) H = (g2}. Again there are three orbits of H but not the same ones: Q1 = {v1}, Q2 = {v2}, and Q3 = {v3,v4}. This gives ^fc(j) = 2Sj,k for k e {1,2} and M3(j) = Sj,3. (iv) H = (g1} x (g2}. Now there are just two orbits: Q1 = {v1, v2} and Q2 = {v3, v4}. Finally, ^1(1) = 2, ^2(2) = 1, and all other functional values are zero. All four choices for H, the partition Q, and the functions clearly yield the same lobetransitive graph r of connectivity 1 by the construction of Theorem 3.5 4 Edge-transitivity Lemma 4.1. If r is an edge-transitive (respectively, arc-transitive) graph, then r is lobetransitive and its lobes are also edge-transitive (respectively, arc-transitive). J. E. Graver and M. E. Watkins: Lobe, edge, and arc transitivity of graphs of connectivity 1 587 Proof. For i = 1,2, let ©i G L(T), and let ei be an edge (respectively, arc) of ©¿. There exists an an automorphism p G Aut(r) such that p(e1) = e2. Since p maps cycles through ei onto cycles through e2, p must map ©1 onto ©2. If e1 and e2 lie in the same lobe ©, then p leaves © invariant, and so its restriction to © is an automorphism of ©. □ Theorem 4.2. Let r be a graph of connectivity 1 with more than one lobe, and let A G L (r). Necessary and sufficient conditions for r to be edge-transitive are the following: (1) The lobes of r are edge-transitive. (2) For each lobe © G L(r), there exists an isomorphism a© : A ^ ©. (3) Exactly one of the following descriptions of r holds: (a) Both r and A are vertex-transitive, in which case every vertex is incident with the same number > 2 of lobes. (b) The graph r is vertex-transitive but A is not vertex-transitive, in which case A is bipartite with bipartition {Q1, Q2}, and there exist constants m1,m2 G N U {Ho} such that for j = 1, 2 and all v G V r, it holds that mj = |{© G L(r): v G a©(Qj)}|. (c) The graph r is not vertex-transitive, in which case r is bipartite with bipartition {P1, P2} and there exist constants m1,m2 G N U {H0}, at least one of which is at least 2, such that for i =1,2, if v G Pi, then |{© G L (r) : v G a © (Pj n V A)}| = mj Sitj. Proof. (Sufficiency) Assume all the conditions in the hypothesis and let e1, e2 G £T be arbitrary edges in lobes A1 and A2, respectively. By condition (2), there exists an isomorphism a: A1 ^ A2. By condition (1), there exists an automorphism a G Aut(A2) such that e2 = aa(e1). Each of the three cases of condition (3) is seen to satisfy the hypothesis of Lemma 2.1, implying that aa: A1 ^ A2 is extendable to an isomorphism of r mapping e1 to e2 . (Necessity) Suppose that r is an edge-transitive graph of connectivity 1. By Lemma 4.1, r is also lobe-transitive and its lobes are edge-transitive, proving condition (1). Condition (2), which establishes notation for the remainder of this proof, also follows from Lemma 4.1. To prove (3), we continue the notation of Theorem 3.2 with I being the index set for the set of orbits of Aut(r) and J being the index set for the orbits of the stabilizer of A that are contained in A. Since both r and all of its lobes are edge-transitive, |I| and |J| equal either 1 or 2. If both r and A are vertex-transitive, then |I| = | J| = 1, and for every vertex v G Vr, t1 (v) = m holds for some m > 2. This is case (3.a). Since any odd cycle in r would be contained in a lobe of r, it holds that r is bipartite if and only if every lobe is bipartite. If either r or A is not vertex-transitive, then each -and hence both - are bipartite, and the sides of the bipartitions (whether or not they are entire orbits of the appropriate automorphism group) are blocks of imprimitivity systems. Let {P1, P2} be the bipartition of Vr, and so {P1 n V©, P2 n V©} is the bipartition of 588 Ars Math. Contemp. 17 (2019) 493-514 any lobe ©. Equivalently, letting {Qi, Q2} denote the bipartition of A, we have Pj = UeeL(r) CTe(Qi) for i = 1,2. Suppose that r is vertex-transitive but A is not, and so |I| = 1 and |J| = 2. By Theorem 2.2, there exist constants m1, m2 e N U {H0} such that for all v e Vr and j e J, we have mj = Tj(v) = |{0 G L(r) : v G ae(Qj)}|. Finally, suppose that r is not vertex-transitive, and so P1 and P2 are the orbits of Aut(r). Also, A is bipartite with bipartition {Q1, Q2}, where Qj = Pj n VA. As no automorphism of r swaps P1 with P2, no automorphism of r swaps Q1 with Q2 (even though A may be vertex-transitive!). Hence |I| = |J| = 2. Since r is lobe-transitive, it follows now from the "necessity" argument of Theorem 3.2 that, for j = 1, 2, the function Tj satisfies the condition Tj(v) > 0 if and only if v e Pj. That means that there exist constants m1, m2 e N U {K0}, at least one of which is greater than 1, such that, if v e Pj, then Tj (v) = mjSj,j. □ Example 4.3. Suppose in the notation of Theorem 4.2 that r is edge-transitive and A is the complete bipartite graph Ks,t with |Q1| = s and |Q2| = t. Suppose that every vertex of r is incident with exactly two lobes isomorphic to A. If s = t, then both r and A are vertex-transitive, and we have case (3.a) of the theorem. If s = t and every vertex lies in one image of Q1 and one image of Q2, then we have the situation of case (3.b). If again s = t but each vertex lies in either two images of Q1 or two images of Q2, then we have the situation described in case (3.c). Remark 4.4. With regard to Example 4.3, we note that having s = t does not assure vertex-transitivity of edge-transitive bipartite graphs. There exist edge-transitive, non-vertex-transitive, finite bipartite graphs where the two sides of the bipartition have the same size. Such graphs are called semisymmetric. The smallest such graph, on 20 vertices with valence 4, was found by J. Folkman [2], who also found several infinite families of semisymmetric graphs. Many more such families as well as forbidden values for s were determined by A. V. Ivanov [4]. Example 4.5. This simple example illustrates how the converse of Lemma 4.1 is false, even though the lobes themselves may be highly symmetric. Let r be a graph of connectivity 1 whose lobes are copies of the Petersen graph (which is 3-arc-transitive!). For each lobe A, let (VA)1 and (VA)2 denote the vertex sets of disjoint 5-cycles indexed "consistently," i.e., if A and © share a vertex v, then v e (VA) j n (V©) j for i =1 or i = 2. (Observe that r is not bipartite.) For i = 1, 2 define Pj = |JeejZ,(r) (V©) j, and suppose that each vertex in Pj belongs to exactly mj lobes. The graph r is lobe-transitive by Theorem 3.2, and r is both vertex- and edge-transitive when m1 = m2, but r is neither vertex- nor edge-transitive when m1 = m2. 5 Arc-transitivity Theorem 5.1. Let r be a graph of connectivity 1. Necessary and sufficient conditions for r to be arc-transitive are the following: (1) The lobes of r are arc-transitive. (2) The lobes of r are pairwise isomorphic. (3) All vertices of r are incident with the same number of lobes. J. E. Graver and M. E. Watkins: Lobe, edge, and arc transitivity of graphs of connectivity 1 589 Proof. (Necessity) Suppose that r is arc-transitive. Conditions (1) and (2) follow from Lemma 4.1. Since arc-transitivity implies vertex-transitivity, condition (3) holds. (Sufficiency) Assume that the three conditions hold. For k =1,2, let ak be an arc of r, and let ©k be the lobe containing ak. By condition (2), there exists an isomorphism a : ©i ^ ©2. By condition (1), there exists an automorphism a e Aut(©2) such that aa(ai) = a2. By condition (3), the functions Tj of Equation (2.1) are constant on Vr. (In fact, since the lobes are vertex-transitive, there is only one such function.) It now follows from Lemma 2.1 that aa is extendable to all of r. □ Remark 5.2. If conditions (1) and (3) of Theorem 5.1 were replaced by the lobes are edge-transitive and r is vertex-transitive, the sufficiency argument would fail. There exist finite graphs [1] and countably infinite graphs with polynomial growth rate [7] that are vertex-and edge-transitive but not arc-transitive. Let A denote such a graph, and consider a graph r whose lobes are isomorphic to A with the same number of lobes incident with every vertex. Then r itself is vertex- and edge-transitive but not arc-transitive. The following proposition is elementary. Proposition 5.3. For all k > 2, the only k-arc-transitive graphs of connectivity 1 are trees of constant valence. References [1] I. Z. Bouwer, Vertex and edge transitive, but not 1-transitive, graphs, Canad. Math. Bull. 13 (1970), 231-237, doi:10.4153/cmb-1970-047-8. [2] J. Folkman, Regular line-symmetric graphs, J. Comb. Theory 3 (1967), 215-232, doi:10.1016/ s0021-9800(67)80069-3. [3] R. Halin, Automorphisms and endomorphisms of infinite locally finite graphs, Abh. Math. Sem. Univ. Hamburg 39 (1973), 251-283, doi:10.1007/bf02992834. [4] A. V. Ivanov, On edge but not vertex transitive regular graphs, in: C. J. Colbourn and R. Mathon (eds.), Combinatorial Design Theory, North-Holland, Amsterdam, volume 149 of North-Holland Mathematics Studies, pp. 273-285, 1987, doi:10.1016/s0304-0208(08)72893-7. [5] H. A. Jung and M. E. Watkins, On the structure of infinite vertex-transitive graphs, Discrete Math. 18 (1977), 45-53, doi:10.1016/0012-365x(77)90005-x. [6] O. Ore, Theory of Graphs, volume XXXVIII of American Mathematical Society Colloquium Publications, American Mathematical Society, Providence, Rhode Island, 1962. [7] C. Thomassen and M. E. Watkins, Infinite vertex-transitive, edge-transitive non-1-transitive graphs, Proc. Amer. Math. Soc. 105 (1989), 258-261, doi:10.2307/2046766. [8] J. Siran and M. E. Watkins, Imprimitivity of locally finite, 1-ended, planar graphs, Ars Math. Contemp. 5 (2012), 213-217, doi:10.26493/1855-3974.213.fcd. [9] M. E. Watkins, Les graphes de Cayley de connectivite un, in: Problèmes combinatoires et théorie des graphes, CNRS, Paris, volume 260 of Colloques Internationaux du Centre National de la Recherche Scientifique, pp. 419-422, 1978, Colloque International CNRS held at the Universite d'Orsay, Orsay, July 9 - 13, 1976. [10] M. E. Watkins, Infinite graphical Frobenius representations, Electron. J. Combin. 25 (2018), #P4.22 (23 pages), https://www.combinatorics.org/ojs/index.php/eljc/ article/view/v25i4p22. [11] M. E. Watkins and J. E. Graver, A characterization of infinite planar primitive graphs, J. Comb. Theory Ser. B 91 (2004), 87-104, doi:10.1016/j.jctb.2003.10.005. ARS MATHEMATICA CONTEMPORANEA ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 17 (2019) 591-615 https://doi.org/10.26493/1855-3974.1801.eb1 (Also available at http://amc-journal.eu) Bipartite edge-transitive bi-p-metacirculants* Yan-QuanFeng t, YiWang Department of Mathematics, Beijing Jiaotong University, Beijing, P.R. China Received 18 September 2018, accepted 18 November 2019, published online 12 December 2019 A graph is a bi-Cayley graph over a group if the group acts semiregularly on the vertex set of the graph with two orbits. Let G be a non-abelian metacyclic p-group for an odd prime p. In this paper, we prove that if G is a Sylow p-subgroup in the full automorphism group Aut(r) of a graph r, then G is normal in Aut(r). As an application, we classify the half-arc-transitive bipartite bi-Cayley graphs over G of valency less than 2p, while the case for valency 4 was given by Zhang and Zhou in 2019. It is further shown that there are no semisymmetric or arc-transitive bipartite bi-Cayley graphs over G of valency less than p. Keywords: Bi-Cayley graph, half-arc-transitive graph, metacyclic group. Math. Subj. Class.: 05C10, 05C25, 20B25 1 Introduction All graphs considered in this paper are finite, connected, simple and undirected. For a graph r, we use V(r), E(r), A(r) and Aut(r) to denote its vertex set, edge set, arc set and full automorphism group, respectively. A graph r is said to be vertex-transitive, edge-transitive or arc-transitive if Aut(r) acts transitively on V(r), E(r) or A(r) respectively, semisymmetric if it is edge-transitive but not vertex-transitive, and half-arc-transitive if it is vertex-transitive, edge-transitive, but not arc-transitive. Let G be a permutation group on a set Q and a e Q. Denote by Ga the stabilizer of a in G, that is, the subgroup of G fixing the point a. We say that G is semiregular on Q if Ga = 1 for every a e Q and regular if G is transitive and semiregular. A group G is metacyclic if it has a normal subgroup N such that both N and G/N are cyclic. Let r be a graph with G < Aut(r). Then r is called a Cayley graph over G if G is regular on V(r) and a bi-Cayley graph over G if G is semiregular on V(r) with two orbits. *The authors acknowledge the partial support from the National Natural Science Foundation of China (11731002) and the 111 Project of China (B16002). 1 Corresponding author. E-mail addresses: yqfeng@bjtu.edu.cn (Yan-Quan Feng), yiwang@bjtu.edu.cn (Yi Wang) Abstract ©® This work is licensed under https://creativecommons.org/licenses/by/4.0/ 592 ArsMath. Contemp. 17(2019)591-615 In particular, if G is normal in Aut(T), the Cayley graph or the bi-Cayley graph r is called a normal Cayley graph or a normal bi-Cayley graph over G, respectively. Determining the automorphism group of a graph is fundamental in algebraic graph theory, but very difficult in general. If r is a connected normal Cayley graph over a group G, then Aut(T) is determined by Godsil [27], and if r is a connected normal bi-Cayley graph over G, then Aut(T) is also determined by Zhou and Feng [55]. Thus a natural problem is to determine normality of Cayley graphs or bi-Cayley graphs over groups. The normality of Cayley graphs over cyclic group of order a prime and over group of order twice a prime was solved by Alspach [1] and Du et al. [19], respectively. Dobson [14] determined all non-normal Cayley graphs over group of order a product of two distinct primes, and Dobson and Witte [16] determined all non-normal Cayley graphs over group of order a prime square. Dobson and Kovacs [15] determined the full automorphism groups of Cayley graphs over elementary abelian group of rank 3. However, it seems still very difficult to obtain normality of Cayley graphs for general valencies. On the other hand, many results on the normality of Cayley graphs with small valencies were obtained, and for example, one may refer to [20, 21, 22] for finite non-abelian simple groups and to [4, 23, 26, 51, 54] for solvable groups. Due to nice properties on automorphism groups of non-abelian p-groups, the normality of Cayley graphs with general valencies over certain non-abelian p-groups was obtained. A connected Cayley graph or bi-Cayley graph over a non-abelian metacyclic p-group, for an odd prime p, is called a p-metacirculant or a bi-p-metacirculant, respectively. Li and Sim [34] proved that a p-metacirculant r is normal except a special case when the non-abelian metacyclic p-group is a Sylow p-subgroup of Aut(r), and Wang and Feng [50] proved that this special case cannot occur. In this paper we prove the following theorem. Theorem 1.1. Let r be a connected bipartite bi-p-metacirculant over a non-abelian metacyclic p-group G. If G is a Sylow p-subgroup of Aut(r), then G is normal in Aut(r). It is well-known that Cayley graphs play an important role in the study of symmetry of graphs. However, graphs with various symmetries can be constructed by bi-Cayley graphs. The smallest trivalent semisymmetric graph is the Gray graph [6], which is a bi-Cayley graph over a non-abelian metacyclic group of order 27, and infinite semisymmetric graphs were constructed in [17, 18, 37]. Boben et al. [5] studied properties of cubic bi-Cayley graphs over cyclic groups and the configurations arising from these graphs. Kovacs et al. [31] gave a description of arc-transitive one-matching bi-Cayley graphs over abelian groups. All cubic vertex-transitive bi-Cayley graphs over cyclic groups, abelian groups or dihedral groups were determined in [39, 52, 54]. Recently, Conder et al. [11] investigated bi-Cayley graphs over abelian groups, dihedral groups and metacyclic p-groups, and using these results, a complete classification of connected trivalent edge-transitive graphs of girth at most 6 was obtained. Furthermore, Qin et al. [41] classified connected edge-transitive bi-p-metacirculants of valency p, and as an application of Theorem 1.1, we prove that there are no such graphs with valency less than p. Theorem 1.2. For any odd prime p, there are no connected arc-transitive or semisymmetric bipartite bi-p-metacirculants of valency less than p. In 1966, Tutte [46] initiated an investigation of half-arc-transitive graphs by showing that a vertex- and edge-transitive graph with odd valency must be arc-transitive. A few years later, in order to answer Tutte's question on the existence of half-arc-transitive graphs Y.-Q. Feng and Y Wang: Bipartite edge-transitive bi-p-metacirculants 593 of even valency, Bouwer [7] constructed a 2k-valent half-arc-transitive graph for every k > 2. One of the standard problems in the study of half-arc-transitive graphs is to classify such graphs for certain orders. Let p be a prime. It is well known that there are no half-arc-transitive graphs of order p or p2, and no such graphs of order 2p by Cheng and Oxley [8]. Alspach and Xu [2] classified half-arc-transitive graphs of order 3p and Kutnar et al. [33] classified such graphs of order 4p. Despite all of these efforts, however, further classifications of half-arc-transitive graphs with general valencies seem to be very difficult, and special attention has been paid to the study of half-arc-transitive graphs with small valencies, which were extensively studied from different perspectives over decades by many authors; see [3, 9, 10, 24, 25, 29, 32, 35, 38, 40,43, 47, 48, 49] for example. The smallest half-arc-transitive graph constructed in Bouwer [7] is a bi-Cayley graph over the non-abelian metacyclic group of order 27 with exponent 9. Zhang and Zhou [56] proved that a half-arc-transitive bi-Cayley graph over cyclic group has valency at least 6, and this was extended to abelian groups by Conder et al. [11]. In fact, half-arc-regular bi-Cayley graphs of valency 6, over cyclic groups, were classified in [56], and two infinite families of bipartite tetravalent half-arc-transitive bi-p-metacirculants of order p3 were constructed in [11], of which one is Cayley and the other is not Cayley. Furthermore, Zhang and Zhou [53] classified tetravalent half-arc-transitive bi-p-metacirculants, and all these graphs are bipartite. This was the main motivation for the research leading to Theorem 1.3, namely the classification of bipartite half-arc-transitive bi-p-metacirculants of valency less than 2p. It was also motivated in part by the classification of half-arc-transitive p-metacirculants of valency less than 2p, given by Li and Sim [35]. For a positive integer n, denote by Zn the cyclic group of order n, as well as the ring of integers modulo n, and by Z*n the multiplicative group of the ring Zn consisting of numbers coprime to n. Theorem 1.3. Let p be an odd prime and let r be a connected bipartite bi-p-metacirculant of valency 2k with k < p over a non-abelian metacyclic p-group G. Then r is half-arc-transitive if and only if k > 2, k | (p - 1), G = Ga,p,7 and r = r^, where 0 < y < a < ^ + 7, m G , 0 < ^ < k with jJ^ \ , and Aut(rm,M) = (Ga,p,7 X Zk).Z2. The groups Ga,^,7 and graphs r^ k t above are defined in Equation (2.1) and Equation (4.3). By Zhang and Zhou [53], the graphs r^ 21 can be Cayley or non-Cayley for certain values m and I, and this implies that the extensions (Ga,p,Y x Z2).Z2 above can be split or non-split. 2 Background results Let G be a finite metacyclic p-group. Lindenberg [36] proved that the automorphism group of G is a p-group when G is non-split. The following proposition describes the automorphism group of the remaining case when G is split. It is easy to show that every non-abelian split metacyclic p-group G for an odd prime p has the following presentation: Ga,ß,Y = (a,b | apa = 1, bp = 1, b-lab = a1+pJ), (2.1) 594 ArsMath. Contemp. 17(2019)591-615 where a, 3, 7 are positive integers such that 0 <7 < a < 3 + 7. Li and Sim characterized the automorphism group Aut(Ga^i7) of the group . Proposition 2.1 ([35, Theorem 2.8]). For an odd prime p, we have | Aut(Ga^r/)| = (p - 1)pmin(«^)+mm(A7) + ^+7-l. Moreover, all Hallp'-subgroups of Aut(Ga,^i7) are conjugate and isomorphic to Zp-1. In particular, the map 9: a ^ a£, b ^ b induces an automorphism of of order p — 1, where e is an element of order p — 1 in Zpa. A p-group G is said to be regular if for any x,y G G there exist dj G (x, y)', 1 < i < n, for some positive integer n such that xpyp = (xy)p n"=1 dp. If G is metacyclic, then the derived subgroup G' is cyclic, and hence G is regular by [30, Kapitel III, 10.2 Satz]. For regular p-groups, the following proposition holds by [30, Kapitel III, 10.8 Satz]. Proposition 2.2. Let G be a metacyclic p-group for an odd prime p. If |G'| = pn, thenfor any m > n, we have (xy)pm = xPm yPm, for any x, y G G. Remark 2.3. For the non-abelian split metacyclic group given in Equation (2.1), we have |G0,,^jT| = pa-Y and a — 7 < 3, and by Proposition 2.2, if (p, m) = 1 then o(bman) = max{o(an),p^}, and if 3 < a andp | n then o(bman) < pa-1. For a finite group G, N < G means that N is a subgroup of G, and N < G means that N is a proper subgroup of G. The following proposition lists non-abelian simple groups having a proper subgroup of index prime-power order. Proposition 2.4 ([28, Theorem 1]). Let T be a non-abelian simple group with H (G) the largest normal subgroup of G whose order is not divisible by p. The next proposition is about transitive permutation groups of prime-power degree. Proposition 2.5 ([34, Lemma 2.5]). Let p be a prime, and let A be a transitive permutation group with p-power degree. Let B be a nontrivial subnormal subgroup of A. Then B has a proper subgroup of p-power index, and Op/ (B) = 1. In particular, Op/ (A) = 1. It is well-known that GL(d, q) has a cyclic group of order qd — 1, the so called Singer-cycle subgroup, which also induces a cyclic subgroup of PSL(d, q). Y.-Q. Feng and Y Wang: Bipartite edge-transitive bi-p-metacirculants 595 Proposition 2.6 ([30, Kapitel II, 7.3 Satz]). The group G = GL(d, q) contains a cyclic subgroup of order qd — 1, and it induces a cyclic subgroup of order (q_i)(--i d) of PSL(d, q). Let G and E be two groups. We call an extension E of G by N a central extension if N lies in the center of E and E/N = G, and if E is further perfect, that is, the derived group E' = E, we call E a covering group of G. Schur [42] proved that for every non-abelian simple group G there is a unique maximal covering group M such that every covering group of G is a factor group of M (also see [30, Chapter 5, Section 23]). This group M is called the full covering group of G, and the center of M is the Schur multiplier of G, denoted by M(G). For a group G, we denote by Out(G) the outer automorphism group of G, that is, Out(G) = Aut(G)/Inn(G), where Inn(G) is the inner automorphism group of G induced by conjugation. The following proposition is about outer automorphism group and Schur multiplier of a non-abelian simple group having a proper subgroup of prime-power index. Proposition 2.7 ([34, Lemma 2.3]). Let p be an odd prime and let T be a non-abelian simple group that has a subgroup H of index pe > 1. Then (1) p { |M(T)|; (2) either p \ | Out(T)|, or T = PSL(2,8) andpe = 32. A group G is said to be a central product of its subgroups H1,..., Hn (n > 2) if G = H1 • • • Hn and for any i = j, H and Hj commute elementwise. A group G is called quasisimple if G' = G and G/Z(G) is a non-abelian simple group, where Z(G) is the centralizer of G. A group G is called semisimple if G' = G and G/Z(G) is a direct product of non-abelian simple groups. Clearly, a quasisimple group is semisimple. Proposition 2.8 ([45, Theorem 6.4]). A central product of two semisimple groups is also semisimple. Any semisimple group can be decomposed into a central product ofquasisim-ple groups, and this set of quasisimple groups is uniquely determined. A subnormal quasisimple subgroup of a group G is called a component of G. By [45, 6.9(iv), p. 450], any two distinct components of G commute elementwise, and by Proposition 2.8, the product of all components of G is semisimple, denoted by E (G), which is characteristic in G. We use F(G) to denote the Fitting subgroup of G, that is, F(G) = OPl (G) x OP2 (G) x • • • x OPt (G), where p1,p2,...,pt are the distinct prime factors of |G|. Set F *(G) = F (G)E (G) and call F *(G) the generalized Fitting subgroup of G. The following is one of the most significant properties of F* (G). For a group G and a subgroup H of G, denote by CG(H) the centralizer of H in G. Proposition 2.9 ([45, Theorem 6.11]). For any finite group G, we have Cg(F*(G)) < F*(G). An action of a group G on a set Q is a homomorphism from G to the symmetric group on Q. We denote by $(G) the Frattini subgroup of G, that is, the intersection of all maximal subgroups of G. Note that for a prime p, OP(G) is ap-groupand OP(G)/$(OP(G)) is an elementary abelian p-group. Thus, OP(G)/$(OP(G)) can be viewed as a vector space over the field ZP. The following lemma considers a natural action of a group G on the vector space OP (G) / $(OP (G)). 596 Ars Math. Contemp. 17 (2019) 493-514 Proposition 2.10 ([50, Lemma 2.9]). For a finite group G and a prime p, let H = Op(G) and V = H/$(H). Then G has a natural action on V, induced by conjugation via elements of G on H. If Cg(H ) < H, then H is the kernel of this action of G on V. Let a group T act on two sets Q and E, and these two actions are equivalent if there is a bijection A: Q ^ E such that (a4)A = (aA)4 for all a G Q and t G T. When the two actions above are transitive, there is a simple criterion on whether or not they are equivalent. Proposition 2.11 ([13, Lemma 1.6B]). Assume that a group T acts transitively on the two sets Q and E, and let W be a stabilizer of a point in the first action. Then the actions are equivalent if and only if W is the stabilizer of some point in the second action. For a group G and two subgroups H and K of G, we consider the actions of G on the right cosets of H and K by right multiplication. The stabilizers of Hx and Ky are Hx and Ky, respectively. By Proposition 2.11, these two right multiplication actions are equivalent if and only if H and K are conjugate in G. 3 Automorphisms of bipartite bi-p-metacirculants Let rN be the quotient graph of a graph r with respect to N < Aut(r), that is, the graph having the orbits of N as vertices with two orbits Oi, O2 adjacent in rN if and only if there exist some u g O1 and v G O2 such that {u, v} is an edge in r. Denote by [O1] the induced subgraph of r by O1, and by [O1,02] the subgraph of [O1 U O2] with edge set {{u,v} G E(r) | u G O1,v G O2}. Proof of Theorem 1.1. Let G a non-abelian metacyclic p-group of order pn for an odd prime p and a positive integer n, and let r be a connected bipartite bi-p-metacirculant over G. Set A = Aut(r), and let G g Sylp(A), where Sylp(A) is the set of Sylow p-subgroups of A. To finish the proof, it suffices to show that G < A. Let W0 and W1 be the two parts of the bipartite graph r. Then {W0, W1} is a complete block system of A on V(r) with | W01 = | W11 = |G| = pn. Let A* be the kernel of A on { Wo, W1}, that is, the subgroup of A fixing Wo and W1 setwise. Then A* < A, A/A* < Z2 and Sylp(A) = Sylp(A*). It follows that G G Sylp(A*). Noting that |G| = pn, we have pn | |A| and pn+1 \ |A|, that is, pn || |A|. The group G has exactly two orbits, that is, W0 and W1, and G is regular on both W0 and W1. By Frattini argument [30, Kapitel I, 7.8 Satz], A* = GAU for any u G V(r), implying that A*u is ap'-group. Clearly, A„ = A;, and so Au is also a p'-group. Let K be the kernel of A* acting on W0. Then K < AV for any v G W0, and K < A*. The orbits of if on W1 have the same length, and so it is a divisor of pn. It follows that if K = 1 then p | |K |, which is impossible because A* is a p'-group. Thus, A* acts faithfully on W0 (resp. W1). Since Sylow p-subgroups of A are conjugate, every p-subgroup of A is semiregular on both W0 and W1. Claim 1. Any minimal normal subgroup N of A* is abelian. Y.-Q. Feng and Y Wang: Bipartite edge-transitive bi-p-metacirculants 597 We argue by contradiction and we suppose that N is non-abelian. Then N = Ti x • • • x Tk with k > 1, where Ti = T is a non-abelian simple group. By Proposition 2.5, p | |N| and sop | |Tj| for each 1 < i < k. Since G G Sylp(A*), we have G n N G Sylp(N), and hence G n N = Pi x • • • x Pk for some P, G Sylp(Tj), where P, = 1 for each 1 < i < k. Since G is metacyclic and G n N < G, G n N is metacyclic and this implies k < 2. Set Q = {Ti,..., Tk} and write B = NA (Ti). Considering the conjugation action of A* on Q, we have B < A* as k < 2, and hence A*/B < S2, forcing B < A*. Thus, Sylp (B) = Sylp(A*) and so B is transitive on both W0 and Wi. Let rTl be the quotient graph of r with respect to Ti. Since Ti < B, all orbits of Ti on W0 have the same length, and the length must be a p-power as | W01 = pn. Since each p-subgroup is semiregular, this length is the order of a Sylow p-subgroup of Ti. Similarly, all orbits of Ti on Wi have the same length and it is also the order of a Sylow p-subgroup of Ti. Thus, V (rTl ) = |Ai,. Ai U • • • U As and W1 = Ai Ai ., As, Ai,..., AS}, the set of all orbits of Ti, with Wo = U • • • U AS. Furthermore, for any 1 < i, j < s we have | Aj | = pm for some 1 < m < n, and hence s = pn-m. Since Ti < B, B has a natural action on V (rTl) and let K be the kernel of this action. Clearly, Ti < K. Recall thatp f |Au | for any u G V(r). Then p f | (Ti)u |, and by Guralnick [28, Corollary 2], Ti is 2-transitive on each Aj or Ai. Sincepn || |A*|,wehavepn || |B|, implying thatpm || |K|. Since (Ti )u is a proper subgroup of Ti of index p-power, Proposition 2.7 implies that either Ti = PSL(2, 8) with pm = 32, or p f | Out(Ti)|. To finish the proof of Claim 1, we will obtain a contradiction for both cases. Case 1. T1 = PSL(2,8) with pm = 32. In this case, |Aj| = |Aj| = 9. If s = 1 then |G| = pm = 32, contradicting that G is non-abelian. Thus s > 2 and s = 3n-2. By Atlas [12], PSL(2, 8) has only one conjugate class of subgroups of index 9, and by Proposition 2.11, Ti acts equivalently on Aj and Aj. «¿1 «¿2 Ai A', aji aj2 JP" Ai Aj Ai Aj Figure 1: The subgraphs [Ai, AJ ]. Set Ai = {«¿i, ai2,..., «¿q} and Aj = { jl,aJ2, ;.q} for 1 < i, j < 3' jQ n-2 Recall that Ti is 2-transitive on Aj and Aj. Since Ti acts equivalently on Aj and Aj, by Proposition 2.11, we may assume that (Ti)au = (Ti)«^, for any 1 < i, j < 3n-2 and 1 < t < 32. The subgraph [A,, Aj] is either a null graph, or one of the three graphs in Figure 3 because (T)^ = (Ti)a^ acts transitively on both A, \ {o^} and Aj \ {aj£}. a ¿P 598 ArsMath. Contemp. 17(2019)591-615 The three graphs have edge sets {{a^, aj£} | 1 < I < 32}, {{ajk, aje} | 1 < k, I < 32} or {{ajk, aj} | 1 < k, I < 32, k = respectively. For any g G S9, define a permutation as on V(T) by (a^)CTs = a^s and (aj£)CTg = aj£s for any 1 < i, j < 3"-2 and 1 < I < 32. Then as fixes each Aj and Aj, and permutes the elements of Aj and Aj in the 'same way' for each 1 < i, j < 3"-2. Since [Aj, Aj] is either a null graph, or one graph in Figure 3, as induces an automorphism of [Aj, Aj], for all 1 < i, j < 3"-2. Also as induces automorphisms of [Aj] and [Aj] for all 1 < i, j < 3"-2 because [Aj] and [Aj] have no edges (T is bipartite). It follows that CTg G Aut(r). Thus, L := {as | g G S9} < Aut(r) and L = S9. Clearly, L < A*. If L < B, there exists x G L such that Tf = Ti, and hence N = Ti x T2 with k = 2 and Tf = T2, which implies that Ti and T2 have the same orbits because x fixes each orbit of Ti, contradicting that Sylow p-subgroups of N are semiregular. Thus L < B. Recall that K is the kernel of B acting on V (rTl) and 32 = pm || |K|. Since L fixes each orbit of Ti and 33 | |L|, we have L < K and 33 | |K|, a contradiction. Case 2. p \ | Out(Ti)|. Since B/TiCb(Ti) < Out(Ti), we havepn || |TiCB(Ti)|. Since Ti is non-abelian simple, Ti n Cb(Ti) = 1 and hence TiCs(Ti) = Ti x CB(Ti). If p | |CB(Ti)|, then G is conjugate to Qi x Q2, where Qi G Sylp(Ti) and Q2 G Sylp(CB(T1i)). Since G is metacyclic, G can be generated by two elements, and since G is a p-group, any minimal generating set of G has cardinality 2. It follows that both Qi and Q2 are cyclic, and so G is abelian, a contradiction. Thus, p \ |CB(T1i)| and hence pn || |Ti|, forcing s = 1. Furthermore, W0 = Ai, Wi = Ai and Ti is 2-transitive on both W0 and Wi. Note that (Ti)u is proper subgroup of T1i of index pn. Since G is a Sylow p-subgroup of A of order pn, all Sylow p-subgroups of T1i are also Sylow p-subgroups of A, and so they are isomorphic to G. Without loss of generality, we may assume G < Ti. By Proposition 2.4, Ti = PSL(2,11), Mii, M23, PSU(4,2), Apn, or PSL(d,q) with ^d-r = p" and d a prime. Suppose Ti = PSL(2,11), Mn or M23. By Proposition 2.4, |Wo| = |Wi| = 11, 11 or 23 respectively, and hence |G| = 11, 11 or 23, contradicting that G is non-abelian. Suppose Ti = PSU(4,2) or Apn. For the former, Ti has one conjugate class of subgroups of index 27 by Atlas [12], and for the latter, Ti has one conjugate class of subgroups of index p". By Proposition 2.11, Ti acts equivalently on W0 and Wi, and since r is connected, the 2-transitivity of Ti on W0 and Wi implies that r = Kp»ipn or Kp»ipn - p"K2. Then A = Spn 1 S2 or Spn x Z2 respectively. Since G is non-abelian, we have n > 3, and so p"+i | |A|, a contradiction. d — i Suppose T1i = PSL(d, q) with ^-¡i = p" and d a prime. By Proposition 2.6, Ti d — i has a cyclic subgroup of order (q-i)(q^i d). Since d is a prime, either (q - 1, d) = 1 or I d -1 (q - 1, d) = d. Note that (q - 1, d) | 1q--ii. If (q - 1, d) = d then d = p andp | (q - 1). Since p > 3 and p2 | (q2 - 1)(q - 1), we have p"+i | (qP-i)(qp-i)(i?( 1, contradicting Claim 1. Thus, A* has no component and E(A*) = 1. It follows that the generalized Fitting subgroup F* (A*) = F(A*). By Proposition 2.5, Op (A*) = 1 and hence F* (A*) = Op(A*). By Proposition 2.9, CA (Op(A*)) < Op(A*), as claimed. Now we are ready to finish the proof. Since | A : A* | < 2 and G has no subgroups of index 2, we only need to show G < A*. Let H = Op(A*). By Claim 1, H =1. Write H = H/$(H) and A* = A*/$(H). Then Op (A*/H) = 1 and H < G as G e Sylp(A*). By Claim 2 and Proposition 2.10, A*/H < Aut(H). Since G is metacyclic, H = Zp or Zp x Zp. Assume H = Zp. Then A*/H < Zp-1, and G = H < A*, as required. Assume H = Zp x Zp. Then A*/H < GL(2,p). If p f |A*/H| then G = H < A*, as required. To finish the proof, we suppose p | |A*/H| and will obtain a contradiction. Since p || | GL(2,p)|, we have p || |A*/H|, and since Sylp(SL(2,p)) = Sylp(GL(2,p)), we have Sylp(A*/H) C Sylp(SL(2,p)). Note that A*/H • SL(2,p) < GL(2,p). Thenp || |A*/H• SL(2,p)| and sop | |(A*/H)nSL(2,p)|. Since Op(A*/H) = 1, A*/H has at least two Sylow p-subgroups, and hence (A*/H) n SL(2,p) has at least two Sylow p-subgroups, that is, (A*/H) n SL(2,p) has no normal Sylow p-subgroups. By [44, Theorem 6.17], (A*/H) n SL(2,p) contains SL(2,p), that is, SL(2,p) < A*/H GL(2,p). In particular, the induced faithful representation of A*/H on the linear space H is irreducible, and hence H is a minimal normal subgroup of A*. Recall that A* < A and H = Op(A*), which is characteristic in A*. Then H < A, and since $(H) is characteristic in H, we have $(H) < A. Let r$(-H) be the quotient digraph of r relative to $(H), and let L be the kernel of A acting on V(r$(H)). Clearly, r$(H) is bipartite. Furthermore, L < A, L < A*, $(H) < L and L = $(H)L„ for any u e V(r) because both $(H) and L are transitive on the orbit of $(H) containing u. Since $(H) < G, $(H) is semiregular on V(r), and hence $(H) n Lu = 1. Since p f |Au|, Lu is a Hall p'-subgroup of L. Since $(H) < L and $(H) e Sylp (L), the Schur-Zassenhaus Theorem implies that all Hall p'-subgroup of L are conjugate. By Frattini argument [30, Kapitel I, 7.8 Satz], A = LNA(L„) = $(H)L„NA(L„) = $(H)NA(L„) and H = Hn A = Hn ($(H)NA(L„)) = $(h)(Hn Na(l„)). SinceFrattini subgroup is generated by nongenerators (see [30, Kapitel III, 3.2 Satz]), H = $(H)(H n NA(L„)) if and only if H = H n NA(Lu), that is, H < NA(Lu). It follows that A = NA(Lu), that is, Lu < A. By taking u e W0, we have Lu = Lv for any v e W0 because A is transitive on W0, and since A* acts faithfully on W0, we have Lu = 1. It follows that L = $(H), that is, the kernel of A acting on V(r$(H)) is $(H). Thus A = A/$(H) is faithful on V(r$(H)), and then A* is faithful on each of the parts of V(r$(H)), that is, 600 ArsMath. Contemp. 17(2019)591-615 A* is a transitive permutation group with p-power degree (the cardinality of each part of Since A*/H = A*/H, we have SL(2,p) < A*/H < GL(2,p). Write R/H = Z(A*/H). Then R < A* and 1 = R/H is a p'-group. Since H < R and H e Sylp(R), the Schur-Zassenhaus Theorem [44, Theorem 8.10] implies that there is a p'-group V < R such that R = HV and all Hall p'-subgroup of R are conjugate. Note that V = 1. By Frattini argument [30, Kapitel I, 7.8 Satz], A* = RN^(V) = HN^(V). Since H is abelian, H n ^nj_(V) ^_A*, aMby the minimality of H,_we have_H n N-^(V) = H or 1. IfH n N-^-V) = H thenJH < N-^ (V) and A* = (V) = ^^ (V ),_that is, V < A*. Thisimpliesthat Op, (A*) = 1, contradicting Proposition 2.5. If H nN-nj-(V )_ = 1 then A* = HN-^(V) implies ASL(2,p) < A*_J AGL(2,p) as SL(2,p) < A*/H < GL(2,p). It follows that a Sylow p-subgroup of A* is not metacyclic. On the other hand, since both normal subgroups and quotient groups of a metacyclic group are metacyclic, any Sylow p-subgroup of A* is metacyclic because each Sylow p-subgroup of A* is metacyclic, a contradiction. This completes the proof. □ 4 Edge-transitive bipartite bi-p-metacirculants A connected edge-transitive graph should be semisymmetric, arc-transitive or half-arc-transitive. In this section, as an application of Theorem 1.1, we prove that there are no connected arc-transitive or semisymmetric bipartite bi-p-metacirculants with valency less than p. Furthermore, we classify the connected half-arc-transitive bipartite bi-p-metacirculants with valency less than 2p. Let G be a group and let R, L and S be subsets of G such that R = R-1, L = L-1, 1 e R U L and 1 e S, where 1 is the identity of G. Let BiCay(G, R, L, S) be the graph having vertex set the union of the right part W0 = {go | g e G} and the left part W1 = {g1 | g e G}, and edge set the union of the right edges {{h0,g0} | gh-1 e R}, the left edges {{h1,g1} | gh-1 e L} and the spokes {{h0,g1} | gh-1 e S}. For g e G, define a permutation g on V(r) = W0 U W1 by the rule hg = (hg)i, Vi e Z2, h,g e G. It is easy to check that g is an automorphism of BiCay(G, R, L, S) and G = {g | g e G} is a semiregular group of automorphisms of BiCay(G, R, L, S) with two orbits W0 and W1. Thus, BiCay(G, R, L, S) is a bi-Cayley graph over G, and BiCay(G, R, L, S) is also called a bi-Cayley graph over G relative to R, L and S. Furthermore, BiCay(G, R, L, S) is connected if and only if G = (R U L U S), and BiCay(G, R, L, S) = BiCay(G, Re, Le,Sd) for any 6 e Aut(G). On the other hand, if r is a Bi-Cayley graph over G then r = BiCay(G, R, L, S) for some subsets R, L and S of G satisfying R = R-1, L = L-1,1 e R U L and 1 e S. For 6 e Aut(G) and x,y,g e G, define two permutations on V(BiCay(G, R, L,S)) = W0 U W1 as following: 5e,x,y: h0 ^ (xhe)1, h1 ^ (yhe)0, Vh e G, °e,g: h0 ^ (he)0, h1 ^ (gh0)1, Vh e G. Y.-Q. Feng and Y Wang: Bipartite edge-transitive bi-p-metacirculants 601 Set I := {d6xy | 6 G Aut(G) s.t. R6 = x-1Lx, L6 = y-1Ry, S6 = y-1S-1x}, F := {a6g | 6 G Aut(G) s.t. Re = R, Le = g-1Lg, Se = g-1S}. The following proposition characterizes the normalizer of G in Aut(r). Proposition 4.1 ([55, Theorem 1.1]). Let r = BiCay(G, R, L, S) be a connected bi-Cayley graph over a group G, where R, L and S are subsets of G with R = R-1, L = L-1, 1 G R U L and 1 G S. If I = 0 then NAut(r)(G) = G x F, and if I = 0, then NAut(r)(G) = G(F,5e ,x,y) for some $8,x,y G 1 Write N = NAut(r)(G). By Proposition 4.1, N10 = F and N^^ = {aeA | 6 G Aut(G) s.t. Re = R, Le = L, S6 = S}. In particular, F is a group. For the special case R = L = 0, it is easy to see that F = {ae,s | 6 G Aut(G), s G S,S6 = s-1S} as 1 G S. Lemma 4.2. Let r = BiCay(G, 0,0, S) be a connected bipartite bi-Cayley graph over G relative to S with 1 G S. Then F = {a6,s | 6 G Aut(G), s G S,S6 = s-1S} is faithful on S1 = {s1 | s G S}. If G is a p-group and F is a p'-group, then F = {6 | a6,s G F}. Proof. Set L = {6 | a6,s}. Since r is connected, G = (S}, and since F1l = {a6j1 | 6 G Aut(G) s.t. S6 = S}, F is faithful on S1. The group F has operation a6xxas,y = &6s,yxs for any a6xx, asy G F, and so the map p: a6,s ^ 6 is an epimorphism from F to L. Let K be the kernel of p. Then a6,s G K if and only if 6 = 1. Let G be a p-group and F a p'-group. If a1jS G K for some 1 = s G S, then s^1'^ = {s1, si,..., so(s)-1,11} because 1a11's = s1 and (s^-1 )ai's = s1 for any positive integer l. Since G is a p-group, o(s) is a p-power and hence p | o(a1jS), which is impossible because F is a p'-group. Thus, s =1 and hence K = 1. Since p is an epimorphism from F to L, we have F = L. □ By Equation (2.1), = (a,b | aP" = 1, bp^ = 1, b-1ab = a1+pY} with 0 2) in Z*pa with k | (p — 1). Then el — 1 G Zpa for any 1 < i < k, and 1 + e +-----+ ek-1 = 0 (mod pa). For i G Zk, let tj = (e — 1)-1(ej — 1) and T = {tj | i G Zk}, where (e — 1)-1 is the inverse of e — 1 in Zpa. For x, y G Zpa, let Tx + y = {tx + y | t G T}. Then Tx + y = T in Zpa if and only if x = e1 (mod pa) and y = (e — 1)-1(e' — 1) (mod pa) for some l G Zk. In particular, Tx = T in Zpa if and only if x = 1 (mod pa). Proof. Suppose ej — 1 G Zpa for some 1 < i < k. Thenp | (ej — 1), and since e has order k, we have ej ^ 1 (mod pa) and (ej)k = 1 (mod pa). Furthermore, p | (ej — 1) implies that there exist l G Z*a (p f l) and 1 < s < a such that ej = 1 + lps. Note that (ej)k — 1 = (1 + lps)k — 1 = klps + C2(lps)2 + • • • + Gfcfc-1(lps)fc-1 + (lps)k. Y.-Q. Feng and Y Wang: Bipartite edge-transitive bi-p-metacirculants 603 Since (e®)fc = 1 (mod pa), we have p | kl, and since 2 < k < p, we have p | l, a contradiction. Thus, p f (e® -1), that is, e® -1 G Zp„. The equation 1 + e +-----+ er-1 = 0 (mod pa) follows from (e-1)(1+e+-----+ek-1) = er-1 = 0 (mod pa) ande-1 G Zp„. Note that T C Zpa and Tx + y C Zpa. Since 1 + e +-----+ er-1 = 0 (mod pa), we have £ t = (e - 1)-1 £ (e® - 1) teT ¿ezfc = (e - 1)-1[(e - 1) + ••• + (ek-1 - 1)] = -k(e - 1)-1 G Zpa. Assume Tx + y = T in Zpa. Then ^teT(tx + y) = ^teT t, and hence ky = (1 - x) ^t£T t = -(1 - x)k(e - 1)-1 in Zpa. It follows y = (e - 1)-1(x - 1) because k G Z;„. Then Tx + (e - 1)-1(x - 1) = T implies x[T(e - 1) + 1] = T(e - 1) + 1. Since T(e - 1) + 1 = {e® | i G Zk} = (e), we have x(e) = (e) in Zp„, that is, x = e1 (mod pa) for some l G Zk. Furthermore, y = (e - 1)-1(e' - 1) (mod pa). On the other hand, let x = e1 (mod pa) and y = (e - 1)-1 (e1 - 1) (mod pa) for some l G Zk. Then in Zpa, we have Tx + y = {e'(e - 1)-1(e® - 1) + (e - 1)-1(e' - 1) | i G Zfc} = (e - 1)-1{eV - 1) + (e1 - 1) | i G Zfc} = (e - 1)-1{ei+1 - 1 | i G Zfc} = {(e - 1)-1(e® - 1) | i G Zfc} = T. Thus Tx + y = T in Zpa if and only if x = e1 (mod pa) and y = (e - 1)-1(e' - 1) (mod pa) for some l G Zk. Applying this with y = 0, we obtain that Tx = T in Zpa if and only if x = 1 (mod pa). □ Lemma 4.6. Lei p be an oddpr/me and let a, 7 be positive integers with 0 < 7 < a. Let e be an element of order k (k > 2) in Z*„ with k | (p - 1). Then for any m G Z*a_Y and any 0 < £ < k - 1, the following equation in Zpa e£(1+ pY)m = [(1+ pY)m - x(1 - e)]2 (4.2) has a solution if and only if pr^y | , and in this case, there are exactly two solutions. Proof. Since e£(1+p7)m G Zpa, Equation (4.2) has a solution if and only if e£(1+p7)m is a square in Z*„. Since Zpa = Zpa-i(p-1), squares in Zpa consists of the unique subgroup of order pa-1 in Zpa, and so Equation (4.2) has a solution if and only if the order of e£(1 + pY)m in Z;„ is a divisor of ip0-1. Clearly, (1 + pY)m has order pa-Y, and ee has order pT^y. Thus, Equation (4.2) has a solution if and only if (k^y | . If e£(1 + pY)m = u2 for some u G Zpa then (1 - e)-1 [(1 + pY)m ± u] are the only two solutions of Equation (4.2) in Zpa. □ Now we construct the half-arc-transitive graphs in Theorem 1.3. Let p be an odd prime, and let a, 7 be positive integers such that 0 2) in Zp* with k | (p - 1). Choose 0 < I < k such that (k^y | ^^. Recall that Gai(8,7 = (a, b | aP" = 1, b^ = 1, b-1ab = a1+pY). Let U = {a4 | t e {(e - 1)-1(e4 - 1) | i e Zk}} and V = {bma4 | i e {(e - 1)-V - 1)(1 + pY)m + e4n | i e Zk}}, where m e and n is a solution of e£(1 + pY)m = [(1 + pY)m - x(1 - e)]2. Define C,m = BiCay(GaA7, 0, 0, U U V). (4.3) By Lemma 4.6, there are exactly two solutions n of equation ee(1 + pY)m = [(1 + pY)m - x(1 - e)]2 in Zpa, and so the notation k t is also written as k e, as used in Theorem 1.3. We first prove the sufficiency of Theorem 1.3. Lemma 4.7. The graphs k e are independent from the choice of element e of order k in Zpa and half-arc-transitive, and Aut(rm k ¿) = (Ga, ^ , Y x Zk).Z2. Proof. Write r = r^ , k ^ and A = Aut(r). Let T = {(e - 1)-1(e4 - 1) | i e Zk} and T7 = {(e - 1)-1(ei - 1)(1 + pY)m + e4n | i e Zk}. Then U = {an | n e T} and V = {bman | n e T'}. Furthermore, r = BiCay(G, 0, 0, S) with G = Ga , ^ , Y and S = U U V. Clearly, 1 e U and Ga ^ 7 = (S), implying that r is connected. Note that T' = T[(1 + pY)m + n(e - 1)] + n. ' ' Since e e Zpa, any element of order k in Zpa can be written as eq with (q, k) = 1 and hence {e4 | i e Zk} = (e) = (eq) = {(eq| i e Zk}. By Lemma 4.5, e - 1 e Zp„ and eq - 1 e Zpa. Let T = {(eq - 1)-1((eq- 1) | i e Zk}, T7 = {(eq - 1)-1((eq- 1)(1 + pY)m + (eq)4n | i e Zk}, U = {an | n e T} and V = {bman | n e T7}. It is easy to see that a ^ a(e-1)(eq-1) 1 and b ^ b induce an automorphism of G, say p. Then Up = {a(e-1)(e'-1)-1(e-1)-1 | i e Zk} = {a(e,-1)-1((eq)i-1) | i e Zk} = {an | n e T} = U, and similarly, Vp = V. Thus, BiCay(G, 0,0, U U V) = BiCay(G, 0,0, U U V), that is, r is independent from the choice of element e of order k in Zpa. To finish the proof, it suffices to prove that r is half-arc-transitive with Aut(r) = (Gaj(Sj7 x Zk).Z2. Claim 1. p f |A1o |. Y.-Q. Feng and Y Wang: Bipartite edge-transitive bi-p-metacirculants 605 We argue by contradiction and we suppose p | |A1o1. Let P is a Sylow p-subgroup of A containing G and let X = NA(G). Then G < P, and hence G < NP(G) < X. In particular, p | |X : G|, and so p | |X1o |. Let t be the automorphism of G induced by a ^ ae and b ^ b. First we prove aT,a G X1o. By Proposition 4.1, it is enough to show ST = a-1S. Clearly, UT = {aen | n G T} = {an | n G Te} and a-1U = {an-1 | n G T} = {an | n G T - 1}. By taking ^ =1 in Lemma 4.5, we have Te = T — 1 and hence UT = a-1U. Similarly, VT = {bmaen | n G T'} = {bman | n G T'e} and a-1V = {a-1bman | n G T'} = {bma-(1+pY)man | n G T'} = {bman | n G T' — (1+ pY)m}. By Equation (4.2), (1 + pY)m + n(e — 1) G Zp*, and hence Te = T — 1 implies T [(1 + pY )m + n(e — 1)]e + ne = T [(1 + pY )m + n(e — 1)] + n — (1 + pY )m. Since T' = T[(1 + pY)m + n(e — 1)] + n, we have T'e = T' — (1 + pY)m, that is, VT = a-1 V. It follows that ST = a-1S, as required. Set U1 = {«1 | u G U}, V1 = {v1 | v G V} and S1 = {s1 | s G S}. Then U1 = 11ffT>a> and V1 = (bma")1ffT>a>. Since G X10, either X10 has two orbits of length k on S1, or is transitive on S1. By Lemma 4.2, X1o acts faithfully on S1, and since p | |X1o |, any element of order p of X1o has an orbit of length p on S1, implying that X1o is transitive on S1 as k < p. From |X1o | = |X1o1l | • |1X1co | = |X1o1l | • 2k, we have p | |X1o111. By Proposition 4.1, X^ = {0-0,1 | 0 G Aut(G) s.t. S® = S}. Let aej1 G X1o1l be of order p with 0 G Aut(G). Then 0 has order p and S® = S. Recall that k > 2. Assume k > 2. Since a G S, we have a® G S® = S = U U V. If a® G V then a® = bma® for some i G T'. Note that a1+e G S as k > 2. Since m G , we have (m,p) = 1, and by Lemma 4.5, (p, 1 + e) = 1. Then (a1+e)® = (bma®)1+e G V, and considering the powers of b, we have m(1 + e) = m (mod p^), that is, e = 0 (mod p^). It follows that p | e, contradicting that e G Z*a. Thus, a® G U, and hence, a® = aj for some j G T. If a® = a then a^0'1 = {a1, a®,..., a1P 1} is an orbit of length p of on S1, which is impossible because there are exactly k < p elements of type aj in S. Thus, a® = a and 0 fixes U pointwise. Furthermore, 0 also fixes V pointwise because |V| = k < p. It follows that 0 = 1 as G = (S), and so = 1, a contradiction. Assume k = 2. Then e = — 1 (mod pa) and S1 = {11,a1, (bman)1, (bma(1+pY)m-n)1}. Since 110'1 = 11 and p > 3, has order 3 and we may assume that a^0'1 = (bman)1, (bma")^r0'1 = (bma(1+pY)m-n)1 and (bma(1+pY)m-n)^0'1 = a1 (replace o^ by ct^ if necessary), that is, a® = bman, (bman)® = bma(1+pY)m-n and (bma(1+pY)m-n)® = a. 606 Ars Math. Contemp. 17 (2019) 591-615 By Lemma 4.3, P < a. It follows that a = (6ma(1+PY = [(6ma>(1+PY )m-2n]6 = 6ma(1+PY )m-"(bman)(1+PY )m-2n, and so 0 = m + m[(1 + pY)m — 2n] (mod p^). Thus, p | (1 — n), which is impossible because otherwisepa = o(a6 ) = o(bma(1+pY)m-n) a, which is impossible because 0 < 7 < a. Thus, A1o = F = (aT,a) = Zk. Since A1o has two orbits on S1, that is U1 and V1, r is not arc-transitive. To prove the half-arc-transitivity of r, we only need to show that A is transitive on V(r) and E(r). Note that 11 € U1 and (bma")1 € V1. By Proposition 4.1, it suffices to construct a A € Aut(G) such that ÎA,bman,1 € I = {¿A,x,y | a € Aut(G), SA = y-1S-1x}, that is SA = S-1bma", because (10,= ((6ma")1,10). Let p = -(1 + pY)m - n(e - 1) and v = -(e - 1)-1p2 - (e - 1)-1p. Then p + 1 + n(e - 1) = 0 (mod pY) and hence v - pn = -(e - 1)-1p2 - (e - 1)-1p - pn = -(e - 1)-1p[p + 1 + n(e - 1)] = 0 (mod pY). By Proposition 2.2, o(bmav-M") = p^. Denote by m-1 the inverse of m in Zp,s. Then (bmav-M")m 1 = bae forsome e in Zpa, and it is easy to check that aM and (bmav-M")m 1 have the same relations as do a and b. Define A as the automorphism of G induced by a ^ aM, b ^ (bmav-M")m-1. Clearly, (bm)A = bmav-^n. Note that S = U U V. First we have UA = {anM | n € T} = {an | n € Tp} and V-1bma" = {(bman)-1bma" | n € T'} = {a-n+n | n € T'} = {an | n € -T' + n}. 608 Ars Math. Contemp. 17 (2019) 591-615 Recall that T' = T[(1 + pY)m + n(e - 1)] + n = -T^ + n. Then —T' + n = TyU, — n + n = Ty«, and so UA = V-1bman. On the other hand, VA = {(bman)A | n e T'} = {bmav-M"aw | n e T'} = {bman | n e T> — Mn + v} and U-1bman = {(an)-1bman | n e T} = {bma-n(1+pY)m+n | n e T} = {bman | n e —T(1 + pY)m + n}. To prove VA = U-1bman, we only need to show T'^ — ^n + v = —T(1 + pY)m + n in Zpa, which is equivalent to show that T(1+ pY)m = T^2 — v + n because T' = —T^ + n. By Equation (4.2), e'(1 + pY )m = [(1 + p7 )m — n(1 — e)]2 = M2, and by Lemma 4.5, T = Te£ + (e — 1)-1(e£ — 1). It follows T (1 + pY )m = Te£(1 + pY )m + (e — 1)-1(e£ — 1)(1 + pY )m = TM2 + (e — 1)-V — (1+ pY )m]. Note that —v + n = (e — 1)-V2 + (e — 1)-V + n = (e — 1)-V + M + n(e — 1)] = (e — 1)-V — (1 + pY )m]. Then T(1 + pY)m = T^2 — v + n, and hence VA = U-1bman. Thus, SA = UA U VA = V-1bman U U-1bman = S-1bman, and so r is half-arc-transitive. Let A* be the subgroup of A fixing the two parts of r setwise. Then A = A* .Z2. Since A1o = Zk and r is normal, we have A* = G x Zk and hence A = (G x Zk).Z2. □ Now we prove the necessity of Theorem 1.3. Lemma 4.8. For an odd prime p, let r be a connected bipartite half-arc-transitive bi-p-metacirculant of valency 2k (k < p) over G. Then k > 2, k | (p — 1), G = Ga^ri and r - rm,fc,£, where m e Z*„_y and 0 < £ < k with ^ | . Proof. Clearly, the two orbits of G are exactly the two parts of r. Then we may assume that r = BiCay(G, 0,0, S), where 1 e S, |S| < 2p and G = (S). Let A = Aut(r). Since r is half-arc-transitive, r has valency at least 4, that is, k > 2, and A1o has exactly two orbits on S1 = {s1 | s e S}, say U1 and V1 with 11 e U1, where U and V are subsets of G with 1 e U. Then S = U U V, |U| = |V| = k > 2 and |S| = 2k. Since k < p, the Orbit-Stabilizer theorem implies that A1o is ap'-group. By Theorem 1.1, G < A, and by Proposition 4.1, A1o = F = {ae,s | 0 e Aut(G), s e S, Se = s-1S}. By Lemma 4.2, A1o is faithful on S1, and F = L := {a | ae,s e F}. Y.-Q. Feng and Y Wang: Bipartite edge-transitive bi-p-metacirculants 609 Suppose that G is non-split. By Lindenberg [36], the automorphism group of G is a p-group. Thus, p | |L| and hencep | |A1o1, a contradiction. Thus, G is split, namely G = Ga p 7, as defined in Equation (2.1). By Proposition 2.1, F is a cyclic subgroup of Zp-1, and hence F = (ct6jS) for some 0 G Aut(G) and s G S with S6 = s-1S. Then ct6jS has order k and (ct6,s) is regular on both U1 and V1. Furthermore, A1o = F = (ct6jS) = Zk, U = l1ffe's> = {11,S1, (ss6 )1,..., (ss6 ••• s6'-2 )1}, and V1 = th'> = {t1, (st6)1, (ss6162)1,. .., (ss6 • • • s6'-2t6'- )1} with (ss6 • • • s6' 1 )1 = 11 for any t G V .It follows U = {1, s, ss6,..., ss6 • • • s6'-2} and V = {t, st6 ,ss6162 ,...,ss6 ••• s6'-216'-1}. In particular, k | (p - 1), 0 has order k, and ss6 ••• s6'-1 = 1. (4.4) By Proposition 2.1, we may assume that 0 is the automorphism induced by a ^ ae, b ^ b, where e G Zpa has order k. Let s = b®aj and t = bman with i, m G Zp,s and j, n G Zpa. Since s6 = b®aej, we have ss6 • • • s6' 1 = bfciae for some e G Zpa. By Equation (4.4), bki = 1, that is, ki = 0 (mod p^). Since k < p, we have i = 0 (mod p^), and hence s = aj. Since ss6 • • • s6i-1 = ajaje • • • ajei-1 = aj(e-1)-1(ei-1), we have U = {1, aj, ajaje, .. ., ajaje • • • a^-} = {aj(e-1)-1 (ei-1) | i G Zfc}. By Lemma 4.3, aj(e-1)-1(ei-1)bm = bmaj(e-1)-1 (e*-1)(1+pY)m , and since (bman)6i = bmaei", we have V = {aj(e-1)-1(ei-1)(bman)6i | i G Zfc} = {bmaj(e-1)-1(ei-1)(1+pY)m+ei" | i G Zk}. By the connectedness of r, G = (S) = (U U V) < (aj, an, bm), forcing G = (aj, an, bm). It follows that p { m and so m G Z*^. Since r is half-arc-transitive, Proposition 4.1 implies that there exists G I such that (1o, 11)^.*,» = ((bman)1,1o) with A G Aut(G) and SA = y-1S-1x. In particular, (bman)1 = 10A,x.» = x1 and 10 = 1^.*.» = y0. It follows that x = bman, y =1 and SA = S-1bman = U-1bman U V-1bman. Furthermore, U-1bman = {a-j(e-1)-1(ei-1)bman | i G Zk} = {bma-j(e-1)-1(ei-1)(1+pY)m+n | i G Zk}, 610 Ars Math. Contemp. 17 (2019) 591-615 and since (bmaj(e-l)-1(ei-l)(l+pY )m+ein)-l ^n = a-j(e-1)-1 (e*-l)(l+pY )m+n(l-ei), we have V-lbman = {a-j(e-l)-1(ei-l)(l+PY)m+n(l-ei) | i G Zk}. Suppose p | j. Since G = (aj, an, bm), we have p f n andp f m. By Proposition 2.2, every element in both V and U-lbman has order max{pa,p^}. Clearly, every element in U has order less than pa, but the element a-j(l+pY)m+n(l-e) g V-lbman has order pa because p f (1 - e) by Lemma 4.5. This is impossible as A G Aut(G) and (U U V)A = SA = S-lbman = U-lbman U V-lbman. Thus, p f j. Now, there is an automorphism of G mapping aj to a and b to b, and so we may assume j = 1 and s = a. It follows that U = {an | n G T}, (4.5) where T = {(e - 1)-l(e4 - 1) | i G Zk}; V = {bman | n G T'}, (4.6) where T' = {(e - 1)-l(ei - 1)(1 + pY)m + e4n | i G Zk}. As (e - 1)-l(e4 - 1)(1+ pY)m + e4n = [(e - 1)-l(e4 - 1)][(1 + pY)m + n(e - 1)] + n, we have T' = T[(1+ pY)m + n(e - 1)] + n. (4.7) Since -(e - 1)-l(e4 - 1)(1 + pY)m + n = [(e - 1)-l(e4 - 1)][-(1 + pY)m] + n and - (e - 1)-l(e4 - 1)(1 + pY)m + n(1 - e4) = [(e - 1)-l(e4 - 1)][-(1 + pY)m + n(1 - e)], we have U-lbman = {bman | n G Tl}, (4.8) where Tl = T[-(1 + pY)m] + n; V-lbman = {an | n G Tl}, (4.9) where T' = T[-(1 + pY)m + n(1 - e)]. Noting that T, T', Tl, T' C Zpa, we have U, V, U-lbman, V-lbman C G. Claim 1. aA G V- lbman. Y.-Q. Feng and Y Wang: Bipartite edge-transitive bi-p-metacirculants 611 Suppose to the contrary that ax / V-1bman. Since ax e Sx = U-1bman UV-1bman, we have ax e U-1bman, that is, ax = bmaM for ^ e T1. By Lemma 4.3, ¡3 < a. Recall k > 2. Let k > 2. Then a1+e e U and (a1+e)x = (bma^)1+e e U-1bman. Note that p \ m and by Lemma 4.5, p \ (1 + e). Considering the power of b of (bmaM)1+e and elements in U-1bman, we have m(1 + e) = m (mod p13) and so e = 0 (mod p13), contradicting e e Zpa. Let k = 2. Then T = {0,1} and e = -1 (mod pa). By Equations (4.5) and (4.6), S = {1, a, bman, bma(1+pY)m-n}, and by Equations (4.8) and (4.9), S-1bman = {1,a-(1+PY )m+2n,bm an,bma-(1+PY )m+n}. Note that ax e U-1bman = {bman, bma-(1+pY)m+n}. Case 1. ax = bman. As Sx = S-1bman, it is easy to see that ((bman)x, (bma(1+pY)m-n)x) = (a-(1+PY)m+2n,b™a-(1+PY)m+n) or (bma-(1+pY)m+n,a-(1+PY)m+2n). For the former, bma-(1+PY )m+n = (b™a(1+PY )m-n)x = [(b™an)a(1+PY )m-2n]x = a-(1+PY)m + 2n ( bm ) (1+PY )m-2n, implying that m = m[(1 + pY)m — 2n] (mod p3), and since p \ m, we have p | n. This is impossible because otherwise pa = o(ax) = o(bman) < pa (3 < a). For the latter, we can verify that a-(1+PY )m+2n = (bma(1+PY )m-n)x = [(bman)a(1+PY )m-2n]x = bma-(1+PY )m+n(bman)(1+PY )m-2n. Thus, 0 = m+m[(1+pY)m —2n] (mod p3), and hencep | (1—n), but it is also impossible because otherwisepa = o(ax) = o(bman) = o((bman)x) = o(bma-(1+pY)m+n) < pa. Case 2. ax = bma-(1+pY)m+n. In this case, we have ((bman)x, (bma(1+pY)m-n)x) = (a-(1+PY)m+2n, bman) or (bman, a-(1+PY)m+2n). For the former, bman = (bma(1+pY )m-n)x = [(bm an)a(1+pY )m-2n]x = a-(1+PY )m + 2n(bma-(1+PY )m+n)(1+pY )m-2n 612 Ars Math. Contemp. 17 (2019) 591-615 implying m = m[(1 + pY)m — 2n] (mod pß), and since p \ m, we have p | n. By Proposition 2.2, 0(6ma-(l+PY )m +n) = 0(6ma(l+PY )m-n) = max{o(a(1+PY) = o(a-(1+PY)m+n), o(bm)}, and it follows that p" = o(aA) = o(6ma-(1+pY )m+" ) = o(6ma(1+PY )m-") = o((6ma(1+pY)m-n)A) = o(6ma") < pa, a contradiction. For the latter, ar(1+PY )m+2" = (6™a(1+PY )m -")A = [(6™a")a(1+PY )m-2"]A = 6ma"(6ma-(1+PY )m+")(1+PY )m-2n. Thus, 0 = m + m[(1 + pY)m — 2n] (mod pß), and hence p | (1 — n), but it is also impossible because otherwise pa = o(aA) = o(6ma-(1+pY)m+n) < pa. This completes the proof of Claim 1. By Claim 1, aA = a^ G V-16ma" for some p G T{. Since pa = o(aA) = o(a^), we have p G Z*a. By Equations (4.5) and (4.9), UA = janM | n G T} = {an | n G Tp} C V-1bman. Then UA = V-1bman = {an | n € T'}, and so Tm = T' in Zpa. By Equation (4.9), Tm = T[-(1+ pY)m + n(1 - e)]. Sincepf M, we have T = T1[-(1 + pY)m + n(1 - e)]M-1. By Lemma 4.5, m = -(1 + PY)m + n(1 - e). Since SA = S-1bman = U-1bman U V-1bman, we have VA = U-1bman. In particular, (bman)A = bmav for some v € T1. For n € T', since (bman )A = [(bman)an-n]A = bmav = , we have that {bman | n € T1} = U-1bman = VA = {(bman)A | n € T'} = {6maw-Mn+v | n € T'} = {bman | n € T'm - Mn + v}. By Equations (4.7) and (4.8), T[-(1 + pY)m] + n = T1 = T'm - Mn + v = T[(1 + pY)m + n(e - 1)]m + Mn - Mn + v in Zpa. Thus, T[(1 + pY)m - n(1 - e)]2(1 + pY)-m - (v - n)(1 + pY)-m = T. By Lemma 4.5, there exists i € Zfc such that e£ = [(1 + pY)m - n(1 - e)]2(1 + pY)-m, that is, n satisfies Equation (4.2). Recall that a - 7 < ft and m € Z*^. Let m = m1 + 1pa-Y with m1 € Zpa-Y. Since (1 + pY) has order pa-Y in Zpa, we have (1 + pY )m = (1 + pY = (1 + pY )m>. Y.-Q. Feng and Y Wang: Bipartite edge-transitive bi-p-metacirculants 613 This implies that replacing m by mi, Equation (4.2) has the same solutions, and T' = {(e - 1)-1(ei - 1)(1 + )mi + e4n | i G Zk} Ç Zpa. The automorphism of G induced by a ^ a and b ^ bmim 1, maps U to U, and V = {bman | n G T'} to {bmian | n G T'}. Thus, we may assume that m G , and therefore, r = rm,M. □ Proof of Theorem 1.3. This is a consequence of Lemmas 4.7 and 4.8. □ References [1] B. Alspach, Point-symmetric graphs and digraphs of prime order and transitive permutation groups of prime degree, J. Comb. Theory Ser. B 15 (1973), 12-17, doi:10.1016/0095-8956(73) 90027-0. [2] B. Alspach and M. Y. Xu, 1/2-transitive graphs of order 3p, J. Algebraic Combin. 3 (1994), 347-355, doi:10.1023/a:1022466626755. [3] I. Antoncic and P. 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B 116 (2016), 504-532, doi:10.1016/j.jctb.2015.10.004. [56] J.-X. Zhou and M.-M. Zhang, The classification of half-arc-regular bi-circulants of valency 6, European J. Combin. 64 (2017), 45-56, doi:10.1016/j.ejc.2017.03.012. ARS MATHEMATICA CONTEMPORANEA ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 17 (2019) 617-626 https://doi.org/10.26493/1855-3974.1891.3b7 (Also available at http://amc-journal.eu) On flag-transitive automorphism groups of symmetric designs* Seyed Hassan AlaviAshraf Daneshkhah, Narges Okhovat Department of Mathematics, Faculty of Science, Bu-Ali Sina University, Hamedan, Iran Received 27 December 2018, accepted 4 November 2019, published online 12 December 2019 In this article, we study flag-transitive automorphism groups of non-trivial symmetric (v, k, A) designs, where A divides k and k > A2. We show that such an automorphism group is either point-primitive of affine or almost simple type, or point-imprimitive with parameters v = A2 (A + 2) and k = A(A +1), for some positive integer A. We also provide some examples in both possibilities. Keywords: Symmetric design, flag-transitive, point-primitive, point-imprimitive, automorphism group. Math. Subj. Class.: 05B05, 05B25, 20B25 1 Introduction A t-design D = (P, B) with parameters (v, k, A) is an incidence structure consisting of a set P of v points, and a set B of k-element subsets of P, called blocks, such that every t-element subset of points lies in exactly A blocks. The design D is non-trivial if t < k < v - t, and is symmetric if |B| = v. By [7, Theorem 1.1], if D is symmetric and non-trivial, then t < 2, see also [12, Theorem 1.27]. Thus we study non-trivial symmetric 2-designs with parameters (v, k, A) which we simply call non-trivial symmetric (v, k, A) designs. A flag of D is an incident pair (a, B), where a and B are a point and a block of D, respectively. An automorphism of a symmetric design D is a permutation of the points permuting the blocks and preserving the incidence relation. An automorphism group G of D is called flag-transitive if it is transitive on the set of flags of D. If G leaves invariant a non-trivial partition of P, then G is said to be point-imprimitive; otherwise G is called * The authors would like to thank anonymous referees for providing us helpful and constructive comments and suggestions. t Corresponding author. E-mail addresses: alavi.s.hassan@basu.ac.ir, alavi.s.hassan@gmail.com (Seyed Hassan Alavi), adanesh@basu.ac.ir, daneshkhah.ashraf@gmail.com (Ashraf Daneshkhah), okhovat.nargeshh@gmail.com (Narges Okhovat) Abstract ©® This work is licensed under https://creativecommons.org/licenses/by/4.0/ 618 Ars Math. Contemp. 17 (2019) 591-615 point-primitive. We here adopt the standard notation as in [8, 23] for finite simple groups of Lie type. For example, we use PSLn(q), PSpn(q), PSUn(q), PQ2n+1 (q) and PQ±„(q) to denote the finite classical simple groups. Symmetric and alternating groups on n letters are denoted by Sn and An, respectively. Further notation and definitions in both design theory and group theory are standard and can be found, for example in [10, 12, 14]. We also use the software GAP [21] for computational arguments. Flag-transitive incidence structures have been of most interest. In 1961, Higman and McLaughlin [11] proved that a flag-transitive automorphism group of a linear space must act primitively on its points set, and then Buekenhout, Delandtsheer and Doyen [5] studied this action in details and proved that a linear space admitting a flag-transitive automorphism group (which is in fact point-primitive) is either of affine, or almost simple type. Thereafter, a deep result [6], namely the classification of flag-transitive finite linear spaces relying on the Classification of Finite Simple Groups (CFSG) was announced. Although, flag-transitive symmetric designs are not necessarily point-primitive, Regueiro [18] proved that a flag-transitive and point-primitive automorphism group of such designs for A < 4 is of affine or almost simple type, and so using CFSG, she determined all flag-transitive and point-primitive biplanes (A = 2). In conclusion, she gave a classification of flag-transitive biplanes except for the 1-dimensional affine case [17]. Tian and Zhou [22] proved that a flag-transitive and point-primitive automorphism group of a symmetric design with A < 100 must be of affine or almost simple type. Generally, Zieschang [25] proved in 1988 that a flag-transitive automorphism group of a 2-design with gcd(r, A) = 1 is (point-primitive) of affine or almost simple type, and this result has been generalised by Zhuo and Zhan [24] for A > gcd(r, A)2. 1.1 Main result In this paper, we study flag-transitive automorphism groups of symmetric (v, k, A) designs, where A divides k and k > A2, and we show that such an automorphism group is not necessarily point-primitive: Theorem 1.1. Let D = (P, B) be a non-trivial symmetric (v, k, A) design with A > 1, and let G be a flag-transitive automorphism group of D. If A divides k and k > A2, then one of the following holds: (a) G is point-primitive of affine or almost simple type; (b) G is point-imprimitive and v = A2 (A + 2) and k = A(A + 1), for some positive integer A. In particular, if G has d classes of imprimitivity of size c, then there is a constant l such that, for each block B and each class A, the size \B n A| is either 0, or l, and (c, d, l) = (A2, A + 2, A) or (A + 2, A2,2). We highlight here that if A divides k, then gcd(k, A)2 = A2 > A which does not satisfy the conditions which have been studied in [24, 25]. Moreover, in Section 1.2, we provide some examples to show that both possibilities in Theorem 1.1 can actually occur. In order to prove Theorem 1.1(a), we apply O'Nan-Scott Theorem [15] and discuss possible types of primitive groups in Section 3. We further note that our proof for part (a) relies on CFSG. To prove part (b), we use an important result by Praeger and Zhou [20, Theorem 1.1] on characterisation of imprimitive flag-transitive symmetric designs. S. H. Alavi et al.: Onflag-transitive automorphism groups of symmetric designs 619 1.2 Examples and comments on Theorem 1.1 Here, we give some examples of symmetric (v, k, A) designs admitting flag-transitive automorphism groups, where A divides k and k > A2. In Table 1, we list some small examples of such designs with A < 3. To our knowledge the design in Line 2 is the only point-primitive example of symmetric designs with v < 2500 satisfying the conditions of Theorem 1.1 and this motivates the authors to investigate symmetric designs admitting symplectic automorphism groups [3]. More examples of symmetric designs admitting flag-transitive and point-imprimitive automorphism groups can be found in [20] and references therein. Line 1. Hussain [13] showed that there are exactly three symmetric (16,6,2) designs, and Regueiro proved that exactly two of such designs are flag-transitive and point-imprimitive [18, p. 139]. Line 2. The symmetric design in this line arises from the study of primitive permutation groups with small degrees. This design belongs to a class of symmetric designs with parameters (3m(3m +1)/2, 3m-1(3m -1)/2, 3m-1(3m-1 -1)/2), for some positive integer m > 1, see [4, 9]. If m = 2, then we obtain the symmetric (45,12, 3) design admitting PSp4(3) or PSp4(3) : 2 as flag-transitive automorphism group of rank 3, see [4]. Lines 3-4. Mathon and Spence [16] constructed 2616 pairwise non-isomorphic symmetric (45,12, 3) designs with non-trivial automorphism groups. Praeger [19] proved that there are exactly two flag-transitive symmetric (45,12, 3) designs, exactly one of which admits a point-imprimitive group, and this example satisfies Line 4, but not Line 3. Table 1: Some symmetric designs satisfying the conditions in Theorem 1.1. Line v k A c d l Case Examples Reference Comments 1 16 6 2 4 4 2 (b) 2 [13], [18] imprimitive 2 45 12 3 - - - (a) 1 [4] primitive 3 45 12 3 5 9 2 (b) None [19] imprimitive 4 45 12 3 9 5 3 (b) 1 [19] imprimitive 2 Preliminaries In this section, we state some useful facts in both design theory and group theory. Lemma 2.1 ([1, Lemma 2.1]). Let D be a symmetric (v, k, A) design, and let G be a flag-transitive automorphism group of D. If a is a point in P and H := Ga, then (a) k(k - 1) = A(v - 1); (b) k divides |H | and Av < k2. Lemma 2.2 ([2, Corollary 4.3]). Let T be a finite simple classical group of dimension n over a finite field Fq of size q. Then 620 Ars Math. Contemp. 17 (2019) 591-615 (a) If T = PSLn(q) with n > 2, then |T| > q"2-2; (b) If T = PSUn(q) with n > 3, then |T| > (1 - q-1)q"2-2; (c) If T = PSpn(q) with n > 4, then |T| > q1 "("+1)/(2a), where a = gcd(2, q - 1); (d) If T = PQ"(q) with n > 7, then |T| > q2"("-1)/(4,0), where £ = gcd(2,n). Lemma 2.3. Let T be a non-abelian finite simple group satisfying |T| < 8 Out(T)|3. (2.1) Then T is isomorphic to A5 or A6. Proof. If T is a sporadic simple group or an alternating group An with n > 7, then | Out(T)| G {1, 2}, and so by (2.1), we must have |T| < 64, which is a contradiction. Note that the alternating groups A5 and A6 satisfy (2.1) as claimed. Therefore, we only need to consider the case where T is a finite simple group of Lie type. In what follows, we discuss each case separately. Let T = PSLn(q) with q = p° and n > 2. If n = 2, then q > 4 and | Out(T)| = a • gcd(2, q - 1), and so by Lemma 2.2(a) and (2.1), we have that q2 < | PSL2(q)| < 8a3 • gcd(2, q - 1)3 ^ 64a3. Thus, q2 < 64a3. This inequality holds only for (p, a) G {(2,1), (2, 2), (2, 3), (2,4), (2, 5), (2, 6), (2, 7), (3,1), (3, 2), (3, 3), (5,1), (7,1)}. Note in this case that q > 4, and hence by (2.1), we conclude that T is either PSL2(4) = PSL2(5) = As, or PSL2(9) = ¿6, as claimed. If n = 3, then by Lemma 2.2(a), we have that q7 < 64a3 • gcd(3, q - 1)3 < 64a3q3, and so q4 < 64a3. If q would be odd, then we would have 34a < 64a3, which is impossible. If q = 2°, then 2° < 64a3 would hold only for a = 1, 2. Therefore, T is isomorphic to PSL3(2) or PSL3(4). These simple groups do not satisfy (2.1). If n > 4, then (2.1) implies that q11 < 64a3, but this inequality has no possible solution. Let T = PSUn(q) with q = p° and n > 3. By Lemma 2.2(b), we have that |T| > (1 -q-1)q"2-2, and so (2.1) implies that (1 - q-1)q"2-2 < 64a3 • gcd(n, q +1)3. If n = 3, then (1 - q-1)q7 < 64a3 • gcd(n, q + 1)3, and so q6 < 27 • 64a3. This inequality holds only for (p, a) G {(2,1), (2, 2), (3,1)}. Note that PSU3(2) is not simple. Therefore, T is isomorphic to PSU3(3) or PSU3(4). These simple groups do not satisfy (2.1). If n > 4, thensince (q +1)3 < 4 • q3(q -1), we would have q" -3 < 64a3 • gcd(n, q +1)3/(q - 1) < 4 • 64a3 (q +1)3/4(q - 1) < 4 • 64a3q3,andso q"2-6 < 4 • 64a3, and hence q10 < 4 • 64a3, which is impossible. Let T = PSp"(q) with q = p° and n > 4. By Lemma 2.2(c), we observe that |T| > q1 "("+1)/2 gcd(2, q-1) > q1 "("+1)/4. By (2.1), we have that q10 < q2"("+1) < 4-64a3, and so q10 < 4 • 64a3, which is impossible. Let T = P^"(q) with q = p° odd and n > 7. Then we conclude by Lemma 2.2(d) that |T| > q2"("-1)/8. Since | Out(T)| = 2a and n > 7, it follows from (2.1) that q21 < 83a3, which is impossible. Let T = P^"(q) with q = p° and n > 8 and e = ±. It follows from Lemma 2.2(d) that |T| > q2"(n-1)/8. Note that | Out(T)| < 6a • gcd(4,qn - e) < 24a. Then (2.1) implies that q28 < 82 • 243a3, which is impossible. S. H. Alavi et al.: Onflag-transitive automorphism groups of symmetric designs 621 Let T be one of the finite exceptional groups F4(q), Ee(q), Eg(q), ^(q) (q = 22m+1), 3D4(q) and 2Ee(q). Then |T| > q20, and so (2.1) implies that q20 < 8 • 23 • 33a3, which is impossible. If T = G2(q) with q = pa = 2. Then by (2.1), we have that q12 < q6(q2 - 1)(q6 - 1) < 8 • 23a3, and so q12 < 8 • 23a3, which is impossible. Similarly, if T is one of the groups 2B2 (q) with q = 22m+1 and 2G2(q) with q = 32m+1, then |T| > q4, and so (2.1) implies that q4 < 8a3, which is impossible. □ 3 Point-primitive designs In what follows, we assume that D = (P, B) is a non-trivial symmetric (v, k, A) design admitting a flag-transitive and point-primitive automorphism group G. Let also A divide k and k > A2 and set t := k/A. Notice that A < k, and so t > 2. We moreover observe by Lemma 2.1(a) that k = ; (3.1) A = ^. (3.2) Since also G is a primitive permutation group on P, by O'Nan-Scott Theorem [15], G is of one of the following types: (a) Affine; (b) Almost simple; (c) Simple diagonal; (d) Product; (e) Twisted wreath product. 3.1 Product and twisted wreath product type In this section, we assume that G is a primitive group of product type on P, that is to say, G < HI Si, where H is of almost simple or diagonal type on the set r of size m := | r | > 5 and I > 2. In this case, P = Lemma 3.1. Let G be a flag-transitive point-primitive automorphism group of product type. Then k divides A^(m — 1). Proof. See the proof of Lemma 4 in [18]. □ Proposition 3.2. If D = (P, B) is a non-trivial symmetric (v,k, A) design admitting a flag-transitive and point-primitive automorphism group G, where A divides k and k > A2, then G is not of product type. Proof. Assume the contrary. Suppose that G is of product type. Then v = m£. Note by Lemma 3.1 that k divides A^(m — 1), and so t = k/A divides ¿(m — 1). We also note by 622 Ars Math. Contemp. 17 (2019) 591-615 Lemma 2.1(b) that Av < k2. Then v < At2, and since A < t, we have that v < t3. Recall that t divides i(m - 1). Hence m£ < i3(m - 1)3. (3.3) Then m£ < i3m3, or equivalently, m£-3 < i3. Since m > 5, it follows that 5£-3 < i3, and this is true for 2 < i < 6. If i = 6, then since m6-3 < 63, we conclude that m = 5, but (m, i) = (5, 6) does not satisfy (3.3). Therefore, 2 < i < 5. Suppose first that i =5. Then by (3.3), we have that m5 < 53(m - 1)3, and so 5 < m < 9. It follows from (3.1) that t divides m5 — 1. For each 5 < m < 9, we can obtain divisors t of m5 - 1. Note by (3.2) that t2 must divide m5 -1 +1. This is true only for m = 7 when t = 2 or 6 for which (v, k, A) = (16807, 8404,4202) or (16807, 2802, 467), respectively. Since A2 < k, these parameters can be ruled out. Suppose that i = 4. Then by (3.3), we have that m5 < 43(m - 1)3, and so 5 < m < 9. By the same argument as in the case where i = 5, by (3.1) and (3.2), we obtain possible parameters (m, t, v, k, A) as in Table 2. Note by Lemma 3.1 that k must divide 4A(m - 1), and this is not true, for all parameters in Table 2. Table 2: Possible values for (m, t, v, k, A) when i = 4. m t v k A 13 51 28561 561 11 31 555 923521 1665 3 47 345 4879681 14145 41 57 416 10556001 25376 61 Suppose now that i = 3. We again apply Lemma 3.1 and conclude that t divides 3(m - 1). Then there exists a positive integer x such that 3(m - 1) = tx, and so m = (tx + 3)/3. By (3.2), we have that m2 +1 - 1 t2x3 + 9tx2 + 27x + 27 A — -- — -. t2 27t Then 27At = t2x3 + 9tx2 + 27x + 27. Therefore, t must divide 27x + 27, and so ty = 27x + 27, for some positive integer y. Thus, _ t(ty - 27)3 + 9 ■ 27(ty - 27)2 + 273y A = 27^ , (3.4) for some positive integers t and y. Since A2 < k, we have that A < t, and so t(ty - 27)3 + 9 ■ 27(ty - 27)2 + 273y < 274t. (3.5) If y > 32, then t(ty - 27)3 + 9 ■ 27(ty - 27)2 + 273y > t(32t - 27)3 + 9 ■ 27(32t - 27)2 + 32 ■ 273 > 274t, S. H. Alavi et al.: Onflag-transitive automorphism groups of symmetric designs 623 for t > 2. Thus 1 < y < 31, and so by (3.5), we conclude that 2 < t < 107. For each such y and t, by straightforward calculation, we observe that A as in (3.4) is not a positive integer. Suppose finally that I = 2. Recall by Lemma 3.1 that t divides 2(m - 1). Then 2(m — 1) = tx for some positive integer x, and so m = (tx + 2)/2. It follows from (3.2) that A = (tx2 + 4x + 4)/4t, or equivalently, 4tA = tx2 + 4x + 4. This shows that t divides 4x + 4, and so ty = 4x + 4, for some positive integer y. Therefore, 43 A = (ty — 4)2 + 16y. Since A2 < k, we have that A < t, and so (ty — 4)2 + 16y < 43t. If y > 6, then (6t — 4)2 + 6 • 16 < 43t, which has no possible solution for t. Thus 1 < y < 5. Since also (t — 4)2 + 16 < 43t, we conclude that 2 < t < 71, and so (3.1) and (3.2) imply that k = t(t2y2 — 8ty + 16y + 16) d A = (ty — 4)2 + 16y = 64 an = 64 ' where 2 < t < 71 and 1 < y < 5. For these values of t and y, considering the fact that m > 5, k > A2 and A divides k, we obtain (v, k, A) = (121,25,5) or (441,56, 7) respectively when (t, y) = (5,4) or (8, 3). These possibilities can be ruled out by [4] or [22, Theorem 1.1]. □ Proposition 3.3. If D = (P, B) is a non-trivial symmetric (v,k, A) design admitting a flag-transitive and point-primitive automorphism group G, where A divides k and k > A2, then G is not of twisted wreath product type. Proof. If G would be of twisted wreath product type, then by [15, Remark 2(ii)], it would be contained in the wreath product H I Sm with H = T x T of simple diagonal type, and so G would act on P by product action, and this contradicts Proposition 3.2. □ 3.2 Simple diagonal type In this section, we suppose that G is a primitive group of diagonal type. Let M = Soc(G) = Ti x • • • x Tm, where T = T is a non-abelian finite simple group, for i = 1,..., m. Then G may be viewed as a subgroup of M • (Out(T) x Sm). Here, Ga is isomorphic to a subgroup of Aut(T) x Sm and Ma = T is a diagonal subgroup of M, and so |P| = |T|m-i. Lemma 3.4. Let G be a flag-transitive point-primitive automorphism group of simple diagonal type with socle Tm. Then k divides Am1h, where m1 < m and h divides |T|. Proof. See the proof of Proposition 3.1 in [22]. □ Proposition 3.5. If D = (P, B) is a non-trivial symmetric (v,k, A) design admitting a flag-transitive and point-primitive automorphism group G, where A divides k and k > A2, then G is not ofsimple diagonal type. Proof. Suppose by contradiction that G is a primitive group of simple diagonal type. Then v = |T|m-1, and so by Lemma 2.1(b), Av < k2. This implies that A|T|m-1 < k2 = A2t2. Since A2 < k, we must have A < t, and hence |T |m-1 60, we must have m < 6. If m = 5, then |T| < 53, and it follows that T = A5. Note that k divides A(v - 1) = A(|T|m-1 - 1). Then t divides |T|m-1 -1 = 604 - 1 = 13 • 59 • 61 • 277. Since t < m|T| = 300 and t > 2, it follows that t G {13, 59, 61,277}. For each such t, we have that A < t and k = tA, and so we easily observe that these parameters does not satisfy Lemma 2.1(a). Therefore m g {2, 3,4}. Note that Ga is isomorphic to a subgroup of Aut(T) x Sm. Then by Lemma 2.1(b), the parameter k divides |Ga|, and so k divides (m!) • |T| • | Out(T)|. On the other hand, Lemma 2.1(a) implies that k divides A(|T|m-1 - 1), and so t divides |T|m-1 - 1 implying that gcd(t, |T|) = 1. Since k divides (m!) • |T| • | Out(T)| and t is a divisor of k, we conclude that t divides (m!) • | Out(T)|. Recall by (3.6) that |T|m-1 < t3. Therefore, |T|m-1 < (m!)3 Out(T)|3, (3.7) where m G {2, 3,4}. If m = 2, then |T| < 8 • | Out(T)|3. If m = 3, then |T|2 < 63| Out(T)|3, and so |T| < 621 Out(T)|. If m = 4, then |T|3 < 243| Out(T)|3, and |T| < 241 Out(T)|. Thus for m < 4, we always have |T| < 8 • | Out(T)|3, where T is a non-abelian finite simple group. We now apply Lemma 2.3 and conclude that T is isomorphic to A5 or A6. If m = 2, then since t divides |T|m-1 - 1 = |T| - 1, we have that t divides 59 or 359 when T is isomorphic to A5 or A6, respectively. Thus (v, k, A) = (60,59A, A) or (v, k, A) = (360,359A,A). Since A > 1, in each case, we conclude that k > v, which is a contradiction. For m = 3,4, since | Out(A5)| = 2 and | Out(A6)| = 4, it follows from (3.7) that |T| < 48 or |T| < 96 when T is isomorphic to A5 or A6, respectively, which is a contradiction. □ 4 Proof of the main result In this section, we prove Theorem 1.1. Suppose that D = (P, B) is a non-trivial symmetric (v, k, A) design with A divides k and k > A2. Suppose also that G is a flag-transitive automorphism group of D. Proof of Theorem 1.1. If G is point-primitive, then by O'Nan-Scott Theorem [15] and Propositions 3.2, 3.3 and 3.5, we conclude that G is of affine or almost simple type. Suppose now that G is point-imprimitive. Then G leaves invariant a non-trivial partition C of P with d classes of size c. By [20, Theorem 1.1], there is a constant l such that, for each B g B and A g C, |B n A| g {0, l} and one of the following holds: (a) k < A(A - 3)/2; (b) (v, k, A) = (A2(A + 2), A(A +1),A) with (c,d,l) = (A2, A + 2, A) or (A + 2, A2, 2); (c) '(A + 2)(A2 - 2A + 2) A2 a + 2 A2 - 2A + 2 (v, k, A, c, d, l) = -, —, A,-,-, 2 k ,,,,, j y 4 '22 2 and either A = 0 (mod 4), or A = 2u2, where u is odd, u > 3, and 2(u2 — 1) is a square; S. H. Alavi et al.: Onflag-transitive automorphism groups of symmetric designs 625 / , ^ , / (A + 6)(A2 +4 A - 1) A(A + 5) _ A2 +4 A - 1 (v,k,A,c,d,l) = ( + /-), ( ■+ ) ,A,A + 6,- + (d) 4 where A = 1 or 3 (mod 6). We easily observe that the cases (a) and (c) can be ruled out as k > A2. If case (d) occurs, then A(A + 5)/2 = k > A2 implying that A < 5. Since A = 1 or 3 (mod 6), it follows that A = 3 for which (v, k, A, c, d, l) = (45,12, 3, 9,5,3) which satisfies the condition in Theorem 1.1(b). Therefore, the case (b) can occur as claimed. □ References [1] S. H. 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Zhan, Flag-transitive automorphism groups of 2-designs with A > (r, A)2 and an application to symmetric designs, Ars Math. Contemp. 14 (2018), 187-195, doi:10.26493/ 1855-3974.1165.105. [25] P.-H. Zieschang, Flag transitive automorphism groups of 2-designs with (r, A) = 1, J. Algebra 118 (1988), 369-375, doi:10.1016/0021-8693(88)90027-0. ARS MATHEMATICA CONTEMPORANEA ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 17 (2019) 627-636 https://doi.org/10.26493/1855-3974.1921.d6f (Also available at http://amc-journal.eu) The symmetric genus spectrum of abelian groups Received 25 January 2019, accepted 6 November 2019, published online 13 December 2019 Let S denote the set of positive integers that appear as the symmetric genus of a finite abelian group and let S0 denote the set of positive integers that appear as the strong symmetric genus of a finite abelian group. The main theorem of this paper is that S = S0. As a result, we obtain a set of necessary and sufficient conditions for an integer g to belong to S. This also shows that S has an asymptotic density and that it is approximately 0.3284. Keywords: Symmetric genus, strong symmetric genus, Riemann surface, abelian groups, genus spectrum, density. Math. Subj. Class.: 57M60, 30F35, 20F38 1 Introduction Let G be a finite group. Among the various genus parameters associated with G, one of the most important is the symmetric genus 1. Our approach utilizes the strong constraints on the Sylow 2-subgroup of a group (not necessarily abelian) acting on a surface of even genus or a surface of genus congruent to 3 (mod 4); here see [7, Theorem 8] and [8, Theorem 5]. In the exceptional case (in which the Sylow 2-subgroup has a special form), we show that there exists an abelian group Ai such that a(A) = a0(Ax). To establish the reverse containment S0 c S, we utilize the characterization of the integers in the spectrum S0 in [3, Theorem 1]. For each integer g satisfying one of the five conditions in that result, we exhibit an abelian group G such that g = a(G). 2 Background results Let A be a non-trivial finite abelian group of rank r. Then A has the standard canonical representation A — Zmi x Zm2 x • • • x Zmr, (2.1) with invariants m1; m2,..., mr subject to m1 > 1 and m; | mi+1 for 1 < i < r. The abelian group A also has another canonical form that is useful in calculating genus parameters of abelian groups. Define the alternate canonical form for A as the direct product of three subgroups T, B and D of A. First, the group D is the subgroup of A C. L. May and J. Zimmerman: The symmetric genus spectrum ofabelian groups 629 generated by the factors Zms, where ms is divisible by 4. Then write A = D x D. Now let T be the Sylow 2-subgroup of D, which is elementary abelian, and B be its direct summand of odd order. Therefore, A = T x B x D. Define t = rank(T), b = rank(B), d = rank(D). It follows that r = rank(A) = d+max(b, t). We point out that this notation differs from that used in [5]. The groups of symmetric genus zero are the classical, well-known groups that act on the Riemann sphere (possibly reversing orientation) [2, §6.3.2]. The abelian group A has symmetric genus zero if and only if A is Zn, Z2 x Z2n, or (Z2)3; see [2, §6.3.2]. The groups of symmetric genus one have also been classified, at least in a sense. These groups act on the torus and fall into 17 classes, corresponding to quotients of the 17 Euclidean space groups [2, §6.3.3]. Each class is characterized by a presentation, typically a partial one. The abelian group A has symmetric genus one if and only if A is Zm x Zmn with m > 3, Z2 x Z2 x Z2n with n > 2, or (Z2)4; see [2, §6.3.3]. Let A be a finite abelian group. The strong symmetric genus of A has been completely determined by Maclachlan [4, Theorem 4], and if A has odd order, then a(A) = a0 (A). The focus of [5] was the determination of the symmetric genus of an abelian group of even order. The approach was to show that, among the various genus actions of A, there is one induced by an NEC group with a signature of one of three types. We established the following result [5, Theorem 3.10]. Theorem A. Let A be an abelian group of even order. Among the NEC groups with minimal non-euclidean area that act on A, there is a group r whose signature has one of the following forms: (I) (g, +, [Ai,...,An], {}); (II) (0, +, [Ai,...,As], {()k }) for some k > 1; (III) (0, +, [], {( )u, (2v)}) for some v > 2. Furthermore, in cases (I) and (II), Aj divides Ai+1 for 1 < i < r — 1. In (III) the notation (2v) means, as usual, a period cycle with v link periods equal to 2. We denote by t(A) (here and in [5]) the minimum genus of any action of A induced by an NEC group of Type II. The size of the largest elementary abelian 2-group factor of A determines whether a (A) is given by an action induced by a group of Type I, II or III. The main result of [5] is the following [5, Theorem 5.7]. Theorem B. Let A be an abelian group of even order with canonical form A = (Z2)a x Zmi x Zm2 x • • • x Zmq , where m1 > 2. If the symmetric genus a(A) > 2, then (i) a(A) = 1 + |A| • (a + 3q — 4)/8, if a > q + 2; (ii) a(A) = t(A), if 1 < a < q + 1; (iii) a(A) = min{a0(A), t (A)}, if a = 0. Thus Theorem B gives the symmetric genus of an abelian group A in terms of the invariants of A and the numbers a0 (A) and t(A). The main result in [3] is the characterization [3, Theorem 1] of the integers in the spectrum S0, and this will be important here. 630 Ars Math. Contemp. 17 (2019) 591-615 Theorem C. Let g > 2. Then g € S0 if and only if g satisfies one of the following conditions: (i) g = 1 (mod 4) or g = 55 (mod 81); (ii) g — 1 is divisible by p4 for some odd prime p; (iii) g — 1 is divisible by a2 for some odd integer a with (a — 1) | g; (iv) g — 1 is divisible by b2a2(a — 1) for some odd integers a,b> 1,with a = 3 (mod 4). 3 So CS To establish the containment S0 C S, we use the characterization of the integers in the spectrum S0 in Theorem C. For each integer g satisfying one of the five conditions in that result, we exhibit an abelian group G such that g = a(G). This is quite easy, as we shall see. In this section, we will assume that A is always written in alternate canonical form, A = T x B x D. We begin by noting some consequences of Theorem B. Directly from part (i) we have the following; this formula was also pointed out in [7, p. 4094]. Proposition 3.1. a(Zf x Z2m) = 1 + 4m for any integer m > 2. Since a((Z2)4) = 1 and a(Z)5) = 5 [7] (the general genus formula is a((Z2)n) = 1 + 2n-3(n — 4) [5, Corollary 5.4]), it follows that the spectrum S contains the entire congruence class g = 1 (mod 4). These odd integers are also in S0 [3, p. 342]. A special case of Theorem B [5, p. 423] will be important here. Theorem 3.2. Let the abelian group A have alternate canonical form A = T x B x D. If T is trivial, then a(A) = a0(A). Let A be a finite abelian group. Then a(A) = a0(A) in another important case. Lemma 3.3. Let A be an abelian group of rank three or more. If the Sylow 2-subgroup S2 of A is cyclic, then a(A) = a0(A). Proof. By Theorem 3.2, we may assume that T is non-trivial. If S2 is cyclic, then we must have S2 = Z2 and D = 1. Now write A = Z2 x B = Z2 x x Z^2 x • • • x Zpb for b > 3, where each ^ is odd. In this case, t = 1, d = 0 and b > 3. Since t < d + 1, [5, p. 416] gives . =1+2|a| (-1+1 (i - s)) (see also (4.1) in Section 4). The group A has canonical form A = Zp1 x Zp2 x • • • x Z2^b, and is the image of a Fuchsian group r with signature (0, +, ..., 2ftb], { }). When we calculate the genus arising from the action of this Fuchsian group on A, we get that it is equal to t(A). Now applying Maclachlan's formula shows that ct°(A) < t(A) and hence 5 and set G = Zp x Zp x Zp x Zmp. Then g = 1, with a = 3 (mod 4). Then g = a°(G), where G is a group of the form Za x Zab x Zabn [3, p. 343], a group with a cyclic Sylow 2-subgroup, so that 1 (the integer k is the number of empty period cycles). Among the signatures of the NEC groups that induce the Type II genus t(A), Lemmas 4.2 and 4.3 of [5] identify one value of k that can be used to calculate t(A). These two lemmas are correct. Unfortunately, there is a mistake in [5, Formula (4.5)], which is used in the final determination of t(A). We correct that here. Let A = T xB xD be the alternate canonical form for A. Remember that t = rank(T), b = rank(B), d = rank(D) and so r = rank(A) = d + max(b, t). The odd order group B is generated by elements with orders ..., pb, where $ divides Pj for i < j. In the case in which t < d + 1, the formula for t(A) [5, p. 416] is correct. Note that k = t gives minimal area by [5, Lemma 4.2]. In the new notation, this formula is T(A) = 1 + 5|A|(k - 2 +1 0 - s) + H' (' - J"))' <41) where the group D is generated by elements with orders J1,..., Sd satisfying S" divides Sj for i < j . Next, we consider the case t > d + 1. Theorem 4.1. Let A be an abelian group in alternate canonical form. Suppose that t > d +1 and k is given by [5, Lemma 4.3]. There are two cases and in each case, define v = b + d - k +1. (i) Suppose that t + d is odd. Then k = (t + d + 1)/2; (a) If b < (k - 1) - d, then t(A) = 1 + 2 |A|(k - 2); 632 Ars Math. Contemp. 17 (2019) 591-615 (b) If b > (k - 1) - d, then v > 1 and - (k - 1) - d, then v > 1 and t (A) = 1 + 2'A|(k - 2+g(1 - S) +(1 - i)). Proof. Among the NEC groups that induce the Type II genus t(A), let r be one with signature (0; +; [Ai,...; As]; ( )k) in which the number k of empty period cycles is given by k = [(t + d + 1)/2] [5, Lemma 4.3, p. 414]. It follows that k > d + 1. The group r has generators xi,..., xs, ei,..., ek, and involutions ci,..., ck. The defining relations for r consist of xi • • • xsei • • • ek = 1, conditions on the order of the elements xj and certain elements commuting. Clearly, one generator is redundant, and, since p(r) is minimal, we may assume that ek is that generator. Now let ai,..., ar be a generating set for A in canonical form so that the orders of these elements satisfy the standard divisibility condition. With p(r) minimal, the elements ei,..., ek-i are mapped onto the subgroup generated by the k -1 elements ar_k+2,..., ar of highest order. In particular, since k -1 > d, the subgroup D of A is contained in the image of (ei,..., ek-i). If t + d is odd, then 2k - 1 = t + d. Since t + d is the rank of the Sylow 2-subgroup S2 of A, we have that S2 is contained in the image of (ci,..., ck, ei,..., ek-i). It follows that (T, D) is contained in the image of (ci,..., ck, ei,..., ek-i). If t + d is even, then 2k = t + d. In this case, there is an additional generator x^ so that (T, D) is contained in the image of (x^, ci,..., ck, ei,..., ek-i). If b < (k - 1) - d, then the images of the generators e» which are not mapped into D generate all of B. Therefore, if t + d is odd, then A is the image of the NEC group with signature (0; +; [ ]; {( )k}) and t(A) = 1 + 2|A|(k - 2). If t + d is even, then we need a generator xi in order to map onto A, and with p(r) minimal, |xi| = 2. Therefore, if t + d is even, then A is the image of the NEC group with signature (0; +; [2]; {( )k}) and t(A) = 1 + i|A|(k - 2). The last case is when b > (k - 1) - d. Let E be the subgroup of A generated by the images of ei,..., ek-i. Then the subgroup B can be decomposed as B = Bi x B2, where B2 = B n E. Let v = b + d - k + 1 so that v is the rank of Bi and v > 1. We need generators xi,..., xv to map onto Bi. If t + d is odd, then A is the image of the NEC group with signature (0; +; [Si,..., ]; {()k}) and t(a)=1+2|a| (k - 2+g(1 - Si ^. Suppose that t + d is even. As in the previous case, we need generators xi,..., xv to map onto Bi. However, the generator xv must map onto an element of order for the NEC group to map onto A. This is because there is the extra involution not in the image C. L. May and J. Zimmerman: The symmetric genus spectrum ofabelian groups 633 of (ci,..., ck). If t + d is even, then A is the image of the NEC group with signature (0;+;k8i,...,&-i, ]; {()k }) and T(A) = 1 + 5|A|(k - 2 + £ (' - s) + (' - n 5 General results Again in this section, we assume that A is written in alternate canonical form, A = T x B x D. Let A be a finite abelian group so that the integer g = a(A) is in the spectrum S. We want to show that g is in S0 as well. This is clearly the case if A has rank one or two or A has a trivial factor T in its alternate canonical form. Thus we may assume that A has rank at least 3 and T is not trivial. In particular, A has even order. Our approach utilizes the Sylow 2-subgroup S2 of A. Lemma 5.1. Let A be an abelian group of rank three or more with a(A) > 2. If the Sylow 2-subgroup of A is isomorphic to (Z2)3, then t (A) = 1 (mod 4). Proof. Since S2 = (Z2)3, we have T = S2, D = 1 and A = T x B. Now t = 3, d = 0, and k = 2. Since b = 1 would imply a (A) = 1, b > 2. We have t > d +1 with t + d odd. The Type II genus is t(A) = 1 + |A| • M/2, where M = (k - 2 + EV=1(1 - )) by Theorem 4.1(i)(b). Since M • |B| is an integer and |A| = 8|B|, we clearly have t(A) = 1 (mod 4). □ Lemma 5.2. Let A be an abelian group of rank three or more with a(A) > 2. If the Sylow 2-subgroup of A is isomorphic to Z2 x Z2 x Z2e for some £ > 2, then t (A) = 1 (mod 4). Proof. In this case we have T = Z2 x Z2 and D is cyclic with order divisible by 2£. Therefore, t = 2 and d =1. This implies that k = t = 2 by [5, Lemma 4.2]. Since b =1 would imply a(A) = 1, b > 2. We have t = d +1 and by (4.1), the Type II genus t(A) = 1 + |A| • M/2 where M = (k - 2 + EbU(1 - ^)). Since M • |B| is an integer and |A| =4 • |D| • |B| with |D| divisible by 2£, again t(A) = 1 (mod 4). □ Lemma 5.3. Let A be an abelian group. If the Sylow 2-subgroup of A is isomorphic to (Z2)4, then t(A) = 1 (mod 4). Proof. Since S2 = (Z2)4, we have T = S2, D = 1 and A = T x B. Now t = 4, d = 0, k = 2 and b > 1. We have t > d +1 with t + d even. We apply Theorem 4.1(ii) and get that k = 2. If b =1, then M = 1/2 by Theorem 4.1(ii)(a). If b > 2, then by Theorem 4.1(ii)(b), M • 2|B| is an integer. In either case, 2M • |B| is an integer. Since |A| = 16|B|, we again have t (A) = 1 (mod 4). □ Theorem 5.4. Let A be an abelian group of rank three or more with a(A) > 2. Then either a(A) = 1 (mod 4) or a(A) = a0 (A) (or both), unless the Sylow 2-subgroup of A has rank 2 and is isomorphic to Z2 x Z2c ,for some £ > 1. Proof. First, if A has odd order, then a (A) = a0 (A). Assume, then, that A has even order so that a (A) is given by Theorem B. Let A have canonical form A = (Z2)0 x Zmi x Zm2 x • • • x Zmq , 634 Ars Math. Contemp. 17 (2019) 591-615 as in Theorem B. It is easy to see in case (i), we always have that a(A) = 1 (mod 4). Suppose a < q + 1. By Theorem B, a(A) is either equal to a0(A) or t(A). Let A act on a surface X of genus g = a(A) > 2, and write |A| = 2n • m, where m is odd. Assume first that g is even. Then by [7, Theorem 9], A is not a 2-group so that m = 1. We consider the possibilities for the Sylow 2-subgroup S2 of A. If S2 is cyclic, then by a (A) = a0 (A) by Lemma 3.3. Assume then that S2 is not cyclic. If A contains an element of order 2n-1 with |S21 = 2n, then S2 is isomorphic to Z2 x Z2n-i, the exceptional case. Assume then that A has no elements of order 2n-1, and apply [7, Theorem 8]. Since A and S2 are abelian, the only possibility is that S2 is isomorphic to (Z2)3. But in this case t(A) = 1 (mod 4) by Lemma 5.1. Therefore, by Theorem B, if g is even, then either a(A) = 1 (mod 4) or a(A) = a0(A), unless S2 = Z2 x Z2c, for some i > 1. Now suppose g = 3 (mod 4), and use [8, Theorem 5]. The Sylow 2-subgroup S2 also acts of the surface X of genus g > 2. By [8, Theorem 5], S2 contains an element of order 2n-3 or larger. Further, if Exp(S2) = 2n-3, then S2 contains a dihedral subgroup of index 4. We consider the possibilities for Exp(S2). If S2 is cyclic, then by a(A) = a0(A) by Lemma 3.3. If S2 is not cyclic and contains an element of order 2n-1, then S2 is isomorphic to Z2 x Z2n-i, the exceptional case. Suppose Exp(S2) = 2n-2. Then S2 is isomorphic to either Z2 x Z2 x Z2n-2 or Z4 x Z2n-2. If S2 = Z2 x Z2 x Z2n-2, then by Lemma 5.2, t(A) = 1 (mod 4). If on the other hand, S2 = Z4 x ^2^-2, then by Theorem 3.2, a(A) = a0(A). Suppose Exp(S2) = 2n-3 and S2 has a dihedral subgroup of index 4. Since S2 is abelian, this forces n = 4 and S2 = (Z2)4. In this case, t(A) = 1 (mod 4) by Lemma 5.3. Therefore, by Theorem B, if g = 3 (mod 4), then either a(A) = t(A) = 1 (mod 4) or a(A) = a0(A), unless S2 = Z2 x Z2t, for some i > 1. □ A consequence of the proof is perhaps worth noting, in connection with the well-known conjecture that "almost all" groups are 2-groups. Theorem 5.5. If A be an abelian 2-group of positive symmetric genus, then a(A) = 1 (mod 4). Proof. Assume A is an abelian 2-group with a(A) > 2. Then a(A) is not even by [7, Theorem 9]. The proof of Theorem 5.4 shows that a (A) cannot be congruent to 3 (mod 4) either, since A = S2 and A is not an abelian group of genus zero or one. □ Next we handle the exceptional case in Theorem 5.4. Theorem 5.6. Let A be an abelian group of rank three or more with a(A) > 2. If the Sylow 2-subgroup of A is isomorphic to Z2 x Z2c, for some i > 1, then there exists an abelian group A1 such that a(A) = a0(A1). Proof. Let A have alternate canonical form A = T x B x D. We consider two cases, i =1 and i > 2. C. L. May and J. Zimmerman: The symmetric genus spectrum ofabelian groups 635 First assume that S2 = (Z2)2. Now T = S2, D = 1 and A = T x B, with t = 2, d = 0, and k =1. Write A = Z2 x Z2 x B = Z2 x Z2 x Z^ x Z^2 x • • • x Z^h for b > 3, where each i is odd. By Theorem 4.1(ii)(b), the Type II genus is T (A) = 1 + 2lAl(-1 + g (1 - i ) + (1 - £ )). By Theorem B, a(A) = min{a0(A),r(A)}. If a(A) = ct°(A), then set Ai = A and we are done. So we assume that t(A) < 1 to obtain 1 as in the calculation of a°(A). By assumption the Fuchsian groups rp for p > 1 give genus larger than t(A). The Fuchsian group r° gives the same genus as t(A). Therefore, t(A) = ct°(A1) and so 2. Now T = Z2 and D is isomorphic to Zm2£, where m is odd. We have alternate canonical form A = Z2 x B x Zm2c = Z2 x Z^1 x Z^2 x • • • x Z^t-1 x Zm2£, where each i is odd, i divides Pj for i < j and m is divisible by ib_1. It follows that t = 1, b > 3, d =1, and k =1 by [5, Lemma 4.2]. By (4.1) the Type II genus is ' 0. The following theorem is simple, but critical in the next two sections. Theorem 4.4. A ribbon graph G is extremal if and only if 7 (Gx) = 0, i.e. Gx is plane. If G, P and Q are ribbon graphs, then we say that G is the join of P and Q, written G = P V Q, if G can be obtained by identifying an arc on the boundary of a vertex of P with an arc on the boundary of a vertex of Q as indicated in Figure 11. The two arcs that are identified do not intersect any common line segments. (a) P (b) Q (c) P V Q Figure 11: The join P V Q of P and Q. Lemma 4.5. Let G = B1 V B2 V B3 V • • • V Bk. Then k MG) = ^ MB*) - k + 1. i=1 Q Q Q. Yan and X. Jin: Extremal embedded graphs 645 Proof. It suffices to show that Lemma 4.5 holds for k = 2. Let v be the new vertex of G formed by merging a vertex of Bi and a vertex of B2. Since v is a cut vertex, the arcs c\ and c2 must belong to a single component, and ci and c'2 must belong to different components, that is, splitting B1 V B2 at v into B1 and B2 increases the number of components by one, as in Figure 12. Thus ^(G) = ^(Bi) + ^(B2) - 1. □ c2 (a) Bi V B2 (b) Bi (c) B2 Figure 12: A component of Bi V B2 is split into a component of Bi and a component of B2. Lemma 4.6. Let G be a ribbon graph and e be a bridge of G. Then ^(G) = ^(G/e). Proof. This follows immediately from Figure 13. Since e is a bridge, the arcs ci and c2 must belong to a single component, and so do the arcs ci and c'2. We have ^(G) = ^(G/e). □ (a) G (b) G/e Figure 13: The medial graphs of G and G/e. Lemma 4.7. Let G be a ribbon graph and ei and e2 be two distinct edges of G (see Figure 4.7). 1. If the 2-cycle given by {ei, e2 } is orientable and the common line segments of ei and e2 are adjacent as in Case 1, then ^(G) = ^(G — ei — e2). 2. If the 2-cycle given by {ei, e2} is non-orientable and the common line segments of ei and e2 are adjacent as in Case 2, then ^(G) = ^(G/ei/e2). 3. If ei and e2 are not parallel edges, but are incident with a common vertex of degree 2 as in Case 3, then ^(G) = ^(G/ei/e2). 646 Ars Math. Contemp. 17(2019) 637-652 (a) Case 1 (b) Case 2 ei e2 (c) Case 3 Figure 14: The three cases of Lemma 4.7. Proof. We illustrate the proof with the following figures. Case 1: Gm G (G - ei - e2)m Case 2: Gm G (G/e i/e2 )m Case 3: Gm ei e2 G ^_^_^ (G/ei/e2)m Theorem 4.8. Let G be a ribbon graph. (1) If G is extremal and e is not a bridge of G, then G — e is extremal. □ Q. Yan and X. Jin: Extremal embedded graphs 647 (2) If G = B1 V B2 V B3 V ■ ■ ■ V Bk, then G is extremal if and only if each Bi is extremal. (3) If e is a bridge of G, then G/e is extremal if and only if G is extremal. (4) Let v be a vertex of degree 2 with exactly one adjacent vertex, the two edges joining these vertices are as in Case 1 of Lemma 4.7. Then G — v is extremal if and only if G is extremal. (5) Let v be a vertex of degree 2 with two different adjacent vertices x and y. Then G/{v, x}/{v, y} is extremal if and only if G is extremal. Proof. (1): Since G is extremal, ^(G) = f (G) + y(G). By Lemma 4.1, we have MG — e) > f (G) + y(G) — 1. Moreover, ^(G — e) < f (G — e) + y(G — e) by Theorem 4.2 and f (G — e) + y (G — e) = f (G) + y(G) — 1. Therefore ^(G — e) = f (G — e) + y(G — e). Hence G — e is extremal. (2): Suppose that G is extremal. Then from Lemma 4.5 we have k m(G) = Y MBi) — k +1 i=1 = f (G)+ y(G) kk = Y f (Bi) — k + 1 + Y Y (Bi) i=1 i=1 k Therefore = Y f (Bi) + Y (Bi) — k +1. Y MBi) = Y f (Bi) + Y (Bi), and so each Bi is extremal. The converse is proved similarly. (3): This follows from Lemma 4.6: MG/e)= m(G), f (G/e) = f (G) and Y(G/e) = y(G). (4): This follows from Case 1 of Lemma 4.7: m(G — v) = m(G) — 1, f (G — v) = f (G) — 1 and y (G — v) = y(G). (5): This follows from Case 3 of Lemma 4.7: M(G/{v,x}/{v,y}) = M(G), f (G/{v,x}/{v,y}) = f (G) and Y (G/{v,x}/{v,y}) = y(G). □ 648 ArsMath. Contemp. 17(2019)591-615 5 Excluded extremal minor characterization Definition 5.1. Let G = (V, E) be a ribbon graph and v G V and e G E. A deletion G - e or G - v of G is admissible if e is not a bridge of G or v is an isolated vertex of G; a contraction G/e or G/v is admissible if e is a bridge of G or v is a vertex of degree 2 with two different adjacent vertices u,w and G/v = G/{v,u}/{v, w}. Definition 5.2. Let G be a ribbon graph. We say that a ribbon graph H is an extremal minor of G, denoted H -< G, if there is a sequence of ribbon graphs G = G0, Gi,..., Gt = H where for each i, Gi+i is obtained from Gi by either an admissible deletion or an admissible contraction. Lemma 5.3. Let G be an extremal ribbon graph and H -< G. Then H is extremal. Proof. It suffices to prove that admissible deletions and admissible contractions preserve extremity. Cases of G - e, G/e and G/v follow from Statements (1), (3), and (5) of Theorem 4.8, respectively. Let v be an isolated vertex of G. Then ^(G - v) = ^(G) - 1, f (G - v) = f (G) - 1 and 7(G - v) = 7(G). Thus if G is extremal, G - v is also extremal. □ Lemma 5.4. Hx -< Gx if and only if H -< G. Proof. We only need to prove the sufficiency since G = (Gx)x. Then by Definition 5.2, it suffices to prove that Lemma 5.4 holds for H = G - e and H = G - v (admissible deletions), and H = G/e and H = G/v (admissible contractions). For clarity, we denote by ex (respectively, vx ) the edge (respectively, the vertex) of Gx corresponding to the edge e (respectively, the vertex v) of G. Then e is a bridge if and only if ex is a bridge of Gx , v is an isolated vertex if and only if vx is an isolated vertex of Gx , and v is a vertex of degree 2 with two different adjacent vertices of G if and only if vx is a vertex of degree 2 with two different adjacent vertices of Gx. Hence H -< G implies Hx -< Gx. □ Theorem 5.5. Let G be a ribbon graph. Then G is extremal if and only if it contains no extremal minor equivalent to Bi, B^, /3, % T or T2 (see Figure 15). Proof. It is not difficult to obtain that 7(5^) = 7(/|) = 1 and 7(B£) = y(4x ) = Y(Tix) = 7(T2x) = 2. By Theorem 4.4, Bi, B2, /3, /2, Ti and T2 are not extremal. It follows from Lemma 5.3 that G contains no extremal minor equivalent to Bi, B2, /3, %, T or T2. To prove the converse we suppose that G is not extremal. Without loss of generality, we assume that G is connected. Then, by Theorem 4.4, y(Gx ) > 0. Case 1: If Gx is non-orientable, then Gx contains a non-orientable cycle C. If C is odd, then C in G is orientable. The edges of G not on C can be deleted or contracted admissibly, and thus C -< G. Note that Bi -< C and hence Bi -< G, a contradiction. If C is even, then C in G is non-orientable. It follows that C -< G, /2 -< C and thus /2 -< G, a contradiction. Case 2: If Gx is orientable, then g(Gx) > 1. Step 1: If f (Gx) > 1, we can take an edge e of Gx whose two edge line segments belong to different boundary components of Gx. Such an edge e must be not a bridge, we can Q. Yan and X. Jin: Extremal embedded graphs 649 (a) Bi (b) B2 (c) h (d) In (e) Ti (f) Tn Figure 15: The "forbidden" minors in Theorem 5.5. delete it in Gx. Deleting such an e will decrease boundary component number by 1 and preserve orientability and the genus. Repeat Step 1 we obtain an orientable ribbon graph (Gx)' with f ((Gx)') = 1 and g((Gx)') = g(Gx). Step 2: If g((Gx)') > 1, (Gx)' must have non-bridges. Take a non-bridge e of (Gx)', deleting e will preserve orientability and connectedness, and must increase boundary components by 1 since f ((Gx)') = 1 and hence decrease the genus by 1. Now go to Step 1. Then carry out Step 2 if the genus is greater than 1. Repeat above process, finally we obtain an orientable ribbon graph (Gx)'' with f ((Gx)'') = 1 and g((Gx)'') = 1. Note that (Gx)'' ^ Gx and (Gx)'' has the cyclomatic number 2. After contracting all bridges and vertices of degree 2 with two distinct adjacent vertices in (Gx)'', we obtain four possible (Gx)''' as shown in Figure 16. Note that ax, bx, cx and dx in Figure 16 are B^, T1, Tn and I3, respectively. By Lemma 5.4, we have B^ ^ G, I3 ^ G, T1 ^ G or T2 ^ G, a contradiction. □ 650 Ars Math. Contemp. 17 (2019) 591-615 As a corollary, we obtain an excluded extremal minor characterization of extremal plane graphs. We need the following lemma and leave its proof to readers. Lemma 5.6. If H ^ G, then 7 (H ) < 7 (G). Corollary 5.7. A plane graph G is extremal if and only if it contains no extremal minor equivalent to Bi or I3. Proof. By Lemma 5.6, an extremal plane graph can not contain B^, I^, Ti and T2 as extremal minors. It then follows from Theorem 5.5. □ 6 Two conjectures and their generalizations We first prove Conjecture 1.2 holds for any extremal cellularly embedded graphs on orientable surfaces. Theorem 6.1. If G is an orientable extremal ribbon graph, then G is bipartite. Proof. By Theorem 4.4, 7(Gx ) = 0 which implies that Gx is a plane graph and thus orientable. If G is not bipartite, then it contains an odd cycle C. C, as a subgraph of the orientable ribbon graph G, is also orientable which will become a non-orientable cycle of Gx .It follows that Gx is non-orientable, a contradiction. □ This theorem is not true for non-orientable extremal ribbon graphs. For example, the non-orientable loop, as in Figure 3, is extremal, but not bipartite. By Theorem 6.1, any orientable extremal ribbon graph with non-zero Euler genus (including the Gx in Example 3.3) is a counterexample of the second claim of Proposition 3.2. To show that Conjecture 1.1 holds for any extremal cellularly embedded graphs we need the following lemma. Lemma 6.2. Let G be a ribbon graph. If Gx is orientable, then Gm has an all-crossing direction such that each of the edges of G is a d-edge. Proof. Since Gx is orientable, it follows that Gx can be drawn on the plane such that each of its edge discs is untwisted. We orient each edge disc anticlockwise as illustrated in Figure 17(a). Note that straight-ahead walks of Gm correspond exactly to boundary components of Gx. Then Gm is oriented in Figure 17(b) as boundary components of Gx. This is an all-crossing direction such that each of the edges of G is a d-edge. □ (a) Orient each edge disc anticlockwise. (b) Get an all-crossing direction of G, and e is a d-edge. m Figure 17: The all-crossing direction of G m- Q. Yan and X. Jin: Extremal embedded graphs 651 Theorem 6.3. If G is an extremal ribbon graph, then each face of G is even. Proof. Since G is an extremal ribbon graph, it follows that 7(Gx) = 0 by Theorem 4.4, and thus Gx is orientable. By Lemma 6.2, Gm has an all-crossing direction such that each of the edges of G is a d-edge. Since each edge is d-edge, we call one of edge line segments in-edge line segment if two marking arrows are all "in" arising from the all-crossing direction of Gm. Otherwise, we call the other out-edge line segment, as in Figure 18(a). 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ARS MATHEMATICA CONTEMPORANEA 17 (2019) 653-698 https://doi.org/10.26493/1855-3974.1813.7ae (Also available at http://amc-journal.eu) Reconfiguring vertex colourings of 2-trees Michael Cavers, Karen Seyffarth Department of Mathematics and Statistics, University of Calgary, Calgary, AB T2N1N4 Canada Received 4 October 2018, accepted 8 October 2019, published online 16 December 2019 Let H be a graph and let k > x(H) be an integer. The k-colouring graph of H, denoted Gk (H), is the graph whose vertex set consists of all proper k-vertex-colourings (or simply k-colourings) of H using colours {1, 2,..., k}; two vertices of Gk (H) are adjacent if and only if the corresponding k-colourings differ in colour on exactly one vertex of H. If Gk (H) has a Hamilton cycle, then H is said to have a Gray code of k-colourings, and the Gray code number of H is the least integer k0(H) such that Gk (H) has a Gray code of k-colourings for all k > k0(H). Choo and MacGillivray determine the Gray code numbers of trees. We extend this result to 2-trees. A 2-tree is constructed recursively by starting with a complete graph on three vertices and connecting each new vertex to an existing clique on two vertices. We prove that if H is a 2-tree, then k0(H) = 4 unless H is isomorphic to the join of a tree T and a vertex u, where T is a star on at least three vertices, or the bipartition of T has two even parts; in these cases, k0(H) = 5. Keywords: 2-trees, graph colouring, Gray codes, Hamilton cycles, reconfiguration problems. Math. Subj. Class.: 05C15, 05C45 1 Introduction Let H be a graph and k a positive integer. We define a proper k-vertex-colouring of H as a function f: V(H) ^ {1,2,..., k} for which f (x) = f (y) for any xy e E(H). Since we are concerned only with proper k-vertex-colourings, we use the simpler term k-colouring, and refer to f (x) as the colour of x. For notation and terminology not defined here, the reader is referred to Bondy and Murty [1]. A graph H has a Gray code of k-colourings if it is possible to list all the k-colourings of H in such a way that consecutive colourings in the list (including the last and the first) differ on exactly one vertex of H, and the Gray code number of H is the least integer E-mail addresses: mcavers@ucalgary.ca (Michael Cavers), kseyffar@ucalgary.ca (Karen Seyffarth) Abstract ©® This work is licensed under https://creativecommons.org/licenses/by/4.0/ 654 Ars Math. Contemp. 17 (2019) 591-615 k0(H) for which H has a Gray code of its k-colourings for all k > k0(H). Equivalently, we may define the k-colouring graph of H, denoted Gk (H), to be the graph whose vertices correspond to all k-colourings of H, and whose edges connect two k-colourings of H that differ on exactly one vertex of H. In this context, H has a Gray code of k-colourings if and only if Gk (H) has a Hamilton cycle, and the Gray code number of H is the the least integer k0(H) for which Gk(H) has a Hamilton cycle for all k > k0(H). The k-colouring graph is an example of a reconfiguration graph. Such graphs are often used in the study of what are known as reconfiguration problems. Generally, a reconfiguration problem asks: given two (different) feasible solutions to a problem, can one solution be transformed to the other through a sequence of allowable moves, while maintaining feasibility at each stage? In the context of k-colourings, the k-colouring graph is connected if and only if any k-colouring can be reconfigured into any other k-colouring by re-colouring one vertex at a time in such a way that each intermediate colouring is a k-colouring. Recently, reconfiguration problems have been receiving wide attention, and have been studied for various graph colouring problems [2, 3, 4, 10, 15], for dominating sets [12, 13, 18], and for various other graph problems including vertex covers, cliques, and independent sets [14]. The k-colouring graph arises in the context of theoretical physics, where it is the graph of the Glauber dynamics Markov chain; the goal is to find efficient algorithms for almost uniform sampling of k-colourings of graphs [16]. The Glauber dynamics Markov chain converges to the uniform distribution provided that the k-colouring graph is connected, prompting Cereceda, van den Heuvel and Johnson [4] to ask the question: given a graph H and a positive integer k, is Gk (H) connected? They prove that if H has chromatic number k G {2, 3}, then Gk (H) is never connected, whereas for k > 4, there are k-chromatic graphs H for which Gk (H) is connected, and other k-chromatic graphs H for which Gk (H) is not connected. In general, they prove that Gk(H) is connected for all k > col(H) + 1, where col(H), the colouring number of H, is defined as col(H) := max{^(G) | G C H} + 1. A slightly weaker version of this result is proven in [8]. Choo and MacGillivray [6] initiated the study of Hamilton cycles in k-colouring graphs by proving that the Gray code number of H is well defined, i.e., Gk (H) has a Hamilton cycle for all k > col(H) + 2. This gives the upper bound k0 (H) < col(H) + 2. Note that if T is a tree, then col(T) = 2 and G2(T) is disconnected; hence 3 < ko(T) < 4. Choo and MacGillivray [6] determine which trees have k0 (T) = 3 and which have k0(T) = 4. In particular, they prove that if T is a tree, then k0(T) = 3 unless T = K1j2£, for I > 1, in which case k0(T) = 4. They also determine the Gray code numbers of complete graphs and cycles. Celaya, Choo, MacGillivray and Seyffarth [3] establish the Gray code numbers of complete bipartite graphs. Haas [11] studies a variation of k-colouring graphs, namely, canonical colouring graphs, and uses techniques developed in [6] to show that canonical colouring graphs of trees have Hamilton cycles. Given the results for trees and complete graphs, a natural question is to determine the Gray code numbers of k-trees. We use the definition of k-tree given in [9], that is, a k-tree is constructed recursively by starting with a complete graph on k + 1 vertices and connecting each new vertex to an existing clique on k vertices (hence a 1-tree is simply a connected acyclic graph). A vertex of degree k in a k-tree is called a leaf. Let H be a k-tree. Then it is clear that the chromatic number of H is x(H) = k + 1, and that Gk+1(H) is disconnected. Since every induced subgraph of H has a leaf, col(H) = k + 1. Thus it follows from [6, Theorem 3.4] that k0(H) < k + 3, and therefore k + 2 < k0(H) < k + 3. The problem M. Cavers and K. Seyffarth: Reconfiguring vertex colourings of 2-trees 655 is therefore reduced to classifying k-trees into those with Gray code number k + 2 and those with Gray code number k + 3. The answer appears to be far from trivial. In the current paper, we provide a complete solution for 2-trees; the characterization can be stated in fairly non-technical language, but the proof involves numerous cases and generalizations of the techniques used in [3, 6]. The join of graphs G1 and G2, denoted by G1 V G2, is obtained from the disjoint union of Gi and G2 by adding all edges between vertices of Gi and vertices of G2. Theorem 1.1. If H is a 2-tree then k0(H) = 4, unless H = T V {u} for some tree T and vertex u, where T is a star on at least three vertices or the bipartition of V (T) has two even parts; in these cases, k0(H) = 5. The remainder of the paper is devoted to proving this theorem. We first characterize 2-trees of diameter two (Lemma 3.2). We then determine the 2-trees, H, of diameter two for which G4(H) has a Hamilton cycle (Lemmas 3.3 and 3.5). This is done by considering the structure of G3(T), where T is a tree (Lemmas A.2 and 3.6). We show that if H is a 2-tree with diameter at least three, then G4 (H) has a Hamilton cycle (Lemmas 6.3 and 6.4). To do so we describe a specific recursive procedure for constructing 2-trees of diameter at least three (Theorem 4.3). 2 Preliminaries Definition 2.1. Let H be a graph, and let X and Y be disjoint subsets of V(H). We denote by [X, Y] the set of edges of H that have one end in X and the other end in Y. Definition 2.2. Let H be a graph and L a function that assigns to each vertex v e V(H) a set of positive integers, L(v), called the list of v. An L-colouring of H (also called a list colouring of H with respect to L) is a (proper) colouring c: V(H) ^ N such that c(v) e L(v) for each v e V(H). We define the L-colouring graph of H, denoted GL(H), to be the graph whose vertex set consists of all L-colourings of H; two L-colourings are joined by an edge of GL(H) if they differ in colour on just one vertex of H. Remark 2.3. If k > x(H) and L(v) := {1,2,..., k} for each v e V(H), then GL(H) = Gk (H), the k-colouring graph of H. We use G □ H to denote the Cartesian product of graphs G and H, and Qn to denote the n-dimensional hypercube, defined as the Cartesian product of n copies of K2. Remark 2.4. Let H1 and H2 be vertex disjoint graphs and let L be an assignment of lists to the vertices of H1 U H2. Then GL(Hi U H2) = GL(Hi) □ Gl(H2). Lemma 2.5. Let H be a 2-tree with clique X = {x1; x2,..., xi} where t < 3, and let L be an assignment of lists to the vertices of H so that 1. |L(xi)| = 1 and L(x,) C {1, 2, 3, 4}, 1 < i < t; 2. L(xj) = L(xj) for all 1 < i = j < t; 3. L(v) = {1,2,3,4} for each v e V(H) \ X. 656 Ars Math. Contemp. 17 (2019) 591-615 Then GL(H) has a spanning tree T with A(T) < 4. Proof. The proof is by induction on |V(H)|. For the basis, H = K3 with V(H) := jvi, v2,v3}. When I = 0, GL(H) = G4(K3), which has a Hamilton cycle [6, Theorem 4.1], so take T to be a Hamilton path in G4(K3). When I = 1, we may assume, without loss of generality, that L(v1) := {1} and L(v2) = L(v3) := {1,2,3,4}. Then Gl(H) = G3(K2), which has a Hamilton cycle [6, Theorem 4.1], so take T to be a Hamilton path in G3(K2). When I = 2, we may may assume, without loss of generality, that L(vi) := {1}, L(v2) := {2}, and L(vj) := {1, 2,3,4}. Then GL(H) = K2, and the result holds. Finally, when I = 3, we may assume, without loss of generality, that L(v1) := {1}, L(v2) := {2}, and L(v3) := {3}. Then GL(H) = K1, and the result holds. Now suppose H is a 2-tree with |V(H)| > 3 and clique X = {x1;..., x.«}. Since I < 3, there is a leaf v e V(H) with v e X, and thus H - v is a 2-tree containing the clique X. It follows by induction that GL(H - v) has a spanning tree T with A(T) < 4. Let V(Gl(H - v)) := {/q, /1,..., fN-1}, and for each / e V(GL(H - v)), let Fj C V(Gl(H)) be the set of L-colourings of H that agree with f on V(H - v). Then {Fq, F1,..., Fn-1} is a partition of the vertices of GL(H). Since v is a leaf of H, there are two ways to extend an L-colouring of H - v to an L-colouring of H, and hence for each j, 0 < j < N - 1, Fj = {aj, bj} is a clique. For each edge // e E(T), there is a unique vertex w e V(H - v) for which fj (w) = fj(w). If wv e E(H), then [Fj, Fj] consists of two disjoint edges, and the subgraph of Gl(H) induced by Fj U Fj is a 4-cycle. Otherwise, wv e E(H), so [Fj, Fj] has only one edge, and the subgraph of GL(H) induced by Fj U Fj is a path of length three. In both cases, label the edge fjfj in T with | [Fj, Fj] |. Let S denote the spanning subgraph of GL(H) corresponding to the spanning tree T of Gl(H - v) as described above, that is, S has edge set Since [Fj, Fj] is nonempty for each fjfj e E(T), S is connected. Also, since A(T) < 4 and [Fj, Fj] contains either one edge or two disjoint edges, A(S) < 5. Furthermore, since there are only two vertices adjacent to v in H, at most two edges incident to fj in GL(H-v) have label '1'. Let S' be the graph obtained from S by deleting the edges aj bj for each fj e V (T) with dT(fj) = 4. Then A(S') < 4, since if aj e V(S) has ds(aj) = 5, then dT(fj) = 4, and thus dS/ (aj) =4. We also claim that S' is connected. To prove this, it suffices to show that there is a path in S' from ap to bp for each fp e V(T) with dT(fp) = 4. Suppose fp e V(T) has dT(fp) = 4. Construct a path, P, in T starting at fp, using edges labelled '2', and whose final vertex fq has dT(fq) < 4. Such a path exists since T has no cycles, and each degree four vertex in T is incident to at least two edges labelled '2'. The union (J [Fj, Fj]) U {ajbj | 0 < j < N - 1}. fifj £E(T) / gives us a path in S' from ap to b. M. Cavers and K. Seyffarth: Reconfiguring vertex colourings of 2-trees 657 Therefore, S' is a connected spanning subgraph of GL(H), and thus S' has a spanning tree T that is also a spanning tree of GL(H). Since A(S') < 4, A(T) < 4, thus completing the proof of the lemma. □ The result in Lemma 2.5 is best possible in that A(T) cannot be reduced from four to three, as illustrated in the following example. Let D denote the unique 2-tree of diameter three on six vertices, with vertices labelled as shown in Figure 1(a). u 2 U1 u4 u 3 U 6 U 5 (a) D 3412 3413 3423 3421 3123 3124 3143 3142 (b) GL(D) 4312 4314 4324 4321 4123 4124 4134 4132 Figure 1: The 2-tree D and GL(D). Let L: V(D) ^ {1, 2,3,4} be defined as follows. L(U1) := {1}, L(u2) := {2}, and L(ui) := {1, 2, 3, 4} for 3 < i < 6. If f is an L-colouring of D, then f (u1) = 1 and f (u2) = 2, and thus we may denote the vertices of GL(D) by strings ijki where f (u3) = i, f (u4) = j, f (u5) = k and f (u6) = I. Using this convention, GL(D) is depicted in Figure 1(b). Notice that GL(D) is unicyclic, and has exactly two nonadjacent vertices of degree four, '3123' and '4124'. Thus every spanning tree of GL(D) has a vertex of degree four. As part of their proof [6, Theorem 5.5], Choo and MacGillivray prove the following. Remark 2.6. Let G be a graph with vertex partition {F0, F1,..., FN_1}, such that (i) G[Fi] is a 4-cycle for each i, 0 < i < N - 1; (ii) G[Fi_1 U Fi] is isomorphic to either P4 □ K2 or Q3 for each i, 1 < i < N — 1; (iii) if G[Fi_1 U Fi] and G[Fi U Fm] are both isomorphic to P4 □ K2, then G[Fi_1 U Fi U Fi+1] is not isomorphic to the graph in Figure 2. Then G has a Hamilton cycle. The conditions imply that [Fi_1, Fi] = 0,1 < i < N — 1, and hence there is a function, h, from a spanning subgraph of G to a path f0 f1 ■ ■ ■ fN_1 of length N — 1 defined by h(u) = fi for all u G Fi, 0 < i < N — 1. In our next lemma, we adapt the result of Choo and MacGillivray to a more general scenario. Suppose G is a graph with vertex partition {F0, F1,..., FN_1} where G[Fi] 658 Ars Math. Contemp. 17 (2019) 591-615 Figure 2: Forbidden subgraph. contains a spanning cycle for each i, 0 < i < N — 1 (these cycles form a 2-factor of G). Further, assume that there is a function, h, from a spanning subgraph of G to a tree with vertex set {f0, f1,..., fn-i} such that h(u) = fi for all u G Fi, 0 < i < N — 1. The general idea is to choose, for each edge fifj of the spanning tree, appropriate edges from the set [Fi; Fj ] of G so that we are able to construct a Hamilton cycle from among these edges and edges of the 2-factor. See Figure 3 for an illustration of this result. Lemma 2.7. Let G be a graph with vertex partition {F0, F1y..., FN_i}, and let T be a tree with V (T) := {f0, fi,..., fN _i}. Suppose there is a function, h, from a spanning subgraph of G to T such that h(u) = fi for all u G Fi, 0 < i < N — 1. Furthermore, suppose that for each fifj G E (T), 0 < i,j < N — 1, there exist edges ei,j in G[Fi] and ej,i in G[Fj] such that (i) if j = k and fifj,fifk G E(T), then ei,j = ei,k; (ii) if ei,j = ac and ej,i = bd, then G[{a, b, c, d}] contains a 4-cycle; (iii) G[Fi] has a Hamilton cycle Ci such that Proof. The proof is by induction on N. The result is trivial when N =1. Let N > 1. For each fi fj G E(T), 0 < i < N — 1, suppose eij G E (G[Fi]), eji G E(G[Fj]), and Ci (a Hamilton cycle in G[Fi]) satisfy conditions (i), (ii) and (iii). Without loss of generality, we may assume that fN_1 is a leaf of T, and that fN_1fN_2 G E(T). Let G' := G — FN_1 and T' := T — fN_1. Using eitj, ej^ and Ci as previously defined, 0 < i < N — 2, and Mi as defined in (iii) except with MN_2 replaced by M'N_2 := Mn_2 \ {eN_2n_1}, we apply the induction hypothesis to G'. The result is a Hamilton cycle C' in G' such that and eN_2,N_1 G E(C'). Let eN_2,N_1 := ac, eN_1,N_2 := bd; without loss of generality, abdca is a 4-cycle in G[{a, b, c, d}], and hence Mi := {ei,j | fifj G E(T)} C E(Ci). Then G has a Hamilton cycle C such that N_1 lJ(E(Ci) \ Mi) C E(C). i=0 C := (C' — eN_2,n_1) U (Cn_1 — eN_1,n_2) + {ab, cd} is a Hamilton cycle in G with the required property. □ M. Cavers and K. Seyffarth: Reconfiguring vertex colourings of 2-trees 659 We remark that if dT(fi) = 1 for some i, then the Hamilton cycle C constructed in Lemma 2.7 contains all except one edge of E(Ci). 3 2-trees of diameter two In this section we characterize 2-trees H of diameter two in which G4(H) has a Hamilton cycle. We begin by defining a class of 2-trees that we denote by T(p, q, r) (see Figure 4). P Q Figure4: The 2-tree T(p, q, r). Definition 3.1. Let P, Q, and R be pairwise disjoint independent sets of vertices with |P| := p, |Q| := q, and |R| := r. The graph T(p, q, r) is the graph onp + q + r + 3 vertices defined on vertex set {x, y, z} U P U Q U R where the subgraph induced by P U Q U R is an independent set, and • the subgraph induced by {x, y} U P is isomorphic to K1j1jP; • the subgraph induced by {y, z} U Q is isomorphic to K1j1jq; • the subgraph induced by {z, x} U R is isomorphic to K1,1,r. A dominating vertex in a graph is a vertex adjacent to all other vertices of the graph. 660 Ars Math. Contemp. 17 (2019) 591-615 Lemma 3.2. A graph H is a 2-tree of diameter two if and only if H has a dominating vertex or H = T(p, q, r) for p, q, r > 0. Proof. Any 2-tree with a dominating vertex has diameter two, and one can easily verify that T(p, q, r) is a 2-tree of diameter two for any p, q, r > 0. For the converse, suppose H is a 2-tree of diameter two. We proceed by induction on n := |V(H)|. When n = 4, H is isomorphic to the graph obtained from K4 by deleting one edge, and has a dominating vertex. Now suppose that n > 5, and let u G V(H) be a leaf of H, i.e., (u) = 2. By the induction hypothesis, H - u has a dominating vertex, or H — u = T(p', q',r') for some p',q',r' > 0. If H — u = T(p', q',r') for some p', q', r' > 0, then let V(H — u) := {x, y, z} U P' U Q' U R', where |P'| := p', |Q'| := q', and |R'| := r'. Since H has diameter two, u must be adjacent to at least two vertices from {x, y, z}. However (u) = 2, and thus NH(u) = {x, y}, NH(u) = {y, z}, or NH(u) = {z,x}. It follows that H = T(p' + 1,q',r'), H = T(p',q' + 1,r'), or H = T (p', q', r' + 1), respectively. Now suppose H — u has a dominating vertex, x. If ux G E(H), then x is a dominating vertex in H. Otherwise, let y and z denote the neighbours of u in H, and note that yz G E(H). Since H has diameter two, every vertex in V(H — u) is adjacent to x and at least one of y or z. Let P' be the set of vertices in H — u adjacent to both x and y, R' be the set of vertices in H — u adjacent to both x and z, and suppose |P'| := p' and |R'| := r'. Since H — u is a 2-tree, P' U R' is an independent set. Therefore, P' U R' U {u} is an independent set in H, and thus H = T (p', 1, r'). □ In what follows, we first prove that if p, q, r > 0, then G4(T(p, q, r)) has a Hamilton cycle. Let G be a graph with vertex partition {F0, Fi, • • •, FN-i}. For each i, 1 < i < N — 1, let Si -i C Fi-i and S' C Fi denote the vertices incident to the edges of [Fi-i, Fi]. Lemma 3.3. If p, q, r > 0, then G4(T(p, q, r)) has a Hamilton cycle. Proof. Let V(K3) := {x, y, z}. Suppose f: V(K3) ^ {1, 2, 3,4} is a 4-colouring of K3 and V(G4(K3)) := {f0, fi, • • •, fN-i}. Since G4(K3) has a Hamilton cycle by [6, Theorem 4.1], we may assume that f0fi • • • fN-i is a Hamilton path in G4(K3). For 0 < i < N — 1, let F' be the set of 4-colourings of T(p, q, r) that agree with fi on {x, y, z}. In order to simplify notation, we define G := G4(T(p, q, r)). Then {F0, F, • • •, Fn-i} is a partition of the vertices of G, and G[F'] = Qp+q+r, 0 < i < N — 1. Let st G [Fi-i,Fi], where s G Si-i and t G S' for some 1 < i < N — 1. Then s(u) = t(u) for all u G V(G) \ {v}, where v is one of {x, y, z}, and s(v) = t(v). Thus, [Fi-i, F'] is a set of independent edges. If v = x, then s(u) = s(w) for all u, w G P and for all u, w G R, and t(u) = t(w) for all u, w G P and for all u, w G R. Thus G[Si-i] = Qq ^ G[S'], and G[Si-i U S'] = Qq+i. Analogously, if v = y, G[Si-i] = Qr = G[S'] and G[Si-i U S'] = Qr+i; and if v = z, G[S'-i] = Qp = G[S'], G[S'-i U S'] = Qp+i. Consider the path fi-ififi+i in G4(K3), 1 < i < N — 2. Note that fi-i(x),fi(x),fi+i(x) use at most two different colours; otherwise, there would be only one colour available for y and z, which is impossible since y and z always receive different colours. Analogously, fi-i, f', fi+i assign at most two different colours to each of y and z. It follows that if |[Fi-i,Fi]| = |[Fi,Fi+i]| = 2, then [F'-i,Fi] U [Fi,F'+i] = 2P3. M. Cavers and K. Seyffarth: Reconfiguring vertex colourings of 2-trees 661 For each edge /¿-if with 1 < i < N - 1, we choose a 4-cycle, Bj_i, containing exactly two edges of [Fi-1, Fj] as follows. • For each |[Fj_i,Fj]| = 2, G[Sj_i U Sj] isa 2-cube, so we choose Bj_i := G[Sj_i U Sj] = ai_ici_idi6iai_i, where aj_i, cj_i e Fj_i and bj, dj e Fj. • For i = 1,2,..., N - 1, and |[Fj_i, Fj]| > 2, first note that |[Fj_i,Fj]| > 4 since G[Sj_i U Sj] = Qn for some n > 3. Thus it is possible to choose edges a^bj and cj_idj of [Fj_i, Fj] so that Bj_i := a^c^d.jbjaj_i is edge disjoint from Bj_2, and also edge disjoint from B, if | [Fj, Fj+1] | = 2 and i < N - 1. Let eM+i := G[S^ n B, and ej+1jj := G[Sj+i] n B^ 0 < i < N - 2. Observe that G and the path /0/i • • • /N_i satisfy conditions (i) and (ii) of Lemma 2.7. Recall that G[Fj] = Qp+q+r, 0 < i < N -1, and p + q + r > 3. Since any pair of edges of Qn, n > 2, is contained in a Hamilton cycle (see [7, Theorem 4.1]), there exist Hamilton cycles C0 in G[F0] containing e0ji, CN_i in G[FN_i] containing eN_1jN_2, and, for 1 < i < N - 2, Gj in G[Fj] containing e^^ and ejjj+1. Therefore, by Lemma 2.7, G has a Hamilton cycle. □ In the case where p > 0 and q = r = 0, T(p, q, r) has a dominating vertex, and is isomorphic to K2 V Kn. Lemma 3.4. For n > 2, G4(K2 V Kn) has no Hamilton cycle. Proof. Let H := K2 V Kn for n > 2, let H := G4(H), and let u, v be the two vertices of H of degree n + 1. For each 1 < i = j < 4 let Vjj := {c e V(H) | c(u) = i and c(v) = j}. Then {V12, V13, Vi4, V21, V23, V24, V31, V32, V34, V41, V42, ^43} is a partition of V(H). Note that [Va^, V7a] = 0 if and only if a = 7 or ft = For 1 < i = j < 4, let Lj be an assignment of lists to the vertices of H such that Lj (u) := {i}, Ljj(v) := {j} and Ljj(x) := {1, 2,3,4} for x e V(H - {u,v}). Note that GLj(H) ^ H[Vjj] = Qn, for each 1 < i = j < 4. Define the three-coloured vertices of H (that is, the colourings of H with three colours) by c^ e Vj such that i, if x = u, cjjfc(x) := ^ j, if x = v, k, otherwise. Each Vjj has two such vertices, cjjfcl and cjjfc2, where k1,k2 e {1, 2, 3,4}\{i,j}. Furthermore, H - {cjjkl, cjjfc2} is disconnected so that any Hamilton cycle in H must contain the edges of a Hamilton path of H[Vjj ] with endpoints cjj fcl and cjj fc2, for each 1 < i = j < 4. By [6, Lemma 2.1] there is no Hamilton path in the n-dimensional cube from 00 • • • 0 to 11 • • • 1 whenever n is even. Thus, for n even, there is no Hamilton path in H[Vj ] with endpoints cjj fcl and cjj fc2. Therefore, H has no Hamilton cycle when n is even. For n odd, such Hamilton paths exist and must be used in any Hamilton cycle of H, if one exists. We construct an auxiliary graph A (see Figure 5(a)) where the vertex (i, j) 662 Ars Math. Contemp. 17 (2019) 591-615 represents a Hamilton path in H[Vj] from cijkl to cijk2. There is an edge in A between (i1,j1) and (i2,j2) whenever cilj1 k is adjacent to ci2j2k in H. We label the edge between (ii,ji) and (i2,j2) by the unique element of {1, 2,3,4} \ ({ii,ji} U {i2, j'2}) (see Figure 5(b)). Notice that the edges labelled i, 1 < i < 4, induce a 6-cycle, and the edges of these four 6-cycles partition E(A). (1,4) (3,4) 3 4 3 .. 4 4 4 1 1 1 ' 2 1 2 (a) (b) Figure 5: The auxiliary graph A. 3 2 3 A Hamilton cycle in H corresponds to a Hamilton cycle, C, in A in which any two consecutive edges of C have different labels. Such a cycle C uses a matching of size three from each labelled 6-cycle in A. Without loss of generality, we may assume C contains horizontal edges of the 6-cycle labelled 1. Now, C uses horizontal edges from one of the remaining labelled 6-cycles, otherwise, C contains a K3. Regardless of whether C uses horizontal edges of the 6-cycle labelled by 2, 3 or 4, using vertical edges of the two remaining 6-cycles gives C = C4 U C8, a contradiction. Therefore H has no Hamilton cycle. □ Observe that if H is a 2-tree of diameter two having a dominating vertex u, then H = T V {u} for some tree T. Lemma 3.5. Let T be a tree on at least two vertices. Then G4(T V {u}) has a Hamilton cycle unless T is a star on at least three vertices or the bipartition of V (T) has two even parts. The proof of this lemma requires a result that we state here, but whose proof is technical and is postponed to the Appendix A. Lemma 3.6. Let T be a tree with bipartition (A, B), where |A| := £ and |B| := r, and let G3(T) be the 3-colouring graph of T with colours C = {1, 2,3}. Define cij to be the vertexof G3(T) with cij (a) = i for all a G A and cij (b) = j for all b G B. (1) If £,r > 0 are both even, then G3 (T) has no spanning subgraph consisting only of paths whose ends are in {c12, c13, c21,c23, c31,c32}. (2) If £ > 1 is odd and r > 0 is even, then G3(T) has a Hamilton path from c12 to c23. (3) If £ > 1 and r > 1 are both odd, then G3(T) has a Hamilton path from c12 to c13. M. Cavers and K. Seyffarth: Reconfiguring vertex colourings of 2-trees 663 Proof of Lemma 3.5. Let T be a tree with | V (T) | > 2 and bipartition (A, B), where A := {xL, x2,..., x^} and B := {y1? y2, • • •, yr}• Let H := T V {u}, and let H := G4(H) be the 4-colouring graph of H with colours {1, 2,3,4}. For each 1 < k < 4, define Vk := {c e V(H) | c(u) = k}. Then {Vl, V2, V3, V4} is a partition of V (H). Define Lk to be an assignment of lists with Lk(u) := {k} and Lk(w) := {1,2,3,4} for w e V(t). Note that GLfc (H) = H[Vk] = G3(T). Define the three-coloured vertices of H (that is, the colourings of H with three colours) by cijk e Vk so that cijk (x) := i for all x e A, cijk(y) := j for all y e B. Observe [Vi, V2] = {c34ic342, c43ic432}, [Vi, V3] = {c24ic243, c42ic423}, [Vi, V4] = {c23ic234, c32ic324}, [V2, V3] = {ci42ci43, c4i2c4i3}, [V2, V4] = {ci32ci34, c3i2c3i4}, [V3, V4] = {ci23ci24, c2i3c2i4}. Case 1. Suppose |V(T)| = 2. Then T V {u} = K3 and G^Ks) has a Hamilton cycle by [6, Theorem 4.1]. Case 2. Suppose T is a star with |V(T)| > 3. Then G4(T V {u}) = K V K„ for some n > 2. By Lemma 3.4, G4(T V {u}) has no Hamilton cycle. Case 3. Suppose both |A| and |B| are odd. By Lemma 3.6, there is a Hamilton path in H[Vi] from c42i to c43i, in H[V2] from c432 to ci32, in H[V3] from ci34 to ci24 and in H[V4] from ci23 to c423. The union of these paths with edges {c43ic432, ci32ci34, ci24ci23, c423 c42i} gives a Hamilton cycle in H. Case 4. Suppose one of | A| and |B| is even and the other is odd. Without loss of generality, |A| is odd and |B| is even. By Lemma 3.6, there is a Hamilton path in H[Vi] from c23i to c34i, in H[Vi] from c342 to c4i2, in H[V3] from c4i3 to ci23 and in H[V4] from ci24 to c234. The union of these paths with edges {c34ic342, c4i2c4i3, c^cm, c234c23i} gives a Hamilton cycle in H. Case 5. Suppose both |A| and |B| are even, and suppose H contains a Hamilton cycle C. Then C [Vi] is a spanning subgraph of H[Vi ] = G3 (T) consisting of a union of paths whose endpoints are three-coloured vertices of Vl, contradicting Lemma 3.6. Thus in this case, H has no Hamilton cycle. □ 4 Constructing 2-trees of diameter at least three To complete the proof of our main result, we must show that if H' is a 2-tree with diameter at least three, then k0(H') =4. Naively deleting two leaves with the intention of applying Remark 2.6 may be problematic. For example, let H' be a 2-tree with diameter at least three having leaves x and y of distance at least three. Let H = H' - {x, y}, NH> (x) = {xL, x2}, and suppose G4(H) has a Hamilton cycle /0/l • • • fN_l/0. Let Fi be the set of 4-colourings of H' that agree with / on V(H), 0 < i < N - 1. In the case that fi_1/i arises from a colour change on xL and /i/i+1 arises from a colour change on x2, the subgraph G[Fj_ l U Fj U Fi+1] is isomorphic to the forbidden subgraph in Figure 2. Because of this we take a more general approach. Suppose a 2-tree H' is obtained from a 2-tree H by repeatedly adding vertices of degree two. Instead of a Hamilton cycle in G4(H) we take 664 Ars Math. Contemp. 17 (2019) 591-615 a spanning tree satisfying certain properties providing the flexibility needed to construct a Hamilton cycle in G4(H'). Our approach does not require G4(H) to have a Hamilton cycle. To facilitate this procedure, we define nine operations (see Figure 6). Definition 4.1. Let H be a 2-tree. Then H' is obtained from H by Operation I if {«,3,7} := V(H') \ V(H), {0102, 6162, cic2} C E(H), and Nh'(a) = {01,02}, Nh>(3) = {61,62}, Nh>(7) = {01,02}. Operation II if {a, 3,7} := V(H') \ V(H), {xa, x6, c1c2} C E(H), and NH, (a) = {x, a}, Nh'(3) = {x, 6}, Nh'(7) = {01,02}. Operation III if {a, 3, 7} := V(H') \ V(H), {ax,xy,yc} C E(H), and NH(a) = {a, x}, Nh'(3) = {x,y}, Nh'(7) = {c,y}. Operation IV if {a, 3,7, 5} := V(H') \ V(H), {xy, zw} C E(H), and NH (a) = Nh' (3) = {x,y}, Nh' (7) = Nh' (5) = {w,z}. Operation V if {a, 3,7, 5} := V(H') \ V(H), {xy, zw} C E(H), and Nh' (a) = {3, x, y}, Nh' (3) = {a, y}, Nh' (7) = Nw (5) = {w,z}. Operation VI if {a, 3,7, 5} := V(H') \ V(H), {xy, zw} C E(H), and NH' (a) = {3, x, y}, Nh' (3) = {a, y}, Nh' (7) = {5, w, z}, Nh' (5) = {7, z}. Operation VII if {a, 3,7,5} := V(H') \ V(H), {xy,xz} C E(H), and NH (a) = {3, x, y}, Nh' (3) = {a, y}, Nh' (7) = {5, x, z}, Nw (5) = {7, z}. Operation VIII if {a, 3,7,5} := V(H') \ V(H), {xy,xz} C E(H), and NH (a) = {3, x, y}, Nh' (3) = {x, a}, Nw (7) = {5, x, z}, Nh' (5) = {7, z}. Operation IX if {a, 3,7,5} := V(H') \ V(H), {xy,xz} C E(H), and NH' (a) = {3, x, y}, Nh' (3) = {a, y}, Nh' (7) = Nh' (5) = {x,z}. Remark 4.2. Since each of Operations I through IX can be performed by sequentially adding simplicial vertices of degree two to H, H' is a 2-tree. Recall that D denotes the unique 2-tree of diameter three on six vertices (Figure 1(a)). Theorem 4.3. A graph H' is a 2-tree of diameter at least three if and only if H' = D or H' can be obtained from a 2-tree H by applying one of Operations I through IX. Proof. (^): If H' = D, the result is trivially true. Otherwise, it follows from Remark 4.2 that H' is a 2-tree. Since each operation produces two leaves that are distance at least three apart, H' has diameter at least three. (^): As already observed, D is the unique 2-tree of diameter three on six vertices. The 2-trees on three, four and five vertices all have diameter less than three. Thus, we may assume that H' is a 2-tree of diameter three and |V(H')| > 7. Since H' has diameter at least three, there are at least two leaves whose neighbourhoods are vertex disjoint. We consider cases based on the number of edges induced by the neighbourhoods of the leaves of H', and the number of leaves of H'. Case 1. First assume that the neighbourhoods of the leaves of H' induce at least three distinct edges. M. Cavers and K. Seyffarth: Reconfiguring vertex colourings of 2-trees 665 a • ^ a,2 bi - ?ß ai Ci N / C2 \ / •7 (a) Operation I ß «^a z .....» x Y • ^ 5 w (d) Operation IV Ci C2 ;• y ß b (b) Operation II ß ^ y z fx I I I a y • 5 xw (e) Operation V y a mC * a > Y C (c) Operation III ß f. i i i • ; a * x 5 A z m" 1 i Y w (f) Operation VI a ß a x 2 y a «- - / \ / \ ß* x •----¥y z¥-----* ß5 (g) Operation VII y z V----o ß y Y ' s '/ / (h) Operation VIII (i) Operation IX Figure 6: Operations I through IX applied to a graph H by the addition of vertices a, ft, 7, and S where applicable. The solid edges belong to H and the dotted edges are added to construct H'. 5 x Y x a Y a z If H' has three leaves whose neighbourhoods are pairwise vertex disjoint, then H' contains vertex disjoint edges aia2, 6162, cic2, and leaves a, ft, 7 with N(a) = {ai; a2}, N(ft) = {61,62}, N(7) = {ci, C2}. Letting H := H' - {a, ft, 7} results in a 2-tree, and applying Operation I to H produces H'. We may now assume that no three leaves of H' have neighbourhoods that are pairwise vertex disjoint. Choose two leaves a and 7 whose neighbourhoods are vertex disjoint, and let ft £ {a, 7} be a leaf such that N(ft) £ {N(a), N(7)}. If N(ft) intersects exactly one of N(a) and N(7), then we may assume without loss of generality that |N(ft) nN(a) | = 1 and N(ft) n N(7) = 0. Then H' contains a path of length two, ax6 and an edge c1c2 that is vertex disjoint from ax6, such that N(a) = {a, x}, N(ft) = {x, 6}, and N(7) = {c1, c2}. Letting H := H' - {a, ft, 7} results in a 2-tree, and applying Operation II to H produces H'. Otherwise, |N(ft) n N(a)| = |N(ft) n N(7)| = 1, and H' contains a path of length three, c1xyc2 such that N(a) = {c1,x}, N(ft) = {x, y} and N(7) = {y, c2}. Letting H := H' - {a, ft, 7} results in a 2-tree, and applying Operation III to H produces H'. Case 2. We may now assume that the neighbourhoods of the leaves of H' induce exactly two edges. 666 Ars Math. Contemp. 17 (2019) 591-615 Case 2(a). Suppose that H' has at least three leaves. Let leaves P and 7 have neighbourhoods that are vertex disjoint, and let S G IP, 7} be a leaf. Without loss of generality, N (S) = N (7). If there exists a leaf a with N(a) = N(P), then H := H' - {a, P, 7, S} is a 2-tree, and applying Operation IV to H produces H'. Otherwise, no other leaf of H' has the same neighbourhood as p. If we let N(P) := {a,y}, then at least one of {a,y} is a leaf of H' - p; without loss of generality, a is a leaf of H' - p, and so d(a) = 3. It follows that N(a) = {P, y, x} for some x G {a, p, 7, S, y}, and that xy G E(H'). Let N(7) = N(S) = {w, z}. We consider two cases depending on |{x} n {w, z}|. If |{x} n {w, z}| =0, then H := H' - {a, P, 7, S} is a 2-tree, and applying Operation V to H produces H'. If |{x} n {w, z}| = 1, then y = w or y = z, and the two cases are analogous. The graph H := H' - {a, P, 7, S} is a 2-tree, and applying Operation IX to H produces H'. Case 2(b). Finally, assume that H' has exactly two leaves, P and S, with N(P) = {a, y} and N(S) = {7, z}. Since dH> (P, S) > 3, {a, y} n {7, z} = 0. We may assume, without loss of generality, that a and 7 are leaves in H' - {P, S}, and so d(a) = 3 and d(Y) = 3. It follows that N (a) = {P, y,p} and N (7 ) = {S, z, q} for some p, q G {P, a, S, 7} with P = y, q = z andpy, qz G E(H'). We consider three cases depending on |{p, y} n {q, z}|. If |{p, y} n {q, z}| = 0, then H := H' - {a, P, 7, S} is a 2-tree, and applying Operation VI to H produces H'. If |{p, y} n {q, z}| = 1, then either p = q, p = z, or q = y. In the case p = q, H := H' - {a, P, 7, S} is a 2-tree, and applying Operation VII to H produces H'. In the case p = z, H := H' - {a, P, 7, S} is a 2-tree, and applying Operation VIII to H produces H'. The case q = y is analogous to the case p = z. Finally if |{p, y} n {q, z}| =2, then the fact that y = z implies that p = z and q = y, and hence H' = D, contradicting the fact that |V(H')| > 7. □ 5 Operations and the 4-colouring graph Let H be a 2-tree, and let H' be the 2-tree obtained from H by applying one of the Operations I through IX. As before, let V(G4(H)) := {/0, /1,..., fN-1}, and let Fj Ç V(G4(H')) be the set of 4-colourings of H' that agree with / on the vertices of H, 0 < j < N - 1. For each Operation I through IX, what follows is a description of the subgraph of G := G4(H') induced by Fj, 0 < i < N - 1, and also a description of the subgraph of G induced by Fi U Fj when // G E(G4(H)), 0 < i, j < N - 1. Each edge // of G4(H) is also assigned a label to indicate the structure of G[Fj U Fj]. We remark that for a path ///k of length two in G4(H), if /¿(u) = /j (u) for some u G V(H), then /j(u) = /k(u). 5.1 Operation I If H' is obtained from H using Operation I, then there are two choices of colour for each of the vertices a, P, and 7, so for each i, 0 < i < N - 1, G[Fj] = To simplify the labelling of the vertices of G[Fj], we label the faces of a plane drawing of Q3 as shown in Figure 7(a), where a1 and a2 are the possible colours of vertex a, P1 and P2 are the possible colours of vertex P, and y1 and y2 are the possible colours of vertex 7. Without loss of generality, assume these colour choices are as shown in Figure 7(b). A vertex u of M. Cavers and K. Seyffarth: Reconfiguring vertex colourings of 2-trees 667 Q3 is assigned label «¿p Yk where i, j, k € {1,2}, and u is incident with the faces labelled Oj, Pj and jk (see Figure 7(c)). «2 \ Yi / ßl ai ß2 7 Y2 (a) 2 \ 3 / 2 1 4 / 223 224 (b) Figure 7: Labelling G [Fi]. 243 244 (c) The following arguments can be made with sets of colours {/¿(ai), /¿(o^)}, j/i(bi), /j(62)} and {/i(c1), /¿(02)} each chosen independently as a subset of {1, 2,3,4}. To ease notation, we assume that /i(a1) = 3, /i(a2) = 4, /i(b1) = 1, /i(b2) = 3, /i(c1) = 1, and /i(c2) = 2. Then the colour choices for a are {1,2}, for fi are {2,4}, and for 7 are {3,4}. As already noted, G [Fi] = Q3; assume that G[Fi] has been drawn in the plane and labelled as in Figure 8(a). If // G E(G4(H)), then / is obtained from /i by changing the colour of a single vertex in V(H). We label each edge /i/ G E(G4(H)) according to the structure of G[Fi U Fj]. (a) 2 2 2 \ 3 / 2 1 4 / (b) (c) 2 \ 3 f 3 (4 / \ (d) Figure 8: G [Fi] and G [Fi U Fj ] for Operation I, II and III. (i) /i(v) = fj(v) for v € {ai, a2}. Without loss of generality, suppose v = ai. Since H is a 2-tree, there is a vertex a3 € V(H) such that H[{a1, a2, a3}] = K3. Observe /i(a3) € {1,2}; we may assume /¿(a3) = 1. Then /j(a1) = 2.1 Since /j(a2) = 1 Since fj (ai) is uniquely determined there is no fp with £ = j such that f and fp differ on ai. 668 Ars Math. Contemp. 17 (2019) 591-615 /¿(a2) = 4, the colours available for a are {1, 3}; the colours available for ft and Y are unchanged. Therefore [Fi, Fj ] is a matching of size four between the 4-cycle bounding the ai face in G[Fj] and the 4-cycle bounding either the ai face or the a2 face in G[Fj]. Thus G[Fj UFj] is isomorphic to the graph in Figure 8(b). We label the edge / /j by a-sq. It follows from Footnote 1 that there are at most two edges incident to / having label a-sq. Furthermore, we remark that if //, // G E(G4(H)) both have label a-sq, then the four vertices of Fj incident to the edges of [Fj; Fj ] are the same four vertices of Fj that are incident to the edges of [Fj, Fj2], and induce a 4-cycle in G[Fj] that bounds a face with label a1 or a2. (ii) /¿(v) = /,(v) for v G {b1,b2}. As in (i), we label the edge //, by b-sq. Note that there are at most two edges incident to /j having label b-sq. If //, /j/j2 G E(G4(H)) both have label b-sq, then the four vertices of Fj incident to the edges of [Fj; Fj ] are the same four vertices of Fj that are incident to the edges of [Fj; Fj2 ], and induce a 4-cycle in G[Fj] that bounds a face with label or ,02. (iii) /¿(v) = /,(v) for v G {c1;c2}. As in (i) and (ii), we label the edge /¿/j by c-sq. Note that there are at most two edges incident to /j having label c-sq. If /¿/j, /¿/72 G E(G4(H)) both have label c-sq, then the four vertices of F; incident to the edges of F Fj ] are the same four vertices of F; that are incident to the edges of [F;, Fj2], and induce a 4-cycle in G[F^ that bounds a face with label y1 or y2. (iv) /¿(u) = /j(u) for u G V(H) \ {a1; a2, b1, b2, c1; c2}. In this case, the vertex labels on G[F;] and G[Fj] are identical. Thus [Fj , Fj] is a perfect matching, and G[Fj U Fj] is isomorphic to the graph in Figure 8(c). We label the edge /¿/j by pm. Table 1: Summary of Operation I. Vertex whose colour is changed Subgraph induced by Fj U Fj Label of /¿/j a1 , a2 Figure 8(b) a-sq bi, 62 Figure 8(b) b-sq c1 , c2 Figure 8(b) c-sq u G V(H) \ {ai, a2, 61, 62, ci, c2} Figure 8(c) pm 5.2 Operation II As with Operation I, G[Fj] = Q3 for each i, 0 < i < N - 1. We may assume that /¿(a) = 4, fj(x) = 3, /j(b) = 1, /j(ci) = 1, and /¿(c2) = 2. Then the colour choices for a are {1,2}, for ß are {2,4}, and for 7 are {3,4}. Using the same labelling convention as for Operation I, we assume that G[Fj] is drawn in the plane and labelled as in Figure 8(a). If // G E(G4(H)), then / is obtained from /j by changing the colour of a single vertex in V(H). We label each edge // G E(G4(H)) according to the structure of G[Fj U Fj]. (i) /¿(v) = /j(v) for v G {a, b, c1;c2}. This is analogous to Operation I when the colour of one of {a1; a2, b1, b2, c1; c2} is changed, and thus G[Fj U Fj] is isomorphic M. Cavers and K. Seyffarth: Reconfiguring vertex colourings of 2-trees 669 to the graph in Figure 8(b). We label the edge /¿/j by either a-sq, b-sq or c-sq as in Operation I. (ii) / (x) = /j (x). We may assume that / (x) = 2. Then the colours available for a are {1,3} and the colours available for fi are {3,4}; the colours available for 7 are unchanged. Therefore, [Fj; Fj ] is a matching of size two where G[Fi] and G[Fj ] are edges whose endpoint labels are unchanged, and thus G[F U Fj ] is isomorphic to the graph in Figure 8(d). We label the edge // by e. (iii) / (u) = /j(u) for u € V(H) \ {a, b, x, ci, c2}. In this case, the vertex labels on G[Fi] and G[Fj ] are identical. Thus F, Fj] is a perfect matching, and G[F U Fj ] is isomorphic to the graph in Figure 8(c). We label the edge / / by pm. Table 2: Summary of Operation II. Vertex whose colour is changed Subgraph induced by Fj U Fj Label of ff a Figure 8(b) a-sq b Figure 8(b) b-sq ci , c2 Figure 8(b) c-sq x Figure 8(d) e u G V(H) \ { a, b, x, ci , c2 } Figure 8(c) pm 5.3 Operation III As with Operations I and II, G [Fi] = Q3 for each i, 0 < i < N - 1. We may assume that /¿(x) = 3, / (y) = 1, /¿(a) = 4, and /¿(c) = 2. Then the colour choices for a are {1, 2}, for fi are {2,4}, and for 7 are {3,4}. Using the same labelling convention as for Operation I, we assume that G[Fi] is drawn in the plane and labelled as in Figure 8(a). If /¿/j G E(G4(H)), then /j is obtained from / by changing the colour of a single vertex in V(H). We label each edge //j G E(G4(H)) according to the structure of G[F u Fj]. (i) /¿(v) = /j (v) for v G {a, c}. This is analogous to Operation I when the colour is changed on one of {ai, a2, ci, c2}, and thus G[Fj U Fj] is isomorphic to the graph in Figure 8(b). We label the edge /j/j by either a-sq or c-sq as in Operation I. (ii) /¿(v) = /j (v) for v G {x,y}. This is analogous to Operation II when the colour of x is changed, and thus G[Fj U Fj ] is isomorphic to the graph in Figure 8(d). We label the edge /¿/j by e. (iii) /¿(u) = /j(u) for vertex u G V(H) \ {a, c, x, y}. In this case, the vertex labels on G[Fj] and G[Fj ] are identical. Thus [Fj, Fj] is a perfect matching, and G[Fj U Fj ] is isomorphic to the graph in Figure 8(c). We label the edge /¿/j by pm. Remark 5.1. We note that for Operations I-III, if /¿/j,/¿/j2 G E(G4(H)) have the same label that is not e, and F/, F2 C Fj are incident to the edges of [Fj, Fj], [Fj, Fj2], respectively, then Fii = Fi2. 670 Ars Math. Contemp. 17 (2019) 591-615 Table 3: Summary of Operation III. Vertex whose colour is changed Subgraph induced by Fi U Fj Label of fi fj a Figure 8(b) a-sq c Figure 8(b) c-sq x, y Figure 8(d) e u £ V(H) \ {a, c, x, y} Figure 8(c) pm 5.4 Operation IV We may assume that fi (x) = 1, fi (y) = 2, fi (w) = 2 and fi (z) = 3. Then the pairs of colour available for a and /, respectively, are {(4,3), (3,3), (3,4), (4,4)}, and the pairs of colours available for 7 and 5, respectively, are {(1,4), (1,1), (4,1), (4,4)}. Thus the subgraph of G induced by Fi is isomorphic to C4 □ C4, and we assume that it is drawn as shown in Figure 9(a), with the rows labelled by the pairs of colours available for a and /, respectively, and the columns labelled by the pairs of colours available for 7 and 5, respectively. If fi fj £ E(G4(H)), then f is obtained from fi by changing the colour of a single vertex in V(H); there are three cases to consider. (a) (b) (c) Figure 9: G[F] and G[F U Fj] for Operation IV. (d) (i) fi(v) = fj (v) for v £ {x,y}. We may assume that fj (x) = 4. Then the pairs of colours available for a and / , respectively, are {(1,3), (3, 3), (3,1), (1,1)}, and the pairs of colours available for 7 and 5 are unchanged. Hence, G[Fi U Fj ] is isomorphic to the graph in Figure 9(b). M. Cavers and K. Seyffarth: Reconfiguring vertex colourings of 2-trees 671 (ii) /¿(v) = fj(v) for v e {w, z}. We may assume that fj(w) = 4. Then the pairs of colours available for 7 and respectively, are {(2, 2), (2,1), (1,1), (1, 2)}, and the pairs of colours available for a and 3 are unchanged. Hence, G[F U Fj ] is isomorphic to the graph in Figure 9(c). (iii) /¿(u) = fj (u) for u e V(H) \ {x, y, w, z}. In this case, the vertex labels on G[Fj] and G[Fj] are identical. Thus [Fj,Fj] is a perfect matching, and G[Fj U Fj] is isomorphic to the graph in Figure 9(d). Table 4: Summary of Operation IV. Subgraph induced by Vertex whose colour is changed Fj U Fj Label of /¿/j x, y Figure 9(b) r z, w Figure 9(c) c u e V(H) \ {x, y, z, w} Figure 9(d) pm Remark 5.2. We note that for Operation IV, if fj fj e E (G4(H)) has label r and e e [Fj, Fj], then each colouring corresponding to an end of e assigns the same colour to a and 3. Similarly, if fjfj e E(G4(H)) has label c and e e [Fj, Fj], then each colouring corresponding to an end of e assigns the same colour to 7 and 5.5 Operation V We may assume that /¿(x) = 1, /¿(y) = 2, /¿(w) = 2 and /¿(z) = 3. Then the pairs of colour available for a and 3, respectively, are {(3,4), (3,1), (4,1), (4, 3)}, and the pairs of colours available for 7 and respectively, are {(4,1), (1,1), (1, 4), (4,4)}. Thus the subgraph of G4 (H') induced by Fj is isomorphic to P4 □ C4, and we assume that it is drawn as shown in Figure 10(a), with the rows labelled by the pairs of colours available for a and 3, respectively, and the columns labelled by the pairs of colours available for 7 and respectively. If /¿fj e E(G4(H)), then fj is obtained from fj by changing the colour of a single vertex in V(H); there are five cases to consider. Since H is a 2-tree, there are vertices a, b e V(H) such that H[{x, y, a}] = K3 and H[{w, z, b}] = K3. Observe /¿(a) e {3,4} and /¿(b) e {1,4}. We may assume that /¿(a) = 4 and fj(b) = 1. Even though b (respectively, a) could be equal to x or y (respectively, w or z), this does not affect the argument. 672 Ars Math. Contemp. 17 (2019) 591-615 (a) (b) (c) (d) (e) (f) Figure 10: Subgraphs induced by F U Fj for Operations V and IX. (i) f (y) = fj (y). Then fj (y) = 3 and the pairs of colours available for a and ^, respectively, are {(4,2), (4,1), (2,1), (2,4)}, and the pairs of colours available for 7 and 5 are unchanged. Hence, G[F^ U Fj ] is isomorphic to the graph in Figure 10(b) with appropriate labels. (ii) f (x) = fj (x). Then fj (x) = 3 and the pairs of colours available for a and ^, respectively, are {(4,1), (4,3), (1,3), (1,4)}, and the pairs of colours available for 7 and 5 are unchanged. Hence, G[F^ U Fj ] is isomorphic to the graph in Figure 10(c). (iii) f (z) = fj (z). Then fj (z) = 4 and the pairs of colours available for 7 and 5, respectively, are {(3,3), (1,3), (1,1), (3,1)}, M. Cavers and K. Seyffarth: Reconfiguring vertex colourings of 2-trees 673 and the pairs of colours available for a and ft are unchanged. Hence, G[Fi U Fj ] is isomorphic to the graph in Figure 10(d). (iv) fi(w) = fj(w). Then fj(w) = 4 and the pairs of colours available for y and S, respectively, are {(3, 3), (1, 3), (1,1), (3,1)}, and the pairs of colours available for a and ft are unchanged. Hence, G[Fi U Fj ] is isomorphic to the graph in Figure 10(d). (v) fi(u) = fj(u) u G V(H) \ {x, y, w, z}. In this case, the vertex labels on G[Fi] and G[Fj] are identical. Thus [Fi; Fj] is a perfect matching, and G[Fi U Fj] is isomorphic to the graph in Figure 10(f). Table 5: Summary of Operation V. Vertex whose colour is changed Subgraph induced by Fi U Fj Label of fifj y Figure 10(b) r x Figure 10(c) rr z, w Figure 10(d) c u G V(H) \ {x, y, z, w} Figure 10(f) pm We informally refer to the rows and columns of vertices in G[Fi] according to the drawing in Figure 10(a). For Operations IV and VI through IX we use a similar convention. Remark 5.3. We note that for Operation V, if fifj G E(G4(H)) has label r, then the set of vertices Si j C Fi incident to the edges of [Fi; Fj] consists of row two or three2. If fifj G E(G4(H)) has label rr, then the set of vertices Sitj C Fi incident to the edges of [Fi; Fj] consists of rows one and two, or rows three and four. If fifj G E(G4(H)) has label c and e G [Fi; Fj], then each colouring corresponding to an end of e assigns the same colour to y and S. 5.6 Operation VI We may assume that fi(x) = 1, fi(y) = 2, fi(z) = 3 and fi(w) = 1. Then the pairs of colour available for a and ft, respectively, are {(3,4), (3,1), (4,1), (4, 3)}, and the pairs of colours available for y and S, respectively, are {(2,4), (2,1), (4,1), (4, 2)}. Thus G[Fi] is isomorphic to P4 □ P4, and we assume that it is drawn in the plane as shown in Figure 11(a), with the rows labelled by the pairs of colours available for a and ft, respectively, and the columns labelled by the pairs of colours available for y and S, respectively. 2 Rows are numbered from top to bottom and columns from left to right. 674 Ars Math. Contemp. 17 (2019) 591-615 (a) (b) (2,1) (2,4) (1,4) (1,2) (e) (c) (3,2)(3,1)(2,1)(2,3) (d) (f) (2,1)(2,4)(1,4)(1,2) (g) (2,1) (2,4) (1,4) (1,2) (h) Figure 11: Subgraphs induced by Fj U Fj for Operations VI, VII and VIII. If fjfj e E(G4(H)), then / is obtained from fj by changing the colour of a single vertex in V(H); there are five cases to consider. Since H is a 2-tree, there are vertices a, b e V (H) such that H [{x,y,a}] = K3 and H [{w, z, b}] = K3 .Observe / (a) e {3, 4} and fj (b) e {2, 4}. We may assume that fj (a) = 4 and fj (b) = 2. Even though b (respectively, a) could be equal to x or y (respectively, w or z), this does not affect the argument. (i) fj (y) = fj (y). Then fj (y) = 3, and the pairs of colours available for a and P, respectively, are {(4,2), (4,1), (2,1), (2,4)}, and the pairs of colours available for 7 and S are unchanged. Hence, G[Fj U Fj ] is isomorphic the the graph in Figure 11(b). (ii) /j (x) = /j (x). Then / (x) = 3, and the pairs of colours available for a and P, respectively, are {(4,1), (4, 3), (1,3), (1,4)}, while the pairs of colours available for 7 and S are unchanged. Hence, G[Fj U Fj ] is isomorphic to the graph in Figure 11(c). M. Cavers and K. Seyffarth: Reconfiguring vertex colourings of 2-trees 675 (iii) /j(z) = fj (z). Then fj (z) = 4 and the pairs of colours available for 7 and J, respectively, are {(3, 2), (3,1), (2,1), (2, 3)}, while the pairs of colours available for a and 3 are unchanged. Hence, G[F U Fj ] is isomorphic the the graph in Figure 11(d). (iv) fj(w) = fj (w). Then fj (w) = 4 and the pairs of colours available for 7 and J, respectively, are {(2,1), (2, 4), (1, 4), (1, 2)}, while the pairs of colours available for a and 3 are unchanged. Hence, G[Fj U Fj ] is isomorphic the the graph in Figure 11(e). (v) fj(u) = fj(u) for some u G V(H) \ {x, y, z, w}. In this case, the vertex labels on G[Fj] and G[Fj ] are identical. Thus [Fj, Fj] is a perfect matching, and G[Fj U Fj ] is isomorphic to the graph in Figure 11(f). Table 6: Summary of Operation VI. Vertex whose colour is changed Subgraph induced by Fj U Fj Label of fifj y Figure 11(b) r x Figure 11(c) rr z Figure 11(d) c w Figure 11(e) cc u G V(H) \ {x, y, z, w} Figure 11(f) pm Remark 5.4. We note that for Operation VI, if fifj G E(G4(H)) has label r (respectively, c), then the set of vertices Sijj C Fi incident to the edges of [Fi, Fj] consists of row (respectively, column) two or three. If fifj G E(G4(H)) has label rr (respectively, cc), then the set of vertices Sijj- C Fi incident to the edges of [Fi, Fj ] consists of rows (respectively, columns) one and two or rows (respectively, columns) three and four. 5.7 Operation VII We may assume that fi(x) = 1, fi(y) = 2 and fi(z) = 3. Then the pairs of colours available for a and ß, respectively, are {(3,4), (3,1), (4,1), (4, 3)}, and the pairs of colours available for 7 and respectively, are {(2,4), (2,1), (4,1), (4, 2)}. Thus G[Fi] is isomorphic to P4 □ P4, and we assume that it is drawn in the plane as shown in Figure 11(a) with rows labelled by the pairs of colours available for a and ß, respectively, and the columns labelled by the pairs of colours available for 7 and respectively. If fifj G E(G4(H)), then fj is obtained from fi by changing the colour of a single vertex in V(H). There are four cases to consider. 676 Ars Math. Contemp. 17 (2019) 591-615 (i) /¿(x) = fj (x). We may assume that /j (x) = 4. Then the pairs of colours available for a and ß, respectively, are {(3,1), (3,4), (1, 4), (1, 3)}, and the pairs of colours available for 7 and respectively, are {(2,1), (2,4), (1, 4), (1, 2)}. Hence, G[F U Fj ] is isomorphic to the graph in Figure 11(g). (ii) /¿(y) = /j (y). This is analogous to Operation VI when the colour of y is changed, and thus G[F U Fj ] is isomorphic to the graph in Figure 11(b). (iii) /¿(z) = /j (z). This is analogous to Operation VI when the colour of z is changed, and thus G[Fi U Fj ] is isomorphic to the graph in Figure 11(d). (iv) /¿(u) = /j(u) for some u G V(H) \ {x,y, z}. In this case, the vertex labels on G[Fi] and G[Fj] are identical. Thus [Fj, Fj] is a perfect matching, and G[Fi U Fj] is isomorphic to the graph in Figure 11(f). Table 7: Summary of Operation VII. Vertex whose colour is changed Subgraph induced by Fj U Fj Label of / x Figure 11(g) sq y Figure 11(b) r z Figure 11(d) c u G V(H) \{x,y,z} Figure 11(f) pm Remark5.5. We note that for Operation VII, if // G E (G4(H)) has label r (respectively, c), then the set of vertices Sjj C Fj incident to the edges of [F^Fj] consists of row (respectively, column) two or three. If // G E(G4(H)) has label sq, then the set of vertices Sjj C Fj incident to the edges of [Fi, Fj] induces a four-cycle using a degree two vertex of G[Fj]. 5.8 Operation VIII We may assume that /¿(x) = 1, /¿(y) = 2 and /¿(z) = 3. Then the pairs of colours available for a and ß, respectively, are {(4, 3), (4, 2), (3, 2), (3,4)}, and the pairs of colours available for 7 and respectively, are {(2,4), (2,1), (4,1), (4, 2)}. Thus G[Fj] is isomorphic to P4 □ P4, and we assume that it is drawn in the plane as shown in Figure 11(a) with rows labelled by the pairs of colours available for a and ß, respectively, M. Cavers and K. Seyffarth: Reconfiguring vertex colourings of 2-trees 677 and the columns labelled by the pairs of colours available for 7 and J, respectively (but not the same labels as in the figure). If // € E(G4(H)), then / is obtained from / by changing the colour of a single vertex in V(H), and there are four cases. (i) /i(x) = /j (x). We may assume that / (x) = 4. Then the pairs of colours available for a and ß, respectively, are {(1, 3), (1, 2), (3, 2), (3,1)}, and the pairs of colours available for 7 and J, respectively, are {(2,1), (2, 4), (1, 4), (1, 2)}. Hence, G[Fi U Fj] is isomorphic to the graph in Figure 11(h) with appropriate labels. (ii) /i(y) = /j (y). This is analogous to Operation VI when the colour of x is changed, and thus G[Fi U Fj ] is isomorphic to the graph in Figure 11(c) with appropriate labels. (iii) /i(z) = /j (z). This is analogous to Operation VI when the colour of z is changed, and thus G[Fi U Fj ] is isomorphic to the graph in Figure 11(d) with appropriate labels. (iv) /i(u) = /j(u) for some u € V(H) \ {x, y, z}. In this case, the vertex labels on G[Fi] and G[Fj] are identical. Thus [Fi, Fj] is a perfect matching, and G[Fi U Fj] is isomorphic to the graph in Figure 11(f). Table 8: Summary of Operation VIII. Vertex whose colour is changed Subgraph induced by Fi U Fj Label of // x Figure 11(h) e y Figure 11(c) rr z Figure 11(d) c u € V(H) \{x,y,z} Figure 11(f) pm Remark 5.6. We note that for Operation VIII, if // G E(G4(H)) has label c, then the set of vertices Si j C Fi incident to the edges of [Fi, Fj ] consists of column two or three. If /¿/j G E(G4(H)) has label rr, then the set of vertices C Fj incident to the edges of [Fj, Fj] consists of rows one and two or rows three and four. If // G E(G4 (H)) has label e, then the set of vertices Sjj C Fj incident to the edges of [Fj, Fj ] induces an edge that is the first or last edge of row two or row three. 5.9 Operation IX We may assume that /¿(x) = 1, /¿(y) = 2 and /¿(z) = 3. Then the pairs of colours available for a and respectively, are {(4, 3), (4,1), (3,1), (3,4)}, 678 Ars Math. Contemp. 17 (2019) 591-615 and the pairs of colours available for 7 and J, respectively, are {(2, 2), (2,4), (4, 4), (4, 2)}. Thus G[Fi] is isomorphic to P4 □ C4, and we assume that it is drawn in the plane as shown in Figure 10(a) with rows labelled by the pairs of colours available for a and ß, respectively, and the columns labelled by the pairs of colours available for 7 and J, respectively (but not the same labels as in the figure). If fifj € E(G4(H)), then fj is obtained from f by changing the colour of a single vertex in V(H); there are four cases. (i) fj(x) = fj (x). We may assume that fj (x) = 4. Then the pairs of colours available for a and ß, respectively, are {(3,1), (3,4), (1, 4), (1, 3)}, and the pairs of colours available for 7 and J, respectively, are {(2,1), (2, 2), (1, 2), (1,1)}. Hence, G[F U Fj] is isomorphic to the graph in Figure 10(e) with appropriate labels. (ii) fj(y) = fj (y). This is analogous to Operation V when the colour of y is changed, and thus G[Fj U Fj ] is isomorphic to the graph in Figure 10(b) with appropriate labels. (iii) fj(z) = fj (z). This is analogous to Operation V when the colour of z is changed, and thus G[Fj U Fj ] is isomorphic to the graph in Figure 10(d) with appropriate labels. (iv) fj(u) = fj(u) for some u € V(H) \ {x,y, z}. In this case, the vertex labels on G[Fj] and G[Fj] are identical. Thus [Fj, Fj] is a perfect matching, and G[Fj U Fj] is isomorphic to the graph in Figure 10(f). Table 9: Summary of Operation IX. Vertex whose colour is changed Subgraph induced by Fj U Fj Label of fjfj x Figure 10(e) e y Figure 10(b) r z Figure 10(d) c u € V(H) \{x,y,z} Figure 10(f) pm Remark 5.7. We note that for Operation IX, if // G E(G4(H)) has label r and e G [Fj, Fj], then each colouring corresponding to an end of e assigns the same colour to a and ft. Similarly, if // G E(G4(H)) has label c and e G [Fj, Fj], then each colouring corresponding to an end of e assigns the same colour to 7 and J. If //j G E(G4(H)) has label e and e G [Fj, Fj], then the set of vertices Sjj C Fj incident to the edges of [Fj, Fj] induces an edge that is either the first or last edge in a column, and each colouring corresponding to an end of e assigns the same colour to 7 and J. M. Cavers and K. Seyffarth: Reconfiguring vertex colourings of 2-trees 679 6 4-colouring graphs of 2-trees of diameter at least three Let H be a 2-tree, and let H' be a 2-tree obtained from H by applying one of the Operations I through IX. We prove G4(H') has a Hamilton cycle. 6.1 Operations I, II and III We first prove a result about Hamilton cycles in a cube Q3 that will later be used to show the existence of edges satisfying Lemma 2.7. In this section, we let each face label in Figure 12(a) denote the 4-cycle bounding that face. h-4 (a) (b) Figure 12: Labelling G[F]. We label the six Hamilton cycles of a plane drawing of Q3 as shown in Figure 13. @010@l (a) C (b) n (c) □ (d) U (e) H (f) I Figure 13: Labels for the Hamilton cycles of Q3. To simplify notation for multisets, we write nZ to mean n copies of Z. Lemma 6.1. Let Q = Q3 be drawn as in Figure 12(a), and let e be an edge of Q. Let Z = {Z1, Z2,..., Zn} be a multiset such that Z C {2a1,2ft1, 2y1, 5Q3} and n < 5. That is, each Zj is an induced subgraph of Q and is either the entire 3-cube or one of the 4-cycles of Q labelled by a1, ft1, or y1. Then there exists a Hamilton cycle in Q containing distinct edges {e, e1, e2,..., en} such that ej £ E(Zj), for 1 < i < n. Proof. It is enough to prove the result when n = 5 and Z = {2a1, 2ft1, y1}. Note that the Hamilton cycles U and n in Q each contain three edges of a1 , three edges of ft1 , and two edges of y1 . When any single edge is deleted from U or n, we see that the resulting Hamilton path contains five distinct edges, two from a1, two from ft1 and one from y1. Since at least one of U and n contains the edge e, either U or n contains distinct edges {e, e1, e2,..., e5} such that ej £ E(Zj), for 1 < i < 5. □ 680 Ars Math. Contemp. 17 (2019) 591-615 Lemma 6.2. Let Q = Q3 be drawn as in Figure 12(a) with edges labelled as in Figure 12(b). Assume {e, e'} C E (Q) with e G {w1, w2,w3, w4}. Let Z = {Z1,Z2,..., Zn} be a multiset such that Z C {a1,^1, 2^1,4Q3} and n < 4. That is, each Zi is an induced subgraph of Q and is either the entire 3-cube or one of the 4-cycles of Q labelled by a1, or y1. Then there exists a Hamilton cycle in Q containing distinct edges {e, e', e1,e2,..., en} such that ei G E(Zi),for 1 < i < n. Proof. It is enough to prove the result when n = 4 and Z = {a1,p1, 2^1} with Z1 := a1, Z2 := A, Z3 := Y1 and Z4 := 71. Let H := {h1, h2, h3, h4}, W := {w1, W2, w4} and D := {d1,d2,d3,d4}. For each Zi, we designate a set of candidate edges for ei as follows (see Table 10). Table 10: Cases in the proof of Lemma 6.2. Cycle Edges et assigned to Zi e', e in Q Zi Z2 Z3 Z4 e' e H, e e {wi, w2} c {h2,h3}\{e'} {wi,w2} \ {e} d2 {hi,h2}\{e'} e' e H, e e {w3, w4 } □ {h2,h3}\{e'} d4 di {hi,h2}\{e'} {e, e'} = {wi, W2} n W3 d4 hi h2 e,e' e W, {e,e'} = {w1,w2} u h3 {wi, W2} \ {e, e'} di d2 e' e D, e e {w1, w4} H h3 {di ,d4}\{e'} hi {di,d2}\{e'} e' e D, e e {w2, W3} I {w2, W3} \ {e} {di ,d4}\{e'} hi {di,d2}\{e'} As indicated in Table 10, the first case has e' G H and e G w2}. We claim that C has the required property. We take e3 := d2, and as |{w1; w2} \ {e}| = 1, we take {e2} := {w1, w2} \ {e}. Observe that the sets {h2, h3} \ {e'} and {h1, h2} \ {e'} are distinct and non-empty. Thus, we may take e1 G {h2, h3} \ {e'} and e4 G {h1, h2} \ {e'} so that e1 = e4. The remaining five cases follow using analogous arguments. □ Lemma 6.3. Suppose H' is obtained from a 2-tree H by applying one of Operations I, II or III. Then G4(H') has a Hamilton cycle. Proof. Case 1. Suppose H' is obtained from H by applying Operation I. By Lemma 2.5, G4(H) has a spanning tree T with A(T) < 4. Let V(T) := {fo, fu ..., fN_1} such that f0 is a leaf, and root T at f0, turning T into a branching, T, by directing all arcs away from fo. Let G := G4(H'), and let Fi be the set of 4-colourings of H' that agree with fi on V (G4(H)), 0 < i < N — 1. Label each ——j g A(—) with the label of fifj G E (G^(H)), as described in Section 5, and let Sitj C Fi denote the vertices incident to the edges of [Fi,Fj]. For each arc —/j G A(7T!), we choose edges eij in G[Fi] and e^ in G[Fj] satisfying conditions (i) and (ii) of Lemma 2.7 as follows. Suppose fofl G A(7^). The fact that G[F0 U F1] is isomorphic to the graph in Figure 8(b) or 8(c) gives us the flexibility to choose e0t1 g E(G[F0]) and e1j0 G E(G[F1 ]) satisfying (ii) of Lemma 2.7. Edge choosing procedure. Now suppose for some i, ei k has been chosen in G[Fi] but eiij has not yet been chosen for each j where —if] G A(T). For this i, let J := {j | M. Cavers and K. Seyffarth: Reconfiguring vertex colourings of 2-trees 681 —fj G A(—)}. We choose edges eij and ej,i for j G J as follows. Assume that G[Fi] is drawn as in Figure 12(a). By Remark 5.1, Sij1 = Sij2 whenever fifj1 and fifj2 have the same label. Without loss of generality, we may assume that an arc fifj with label a-sq has G[Sitj] isomorphic to the 4-cycle a1, an arc fifj with label b-sq has G[Sitj] isomorphic to the 4-cycle and an arc fifj with label c-sq has G[Sitj] isomorphic to the 4-cycle j1. For each j G J, let Zj := G[Sijj], and define the multiset Z := {Zj | j G J}. Then each Zj is either a 3-cube or one of the 4-cycles a1, l31, or j1. Since fi is incident to at most two edges with label a-sq, at most two edges with label b-sq, and at most two edges with label c-sq, Z C {2a1,2/3h 2^1,5Q3}. Observe |Z| < 4 since A(T) < 4. By Lemma 6.1, using e := ei 4 having bipartition (A, B). Fix u G A, v G B with u adjacent to v in T. Define H := G3(T). For each 1 < i = j < 3 let Vîj := {c G V(H) | c(u) = i and c(v) = j}. Then {Vi2, Vi3, V23, V2i, V3i, V32} is a partition of V(H). Observe each H[Vj] is connected (this can be shown by induction on |V(T)|), and that [Va^, V7(5] = 0 if and only if a = y or fi = J. It follows that Hi = H[Vi2 U V13 U V23 U V21] is connected, and by Lemma A.1 is bipartite. Similarly, H2 = H[V23 U V2i U V3i U V32] and H3 = H[V3i U V32 U Vi2 U Vi3] are connected and bipartite. Denote the two-coloured vertices of H (that is, the colourings of T with two colours) by cj G Vj such that , I i, if x G A, c,-o(x) := < 3K ' \j, if x G B. Suppose the bipartition of each H[Vj ] is (Aj, Bjj ) where c^j G Bjj. If |A| and |B| are even, then dH(cjj, c^/j/) is even. It follows that (Ai2 u Ai3 u A23 U A2i, Bi2 u Bi3 U B23 U B2i) is a bipartition of Hi, (A23 U A2i u A3i U A32, B23 U B2i u B3i U B32) is a bipartition of H2 , and (A3i U A32 U Ai2 U Ai3, B3i U B32 U Bi2 U Bi3) is a bipartition of H3. Hence, A := (J Aij and B := J Bij i 2 and kj > 1, 1 < i < n [17]. 2. Q„, for n > 1 [5]. Definition A.5. A B-graph with vertex partition {F0, Fi,..., FN _i} is a bipartite graph G with bipartition (A, B) together with a partition {F0, F1,..., FN_1} of V(G) so that, for i = 0,1,..., N — 1, G [Fi] is Hamilton laceable. Lemma A.6. Let G be a B-graph with vertex partition {F0, F1,..., FN_1} and bipartition (A, B). Suppose for each i = 1,2,..., N — 1, there is an edge bi_1ai with 6i_1 G B n Fi_1 and G An Fj. Then G has a Hamilton path between any vertex in An F0 and any vertex in B n FN _1. Proof. Let a0 G An F0 and bN_1 G B n FN_1. For each Fj, i = 0,1,..., N — 1, choose a Hamilton path Pi in G[Fj] between bj and aj. Then Corollary A.7. Let G be a B-graph with vertex partition {F0, F1,..., FN_1} and bipartition (A, B) such that [Fi_1; Fj] is a set of independent edges and | [Fj_1, Fj]| > 2, i = 1,2,..., N — 1. If for each i = 1,2,..., N — 1, the endpoints of any pair of edges in [Fi_1, Fj] induces a 4-cycle in G, then G has a Hamilton path between any vertex in An Fo and any vertex in B n FN _1. Proof. Let a0 G An F0 and _1 G B n FN _1. For each [Fi_1, F, ], i = 1, 2,..., N — 1, choose two edges 6i_1ai and bj_1ai. Then G[{a,, aj, bi_1, bi_1}] induces a 4-cycle ajajbj^bj^aj. Note that either 6j_1 G B or 6j_1 G B. Without loss of generality, suppose 6j_1 G B. Then aj G A. The result follows by Lemma A.6. □ Definition A.8. An odd flare is a tree obtained from K1ji, t > 3 and odd, by a single subdivision of one edge. Lemma 3.6. Let T be a tree with bipartition (A, B), where |A| := £ and |B| := r, and let G3(T) be the 3-colouring graph of T with colours C = {1, 2, 3}. Define cj to be the vertex of G3(T) with Cj (a) = i for all a G A and Cj (b) = j for all b G B. (1) If £, r > 0 are both even, then G3(T) has no spanning subgraph consisting only of paths whose ends are in {c12, c13, c21, c23, c31, c32}. (2) If £ > 1 is odd and r > 0 is even, then G3(T ) has a Hamilton path from c12 to c23. (3) If £ > 1 and r > 1 are both odd, then G3(T ) has a Hamilton path from c12 to c13. are the edges of a Hamilton path in G between a0 and _i. □ 692 Ars Math. Contemp. 17 (2019) 591-615 Proof. (1): Suppose r > 0 are both even. By Lemma A.2, H := G3(T) is bipartite. Suppose H has bipartition (A, B). Without loss of generality, assume c12 £ A. Since dH(c12, cjj) is even for each cjj £ {c12, c13, c21, c23, c31, c32} and H is bipartite, we have {c12,c13, c21, c23, c31, c32} C A. By [6, Theorem 5.5] there is a Hamilton cycle in H, and thus |A| = |B|. It follows that there is no spanning subgraph of H consisting only of paths whose ends are in {c12, c13, c21, c23, c31, c32} (otherwise |A| > |B|). (2): Suppose I > 1 is odd and r > 0 is even. Then I + r > 5. We first prove that G3 (T) has a Hamilton path between c12 and c23 whenever T is P5 or any odd flare. If T = P5 (with |A| = 3, |B| = 2), then there is a Hamilton path between c12 to c23 in G3(P5), as described in Figure 15. A 3-colouring f of P5 = x1x2x3x4x5 is represented by the string f (X1)f (X2)f (X3)f (X4)f (X5); for example, c12 = 12121 and c23 = 23232. 12121 21313 32321 13131 23231 31212 12321 21323 32121 12131 13231 31232 12323 21321 32123 32131 13232 31231 12313 31321 12123 32132 13212 21231 12312 31323 13123 12132 13213 21232 32312 31313 23123 13132 23213 21212 31312 32313 23121 23132 21213 23212 21312 32323 13121 23131 31213 23232 Figure 15: A Hamilton path in G3(P5) from ci2 to c23. Let T be an odd flare on n vertices with u denoting the unique vertex of degree two and let NT(u) = {v, v'} where v is the unique vertex of degree n - 2 (Figure 16). v u (n - 3) Figure 16: An odd flare T. Here |A| = n - 2 and |B| = 2. We partition H := G3(T) according to the colours of u and v. For each 1 < i = j < 3, let Vjj := {c £ V(H) | c(u) = i and c(v) = j} and let Ljj be an assignment of lists to the vertices of H such that Ljj (u) := {i}, Ljj (v) := {j} and Ljj(w) := {1,2, 3} for w £ V(H - {u, v}). Note that GLjj (H) = H[Vjj] = Qn-2, for each 1 < i = j < 3. Let H1 := H[Vi2 U V12] = Qn-3 □ PI H4 := H[V21 U V31] = Qn-3 □ P4, H2 := H[V13], and H3 := H[V23]. For {i,j, k} = {1,2,3}, let djjk £ V(H) denote the vertex with djjk(v) = j, djjk(v') = k, and djjfc(w) = i for all w £ Nt(v). Then [V12, V13] = {c^d^, d123c13} and [V23, V21] = {c21d231, d213c23}. Claim. For m > 2, every edge of Qm □ P4 is in a Hamilton cycle of Qm □ P4. The claim follows by induction and the fact that any pair of distinct edges of Qm-1 (m > 3) belongs to a Hamilton cycle of Qm-1 (for example, see [7, Theorem 4.1]). In what follows, we define cycles and paths to construct a Hamilton path from c12 to c23 in G3(T) when T is an odd flare on n vertices. See Figure 17 for the case n = 5; here, M. Cavers and K. Seyffarth: Reconfiguring vertex colourings of 2-trees 693 the labels of the two columns of Vj represent the colour choices for v' and the labels of the rows of Vj represent the colour choices for the two vertices in NT(v) \ {u}. Let C1 be a Hamilton cycle of H containing c12d123 and C4 be a Hamilton cycle of H4 containing d213c21; these exist by the previous claim. Let c G NH3 (c23) have c(v') = 3 and d G V13 be the unique vertex of H2 adjacent to c. By [7, Theorem 4.1], there is a Hamilton cycle C3 in H3 containing the edges cc23 and c23d231. Observe dH2 (d, c13) is odd. Since H2 is Hamilton laceable, there is a Hamilton path P between d and c13 (see Figure 17). Figure 17: H = G3(T), T an odd flare on five vertices, along with C1, P, C3, and C4. Now, (C1 - {C12d123}) u {d123C13} U P U {dc} U (C3 - {C23C, C23d231}) u {d231C21} u (C4 - {C21d213}) u {d213C23} is a Hamilton path in H between c12 and c23 (see Figure 18 for n = 5). Figure 18: A Hamilton path between c12 and c23 in H = G3(T), T an odd flare on five vertices. Now suppose T is a tree on n > 5 vertices with n odd that is not isomorphic to a star or an odd flare. Then there are leaves x, y G V(T) with dT(x, y) > 3. By choosing leaves x and y so that dT(x, y) > 3 is minimum, T' := T - {x, y} is not a star. Let NT(x) := {x'} and Nt(y) := {y'}; since dT(x, y) > 3, x' = y'. Let (A', B') denote the bipartition of T' with A' C A, B' C B, and define cj to be the vertex of H' := G3(T') with ci^. (a) = i for all a G A' and ci^. (b) = j for all bG B'. By the inductive hypothesis, H' has a Hamilton path between c'12 and c23. Let V (H') := {/0, f1,..., _1}. Since H' has a Hamilton path, we may assume that /o/1 • • • -1 is a Hamilton path in H' between /0 = c'12 and /N_1 = c23. Since c'12 and c23 differ in colour on at least two vertices, /0 is not adjacent to /N_1 in H'. 694 Ars Math. Contemp. 17 (2019) 591-615 For 0 < i < N — 1, let Fi be the set of 3-colourings of H that agree with fi on V(T'). Then {Fo, Fu ..., FN_i} is a partition of the vertices of H, and H[Fi] = C4, 0 < i < N — 1. If fi_1 and fi differ on the colour of a vertex of V(T') \ {x', y'}, then |[Fi_1, Fi]| = 4 and H[Fi_1 U Fi] = Q3. Otherwise, fi_1 and fi differ on the colour of x' or y', implying that | [Fi_1, Fi] | = 2, the subgraph of H induced by the endpoints of the edges of [Fi_1, Fi] is a 4-cycle, and H[Fi_1 U Fi] = P4 □ K. Consider the spanning subgraph G of H with edge set /N_1 \ /N_1 \ \JE(H[Fi]) U (J[Fi_1,Fi] Note that G is a connected B-graph. Let (A, B) be the bipartition of G and assume C12 € A. First suppose c23 € B. As c12 € F0 and c23 € FN_1, it follows from Corollary A.7 that there is a Hamilton path in G between c12 and c23. Now suppose c23 € A. Since dG(c12, c23) is odd, H must have an edge e with both endpoints in A or both endpoints in B. Suppose e € [Fp, Fq], where 0 < p < q < N — 1. Since f0 is not adjacent to fN_1 in H', either p = 0 or q = N — 1. Without loss of generality we assume p = 0. Then fpfq € E(H'), and either (i) | [Fp, Fq] | =4 and H[Fp U Fq] = Q3, or (ii) | [Fp, Fq] | = 2, the subgraph of H induced by the endpoints of the edges of [Fp, Fq] is a 4-cycle, and H[Fp U Fq] = P4 □ K2. In either case, there exists another edge e' € [Fp, Fq] such that e := uv, e' := u'v', u, u' € Fp, and uvv'u'u is a 4-cycle. The choice of e = uv ensures that u,v € A or u, v € B. Consider the spanning tree T of H' with edge set E(T) := {fi_1fi | 1 < i < q — 1} U {fi_1fi | q + 1 < i < N — 1} U {fpfq}. We define J to be the spanning subgraph of H with edge set m E(H[Fi])j U I (J [Fi, Fj] ^i=0 ' \fifj e-E(T) Then J is a connected B-graph. Let (K, L) be the bipartition of J; we may assume that c12 € K. Since fpfq € E(T), [Fp, Fq] C E( J); in particular, e = uv and e' = u'v' are edges of J, so u and v are in different parts of the partition (K, L), and thus c23 € L. Case 1. If |[Fp, Fp+1]| = |[Fp, Fq]| = 2, then fpfq, fpfp+1 € E(H') arise from colour changes on x' and y'. Since there are only three possible vertex colours, there is only one possible colour that x' could change to, and only one possible colour that y' could change to; i.e., one of fpfq, fpfp+1 arises from a colour change on x' and the other from a colour change on y'. Assuming that H[Fp] is the 4-cycle uu'ww'u, it follows that u, u' are incident to the edges of [Fp, Fq] and that without loss of generality u', w are incident to the edges of [Fp, Fp+1]. M. Cavers and K. Seyffarth: Reconfiguring vertex colourings of 2-trees 695 Let [Fp,Ff] = and [Fp, Fp+1] = {wz,u'z'}. Since uu' G E(J), exactly one of u, u' is in L. Case 1(a). First suppose that u G L. Let J1 denote the subgraph of J induced by uf=0Fj. Then J1 is a B-graph with vertex partition {F0, F1;..., Fp} and bipartition (K, L) satisfying the conditions of Corollary A.7, with c12 G K n F0 and u G L n Fp. Thus J1 has a Hamilton path R1 between c12 and u. Note that the proof of Corollary A.7 implies that R1 can be constructed so as to contain the edge u'w. Let J2 be the subgraph of J induced by uN_ 1Fi. Then J2 is a B-graph with vertex partition {Ff, Ff+1,..., FN_1} and bipartition (K, L) satisfying the conditions of Corollary A.7, with v G K n Ff and c23 G L n FN_1. Thus J2 has a Hamilton path R2 between v and c23. Finally, let J3 be the subgraph of J induced by ufrpFj. Then by Lemma 2.7, J3 has a Hamilton cycle C containing the edges uu', ww' and uw'. Let R3 be the path between u' and w obtained by deleting u and w' from C. Now concatenate paths R1, R2, R3 and delete edge u'w G R1 to form a Hamilton path between c12 and c23. Case 1(b). Now suppose that u G K. Then u' G L. Since p > 1 and |[Fp,Fp+1]| = |[Fp,Ff ]| = 2, we have |[Fp_1;Fp]| = 4. Let t G be such that tu G [Fp_1;Fp]. Then t G L. Let J1 denote the subgraph of J induced by up=_01Fi. Then J1 is a B-graph with vertex partition {F0, F1,..., Fp_1} and bipartition (K, L) satisfying the conditions of Corollary A.7, with c12 G K n F0 and t G L n . Thus J1 has a Hamilton path R0 between c12 and t. Let R1 be the concatenation of paths R0 and tuw'wu'. We define J2 and R3 as in Case 1(a). The same argument with v' in place of v gives a Hamilton path R2 between v' and c23. Now concatenate paths R1, R2, R3 and delete edge u'w G R1 to form a Hamilton path between c12 and c23. Case 2. Suppose |[Fp, Ff]| = 4 and label the 4-cycle of H[Fp] as uu'ww'u. Let J3 be the subgraph of J induced by uflpFj. Then by Lemma 2.7, J3 has a Hamilton cycle C. Without loss of generality, suppose C contains the edges uu', ww' and uw'. Let R3 be the path between u' and w obtained by deleting u and w' from C. Let J1 denote the subgraph of J induced by Uf=0Fj. Then J1 is a B-graph with vertex partition {F0, F1,..., Fp} and bipartition (K, L) satisfying the conditions of Corollary A.7, with c12 G K n F0. Thus J1 has Hamilton paths R and R'/ between c12 and the two vertices in L n Fp. Let R1 be one of R1 and R1' such that R1 contains edge u'w. Suppose R1 is between c12 and t. Then t g L. Let t' g Ff be such that tt' g [Fp, Ff]. Then t' G K. We define J2 as in Case 1(a). The same argument with t' in place of v gives a Hamilton path R2 between t' and c23. Now concatenate paths R1, R2, R3 and delete edge u'w g R1 to form a Hamilton path between c12 and c23. Case 3. Suppose | [Fp, Fp+1] | = 4. Let J1 denote the subgraph of J induced by (up=0Fj) U (u^-1 Fj). Then J1 is a B-graph with vertex partition {F0, F1,..., Fp, Fq, Fq+1,..., Fn-1} and bipartition (K, L) satisfying the conditions of Corollary A.7, with c12 G K n F0 and c23 G L n Fn_ 1. Thus J1 has a Hamilton path R between c12 and c23. Note that the proof 696 Ars Math. Contemp. 17 (2019) 591-615 of Corollary A.7 implies that R can be constructed so as to contain three edges of H[Fp]. Let J3 be the subgraph of J induced by uq-p+Fj. Then by Lemma 2.7, J3 has a Hamilton cycle C. Note that C contains three edges of H[Fp+1]. By the Pigeonhole Principle there are s, t G Fp and s', t' G Fp+1 with st G R, s't' G C such that ss'tt's is a 4-cycle in H[Fp U Fp+1 ]. Now (R U C) - {st, s't'} is a Hamilton path in J between c12 and c23. (3): Finally suppose, I > 1 and r > 1 are both odd. We define £sfct to be the tree obtained from Pk with ends u and v by appending s leaves to u and t leaves to v. The proof is by induction and has the following base cases. Base Case 1. We first prove that H := G3(T) has a Hamilton path between c12 and c13 when T = £4,0+2 for k > 0, that is, when T = P4fc+2. If T = P2 then c12c32c31c21c23c13 is such a Hamilton path. Suppose k > 0 and T = P4k+2. Let u and v be the leaves of T, NT(u) := {u'}, NT(v) := {v'}, NT(u') := {u,u''} and NT(v') := {v,v''}. Then T' := T - {u, u', v, v'} is isomorphic to P4(fc_1)+2. Let (A', B') denote the bipartition of T' with A' C A, B' C B, and define cj to be the vertex of H' := G3(T') with cj(a) = i for all a G A' and cj (b) = j for all b G B'. By the inductive hypothesis, H' has a Hamilton path between c'12 and c'13. Let V(H') := {/0, f1,..., /Nt1}. Since H' has a Hamilton path we may assume that /0/1 • • • /Nt1 is a Hamilton path in H' between /0 := c'12 and _1 := c13. For 0 < i < N - 1, let F; be the set of 3-colourings of H that agree with / on V(T'). Then {Fo, F1,..., Fn-1} is apartition of the vertices of H, and H[Fj] = P4 □ P4, 0 < i < N - 1. If /jt1 and /, differ on the colour of a vertex of V(T') \ {u'', v''}, then | [Fj_1, Fi] | = 16 and H[Fi_1 U F,] = (P4 □ P4) □ K2. Otherwise, /jt1 and /, differ on the colour of u'' or v'', implying that | [Fi—1, F,] | = 8, and the subgraph of H induced by the endpoints of the edges of [Fi—1, F,] is C4 □ P4. Consider the spanning subgraph G of H with edge set (N _1 \ / N _1 U E(H[Fj]H u( U [Fj_1,Fj] i=0 J \i=1 Note that by Remark A.4, G is a connected B-graph. Let (A, B) be the bipartition of G and assume c12 G A. Since H is bipartite by Lemma A.2, and dH(c12, c13) is odd, c13 G B. As c12 G F0 and c13 G FN_1, it follows from Corollary A.7 that there is a Hamilton path in G between c12 and c13. Base Case 2. Let T = £24sfc2 2 or T = f4sfc+1j2i+1 with s, t > 1, k > 0, and T' = P4fc+2 be obtained from T by deleting 2s leaves adjacent to u and 2t leaves adjacent to v. Let (A', B') denote the bipartition of T' with A' C A, B' C B, and define cj to be the vertex of H' := G3(T') with cj(a) = i for all a G A' and cj (b) = j for all b G B'. By Case 1, H' has a Hamilton path between c'12 and c'13. Let V(H') := {/0, /1,..., /N_1}. Since H' has a Hamilton path we may assume that /0/1 • • • /N_1 is a Hamilton path in H' between /0 := c'12 and /N_1 := c'13. For 0 < i < N -1, let Fj be the set of 3-colourings of H := G3(T) that agree with /, on V (T'). Then {F0, F1,..., Fn _1} is apartition of the vertices of H, and H[Fj] = Q2S+2i, 0 < i < N - 1. If /j_1 and /, differ on the colour of a vertex of V(T') \ {u, v}, then |[Fj_1, Fj] | = 22s+2t and H[Fj_1 U Fj] = Q2S+2t □ K2 = Q2,+2t+1. If /i_1 and /j differ on the colour of u then |[Fj_1, F,]| = 22t, and the subgraph of H induced by the M. Cavers and K. Seyffarth: Reconfiguring vertex colourings of 2-trees 697 endpoints of the edges of [ii-i, Fi] is Q2i+i. Otherwise /j_i and / differ on the colour of v implying that | [Fj_1, Fj] | = 22s, and the subgraph of H induced by the endpoints of the edges of [Fj_i, Fj] is Q2s+i. Consider the spanning subgraph G of H with edge set Note that by Remark A.4, G is a connected B-graph. Let (A, B) be the bipartition of G and assume ci2 G A. Since H is bipartite by Lemma A.2, and dH(ci2, ci3) is odd, ci3 G B. As ci2 G F0 and ci3 G FN_i, it follows from Corollary A.7 that there is a Hamilton path in G between ci2 and ci3. Base Case 3. Let T = f2sfc+iji with s, k > 1, and let T' ^ P4k_2 be obtained from T by deleting the 2s + 2 leaves and the vertices u and v. Define u', v' as the leaves of T' so that in T, u' is adjacent to u and v' is adjacent to v. Let (A', B') denote the bipartition of T' with A' C A, B' C B, and define cj to be the vertex of H' := G3(T') with cj^ (a) = i for all a G A' and cj (b) = j for all bG B'. By Case 1, H' has a Hamilton path between c'i2 and c'i3. Let V(H') := {/0, /i;..., /N_i}. Since H' has a Hamilton path we may assume that /0/i • • • /N_i is a Hamilton path in H' between /0 := c'i2 and /N_i := c'i3. For 0 < i < N - 1, let Fj be the set of 3-colourings of H := G3(T) that agree with /i on V(T'). Then {F0, Fi,..., Fw_i} is a partition of the vertices of H, and H[Fj] D P4 □ P22s+2, 0 < i < N - 1. This follows since GL(Kij2s+i) has a Hamilton path P22s+2 where L is an assignment of lists in which vertices of degree one have lists {1, 2, 3} and the remaining vertex has list {1, 2}. If /j_i and /j differ on the colour of a vertex of V(T') \ {u', v'}, then |[Fj_i,Fj]| = 4• 22s+i andH[Fj_iUFj] D (P4 □ P22=+i) □ K2. If /j_i and /j differ on the colour of v' then | [Fj_i, Fj] | =2 • 22s+i, and the subgraph of H induced by the endpoints of the edges of [Fj_i, Fj] contains (P2 □ P22s+i) □ K2. Otherwise /j_i and /j differ on the colour of u' implying that | [Fj_i, Fj] | =4 • 22s, and the subgraph of H induced by the endpoints of the edges of [Fj_i, Fj] contains (P4 □ P22s) □ K2. Consider the spanning subgraph G of H with edge set Note that by Remark A.4, G is a connected B-graph . Let (A, B) be the bipartition of G and assume ci2 G A. Since H is bipartite by Lemma A.2, and (ci2, ci3) is odd, ci3 G B. As ci2 G F0 and ci3 G FN_i, it follows from Corollary A.7 that there is a Hamilton path in G between ci2 and ci3. Induction Step. Now suppose T is a tree with bipartition (A, B), where |A| := I > 1 and |B| := r > 1 are both odd and T is not isomorphic to any of the graphs in Base Cases 1 to 3. If every pair of leaves x, y G A or x, y G B satisfy (x, y) < 2, then T = fsfct; since r > 1 are both odd, T is isomorphic to one of the graphs in Base Cases 1 to 3. Thus, there are leaves x, y G A (or x, y G B) with (x, y) > 3. Case 1. If T - {x,y} is a star, then T is the graph obtained from Kij2s+i, s > 1, by subdividing two of its edges. Let u g V(T) be the vertex of degree 2s + 1 and v G V(T) a leaf adjacent to u. Define T' := T[{u, v}] and observe T' = K2. 698 Ars Math. Contemp. 17 (2019) 591-615 Let W' := G3(T') have Hamilton path /0/1/2/3/4/5 := c'12c'32c'31c'21c'23c'13, where cij(u) = i and cjj(v) = j. For 0 < i < 5, let Fj be the set of 3-colourings of W := G3(T) that agree with / on V(T'). Then {F0, F1;..., F5} is a partition of the vertices of W, and W[Fj] = P4 □ P4 □ Q2s-2, 0 < i < 5. If /i-1 and / differ on the colour of vertex v, then |[Fi-1, Fj]| = 4 • 4 • 22s-2 and W[Fj-1 U Fj] = (P4 □ P4 □ Q2s-2) □ K2. If /j-1 and / differ on the colour of u then | [Fj-1; Fj] | = 4, and the subgraph of W induced by the endpoints of the edges of [Fj-1, Fj] is C4. Consider the spanning subgraph G of W with edge set Note that by Remark A.4, the graph P4 □ P4 □ Q2s-2 is Hamilton laceable since P4 □ P4 □ P22s-2 is a Hamilton laceable spanning subgraph. Thus, G is a connected B-graph. Let (A, B) be the bipartition of G and assume c12 6 A. Since W is bipartite by Lemma A.2, and dH(c12, c13) is odd, c13 6 B. As c12 6 F0 and c13 6 F5, it follows from Corollary A.7 that there is a Hamilton path in G between c12 and c13. Case 2. Suppose NT(x) := {x'} and NT(y) := {y'}, and that T' := T - {x, y} is not a star. Let (A', B') denote the bipartition of T' with A' C A, B' C B, and define cj^ to be the vertex of W' := G3(T') with cj^ (a) = i for all a 6 A' and cj^(b) = j for all b €E B'. By the inductive hypothesis, W' has a Hamilton path between c'12 and c'13. Let V(W') := {/0, /1;..., /N-1}. Since W' has a Hamilton path we may assume that /0/1 • • • /N-1 is a Hamilton path in W' between /0 := c'12 and /N-1 := c'13. For 0 < i < N - 1, let Fj be the set of 3-colourings of W := G3 (T) that agree with /j on V(T'). Then {F0, F1;..., FN-1} is a partition of the vertices of W, and W[Fj] = C4, 0 < i < N — 1. If /j-1 and /j differ on the colour of a vertex of V(T') \ {x', y'}, then |[Fj-1; Fj]| =4 and W[Fj-1 U Fj] = Q3. Otherwise, /j-1 and /j differ on the colour of x' or y', implying that | [Fj-1; Fj] | = 2, the subgraph of W induced by the endpoints of the edges of [Fj-1, Fj] is a 4-cycle, and W[Fj-1 U Fj] = P4 □ K2. Consider the spanning subgraph G of W with edge set Note that G is a connected B-graph. Let (A, B) be the bipartition of G and assume ci2 g A. Since H is bipartite by Lemma A.2 and dH(c12, c13) is odd, c13 g B. As c12 g F0 and c13 g Fn_1, it follows from Corollary A.7 that there is a Hamilton path in G between c12 and c13. □ ARS MATHEMATICA CONTEMPORANEA Author Guidelines Before submission Papers should be written in English, prepared in etex, and must be submitted as a PDF file. The title page of the submissions must contain: • Title. The title must be concise and informative. • Author names and affiliations. For each author add his/her affiliation which should include the full postal address and the country name. If avilable, specify the e-mail address of each author. Clearly indicate who is the corresponding author of the paper. • Abstract. A concise abstract is required. The abstract should state the problem studied and the principal results proven. • Keywords. 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