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). t 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/
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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
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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,g,7 and r = r^,
where 0 < 7 < a < P + 7, m G , 0 < ^ < k with jJ^ \ , and Aut(rm,M) = (Ga,p,7 X Zk).Z2.
The groups Ga,g,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)
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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 <T, and let |T : H | = pa for a prime p. Then one of the following holds.
(1)	T = PSL(n, q) and H is the stabilizer of a line or hyperplane. Furthermore, |T : H| = (qn — 1)/(q — 1) = pa and n must be a prime.
(2)	T = An and H = An-i with n = pa.
(3)	T = PSL(2,11) and H = A5.
(4)	T = M23 and H = M22 or T = M11 and H = M10.
(5)	T = PSU(4, 2) = PSp(4, 3) and H is the parabolic subgroup of index 27.
For a group G and a prime p, denote by Op (G) the largest normal p-subgroup of G, and by Op> (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).
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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) / $(OP (G)).
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Proposition 2.10 ([50, Lemma 2.9]). For a finite group G and a prime p, let H = Op(G) and V = H/$(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.
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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 | |T,| 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, Oi2,..., «¿q} and Aj = {
jl,aJ2,
;.q} for 1 < i, j < 3'
jQ
n-2
Recall that Ti is 2-transitive on A, and Aj. Since Ti acts equivalently on A, 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
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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?(<ii)P-qP"1), that is, p"+i | |Ti|, a contradiction. If (q - 1, d) = 1 then Ti has a cyclic subgroup of order
id-q—i
qd1
1-= p", contradicting that G is non-abelian. This completes the proof of Claim 1.
Claim 2. CA. (Op (A*)) < Op (A*).
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Suppose that D is a component of A*, that is, a subnormal quasisimple subgroup of A*. Then D = D' and D/Z(D) = T, a non-abelian simple group. By Proposition 2.5, D has a proper subgroup C of p-power index and Z(D) is a p-group. Since |D : C| = |D : CZ(D)| • |CZ(D) : C|, we have that |D : CZ(D)| is a p-power. If D = CZ(D) then D = D' = C', contradicting that C is a proper subgroup of D. Thus, CZ(D) = D. Since |D/Z(D) : CZ(D)/Z(D)| = |D : CZ(D)|, we have that D/Z(D) has a proper subgroup CZ (D)/Z (D) of p-power index. By Proposition 2.7(1), p f |M (D/Z (D))| and hence p f |Z(D)|. Since Z(D) is a p-group, we have Z(D) = 1 and so D = T. Recall that E(A*) is the product of all components of A*. Then D < E(A*) and since D = T, D is a direct factor of E(A*). Clearly, D° is also a direct factor of E(A*) for any a e A*. It follows that A* contains a minimal normal subgroup which is isomorphic to Te with e > 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,
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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* = HN^ (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,Se) 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.
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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,Se ,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 = {aejS | 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 <Y < a < [ + 7.
Lemma 4.3. In , the following properties hold:
(1)	For any non-negative integers i,j, we have albj = bjal(1+pJ)j;
(2)	Let 6 G Aut(G ) such that a6 = bman with (m,p) = 1. Then [ < a.
Proof. From b-1ab = a1+pY, we have b-1alb = ai(1+pY) and hence b-jalbj = ai(1+pY)j. Part (1) follows. Since a6 = bman, we have o(bman) = o(a) = pa, and since (a)<Ga^n, we have (bman) < Ga^x/. Then (p,m) = 1 implies Ga^r/ = (a,bman) = (a}(bman), and hence
a+ß = |G , = Ka)|-Kbma")l = Pa ■ Pa < pa pa 1 a'ß'Y 1 |(a)n(bma«)| |(a) n (bma«)| " P P
that is, 3 < a. If [ = a, then |(a) n (bman}| = 1 and hence	= (a) x (bman),
contradicting that	is non-abelian. Thus, [ < a and part (2) follows.	□
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A graph r is called locally-transitive if the stabilizer Aut(T)„, for any u G V(T), is transitive on the neighborhood of u in V(T).
Theorem 4.4. There are no connected locally-transitive bipartite bi-p-metacirculants of valency less than p for any odd prime p.
Proof. Suppose to the contrary that r is a connected locally-transitive bipartite bi-Cayley graph of valency less than p over a non-abelian metacirculant p-group G. Since p is odd, the two orbits of G are exactly the parts of r, and we may assume that r = BiCay(G, 0,0, S), where 1 G S, |S| < p and G = (S}. Let A = Aut(r). Since r has valency less thanp, A1} is ap'-group, and by Theorem 1.1, G < A. Write F = {ct6,s | 0 G Aut(G), S6 = g-1S} and L = {0 | ct6,s G F}. By Proposition 4.1, A i0 = F, and by Lemma 4.2, F = L.
Assume that G is non-split. By Lindenberg [36], the automorphism group of G is a p-group. Thus, p | |L| and hence p | |A1} |, a contradiction.
Assume that G is split. Then G = Ga ^i7, as defined in Equation (2.1). Since F is a p'-group and F = L, Proposition 2.1 implies that F is cyclic and |F| | (p - 1). Let |S1| = k and F = (ct6,s}, where 0 G Aut(G), s G S and S6 = s-1S. Since F is transitive on S1, ct6,s permutes all elements in S1 cyclically, and so ct^ s fixes all elements in S1. By Proposition 4.1, F is faithful on S1, implying that ct^ s = 1. It follows that ct6,s has order k and is regular on S1. Since F = L, 0 also has order k. Furthermore, S1 = l^ = {11,S1, (ss6 )1,..., (ss6 ••• s6'-2 )1} and I1 = (ss6 ••• s6'-1 )1, that is, S = {1, s, ss6,..., ss6 • • • s6'-2} and
û	ÛK—1
ss6 ••• s6 =1.	(4.1)
Note that for any t g Aut(G), we have r = BiCay(G, 0, 0, S) = BiCay(G, 0, 0, ST),
where ST = {1,t,tt6T , ...,tt6T ••• t(6T )k-2} and tt6T ••• t(6T )k-1 = 1 with t = sT. By
Proposition 2.1, all cyclic groups of order k in Aut(G) are conjugate, and so 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 G Ga,^i7. By Lemma 4.3, a®bj = bjai(1+pY, and since a6 = ae and
b6 = b, we have ss6 • • • s6' 1 = bfciae for some e G Zpa. By Equation (4.1), bki = 1, that
is, ki = 0 (mod p^ ). Since k < p, we have i = 0 (mod p^ ), and hence G = (S} = - - - - - - k — 2 (1,aj ,aj aje,...,aj aje ••• aje } < (a}, a contradiction. This completes the proof. □
Theorem 1.2 is a direct corollary of Theorem 4.4.
To prove Theorem 1.3, we need two technical lemmas on integer numbers.
Lemma 4.5. Let p be an odd prime and a a positive integer. Let e be an element of order k (k > 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 Z*pa. 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*„ (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.
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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
E t = (e - 1)-1 E (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 J2teT(tx + y) = ^teT t, and hence ky = (1 - x) ^teT 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 Zr} = (e), we have x(e) = (e) in Zp„, that is, x = e1 (mod pa) for some l G Zr. 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 Zr. 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 Zr. 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 oddprime 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 -p^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 = Zp«-i(p-1), squares in Zpa consists of the unique subgroup of order (p-1ypa-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 | (p-1y. 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, p, 7 be positive integers such that 0 <7<a < ft + 7. Let e be an element of
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order k (k > 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 Zfc} 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 |.
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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 Se = S. Recall that
k > 2.
Assume k > 2. Since a G S, we have ae G Se = S = U U V. If ae G V then ae = 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)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, ae G U, and hence, ae = aj
for some j G T. If ae = 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 aej1 by if necessary), that is, a® = bman, (bman)® = bma(1+pY)m-n and (bma(1+pY)m-n)® = a.
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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) <pa.
Summing up, we have proved p \ |A1o |, and this completes the proof of Claim 1.
By Claim 1, A1o is a p'-group, and by Theorem 1.1, r is normal. By Proposition 4.1,
A1o = X1o = F = {ct6jS | a G Aut(G), s G S, S6 = s-1S}, and by Lemma 4.2, F = L < Aut(G), where L = {0 | ct6jS G F}. By Proposition 2.1, F is cyclic, and since aT,a G A1o, F is transitive on S1 or has two orbits. By Lemma 4.2, F is faithful on S1, and since F is cyclic, either F is regular on S1, or F = (aT,a).
We suppose that F is regular on S1 and will obtain a contradiction. Note that F = Z2k and |F : (aT,a)| = 2. Then (aT,a) < F, and the two orbits U1 and V1 of (aT,a) consist of an imprimitive block system of F on S1. By the regularity of F, there exists ct6jS G F such that = (bman)1, implying that s = bman and S6 = s-1S = (6ma")-1S, and hence Ule,s = V1 because 11 G U1 and (bman)1 G V1. It follows that a6 G (6ma")-1V. It is easy to see that
where
(bman)-1S = (bman)-1U U (6ma")-1V,
(bman)-1U = {(bman)-1an | n G T} = {6-ma-n(1+pY) | n G T}
= |6-man | n e T - n(1 + pY)-m} and (6ma")-1V = j(6ma")-16man | n e T'}
= ja-n+n | n e T'} = jan | n e T' - n}.
Since T' = T[(1 + pY)m + n(e - 1)] + n, we have
(6man)-1 V = jan | n e T[(1 + pY)m + n(e - 1)]}.
Let a8 = ar e (6ma")-1V for some r e T[(1 + pY)m + n(e - 1)]. Since pa = o(a8) = o(ar), we have r e Zp„. Note that
U8 = janr | n e T} = jan | n e Tr} C (6ma")-1V.
Then
U6 = (6man) V = {an | n G T[(1 + pY)m + n(e — 1)]},
and so Tr = T[(1 + pY)m + n(e - 1)] in Zpa. By Lemma4.5, r = (1 + pY)m + n(e - 1).
Since S8 = (6ma")-1S = (6ma")-1U U (6ma")-1V, we have V8 = (6ma")-1U. In particular, (6man)8 = b-ma' for some t e T - n(1 + pY)-m. For n e T', since
(6man )8 = [(6man)an-n]8 = 6-maiar(n-n) =
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we have
{b-man | n G T - n(1 + pY)-m}
= (bma")-1U = Ve = {(bman)e | n G T'} =	| n G T'} = {b-man | n G T'r - rn +t}.
This implies that
T - n(1+ pY)-m
= T 'r — rn + t = Tr [(1 + pY )m + n(e — 1)] + rn — rn + t = Tr[(1 + pY)m + n(e - 1)] + t = T[(1 + pY)m + n(e - 1)]2 + t
in Zpa. By Equation (4.2), e£(1 + pY)m = [(1 + pY)m + n(e - 1)]2. It follows that
T - n(1 + pY)-m = Te£(1 + pY)m + t,
and hence
T = Te£(1 + pY )m + t + n(1 + pY )-m.
By Lemma 4.5, there exists £' G Zk such that e£ = e£(1 + pY)m (mod pa), that is, e£ — = (1 + PY)m (mod pa). Since e is an element of order k, we have (1 + pY)mk = 1 (mod pa) and since (mk,p) = 1, we havepY = 0 (mod pa), implying that 7 > 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 G U1 and (bma")1 G V1. By Proposition 4.1, it suffices to construct a A g Aut(G) such that
ÎA,bman,1 G I = {¿A,x,y | a G Aut(G), SA = y-1S-1x},
that is SA = S-1bma", because (10,11)<^,bm»n,i = ((6ma")1,10).
Let ^ = -(1 + pY)m - n(e - 1) and v = -(e - 1)"V - (e - 1)-V Then ^ + 1 + n(e - 1) = 0 (mod pY) and hence
v - ^n = -(e - 1)-1^,2 - (e - 1)-V - ^n
= -(e - 1)-V[m + 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 = {aw | n G T} = {an | n G T^} and V-1bma" = {(bman)-1bma" | n G T'}
= {a-n+n | n G T'} = {an | n G -T' + n}.
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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 € T'} = {bmav-M"aw | n € T'} = {bman | n € T> — Mn + v} and U-1bman = {(an)-1bman | n € T} = {bma-n(1+pY)m+n | n € T} = {bman | n € —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,^,7 and r - rm,fc,£, where m € Z*„_y and 0 < t < 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 € 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 € S}, say U1 and V1 with 11 € U1, where U and V are subsets of G with 1 € 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 € Aut(G), s € S, Se = s-1S}. By Lemma 4.2, A1o is faithful on S1, and F = L := {a | ae,s € F}.
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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 ^ 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 = t^ = {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},
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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.
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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
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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
pa = 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>.
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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.	□
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