ARS MATHEMATICA CONTEMPORANEA ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 15 (2018) 499-522 https://doi.org/10.26493/1855-3974.1409.e54 (Also available at http://amc-journal.eu) Arc-transitive cyclic and dihedral covers of pentavalent symmetric graphs of order twice a prime* Yan-Quan Feng t, Da-Wei Yang , Jin-Xin Zhou Mathematics, Beijing Jiaotong University, Beijing, 100044, P.R. China Received 23 May 2017, accepted 10 May 2018, published online 9 September 2018 A regular cover of a connected graph is called cyclic or dihedral if its transformation group is cyclic or dihedral respectively, and arc-transitive (or symmetric) if the fibre-preserving automorphism subgroup acts arc-transitively on the regular cover. In this paper, we give a classification of arc-transitive cyclic and dihedral covers of a connected pentava-lent symmetric graph of order twice a prime. All those covers are explicitly constructed as Cayley graphs on some groups, and their full automorphism groups are determined. Keywords: Symmetric graph, Cayley graph, bi-Cayley graph, regular cover. Math. Subj. Class.: 05C25, 20B25 1 Introduction All groups and graphs considered in this paper are finite, and all graphs are simple, connected and undirected, unless otherwise stated. Let G be a permutation group on a set Q and let 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. Denote by Zn, Z*n, Dn, An and Sn the cyclic group of order n, the multiplicative group of units of Zn, the dihedral group of order 2n, the alternating and symmetric group of degree n, respectively. For two groups M and N, we use MN, M.N, M x N and M x N to denote the product of M and N, an extension of M by N, a split extension of M by N and the direct product of M and N, respectively. For a subgroup H *This work was supported by the National Natural Science Foundation of China (11731002, 11571035, 11711540291, 11671030) and by the 111 Project of China (B16002). t Corresponding author. E-mail addresses: yqfeng@bjtu.edu.cn (Yan-Quan Feng), dwyang@bjtu.edu.cn (Da-Wei Yang), jxzhou@bjtu.edu.cn (Jin-Xin Zhou) Abstract ©® This work is licensed under https://creativecommons.org/licenses/by/3.0/ 500 Ars Math. Contemp. 15 (2018) 441-466 of a group G, CG(H) means the centralizer of H in G and NG(H) means the normalizer of H in G. For a graph r, we denote its vertex set, edge set and full automorphism group by V(r), E(r) and Aut(r), respectively. An s-arc in r is an ordered (s + 1)-tuple (v0, vi,..., vs) of vertices of r such that {vi-1, v¿} G E(r) for 1 < i < s, and vi-1 = vi+1 for 1 < i < s. A 1-arc is just an arc. A graph r is (G, s)-arc-transitive for a subgroup G of Aut(r) if G acts transitively on the set of s-arcs of r, and (G, s)-transitive if r is (G, s)-arc-transitive but not (G, s + 1)-arc-transitive. A graph r is said to be s-arc-transitive or s-transitive if it is (Aut(r), s)-arc-transitive or (Aut(r), s)-transitive, respectively. In particular, 0-arc-transitive means vertex-transitive, and 1-arc-transitive means arc-transitive or symmetric. A graph r is edge-transitive if Aut(r) is transitive on the edge set E(r). Let r be a graph and N < Aut(r). The quotient graph rN of r relative to the orbits of N is defined as the graph with vertices the orbits of N on V(r) and with two orbits adjacent if there is an edge in r between those two orbits. In particular, for a normal subgroup N of Aut(r), if r and rN have the same valency, then rN is a normal quotient of r, and if r has no proper normal quotient, then r is basic. To study a symmetric graph r, there is an extensive used strategy consisting of two steps: the first one is to investigate normal quotient graph rN for some normal subgroup N of Aut(r) and the second one is to reconstruct the original graph r from the normal quotient rN by using covering techniques. This strategy was first laid out by Praeger (see [31]), and it is usually done by taking the normal subgroup N as large as possible and then the graph r is reduced a basic graph. In the literature, there are many works about basic graphs (see [1, 14, 16] for example), while the works about the second step, that is, covers of graphs, are fewer. An epimorphism n: f ^ r of graphs is called a regular covering projection if Aut(f) has a semiregular subgroup K whose orbits in V (f) coincide with the vertex fibres n-1 (v), v G V (r), and whose arc and edge orbits coincide with the arc fibres n-1((u, v)) and the edge fibres n-1({u,v}), {u, v} G E(r), respectively. In particular, we call the graph f a regular cover or a K-cover of the graph r, and the group K the covering transformation group. If K is dihedral, cyclic or elementary abelian, then T is called a dihedral, cyclic or elementary abelian cover of r, respectively. An automorphism of r is said to be fibre-preserving if it maps a vertex fibre to a vertex fibre, and all such fibre-preserving automorphisms form a group called the fibre-preserving group, denoted by F. It is easy to see that F = NAut(p) (K). If r is F-arc-transitive, we say that T is an arc-transitive cover or a symmetric cover of r. For an extensive treatment of regular cover, one can see [3, 26, 27]. Covering techniques have long been known as a powerful tool in algebraic and topo-logical graph theory. Application of these techniques has resulted in many constructions and classifications of certain families of graphs with particular symmetry properties. For example, by using covering techniques, Djokovic [10] constructed the first infinite family of 5-arc-transitive cubic graphs as covers of Tutte's 8-cage, and Biggs [4] constructed some 5-arc-transitive cubic graphs as covers of cubic graphs that are 4-arc-transitive but not 5-arc-transitive. Gross and Tucker [18] proved that every regular cover of a base graph can be reconstructed as a voltage graph on the base graph. Later, Malnic et al. [26] and Du et al. [12] developed these ideas further in a systematic study of regular covering projections of a graph along which a group of automorphisms lifts. Based on the approaches studied in [12, 26], many arc-transitive covers of symmetric graphs of small orders and small valencies have been classified. For example, Pan et al. [29] studied arc-transitive cyclic covers of some complete graphs of small orders. One Y.-Q. Feng et al.: Arc-transitive cyclic and dihedral covers of pentavalent symmetric graphs 501 may see [2, 27] for other works. Moreover, a new approach was proposed by Conder and Ma [7, 8] by considering a presentation (quotient group) of a universal group, which can be obtained from Reidemeister-Schreier theory, and representation theory and other methods are applied when determining suitable quotients. As an application, arc-transitive abelian covers of the complete graph K4, the complete bipartite graph K3 3, the 3-dimensional hypercube Q3, the Petersen graph and the Heawood graph, were classified. Later, arc-transitive dihedral covers of these graphs were determined by Ma [25]. For arc-transitive covers of infinite families of graphs, Du et al. studied 2-arc-transitive elementary abelian and cyclic covers of complete graphs Kn in [11, 13] and Kn n - nK2 in [32, 33]. Recently, Pan et al. [28] determined arc-transitive cyclic covers of the complete bipartite graph Kp,p of order 2p for a prime p. Compared with symmetric covers of graphs of small orders and valencies, there are only a few contributions on symmetric covers of infinite families of graphs. Arc-transitive covers of non-simple graphs were also considered in literature. For example, regular covers of the dipole Dipk (a graph with two vertices and k parallel edges) were extensively studied in [2, 16, 26, 27, 34]. Such covers are called Haar graphs, and in particular, cyclic regular covers of dipoles are called cyclic Haar graphs, which can be regarded as a generalization of bipartite circulants and were studied in [21] (also see [15]). Construction of Haar graphs have aroused wide concern. Marusic et al. [26] studied elementary abelian covers of the dipole Dipp for a prime p. In particular, symmetric elementary abelian covers and Zp x Zp-covers for a prime p of the dipole Dip5 were classified completely in [16] and [34], respectively. Let p be a prime. Pentavalent symmetric graphs of order 2p were classified by Cheng and Oxley in [6], which are the complete graph K6 of order 6 and a family of Cayley graphs CDp with p = 5 or 5 | (p - 1) on dihedral groups (see Proposition 3.4). It has been shown that many pentavalent symmetric graphs are regular covers of them, see [16, 34]. In this paper, we consider arc-transitive cyclic and dihedral covers of these graphs. For K6, the cyclic covers have been classified in [29], which should be the complete bipartite graph K6,6 and the Icosahedron graph I12 (note that Ii2 is missed in [29]). For CDp, the cyclic covers consist of six infinite families of graphs, which are Cayley graphs on generalized dihedral groups. In particular, one family of graphs are cyclic Haar graphs and the other five families are non-cyclic Haar graphs. What is more, the full automorphism groups of them are determined. Arc-transitive dihedral covers of K6 and CDp are also classified, and there are only four sporadic graphs of order 24, 48, 60 and 120, respectively. A similar work about cubic graphs was done by Zhou and Feng [37]. Different from regular covers of graphs mentioned above, the method to classify arc-transitive cyclic covers used in this paper is related to the so called bi-Cayley graph. A graph r is a bi-Cayley graph over some group H if Aut(r) has a semiregular subgroup isomorphic to H having exactly two orbits on V(r). Clearly, a Haar graph is a bipartite bi-Cayley graph. Recently, Zhou and Feng [38] gave a depiction of the automorphisms of bi-Cayley graphs (see Section 4), and based on this work, we classify the cyclic covers. In particular, all these covers are bi-Cayley graphs over some abelian groups. Note that vertex-transitive bi-Cayley graphs of valency 3 over abelian groups were determined in [36], while the case for valency 5 is still elusive. Indeed, even for arc-transitive pentavalent bi-Cayley graphs over abelian groups, it seems to be very difficult to give a classification, and one may see [2, 16, 34] for partial works. The paper is organized as follows. After this introductory section, in Section 2 we give 502 Ars Math. Contemp. 15 (2018) 441-466 some notation and preliminary results. In Section 3, several infinite families of connected pentavalent symmetric graphs are constructed as Cayley graphs on generalized dihedral groups Dih(Zmpe x Zp), where e, m are two positive integers and p is a prime such that (m, p) = 1. In Section 4, it is proved that these Cayley graphs include all arc-transitive normal bipartite bi-Cayley graphs over Zmpe x Zp, and using this result, arc-transitive cyclic and dihedral covers of connected pentavalent symmetric graphs of order 2p are classified in Sections 5 and 6, respectively. In Section 7, the full automorphism groups of these covers are determined. 2 Preliminaries In this section, we describe some preliminary results which will be used later. The following result is important to investigate symmetric pentavalent graphs. Proposition 2.1 ([19, Theorem 1.1]). Let r be a connected pentavalent (G, s)-transitive graph for some G < Aut(r) and s > 1, and let v G V(r). Then one of the following holds: (1) s = 1 and Gv = Z5, D5 or D\0; (2) s = 2 and Gv = F20, F20 x Z2, A5 or S5, where F20 is the Frobenius group of order 20; (3) s = 3 and Gv = F20 x Z4, A4 x A5, S4 x S5 or (A4 x A5) x Z2 with A4 x Z2 = S4 and A5 xZ2 = S5; (4) s = 4 and Gv = ASL(2,4), AGL(2,4), AEL(2,4) or ArL(2,4); (5) s = 5 and Gv = Z2 x rL(2,4). From [24, Theorem 9], we have the following proposition. Proposition 2.2. Let r be a connected G-arc-transitive graph of prime valency, and let N be a normal subgroup of G. If N has at least three orbits, then it is semiregular on V(r) and the kernel of G on the quotient graph rN. Furthermore, rN is G/N-arc-transitive, and r is a regular cover of rN with N as the covering transformation group. Let G and E be two groups. We call an extension E of G by N a central extension of G if E has a central subgroup N such that E/N = G, and if further E is perfect, that is, if it equals its derived group E', we call E a covering group of G. Schur 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 (see [22, V, § 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 Mult(G). Lemma 2.3. Let G be a group, and let N be an abelian normal subgroup of G such that G/N is a non-abelian simple group. If N is a proper subgroup of CG(N), then G = G'N and G' n N < Mult(G/N). Proof. Since N is aproper subgroup of CG(N), we have 1 = CG(N)/N < G/N, forcing CG (N)/N = G/N because G/N is simple. Thus G = CG (N) and it is a central extension of G/N by N. Since G/N = (G/N)' = G'N/N = G'/(G' n N), we have G = G'N, Y.-Q. Feng et al.: Arc-transitive cyclic and dihedral covers of pentavalent symmetric graphs 503 and since G = {GN)' = (G')', G is a covering group of G/N. Hence G n N < Mult(G/N). □ Denote by soc(G) the socle of G, that is, the product of all minimal normal subgroups of G. A list of all proper primitive permutation groups of degree less than 1000 was given by Dixon and Mortimer [9, Appendix B], and based on the list, we have: Lemma 2.4. Let G be a primitive permutation group on a set Q and let a G Q, where |Q | G {2, 4, 6, 8,12,16, 24, 72,144, 288, 576}. If Ga is solvable, then either G < AGL(n, 2) and |Q| = 2n with 1 < n < 4, or soc(G) = PSL(2,p), PSL(3, 3) or PSL(2,q) x PSL(2, q) with |Q| = p + 1, 144 or (q + 1)2 respectively, wherep G {5,7,11,23,71} and q G {11, 23}. Proof. If |Q| = 2 or 4, then G < S2 ^ AGL(1, 2) or G < S4 = AGL(2,2), respectively. Let |Q| > 6 and write N := soc(G). Then N < G and Na < Ga. Since Ga is solvable, Na is solvable. By [9, Appendix B, Tables B.2 and B.3], G is an affine group, N = A|fi|, or G is isomorphic to one group listed in [9, Tables B.2 and B.3]. If G is affine, then |Q| is a prime power and thus |Q| = 2n with n =3 or 4. By [9, Theorem 4.1A (a)], we have G < AGL(n, 2). If N = A|q| then Na = A|q|-i, which is insolvable because |Q| — 1 > 5, a contradiction. In what follows we assume that G is isomorphic to one group listed in [9, Tables B.2 and B.3]. Note that all groups in the tables are collected into cohorts and all groups in a cohort have the same socle. Assume that |Q| = 144. By [9, Table B.4, pp. 324], there are one cohort of type C, two cohorts of type H and four cohorts of type I (see [9, Table B.1, pp. 306] for types of cohorts of primitive groups) of primitive groups of degree 144. For the cohort of type C, by [9, Table B.2, pp. 314], N = PSL(3, 3) and Na = Z13 x Z3. For the two cohorts of type H, by [9, Table B.2, pp. 321], they have the same socle N = M12 and Na = PSL(2,11). For the four cohorts of type I, by [9, Table B.3, pp. 323], N = A12 x A12, PSL(2,11) x PSL(2,11), M11 x M11 or M12 x M12 and Na = A11 x A11, (Z11 x Z5) x (Z11 x Z5), M10 x M10 or M11 x M11, respectively. Since Na is solvable, we have N = PSL(3, 3) or PSL(2,11) x PSL(2,11). For |Q| G {6, 8,12,16,24,72,288, 576}, by [9, Tables B.2, B.3 and B.4], a similar argument to the above paragraph implies that either N = PSL(2,23) x PSL(2,23) with degree 232 = 576 and Na = (Z23 x Zn) x (Z23 x Zn), or N = PSL(2,p) with degree p + 1 and Na = Zp x Zp-l wherep G {5,7,11,23,71}. □ 3 Graph constructions as Cayley graphs Let G be a finite group and S a subset of G with 1 G S and S-1 = S. The Cayley graph r = Cay(G, S) on G with respect to S is defined to have vertex set V(r) = G and edge set E(r) = {{g, sg} | g g G, s g S}. It is well-known that Aut(r) contains the right regular representation R(G) of G, the acting group of G by right multiplication, and r is connected if and only if G = (S), that is, S generates G. By Godsil [17], NAut(r)(R(G)) = R(G) x Aut(G, S), where Aut(G, S) = {a G Aut(G) | Sa = S}. A Cayley graph r = Cay(G, S) is said to be normal if R(G) is normal in Aut(r), and in this case, Aut(r) = R(G) x Aut(G, S). For an abelian group H, the generalized dihedral group Dih(H) is the semidirect product H x Z2, where the unique involution in Z2 maps each element of H to its inverse. In particular, if H is cyclic, then Dih(H) is a dihedral group. In this section, we introduce 504 Ars Math. Contemp. 15 (2018) 441-466 several infinite families of connected pentavalent symmetric graphs which are constructed as Cayley graphs on generalized dihedral groups. Example 3.1. Let Dih(Zf) = {a, b,c,h | a5 = b5 = c5 = h2 = [a, b] = [a, c] = [b, c] = 1, ah = a-1, bh = b-1,ch = c-1}, and define CgV53 = Cay(Dih(Zf), {h, ah, bh, ch, a-1b-1c-1h}). By [34, Theorem 1.1], Aut(C£D53) ^ Dih(Zf) x S5 and CGD53 is the unique connected pentavalent symmetric graph of order 250 up to isomorphism. Let m be a positive integer. Consider the following equation in Zm x4 + x3 + x2 + x +1 = 0. (3.1) In view of [15, Lemma 3.3], we have the following proposition. Proposition 3.2. Equation (3.1) has a solution r in Zm if and only if (r, m) e {(0,1), (1,5)} or m = 5tpe11 p22 ■ ■ ■ pess and r is an element in Z*m of order 5, where t < 1, s > 1, ei > 1 and pi's are distinct primes such that 5 | (p^ — 1). The following infinite family of Cayley graphs was first constructed in [23]. Example 3.3. Let m > 1 be an integer such that Equation (3.1) has a solution r in Zm. Then m = 5, 11 or m > 31. Let CDm = Cay(Dm, {b, ab, ar+1b, ar'2+r+1b, ar'3+r'2 +r+1b}) be a Cayley graph on the dihedral group Dm = {a,b | an = b2 = 1, ab = a-1}. For m = 5 or 11, by [6], Aut(CDm) = (S5 x S5) x Z2 or PGL(2,11), respectively. In particular, CV5 = K55. For m > 31, by [23, Theorem B and Proposition 4.1], Aut(CDm) = Dm x Z5, and obviously, if m has a prime divisor p with p < m, then Aut(CDm) has a normal subgroup Zm/p, and by Proposition 2.2, CDm is a symmetric Zm/p-cover of a connected pentavalent symmetric graph of order 2p. By [6], we have the following proposition. Proposition 3.4. Let r be a connected pentavalent symmetric graph of order 2p for a prime p. Then r = K6 or CDp with p = 5 or 5 | (p — 1). In the remaining part of this section, we construct five infinite families of Cayley graphs on some generalized dihedral groups, and for convenience, we always assume that G = Dih(Zm x Zpe x Zp) = {a, b,c,h | am = bp° = cp = h2 = [a, b] = [a, c] = [b, c] = 1, ah = a-1 ,bh = b-1,ch = c-1} and r is a solution of Equation (3.1) in Zm, that is, r4 + r3 + r2 + r + 1 = 0 (mod m). By Proposition 3.2, m is odd and 52 { m. Example 3.5. Assume that e > 2 and p is a prime such that (m,p) = 1 and 5 | (p — 1). Let A be an element of order 5 in Zpe. Then A is a solution of Equation (3.1) in Zpe. Set T1 (r, A) = {h, hab, har+1bx+1c, har2+r+1bx2+x+1cx4+x+1, ha^+^+^b^+^+^c}, T2 (r, A) = {h, hab, har+1bx+1c, har'2+r+1bx2+x+1cx3+x+1, har'3+r'2+r+1bx3+x2+x+1cx}, T3 (r, A) = {h, hab, har+1bx+1c, har2+r+1bx2+x+1cx2+x+1, har3+r"+r+1bx3+x2+x+1 cx"}. Y.-Q. Feng et al.: Arc-transitive cyclic and dihedral covers of pentavalent symmetric graphs 505 It is easy to see that each of T\ (r, A), T2 (r, A) and T3 (r, A) generates G. Define CGVi„ipexp = Cay(G, Ti(r, A)), i = 1, 2, 3. The maps X X 4 a ^ ar, b ^ b c, c ^ c , h ^ hab; a ^ ar, b ^ bXc, c ^ cX , h ^ hab; a ^ ar ,b ^ b c, c ^ c ,h ^ hab induce three automorphisms of order 5 of G, denoted by a1, a2 and a3 respectively, and a fixes the set Tj (r, A) and permutes its five elements cyclicly. It follows that for each i = 1,2, 3, a G Aut(G, Tj(r, A)) and (R(G), Oj) = G x Z5, which is arc-transitive on CQVmpe xp" The graphs CQVlmpexp, CQV2mpexp and CQV3mpexp for (m, e) = (1,2) have been introduced in [34, Example 4.4], and they are not isomorphic to each other by [34, Lemma 4.5]. Indeed, we can also prove that the graphs CQVlmpe xp, CQV2mpe xp and CQV3mpe xp for each integers m > 1 and e > 2 are not isomorphic to each other. Since the proof is similar to [34, Lemma 4.5], we omit it, and one can see [35] for a detailed proof. Example 3.6. Let p be a prime such that p = 5 or 5 | (p ± 1). Assume that e =1 and (m,p) = 1. Then G = Dih(Zm x Zp x Zp). For p = 5, let A = 0, and for 5 | (p ± 1), let A G Zp satisfying the equation A2 =5 in Zp. Set S(r, A) = {h, hab, har+1c, ha^+V2"^ X)c2"(1+X), ha^+^+V2-1^X)c}. It is easy to see that S(r, A) generates G. Define CGD^pxp = Cay(G,S(r,A)). The map a ^ ar, b ^ b-1c, c ^ b-2 1(3+X)c2 1(1+X) and h ^ hab induces an automorphism of the group G, denoted by a4, which permutes the elements in S(r, A) cyclicly. Then a4 G Aut(G, S(r, A)) and (R(G), a4) = G x Z5 acts arc-transitive on CQVAmpxp. Moreover, for m = 1 or 5, we have r = 0 or 1 respectively, and the map a ^ a-1, b ^ b-2 (1+X)c, c ^ b-2 (1+X)c2 (1+X), h ^ h induces an automorphism p of G. It is easy to check that p G Aut(G, S(r, A)) and (a4,P) = D5. In particular, by [16, Theorem 6.1], CQT>5x5 is the unique connected pentavalent symmetric graph of order 50 up to isomorphism. Example 3.7. Assume that e = 1 andp is a prime such that (m,p) = 1 and 5 | (p - 1). By [34, Case 2, page 14], x4 + 10x2 + 5 = 0 has a root A in Zp. Set S(r, A) = {h, hab, har+1c, har2 +r+1b8-1( X3-X2+7X+V~1( X+1), har3+r2+r + 1b-8-1( X3 + X 2+7 X-1)c8-1( X 3 + X2 + 11 X+3)} It is easy to check that S(r, A) generates G. Define CGD^pxp = Cay(G, S(r, A)). 506 Ars Math. Contemp. 15 (2018) 441-466 The map a ^ ar, b ^ b-1c, c ^ b8 1(*3-*2+7*-7)c2 1(a+i) and h ^ hab induces an automorphism of the group G, denoted by a5, which permutes the elements in S(r, A) cyclicly. Then a5 G Aut(G, S(r, A)) and (R(G),a5) = G x Z5 acts arc-transitive on Let r = CGVlnpexp withp = 5 or 5 | (p - 1), where 1 < i < 5. By Examples 3.53.7, Aut(r,) contains an arc-transitive subgroup R(G) x (oj) for each 1 < i <5. Let N be a subgroup of R(G) x (oj) as listed in Table 1. In particular, for r4 with 5 | (p - 1), since A2 = 5 in Zp, the equation x4 + 10x2 +5 = 0 has a root t such that t2 = 2A — 5 (see Example 3.7). It is easy to compute that Nj = Zmpe and Nj < R(G) x (aj) for each 1 < i < 5 (see [35] for a detailed computation). By Proposition 2.2, we have the following lemma. Table 1: Subgroups of Aut(CGD mpe xp ) for 1 < i < 5. Tj P Nj CGDmpexp 5 (p — 1) (R(a),R(b5 c3^2*2-^1)) CGDmpexp 5 (p — 1) (R(a),R(b-5c2A3+4A2+A+3)) CGDmpexp 5 (p — 1) (R(a),R(b-5c4A3+3A2+2A+1)) CGDmpxp p = 5 (R(a),R(b2c4)) 5 (p — 1) (R(a), R(bi+1cA-3)) (t2 = 2A — 5) CGDmpxp 5 | (p — 1) (R(a),R(b2(A2+5)-1(A3+10A+5)-(A+3)c4)) Lemma 3.8. Let p be a prime such that p = 5 or 5 | (p — 1). Then for each 1 < i < 5, CGDmpexp is a connected symmetric cyclic cover of a connected pentavalent symmetric graph of order 2p. 4 Pentavalent symmetric bi-Cayley graphs over abelian groups Given a group H, let R, L and S be three subsets of H such that R-1 = R, L-1 = L, and 1 ^ R U L. The bi-Cayley graph over H relative to the triple (R, L, S), denoted by BiCay(H, R, L, S), is the graph having vertex set {h0 | h G H} U {h1 | h G H} and edge set {{ho,go} | gh-1 G R} U {{hi,gi} | gh-1 G L} U {{ho,gi} | gh-1 G S}. For a bi-Cayley graph r = BiCay(H, R, L, S), it is easy to see that R(H) can be regarded as a semiregular subgroup of Aut(r) with two orbits, which acts on V(r) by the rule hR(g) = (hg)j, i = 0,1, h, g G H. If R(H) is normal in Aut(r), then r is a normal bi-Cayley graph over H. Let r = BiCay(H, 0,0, S) be a connected bi-Cayley graph over an abelian group H. Then r is bipartite. By [38, Lemma 3.1], we may always assume that 1 G S. Moreover, r = BiCay(H, 0, 0, Sa) for a G Aut(H), and H = (S). Since H is abelian, there is an automorphism of H of order 2, denoted by 7, induced by g ^ g-1, Vg G H. For Y.-Q. Feng et al.: Arc-transitive cyclic and dihedral covers of pentavalent symmetric graphs 507 a G Aut(H) and x G H, define ¿7,1,1 : h0 ^ (h-1)i, hi ^ (h-1)0, Vh G H; &a,x : h0 ^ (ha)0, h1 ^ (xha)1, Vh G H. Set F = jaa,x | a G Aut(H), Sa = x-1S}. Then ¿Y,1,1 G Aut(r) and F < Aut(r)1o (see [38, Lemma 3.3]). Since r is connected, F acts on Nr(10) faithfully. By [38, Theorem 1.1 and Lemma 3.2], we have the following proposition. Proposition 4.1. Let r = BiCay(H, 0, 0, S) be a connected bi-Cayley graph over an abelian group H, and let A = Aut(r). Then NA(R(H)) = R(H)(F, ¿7,1,1) with vertex stabilizer (NA(R(H)))1o = F, andr is isomorphic to the Cayley graph Cay(Dih(H), 7S), where Dih(H) = H x (7). The following lemma is from [2, Theorem 1.1]. Lemma 4.2. Let n be a positive integer and p a prime such that p > 5. Let r be a connected pentavalent symmetric bi-Cayley graph over Znp. Then r = CVnp, as defined in Example 3.3. Let H = (x) x (y) x (z) = Zm x Zpe x Zp, where m and e are two positive integers and p is a prime such that p > 5 and (m,p) = 1. In the remaining of this section, we always let r = BiCay(H, 0,0, S) be a connected pentavalent bi-Cayley graph over H such that NAut(r)(R(H)) is arc-transitive on r. Assume that S = {1, a, b, c, d}. Then H = (a, b, c, d). By Proposition 4.1, there exists a oag G F of order 5 permuting the neighborhood {I1, a1, b1, C1, d1} of I0 in r cyclicly. One may assume that 1 1 = a1 , which implies that g = a because 1^'® = g1, and that b1 = , c1 = , d1 = and 11 = ¿1a'a. It follows that aa = ba-1, ba = ca-1, ca = da-1, da = a-1. (4.1) For h G H, denote by o(h) the order of h in H. Since aa = ba-1 by Equation (4.1), o(ba-1) = o(aa) = o(a), forcing that o(b) | o(a). Similarly, since da = a-1 and ca = da-1, we have o(d) = o(a) and o(c) | o(a). Since H = (a, b, c, d), we have o(x) | o(a) for any x G H, and since H = Zm x Zpe x Zp, we have o(a) = mpe and |H : (a)| = p. Suppose that b G (a), say b = a® for some integer i. By Equation (4.1), aa = ba 1 = a®-1 G (a) and ca-1 = ba = (a®)a = ai(i-1) G (a), implying that c G (a). Similarly, d G (a) because d = a • ca. Since H = (a, b, c, d), we have H = (a) = Zmpe, a contradiction. Hence b G (a), and since |H : (a)| = p, we have H = (a, b) and p | o(b). Let A = Aut(r). Since r is Na(R(H))-arc-transitive, F = Na(R(H))10 acts transitively on Nr(10). Let ap,g G F for some fi G Aut(H) and g G H such that 1^'9 = 11. Then 11 = (^g)1 = g1, forcing that g =1. Hence Fu = {03,1 | £ G Aut(H), S^ = S}, that is, F1l = Aut(H, S). By Proposition 4.1, |Na(R(H))| = 2|H||F| = 2|H| • |NC1q)||F1i| = 10|H|| Aut(H,S)|. 508 Ars Math. Contemp. 15 (2018) 441-466 Observation 4.3. o(a) = mpe, p | o(6), H = (a, 6} and |na(r(h))| = 10|H|| Aut(H,S)|. In the following two lemmas we consider the two cases: e > 2 and e =1, respectively. Lemma 4.4. If e > 2, then 5 | (p - 1), r = C£Dm„pexp for some 1 < i < 3 and |na(r(h))| = 10|H |. Proof. By Observation 4.3, o(a) = mpe, p | o(6) and H = (a, 6} = (x, y, z} = Zm x Zpe x Zp, where (m,p) = 1. Then H has an automorphism mapping xy to a, and thus we may assume a = xy, which implies that 6 = xr+1yA+1zl for some r +1 G Zm, A +1 G Zpe and 0 = i G Zp because H = (a, 6}. Furthermore, H has an automorphism fixing x, y and mapping z to zl, and so we may assume 6 = xr+1yA+1z. Let c and d = xfcy£z4, where i, k G Zm, j, ^ G Zpe and s, t G Zp. = x®yjzs 1, et > 1 and pt's are distinct primes such that 5 | (pt - 1) for 1 < i < f. Note that e > 2. By Equation (4.3), j = A2 + A+1 (mod pe-1) and by Equations (4.6) and (4.7), I = A3 + A2 + A +1 (mod pe-1) and A4 + A3 + A2 + A +1 = 0 (mod pe-1), implying A5 = 1 (mod pe-1). It follows from Proposition 3.2 that either (A,pe-1) = Y.-Q. Feng et al.: Arc-transitive cyclic and dihedral covers of pentavalent symmetric graphs 509 (1,5), or 5 | (p - 1) and A is an element in Zpe_i of order 5, forcing that A = 0 and i-1 = (-A4)-1 = -A. Furthermore, one may assume that j = A2 + A + 1 + sip6-1 (mod pe), i = A3 + A2 + A + 1 + s2pe-i (mod pe) and A4 + A3 + A2 + A +1 = ipe-1 (mod pe) for some s1, s2, i G Zp. In what follows all equations are considered in Zp, unless otherwise stated. As p | pe-1, the following equations are also true in Zp: j = A2 + A + 1, i = A3 + A2 + A +1, A4 + A3 + A2 + A + 1 = 0, i-1 = -A. By s x (4.9) - t x (4.8), s = i-1(jt - t2) = -A(jt - t2), and by Equation (4.9), we have At3 - (A3 + A2 + A)t2 - (A + 1)t + (A3 + A2 + A +1) =0. Combined with A4 + A3 + A2 + A +1 = 0 and A = 0, we have (t - 1)(t - A)(t - A2) = 0, which implies that t =1, A or A2. Recall that j = A2 + A + 1 and s = -A(jt - t2). Thus (t, s) = (1, A4 + A + 1), (A, A3 + A + 1) or (A2, A2 + A + 1). Since j = A2 + A +1 + s1pe-1 (mod pe) and i = A3 + A2 + A +1 + s2pe-1 (mod pe), by Equations (4.6) and (4.7) we have: I (A + s)sipe 1 = s2pe 1 (mod pe) I tsipe-1 + As2pe-1 = -(A4 + A3 + A2 + A +1) (mod pe) (4.10) Recall that either (A,pe-1) = (1,5) or 5 | (p - 1). Suppose that pe-1 = 5. Then p = 5, e = 2 and (A, s,t) = (1,3,1). By Equation (4.10), we have 5s2 = 20s1 and 52s1 + 5 = 0 in Z52, a contradiction. Hence 5 | (p - 1). Again by Equation (4.10), we have -(t + A2 + As)s1pe-1 = ipe-1 (mod pe), where ipe-1 = A4 + A3 + A2 + A + 1. Furthermore, Ut + a2 + As)s1 = -1 (411) \(t + A2 + As)s2 = -i(A + s) Since (t, s) = (1, A4 + A + 1), (A, A3 + A +1) or (A2, A2 + A +1), we have t + A2 + As = 2A2 + A + 2, A4 + 2A2 + 2A or A3 + 3A2 + A, respectively, and since (2A2 + A + 2)(A4 + 2A2 + 2A) = 6(A4 + A3 + A2 + A) + 1 = -5 and (A3 + 3A2 + A)(A4 - 2A3 + A2) = A4 + A3 + A2 + A - 4 = -5, we have (t + A2 + As)-1 = -5-1 (A4 + 2A2 + 2A), -5-1(2A2 + A + 2) or -5-1(A4 - 2A3 + A2), respectively. By Equation (4.11), (sb s2) = (5-1i(A4+2A2 +2A), 5-1 i(-3A4+A3+2A2)), (5-1i(2A2 +A+2), 5-1i(-3A4+2A3+A)) or (5-1i(A4 - 2A3 + A2), 5-1i(-2A4 + A2 + a)). Note that a = xy, b = xr+1yA+1z, and (c,d) = (xr2+r+1yA2+A+1+SlPe-1 zA4+A+1,xr3+r2+r+1 yA3+A2+A+1+s2Pe-1 z), (Xr2+r+1yA2 + A+1+SlPe-1 ZA3+A+1 xr3+r2+r+1 yA3 + A2+A+1+S2Pe-1 zA) or (xr2+r+1yA2 + A+1+sipe-1 zA2+A+1 xr3+r2+r+1 yA3 + A2+A+1+S2Pe-1 zA2 ) 510 Ars Math. Contemp. 15 (2018) 441-466 we have S = Si, S2 or S3, where 51 = {1,xy,xr+y+1z,xr2+r+y2+A+1+5-1(A4+2A2+2A)lPe-1 zA4+A+1, xr3 +r2 +r+1yA3 + A2 +A+1+5-1 (-3A4+A3 + 2A2 )ipe-1 z} 52 = {1,xy,xr+1yA+1z,xr2+r+1yA2+A+1+5-1(2A2+A+2)lPe-1 za3+a+1, xr3+r2+r+1yA3+A2 + A+1+5-1(-3A4 + 2A3 + A)ipe-1 zA } 53 = {1,xy,xr+y+1z,xr2+r+y2+A+1+5-1(A4-2A3+A2)lPe-1 zA2+A+1, xr3+r2+r+1yA3 + A2 + A+1+5-1(-2A4+A2 + A)ipe-1 zA2 Since x5 = 1 (mod pe) implies that x5 = 1 (mod pe 1), there exists f G Zp such that A1 = A + fpe-1 is an element of order 5 in Z*e. Then A = A1 - fpe-1, A® = 1 and A| + A? + A2 + A1 + 1 =0 in Zpe. Hence ipe-1 = A4 + A3 + A2 + A +1 = (A1 - fpe-1)4 + (A1 - fpe-1)3 + (A1 - fpe-1)2 + (A1 - fpe-1) + 1 = -(4A3 + 3A1 + 2A1 + 1)fpe-1 Upe. 1UC11 A = A1 - fp , A1 Uati/^a i/ne 1 — -•p 2 , n 1\ i 1 _ /n3 , in Zpe, and thus S1 = {1,xy,xr+1yA+1z,xr2+r+1yA2+A+1+5-1(A4+2A2+2A)lPe-1 zA4+A+1, xr3+r2+r+1yA3+A2 + A+1+5-1(-3A4 + A3+2A2)ipe-1 z} = {1, xy, xr+1yA1+1y-fPe-1 z,xr2+r+1 y^+^y-M+^+W"1 zA!+A+1, xr3+r2+r+1 yA3 + A?+A1 + 1y-/pe-1 z}. Let f be the automorphism of H induced by x ^ x, y ^ y and z ^ yfpe 1 z. Then (S1)^ = {1, xy, xr+1 yA1 + 1z, xr2+r+1yA2+A1 + 1zA4+A1+1, xr3+r2+r+1yA3+A1+A1+1z}. Since BiCay(H, 0,0, S1) = BiCay(H, 0, 0, Sf), we may assume that A = A1 is an element of order 5 in Zpe, and S1 = {1, xy, xr+1yA+1z, x^+y^+V^1, xr3+r2 +r+y3+A2 +A+1z}. Similarly, we can also assume that 52 = {1, xy, xr+1yA+1z, x^+y^+y^1, x^+^+V^+^V}, 53 = {1, xy, xr+1yA+1z, x^+y^+V^1, x^+^+y^^+V2}. By Proposition 4.1 and Example 3.5, r = BiCay(H, 0,0, S) = CGD^^ with 1 < i < 3. Note that |NA(R(H))| = 10|H|| Aut(H,S)| (see Observation 4.3). For S1, let ft G Aut(H,S1). Then Sf = S1. Since (y, z) is characteristic in H = (x) x (y) x (z) = Zm x Zpe x Zp, we have {y, yA+1 z, yA2+A+1 zA4+A+1, yA3+A2+A+1z}f = {y, yA+1z, yA2+A+1 za4+a+1, y^+^z}. It follows that yf = ysz4 with (s, t) = (1,0), (A + 1,1), (A2 + A + 1, A4 + A + 1), or (A3 + A2 + A + 1,1). Furthermore, we have (y • yA+1z • yA2+A+1zA4+A+1 • yA3+A2+A+1z)f = y • yA+1 z • yA2+A+1zA4+A+1 • yA3+A2+A+1z Y.-Q. Feng et al.: Arc-transitive cyclic and dihedral covers of pentavalent symmetric graphs 511 and (y8 y-1)A3+2A2+3A+4 = (z-a4-a-3)8 za4+a+3 In particular, (y8y-1)(A +2A +3A+4)p = 1. Note that A4 + A3 + A2 + A +1 = 0 in Zpe implies that A4 + A3 + A2 + A +1 =0 in Zp. If A3 + 2A2 + 3A + 4 = 0 in Zp, then A3 = -2A2 -3A - 4, A4 = A • A3 = A2 +2A + 8, and thus 0 = A4 + A3 + A2 + A +1 = 5, contrary to 5 | (p - 1). Hence A3 + 2A2 + 3A + 4 = 0 in Zp and (y8y-1)p = 1. Suppose that (s,t) = (1,0). Then y8y-1 = ys-1z4 with s - 1 = A, A2 + A or A3+A2+A. Since A4+A3+A2+A+1 =0 in Zp, we have A = 0,-1 and thus (s-1,p) = 1. This implies that y8y-1 = ys-1z4 has orderpe, and since e > 2, we have (y8y-1)p = 1, a contradiction. Hence (s, t) = (1,0), that is, y8 = y. It follows that (yA+1z)8 = yA+1z8 G {yA+1z,yA2 +A+1zA4+A+1,yA3+A2+A+1z}, and thus z8 G {z,yA2zA4+A+1,yA3+A2z}. If z8 = yA2 zA4+A+1 or yA3 +a2 z, then (yA2 )p = 1 or (yA3+A2 )p = 1. It forces that A2 = 0 or A3 + A2 =0 in Zpe-i, and A = 0, -1, a contradiction. Hence z8 = z. Noting that (x) is characteristic in H, we have (xy)8 = x8y G S8 = S1. Then it is easy to check that (xy)8 = xy and thus x8 = x. It implies that ft is the identity automorphism. Hence | Aut(H, S1)| = 1 and |NA(R(H))| = 10|H|. By a similar argument as above, for S2 and S3, we also have | Aut(H,S2)| = | Aut(H,S3)| = 1 and |Na(R(H))| = 10|H|. □ Lemma 4.5. If e =1, that is, H = Zm x Zp x Zp, then one of the following holds: (1) p = 5 or 5 1 (p ± 1) and r = CGDmpxp as defined in Example 3.6. Furthermore, (i) |Na(R(H))| = 10|HI if m =1, 5; (ii) |Na(R(H))| = 20|H| if m = 5; (iii) |Na(R(H))| = 20|H| if m = 1 andp = 5; and (iv) |Na(R(H))| = 40|H| if m = 1 andp = 5; (2) 5 | (p - 1), r = CGD;mpxp as defined in Example 3.7 and |NA(R(H))| = 10|H|. Proof. Note that (m,p) = 1. By Observation 4.3, we have o(a) = mp, p | o(b) and H = (a, b) = (x,y, z) = Zm x Zp x Zp. Then H has an automorphism mapping xy to a, and we may assume a = xy, implying that b = xr+1yAzl for some r +1 G Zm, i, A G Zp and i = 0 because H = (a, b). The group H also has an automorphism fixing x, y and mapping z to yAzl, and we may further assume b = xr+1z. Let c = x®yjzs and d = xky£z4, where i, k G Zm, j, s, t G Zp. By Equation (4.1), aa = ba-1, that is, (xy)a = xry-1z. Since both (x) and (y, z) are characteristic in H, we have xa = xr and ya = y-1z. Again by Equation (4.1), since (xr+1z)a = ba = ca-1 = xi-1yj-1zs, we have za = (x-r-1)a • ba = x-r2-r-1+i • yj-1zs, implying that za = yj-1zs and -r2 - r - 1 + i = 0 (mod m). (4.12) Moreover, we have cfc-y-V = da-1 = ca = (x4 yj zs)a = (xr )4(y-1z)j (yj-1zs)s = xri y-j+s(j-1)zj+s 2 512 Ars Math. Contemp. 15 (2018) 441-466 and x-1 y-1 = a-1 = da = (xk yV)a = (xr )k (y-1z) V'-^)4 = xrk y-i+j-1)izsi+£. Considering the powers of x, y and z, we have Equations (4.13) - (4.18). As shown in these equations, in what follows all equations are considered in Zp, unless otherwise stated: k — 1 = ri (mod m); (4.13) i — 1 = —j + s(j — 1); (4.14) t = j + s2; (4.15) — 1 = rk (mod m); (4.16) —1 = —i + (j — 1)t; (4.17) 0 = st + i. (4.18) By Equation (4.12), we have i = r2 + r +1 (mod m) and by Equations (4.13) and (4.16), k = r3 + r2 + r +1 (mod m) and r4 + r3 + r2 + r +1 = 0 (mod m). It follows from Proposition 3.2 that either (r , m) G {(0 , 1), (1,5)} or r is an element of order 5 in Z^ and the prime decomposition of m is • • • with t < 1, f > 1, et > 1 and 5 | (pt — 1) for 1 < i < f. By Equation (4.15), t = j + s2, and by Equations (4.14), (4.17) and (4.18), t = 1 — j + s(j — 1), t = 1 + (j — 1)t = 1 + (j — 1)(j + s2) and t = —st = —sj — s3. It follows j2 + (s2 — s)j — (s2 — s) = 0; (4.19) (2s — 1)j + s3 — s + 1=0. (4.20) By Equation (4.19), (2s — 1)2j2 + (2s — 1)2(s2 — s)j — (2s — 1)2(s2 — s) = 0, and since (2s — 1)j = —(s3 — s + 1), we have s6 — 3s5 + 5s4 — 5s3 + 2s — 1=0, that is, (s2 — s — 1)(s4 — 2s3 + 4s2 — 3s + 1) =0. Hence, either s2 — s — 1 =0 or s4 — 2s3 +4s2 — 3s +1 = 0. Case 1: s2 — s — 1=0. Let A = 2s — 1. Then s = 2-1(1 + A) and A2 = 5, and thus (A,p) = (0,5) or 5 | (p ± 1) by [34, Example 4.6]. By Equations (4.19) and (4.20), j2 + j — 1 = 0 and (2s — 1)j + (s + 2) = 0. For (A,p) = (0,5), j2 + j — 1 = 0 implies that j = 2 = —2-1(1 + A). For 5 | (p ± 1), we have A = 0, and since 2s — 1 = A and (2s — 1)j + (s + 2) = 0, we have j = — (2s — 1)-1(s + 2) = —A-1 • 2-1(A + 5) = —2-1(1 + A) (note that 5 = A2). Itfollows from Equations (4.15) and (4.18) that t = j + s2 = 1 and i = —st = —2-1(1 + A). Recall that i = r2 + r + 1 (mod m) and k = «3 + «2 + i + 1 (mod m). Hence c = xr2+r+1y-2-1(1+A)z2-1(1+A) and d = xr3+r2+r+1y-2-1(1+A)z. Now, S = {1, xy, xr+1z, xr2+r+1y-2-1(1+A)z2-1(1+A), xr3+r2+r+1y-2-1(1+A)z}. By Proposition 4.1 and Example 3.6, r = BiCay(H, 0,0, S) = CGD^pxp. For (m,p) = (1,5), we have A = 0 and S = {1, y, z, y-3z3,y-3z}. By Magma [5], |Na(R(H))| = 40|H|. Assume that (m,p) = (1, 5), and let ^ G Aut(H, S). Then S^ = S, and since both (x) and (y, z) are characteristic subgroups of H, we have {x xr+1 xr2+r+1 xr3+r2+r+1}^ = {x xr + 1 xr2+r+1 xr3+r2+r + 1} Y.-Q. Feng et al.: Arc-transitive cyclic and dihedral covers of pentavalent symmetric graphs 513 {y,z,y-2-1(1+A)z2-1(1+A),y-2-1(1+A)z}^ = {y, z, y-2"(1+A)z2-1(1+A), y-2"(1+A)z}. Similarly to Lemma 4.4, the two equations imply that for m = 1,5 (r = 0, ±1), ft is the identity automorphism of H, and for m = 1 or 5 (r = 0 or 1), ft has order 2 that are inducedby x ^ xr +r +r+1, y ^ y-2 (1+A)z,and z ^ y-2 (A+1)z2 (A+1) (onemay also see [35] for a detailed computation). It implies that | Aut(H, S)| = 1 for m = 1,5 and | Aut(H,S)| = 2 for m = 1 or 5. By Observation 4.3, we have |NA(R(H))| = 10|H| or 201H |, respectively. Case 2: s 4 - 2s3 + 4s2 - 3s + 1 = 0. By Case 1, we may assume that s2 - s - 1=0. If p = 5, then s4 - 2s3 + 4s2 - 3s + 1=0 implies that s = 3 and thus s2 - s - 1 = 0, a contradiction. Hence p = 5. By [34, Lemma 5.4, Case 2], we have 5 | (p - 1) and s = 2-1(1 + A), where A4 + 10A2 +5 = 0 and A = 0, ±1. Since s4 - 2s3 + 4s2 - 3s +1 = 0, we have (2s - 1)(8s3 - 12s2 + 26s - 11) = -5, and since p = 5, we have (2s - 1)-1 = -5-1(8s3 - 12s2 + 26s - 11). Noting that s4 = 2s3 - 4s2 + 3s - 1, we have s5 = -5s2 + 5s - 2 and s6 = -5s3 + 5s2 - 2s. By Equation (4.20), j = -(2s - 1)-1(s3 - s + 1) = 5-1(8s3 - 12s2 + 26s - 11)(s3 - s + 1) = s3 - 2s2 + 3s - 1 = 8-1(A3 - A2 + 7A + 1) and by Equations (4.15) and (4.18), t = j + s2 = s3-s2 + 3s-1 = 8-1(A3 + A2 + 11A+3) and I = -st = -s3 + s2-2s + 1 = -8-1(A3 + A2 + 7A - 1). It follows that S = {1, xy, xr+1z,xr2+r+1y8-1(A3-A2+7A+1)z2-1(1+A), y xr3+r2+r+1y-8-1(A3 + A2+7A-1)z8-1(A3 + A2 + 11A+3)} By Proposition 4.1 and Example 3.7, r = BiCay(H, 0,0, S) = CGD^pxp. Let ft G Aut(H, S). Then S^ = S. Since (x) and (y, z) are characteristic in H, we have 8-1(A3-A2+7A+1) z2-1(1+A) y-8-1(A3+A2 + 7A-1) z8-1(A3+A2 {y, Z,yS ^+7A+X)(l+A),y-8 (A" +A" + 7A-i)z8 = {y Z y8-1 (A3-A2 + rA+l)Z2-1(l+A) y-8-1(A3 + A2+7A-l)Z8-1(A3 + A2 + llA+3)} and since A = 0, ±1, we have y^ = y and z^ = z (also see [35] for a detailed computation). Since (xy)^ = x^y e S, it is easy to check that (xy)^ = xy and thus x^ = x. Hence ft is the identity automorphism of H and | Aut(H, S)| = 1. By Observation 4.3, |Na(R(H))| = 10|H |. □ 5 Cyclic covers In this section, we classify connected symmetric cyclic covers of connected pentavalent symmetric graphs of order twice a prime. Denote by K6,6 - 6K2 the complete bipartite graph of order 12 minus a one-factor and by Il2 the Icosahedron graph. Edge-transitive cyclic covers of K6 were classified in [29, Theorem 1.1], and by [29, Line 20, pp. 40], such graphs have order 12 and thus isomorphic to K6,6 - 6K2 or Il2 by [20, Proposition 2.7] (note that the graph Il2 is missed in [29, Theorem 1.1]). Theorem 5.1. Let r be a connected pentavalent symmetric graph of order 2p for a prime p, and let T be a connected symmetric Zn-cover of r with n > 2. Then r = K6,6 — 6K2, 514 Ars Math. Contemp. 15 (2018) 441-466 I12, CDnp, or CQVlmpC^p for 1 < i < 5 with n = mpe, (m,p) = 1, 5 | (p — 1) and e > 1, which are defined in Examples 3.3, 3.5, 3.6 and 3.7. Proof. By Proposition 3.4, r = K6 for p = 3, K5,5 for p = 5, or CDp for 5 | (p — 1). If r = K6 then r = K6,6 — 6K2 or Ii2 by [29, Theorem 1.1] (also see the proof in [2, Theorem 3.6]). In the following, we assume that p > 5. Let A = Aut(r). Let K = Zn and F = NA (K). Since r is a symmetric K-cover of r, F is arc-transitive on f and F/K is arc-transitive on = r. Let B/K be a minimal arc-transitive subgroup of F/K. By Proposition 3.4, B/K = Dp x Z5 for p > 11; by Magma [5], B/K ^ D11 x Z5 for p =11, and B/K = Z2 x Z2, Z2 x Z4 or Z2 x Z8 for p = 5. Each minimal normal subgroup of B/K is isomorphic to Zp or Z5 with p = 5 and B/K = Z5 x Z8. Clearly, B is arc-transitive on f and B/K is non-abelian. Set C = Cb(K). Since K is abelian, K < Z(C) < C, where Z(C) is the center of C. Suppose K = C. Then B/K = B/C < Aut(K) = Z*n, which forces that B/K is abelian, a contradiction. Hence K < C and 1 = C/K < B/K. It follows that C/K contains a minimal normal subgroup of B/K, say L/K. Then L < B and L < C < B. Furthermore, L/K = Zp, or L/K = Z5 withp = 5 and B/K = Z2 x Z8. Clearly, L and L/K have two orbits on V(r) and V(rK), and r and rK are bipartite graphs with the two orbits of L and L/K as their bipartite sets, respectively. Since K < Z(C) and L < C, K < Z(L). First, assume L/K = Zp. Since K < Z(L), L is abelian, and so L = Znp or Zn x Zp with p | n. For the latter, L = Zm x Zpe x Zp with n = mpe, (m,p) = 1 and e > 1. Since L/K is semiregular on V (fK ), L is semiregular on V (f) and thus r is a bi-Cayley graph over L. Noting that L < B, we have that NA(L) is arc-transitive on T, forcing that f = BiCay(L, 0,0, S) for some subset S Ç L. Recall thatp > 5. By Lemmas 4.2-4.5, f = CD„p or CGDmpexp (1 < i < 5), as required. Now, assume L/K = Z5. Then p = 5 and B/K = Z5 x Z8. Since K < Z(L) and K = Zn, L = P x H, where P and H are the Sylow 5-subgroup and the Hall 5'-subgroup of L, respectively. Note that H < K is abelian, but P may not. Since (L/K)vk = Z5, we have Lv = Pv = Z5, where v g V(r and vK is an orbit of K on V(r containing v. Note that P < B^s P is characteristic in L and L < B. By Proposition 2.2, P has at most two orbits on V(r) because Pv = 1, and since L has exactly two orbits on V(r), P and L have the same orbits. It follows that L = PLv = PPv = P, forcing that H =1 and K is a 5-group. Suppose |K| = 54 with t > 2. Since K is cyclic, K has a characteristic subgroup N such that |K/N| = 25, and since K < B, N < B. By Proposition 2.2, fN is a connected pentavalent B/N-arc-transitive graph of order 10|K|/|N| = 250, and by Example 3.1, rN = CGD53. Since B/K = Z5 x Z8 and |K/N| = 52, all Sylow 2-subgroups of B/N are isomorphic to Z8 and |B/N| = 8 • 54. However, by Magma [5], Aut(C£D53) has no arc-transitive subgroup of order 8 • 54 that has a Sylow 2-subgroup isomorphic to Z8, a contradiction. Since K = 1, we have |K| = 5 and |V(f)| = 10|K| = 50. By Example 3.6, f = x5, as required. □ Y.-Q. Feng et al.: Arc-transitive cyclic and dihedral covers of pentavalent symmetric graphs 515 6 Dihedral covers In this section, we aim to classify symmetric dihedral covers of connected pentavalent symmetric graphs of order twice a prime. First, we introduce four graphs which are from [30]. Example 6.1. Let I(122) = Cay(D12, {b, 6a, 6a2, ba4, ba9}), G48 = Cay(D24, {b, ba, ba3, ba11, ba20}). be two Cayley graphs on the dihedral groups D12 = (a, b | a12 = b2 = 1,ab = a-1} and D24 = (a, b | a24 = b2 = 1, ab = a-1}, respectively. By Magma [5], Aut(l12)) = A5 x D4 and Aut(G48) = SL(2, 5) x D4, and their vertex stabilizers are isomorphic to F20. Example 6.2. Let Geo = Cay(A5, {(1 4)(2 5), (1 3)(2 5), (1 3)(2 4), (2 4)(3 5), (1 4)(3 5)}) be a Cayley graph on A5. By Magma [5], it is a connected pentavalent symmetric graph of order 60 and Aut(Ge0) = A5 x D5 with vertex stabilizer isomorphic to D5. Example 6.3. Let G be a subgroup of S7 generated by the elements a = (1 4)(2 5)(6 7), b = (1 3)(2 5)(6 7), c = (1 3)(2 4)(6 7), d = (2 4)(3 5)(6 7) and e = (1 4)(3 5)(6 7), and define G120 = Cay(G, {a, b, c, d, e}). By Magma [5], G = A5 x Z2 and G120 is a connected pentavalent symmetric graph of order 120. Moreover, Aut(G120) = A5 x D10 with vertex stabilizer isomorphic to D5. A list of all pentavalent G-arc-transitive graphs on up to 500 vertices with the vertex stabilizer Gv = Z5, D5 or F20 was given in Magma code by Potočnik [30]. Based on this list, we have the following lemma. Lemma 6.4. Let r be a G-arc-transitive graph of order 24, 48, 60, 120 or 240 with vertex stabilizer Gv = Z5, D5 or F20 for some G < Aut(r) and v G V(r). Then r is a connected symmetric dihedral cover of Ke if and only if r = I^, G48, Ge0 or G120. Proof. To show the necessity, let r be a connected symmetric dihedral cover of Ke. Then Aut(r) has an arc-transitive subgroup having a normal dihedral subgroup of order |V(r)|/6. Since r is G-arc-transitive with Gv = Z5, D5 or F20, by [30] r is isomorphic to one of the seven graphs: three graphs of order 24, 48 and 60 respectively, two graphs of order 120 and two graphs of order 240. For the orders 24, 48 and 60, r = I^, G48 or Ge0 by Examples 6.1 and 6.2. For the order 120, by Magma [5] one graph is isomorphic to G120 and the other has no arc-transitive group of automorphisms having a normal dihedral subgroup of order 20; in this case r = G120. For the order 240, again by Magma [5] none of the two graphs has an arc-transitive group of automorphisms having a normal dihedral subgroup of order 40. Now, we show the sufficiency. By Magma [5], Aut(l12)) has a normal subgroup N = D2. Clearly, N has more than two orbits on V(I^), and by Proposition 2.2, the quotient graph (I^ )N is a connected pentavalent symmetric graph of order 6, that is, the (2) complete graph Ke. Thus I12 is a D2-cover of Ke. Similarly, one may show that G48, Ge0 or G120 is a symmetric D3-, D5- or D10-cover of Ke, respectively. □ 516 Ars Math. Contemp. 15 (2018) 441-466 Now, we are ready to classify symmetric dihedral covers of connected pentavalent symmetric graphs of order 2p for any prime p. Clearly, we have p > 3. Theorem 6.5. Let r be a connected pentavalent symmetric graph of order 2p with p a prime, and le Ö60 or Gl20- (2) prime, and let T be a connected symmetric Dn-cover of r with n > 2. Then r = , G48 Proof. Let K = Dn and let F be^the fibre-preserving group. Since r is a symmetric K-cover of r, F is arc-transitive on r and F/K is arc-transitive on rK = r. Assume n = 2. Then |V(f)| = 2n • |V(r)| = 8p. Recall that p > 3. By [20, Proposition 2.9], f = I^ or a graph G248 of order 248 with Aut(G248) = PSL(2, 31). Since PSL(2, 31) has no proper subgroup of order divisible by 248 by Magma [5], Aut(r) is the unique arc-transitive group of automorphisms of T, that is, F = PSL(2, 31). It implies that T C G248 because F has no normal subgroup isomorphic to Dn. Hence f C I(2) r = I12 . Assume n > 2. Let Zn be the cyclic subgroup of K = Dn of order n. Then Zn is characteristic in K and so Zn < F as K < F. By Proposition 2.2, rZn is a connected pentavalent F/Zn-arc-transitive graph of order 4p, and by [20, Proposition 2.7], fZn = I12 or K6,6 - 6K2. Thus r is a symmetric Zn-cover of K6,6 - 6K2 or I12. Note that |V (f)| = 12n. Let rZn = K6,6 - 6K2. Since each minimal arc-transitive subgroup of Aut(K6,6 -6K2) is isomorphic to A5 x Z2 or S5 by Magma [5], F/Zn has an arc-transitive subgroup B/Zn = A5 x Z2 or S5. It follows that |Bv | = 10 forv G V (f), and form Proposition 2.1 that Bv = D5. In particular, B is arc-transitive on f and B/Zn has a normal subgroup M/Zn = A5, which is edge-transitive on f Zn and has exactly two orbits on V (rZn ). Thus M < B is edge-transitive and has two orbits on V(r). Since |B : M| = 2, we have Mv = D5. Clearly, Zn < Cm (Zn). If Zn = Cm (Zn), then A5 = M/Zn = M/Cm(Zn) < Aut(Zn) = Zn, which is impossible. Hence Zn is a proper subgroup of CM(Zn), and since Mult(A5 ) = Z2, Lemma 2.3 implies that either M = M' x Zn = A5 x Zn or M = M'Zn = SL(2, 5)Zn with M' n Zn C Z2. In particular, M/M' is cyclic. Since M' is characteristic in M and M < B, we have M' < B. If M' has at least three orbits on V(f ), by Proposition 2.2, M' is semiregular on V (f ) and f M/ is a connected pentavalent B/M'-arc-transitive graph. The stabilizer of a G V(rM/) in M/M' is isomorphic to Mv = D5, but this is impossible because M/M' is cyclic. Thus M' has at most two orbits on V (f ) and so |V(f)| | 2|M'|, that is, 6n | |M'|. If M = A5 x Zn, then M' = A5 and 6n | |M'| implies that n = 5 or 10 as n > 2. It follows that |V(f )| = 60 or 120. Since Bv = D5, we have r C G60 or G120 by Lemma 6.4. If M = SL(2,5)Zn with M' = SL(2,5) and SL(2,5) n Zn = Z2, then n is even and 6n | |M'| implies that n = 4, 10 or 20. It follows that | V(f ) | = 48, 120 or 240, and from Lemma 6.4 that r C G48 or G120. Let fZn = I12. By Magma [5], under conjugation Aut(I12) has only one minimal arc-transitive subgroup isomorphic to A5, and so F/Zn has an arc-transitive subgroup B/Zn = A5. By a similar argument as the previous paragraph, B = B'Zn and B'nZn < Mult(A5) by Lemma 2.3, forcing that either B = B' x Zn = A5 x Zn or B = B'Zn = SL(2, 5)Zn with SL(2, 5) n Zn = Z2. Furthermore, B is arc-transitive on T with Bv = Z5 for Y.-Q. Feng et al.: Arc-transitive cyclic and dihedral covers of pentavalent symmetric graphs 517 v G V(r), and B/B' is cyclic. If B' has more than two orbits on V(T), then rB is a connected pentavalent B/B'-arc-transitive graph by Proposition 2.2, which is impossible because B/B' is abelian. Thus B' has at most two orbits on V(f) and so 12n | 2|B'|. If B = A5 x Zn, then B' = A5, and 12n | 2|B'| implies that n = 5 or 10. It follows that |V(f)| = 60 or 120. Since Bv = Z5, we have f = Geo or G120 by Lemma 6.4. If B = SL(2, 5)Zn with SL(2, 5) n Zn = Z2, then B' = SL(2, 5) and n is even. Since 12n | 2|B'|, we have n = 4, 10 or 20, and so |V(f)| = 48, 120 or 240. It follows from Lemma 6.4 thatf = G48 or G120. □ 7 Full automorphism groups of covers Let r be a symmetric Dn- or Zn-cover of a connected symmetric pentavalent graph of order 2p, where n > 2 is an integer and p is a prime. In this section, we aim to determine (2) the full automorphism group of r. For Dn, by Theorem 6.5, r = I12, G48, Geo or Gi2o and by Examples 6.1 -6.3, Aut(T) is known. For Zn, by Theorem 5.1, r = K6,6 — 6K2, Ii2, CDnp (see Example 3.3), or CGVlmpexp with 1 < i < 5 (see Examples 3.5, 3.6 and 3.7). In particular, for the graph CGD„pexp, we have mpe = n and m is given by m = 5tp11 • • • pe/ s.t. t < 1, s > 0, ej > 1, 5 | (py — 1) for 0 < j < s, (7.1) where m,p, e satisfy the conditions as listed in the second column in Table 2. Note that m is odd by Equation (7.1). By Magma [5], Aut(K6,6 — 6K2) = S6 x Z2 and Aut(I12) = A5 x Z2, and by Example 3.3, Aut(CDnp) = Dnp xZ5. Hence we only need to determine the full automorphism groups of CGD'lmpexp for 1 < i < 5. All theses graphs are connected symmetric cyclic covers of some pentavalent symmetric graph of order 2p except CGD>mpxp with 5 | (p + 1), which are connected symmetric bi-Cayley graphs over Zmp x Zp. Theorem 7.1. Aut(CGD„pe xp) for 1 < i < 5 is isomorphic to one group listed in Table 2. Table 2: Full automorphism groups of CGD mpe xp for 1 < i< 5. r Conditions: (m,p) = 1, m: Eq. (7.1) Aut(r) CGD^pexp,» =1, 2 5 | (p - 1) and e > 2 Dih(Zmpe x Zp) x Z5 m =1, 5, and p = 5 or 5 | (p ± 1) Dih(Zmp x Zp) x Z5 CGDmpxp m =1 or 5, and 5 | (p ± 1) Dih(Zmp x Zp) x D5 m =1 and p = 5 (Dih(Z5) x f2o).Z4 CGDmpxp 5 | (p - 1) Dih(Zmp x Zp) x Z5 Proof. Let r = CGDmpexp for 1 < i < 5 and A = Aut(r). For (m,p) = (1, 5), we have r = CGD5x5 and by [16, Theorem 4.3 (1)], Aut(r) = (Dih(Z2) x ^2o).Z4. In what follows we assume that (m,p) = (1, 5). By Examples 3.5, 3.6 and 3.7, A has an arc-transitive subgroup F isomorphic to Dih(Zmpe x Zp) x Z5 for CGDmpe xp (i = 1, 2, 3), 518 Ars Math. Contemp. 15 (2018) 441-466 Dih(Zmp x Zp) x Z5 for CQV^p^p with m = 1, 5 andp = 5 or 5 | (p ± 1), Dih(Zmp x Zp) x D5 for CGD^pxp with m =1 or 5 and 5 | (p ± 1), and Dih(Zmp x Zp) x Z5 for CgV^pxp with 5 | (p - 1). Note that Fv = Z5 or D5 for v e V(r). Furthermore, F has a normal semiregular subgroup K = Zmpe x Zp having two orbits on V(r), and hence r is an F-arc-transitive bi-Cayley graph over K. By Lemmas 4.4 and 4.5, |NA(K)| = |F|, implying that NA(K) = F. Note that |F| = 10|K| or 20|K|, that is, |F| = 10mpe+1 or 20mpe+1 with p = 5 or 5 | (p ± 1), and by Equation (7.1), both m and |K| are odd. In particular, |V(r)| = 2|K| = 2mpe+1 is twice an odd integer. Clearly, K = Zmpe x Zp has a characteristic Hall 5'-subgroup, say H. Then H < F as K < F .If H = K, then 5 | mpe+1 and H has at least three orbits. For p = 5, we have 5 | m, and since 52 { m by Equation (7.1), we have |K : H| = 5. For p = 5, by Table 2, r = CGDmpxp with (m, 5) = 1 and K = Zm x Zp x Zp, implying that |K : H| = 52. By Proposition 2.2, rH is a connected pentavalent F/H-arc-transitive graph of order 2 • 5 or 2 • 52. By Proposition 3.4 and Example 3.6, rH = K5,5 or = CGl4x5. Since |F| = 10|K| or 20|K| and |K| is odd, H is the characteristic Hall {2, 5}'-subgroup of F. Thus we have the following claim. Claim 7.2. H is the characteristic Hall {2, 5}'-subgroup of F, and we have H = K, or ^5x5- |K : H| = 5 and Fh = K55, or |K : H| = 25 and Fh = To finish the proof, we only need to show that A = F. Suppose to the contrary that A = F. Then A has a subgroup M such that F is a maximal subgroup of M. Since F is arc-transitive on r, M is arc-transitive, and since NA(K) = F, we have K ^ M. By the definitions of the graphs CGDmpexp (1 < i < 5) in Examples 3.5, 3.6 and 3.7, r has the 6-cycle (1, h, a-r-1b-A-1c-1, ha-rb-Ac-1, a-rb-Ac-1, hab, 1) for 1 < i < 3, and the 6-cycle (1, h, a-r-1 c-1, ha-r bc-1, a-r bc-1, hab, 1) for 4 < i < 5. Suppose that r is (M, 4)-arc-transitive. Then each 4-arc lies in a 6-cycle in r and so r has diameter at most three. It follows that |V(r)| = 2mpe+1 < 1 + 5 + 5 • 4 + 5 • 4 • 4 = 106, that is, mpe+1 < 53. Since p = 5 or 5 | (p ± 1) and e +1 > 2 (see the second column of Table 2), we have p = 5 and m < 2. Since m is odd, (m,p) = (1,5), contrary to assumption. Thus r is at most 3-arc-transitive, and by Proposition 2.1, we have |Mv | G {5,10, 20, 40, 60, 80,120, 720,1440, 2880}. Note that |M : F| = |Mv : Fv | G {2,4, 6,8,12,16,24,72,144, 288, 576} because M = F and |Fv | = 5 or 10. Let [M : F] be the set of right cosets of F in M. Consider the action of M on [M : F] by right multiplication, and let FM be the kernel of this action, that is, the largest normal subgroup of M contained in F. Then M/Fm is a primitive permutation group on [M : F] because F/FM is maximal in M/Fm, and (M/Fm)f = F/FM, the stabilizer of F g [M : F] in m/Fm. It follows that |m/Fm | = |M : F||F/Fm | and so |F/Fm| = |M/FM|/|M : F|. Since |M : F| G {2,4, 6,8,12,16,24,72,144, 288, 576}, by Lemma 2.4 we have M/FM < AGL(t, 2) with |M : F| = 2l and 1 < t < 4, or soc(M/FM) = PSL(2, q), PSL(3, 3) or PSL(2,r) x PSL(2,r) with |M : F| = q + 1, 144 or (r + 1)2 respectively, where q G {5, 7,11, 23, 71} and r G {11,23}. Suppose M/FM < AGL(2,2) and |M : F| = 4. Since a 2-group cannot be primitive on [M : F], we have 3 | |m/Fm| and so 3 | |MM/Fm|/|M : F| = |^F/Fm|. Since |F| = 10mpe+1 or 20mpe+1 with p = 5 or 5 | (p ± 1), we have 3 | m, which is impossible by Equation (7.1). Thus M/FM < AGL(2, 2). Similarly, since 7 { m, we have M/Fm < AGL(3, 2), and if M/Fm < AGL(4,2), then M/Fm is a {2,5}-group. Furthermore, soc(M/FM) ^ PSL(2, q), PSL(3, 3) or PSL(2, r) x PSL(2, r) for q G Y.-Q. Feng et al.: Arc-transitive cyclic and dihedral covers of pentavalent symmetric graphs 519 {7, 23, 71} and r = 23 because otherwise one of 7, 23,13, 23 is a divisor of m. It follows that M/Fm = Z2 with |M : F| = 2, M/Fm < AGL(4, 2) with |M : F| = 24 and M/Fm a {2, 5}-group, soc(M/FM) = PSL(2, q) with |M : F| = q +1 and q G {5,11}, or soc(M/FM) = PSL(2,11) x PSL(2,11) with |M : F| = 144. First assume that M/Fm = Z2 with |M : F| =2. Then F < M and H < M as H is characteristic in F by Claim 7.2. Let C = CM (H). Since K is abelian, H < K < C. Let P be a Sylow 5-subgroup of C containing the unique Sylow 5-subgroup of K. Since H is the Hall 5'-group of K, K < HP = H x P. Clearly, HP/H is a Sylow 5-subgroup of C/H. Recall that |F/K| | 20 and |K/H| | 25 (see Claim 7.2). Since |M| = 2|F|, we have |M/H| | 23 • 53, and by Sylow theorem, M/H has a normal Sylow 5-subgroup. In particular, C/H has a normal Sylow 5-subgroup, that is, HP/H < C/H. This implies H x P < C, and since C < M and P is characteristic in C, we have P < M. Since (m,p) = (1,5) and |V(r)| = 2mpe+1, P has at least three orbits on V(r). By Proposition 2.2, P is semiregular on V(r). Thus |P| | |V(r)| and |P| | |K|. It follows that |HP| = |H||P| | |K|, and since K < HP, we have K = HP < M, a contradiction. Assume that M/Fm < AGL(4,2) with |M : F| = 24 and M/Fm a {2,5}-group. Then M/Fm has a regular normal subgroup of order 24, say L/FM, and hence L < M, 24 | |L| and 5 | |M : L|. If L is semiregular then 24 | |V(r)| = 2mpe+1, which is impossible. Thus L is not semiregular, and so 5 | |Lv |. By Proposition 2.2, L has one or two orbits, yielding that |L| = |V(r)||L„| or |L| = |V"(r)||L„|/2. Since |M| = |V(r)||M„|, we have |M : L| = |Mv : Lv | or 2|Mv : Lv |, and since 52 f |Mv|, we have 5 f |M : L|, a contradiction. Assume that soc(M/FM) = PSL(2, 5) with |M : F| = 6. Then M/Fm = PSL(2, 5) or PGL(2,5), and |F/Fm| = |M/Fm|/|M : F| = 10 or 20. Since H is the unique normal Hall {2, 5}'-subgroup of F, we have H < FM and so H is characteristic in FM. This implies H < M because Fm < M. Since M/Fm = (M/H)/(Fm/H), M/H is insolvable, and since K ^ M, we have H = K .By Claim 7.2, Th = K5,5 or cgd5x5. If rH = cgd4x5 then Aut(rH) = (Dih(Zg) x F2o).Z4 is solvable and so M/H is solvable, a contradiction. If rH = K5 5 then as Aut(K5,5) — (S5 x S5) x Z2, it is easy to show that each insolvable arc-transitive group of Aut(K5 5) contains A5 x A5 (this is also easily checked by Magma [5]), and so |m/h| > 2 • 602. Noting that FM is semiregular on V(r), we have |Fm| | |K|. By Claim 7.2, |K : H| | 52, and hence |Fm : H| | 52. It follows that |M/Fm | = |M/H|/|Fm/H| > 2 • 602/52 > | PGL(2,5) |, a contradiction. Assume that l/Fm := soc(m/Fm) = PSL(2,11) with |M : F| = 12. Then M/Fm = PSL(2,11) or PGL(2,11), and |F/Fm| = |M/Fm|/|M : F| = 55 or 110. Moreover, L < M and K < L as |K| is odd and |M : L| < 2. Since 11 | |L/Fm|, Fm has at least three orbits on V(r), and by Proposition 2.2 FM is semiregular and rFM is a pentavalent F/FM-arc-transitive graph. Thus |FM| | |V(r)| and |V(rfm)| is even. Since |V(rFM)| = |V(r)|/|FM| = 2|K|/|Fm|, |Fm| is odd and |Fm| | |K|. Recall that H is the characteristic Hall {2,5}'-subgroup of F by Claim 7.2. Set N = HnFM. Since FM has odd order, N is the characteristic Hall 5'-subgroup of FM, and since Fm < M, we have N < M. Hence FM/N is a 5-subgroup. By Claim 7.2, 53 f |K|, and since |Fm | | |K|, we have 53 f |Fm |, that is, |Fm/N| | 25. Thus Fm/N is abelian, and Aut(FM/N) is cyclic or Aut(FM/N) ^ GL(2,5). If Fm/N = CL/N(Fm/N), then PSL(2,11) = L/Fm = (L/N)/(Fm/N) < Aut(FM/N), which is impossible. Thus Fm/N is a proper subgroup of CL/N(FM/N), and since Mult(PSL(2,11)) = Z2, Lemma 2.3 implies that L/N = (L/N)' x FM/N with (L/N)' = PSL(2,11). Since 520 Ars Math. Contemp. 15 (2018) 441-466 |V(rw)| = |V(r)|/|N| = 2|K|/|N| with |K| odd, (L/N)' = PSL(2,11) cannot be semiregular on V(rN), implying that 5 | |(L/N)0J for a G V(rN). It follows from Proposition 2.2 that (L/N)' has at most two orbits on V(rN), and so |(L/N)|/|(L/N)'| = |V(rw)||(L/N)a|/(|V(rw)||(L/N)a|) = |(L/N)a|/|(L/Nor2|(L/N)«|/|(L/NU implying that 5 \ (|L/N|/|(L/N)'|). Since |(L/N)/(L/N)'| = |Fm/N| and Fm/N is a 5-group, we have |Fm/N| = 1, that is, L/N = (l/n)' = PSL(2,11) Since K ^ M and N < M, we have N = K, and since K < CL(N) and |N| is odd, Lemma 2.3 implies L = L' x N with L' = PSL(2,11). Note that L' < M. Since r has order twice an odd integer, L' cannot be semiregular on r, yielding 5 | |LV |. By Proposition 2.2, L' has at most two orbits, and so | PSL(2,11) | = |L'| = |V(r)||L(, | or |V(r)||L(, |/2. It implies that |V(r)| | 2| PSL(2,11)|, that is, |V(r)| | 23 • 3 • 11. S ince | V(r) | = 2mpe+1 and it is not divided by 3 or 22 by Equation (7.1), we have | V(r) | = 22, contrary to the fact that e + 1 > 2. Assume that L/FM := soc(M/FM) = PSL(2,11) x PSL(2,11) with |M : F| = 144. Then there exists L1/FM < L/Fm such that L1/FM = PSL(2,11) and 11 | |L : L11. Since 11 | |L : FM |, FM has at least three orbits and so Tfm has order twice an odd integer. This implies that L/FM cannot be semiregular, and by Proposition 2.2, L/FM has one or two orbits. If L/FM has one orbit then L1/FM is semiregular on rfm as 11 | |L : L1| implies that L1 /FM has at least three orbits, and so 4 | |V(rfm)|, a contradiction. If L/FM has two orbits then rfm is bipartite and L/FM is edge-transitive on rfm. Furthermore, L1/FM fixes the bipartite sets setwise. 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