/^creative ^commor ARS MATHEMATICA CONTEMPORANEA ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 16 (2019) 81-95 https://doi.org/10.26493/1855-3974.1333.68f (Also available at http://amc-journal.eu) Pentavalent symmetric graphs of order four times an odd square-free integer* Bo Ling t School of Mathematics and Computer Sciences, Yunnan Minzu University, Kunming, Yunnan 650504, P. R. China Ben Gong Lou School ofMathematics and Statistics, Yunnan University, Kunming, Yunnan 650031, P. R. China Ci Xuan Wu School of Statistics and Mathematics, Yunnan University of Finance and Economics, Kunming, Yunnan 650221, P. R. China Received 24 February 2017, accepted 6 June 2018, published online 18 September 2018 Abstract A graph is said to be symmetric if its automorphism group is transitive on its arcs. Guo et al. in 2011 and Pan et al. in 2013 determined all pentavalent symmetric graphs of order 4pq. In this paper, we shall generalize this result by determining all connected pentavalent symmetric graphs of order four times an odd square-free integer. It is shown in this paper that, for each such graph r, either the full automorphism group Aut r is isomorphic to PSL(2,p), PGL(2,p), PSL(2,p) x Z2 or PGL(2,p) x Z2, or r is isomorphic to one of 9 graphs. Keywords: Arc-transitive graph, normal quotient, automorphism group. Math. Subj. Class.: 05C25, 05E18, 20B25 *The authors are very grateful to the referee for the constructive comments and suggestions. The first author was supported by the National Natural Science Foundation of China (11701503, 11861076, 11761079), Yunnan Applied Basic Research Projects (2018FB003) and the Scientific Research Foundation Project of Yunnan Education Department (2017ZZX086). The second author was supported by the National Natural Science Foundation of China (11861076, 11231008, 11461004, 11301468). t Corresponding author. E-mail addresses: bolinggxu@163.com (Bo Ling), bengong188@163.com (Ben Gong Lou), wucixuan@gmail.com (Ci Xuan Wu) ©® This work is licensed under https://creativecommons.org/licenses/by/4.0/ 100 Ars Math. Contemp. 16 (2019) 97-109 1 Introduction All graphs in this paper are assumed to be finite, simple and undirected. Let r be a graph and denote Vr and Ar the vertex set and arc set of r, respectively. Let G be a subgroup of the full automorphism group Aut r of r. Then r is called G-vertex-transitive and G-arc-transitive if G is transitive on Vr and Ar, respectively. An arc-transitive graph is also called a symmetric graph. It is well known that r is G-arc-transitive if and only if G is transitive on Vr and the stabilizer Ga := {g G G | a9 = a} is transitive on the neighbor set r (a) of the vertex a of r. The cubic and tetravalent graphs have been studied extensively in the literature. In recent years, attention has moved on to pentavalent symmetric graphs and a series of results have been obtained. For example, all the possibilities of vertex stabilizers of pentavalent symmetric graphs are determined in [7, 20]. Also, for distinct primes p, q and r, the classifications of pentavalent symmetric graphs of order 2pq and 2pqr are presented in [9, 19], respectively. A classification of 1-regular pentavalent graph (that is, the full automorphism group acts regularly on its arc set) of square-free order is presented in [13]. Recently, pentavalent symmetric graphs of square-free order have been completely classified in [11]. Furthermore, some classifications of pentavalent symmetric graphs of cube-free order also have been obtained in recent years. For example, the classifications of pentavalent symmetric graphs of order 12p, 4pq and 2p2 are presented in [8, 16, 5]. More recently, symmetric graphs of any prime valency which admit a soluble arc-transitive group have been classified in [14]. The main purpose of this paper is to extend the results in [8, 16] to four times an odd square-free integer case. The main result of this paper is the following theorem. Theorem 1.1. Let n be an odd square-free integer and let r be a pentavalent symmetric graph of order 4n. If n has at least three prime factors, then one of the following statements holds. (1) Aut r = PSL(2,p), PGL(2,p), PSL(2,p) x Z2 or PGL(2,p) x Z2, where p > 29 is a prime. Furthermore, the stabilizer (Aut r)a and the prime p appear in Table 5 or Table 6. (2) The triple (r, n, Aut r) lies in the following Table 1. Remark 1.2 (Remarks on Theorem 1.1). (a) The graphs in Table 1 are introduced in Example 3.2. (b) The graphs C5852 and C380 in Table 1, and the graphs in part (1) with automorphism group PSL(2,p) x Z2 or PGL(2,p) x Z2 can also be constructed from the bipartite double cover (the definition of bipartite double cover see Section 3) of a pentavalent symmetric graph of square-free order (see [11, Example 4.3 and Example 4.5] and [19, Example 3.9 and Example 3.11] for details on these graphs). 2 Preliminaries We now give some necessary preliminary results. The first one is a property of the Fitting subgroup, see [18, p. 30, Corollary]. Lemma 2.1. Let F be the Fitting subgroup of a group G. If G is soluble, then F = 1 and the centralizer CG(F) < F. B. Ling et al.: Pentavalent symmetric graphs of order four times an odd square-free integer 83 Table 1: Nine 'sporadic' pentavalent symmetric graphs of order four times an odd squarefree integer. Row r Aut r (Aut r )a Transitivity Bipartite? 1 r1 C17556 3 • 7 • 11•19 J1 D10 1-transitive No 2 r2 C17556 3 • 7 • 11•19 J1 D10 1-transitive No 3 r3 C17556 3 •7 • 11•19 J1 D10 1-transitive No 4 r4 C17556 3 •7 • 11•19 J1 D10 1-transitive No 5 5 C17556 3 •7 • 11•19 J1 D10 1-transitive No 6 C5852 7 • 11 • 19 J1 x Z2 A5 2-transitive Yes 7 1 C780 3 • 5 • 13 PSL(2,25) x Z2 F20 2-transitive No 8 2 C780 3 • 5 • 13 PSL(2, 25) x Z2 F20 2-transitive No 9 C380 3 • 5 • 13 PSL(2, 25) x Z2 F20 2-transitive Yes The maximal subgroups of PSL(2,p) are known, see [4, Section 239]. Lemma 2.2. Let T = PSL(2,p), where p > 5 is a prime. Then a maximal subgroup of T is isomorphic to one of the following groups: (1) Dp-!, where p = 5,7, 9,11; (2) Dp+i, where p = 7, 9; (3) Zp : Z(p-1)/2; (4) A4, where p = 5 or p = 3,13, 27, 37 (mod 40); (5) S4, where p = ±1 (mod 8) (6) A5, where p = ±1 (mod 5). By [2, Theorem 2], we may easily derive the maximal subgroups of PGL(2,p). Lemma 2.3. Let T = PGL(2,p) with p > 5 a prime. Then a maximal subgroup of T is isomorphic to one of the following groups: (1) Zp : Zp-i; (2) D2(p+1); (3) D2(p-i), wherep > 7; (4) S4, where p = ±3 (mod 8); (5) PSL(2,p). From [6, pp. 134-136], we can obtain the following lemma by checking the orders of nonabelian simple groups. Lemma 2.4. Let n be an odd square-free integer such that n has at least three prime factors. Let T be a nonabelian simple group of order 2® • 3j • 5 • n, where 1 < i < 11 and 0 < j < 2. Let p be the largest prime factor of n. Then T is listed in Table 2. n 100 Ars Math. Contemp. 16 (2019) 97-109 Table 2: Nonabelian simple groups of order 2® • 3j • 5 • n with 1 < i < 11 and 0 < j < 2. T |T| n M22 27 • 32 • 5 • 7 • 11 3 7 • 11 M23 27 32 • 5 • 7 • 11 • 23 7 11 • 23 J1 23 • 3 • 5 • 7•11 • 19 7 11 • 19 J2 27 • 33 • 52 • 7 3 5 • 7 Sz(32) 210 • 52 • 31 • 41 5 31 • 41 PSU(3, 4) 26 3 • 52 • 13 3 5 • 13 PSp(4, 4) 28 32 • 52 • 17 3 5 • 17 PSL(2, 25) 23 3 • 52 • 13 3 5 • 13 PSL(2, 28) 28 • 3 • 5 • 17 • 257 3 17 • 257 PSL(5, 2) 210 • 32 • 5 • 7 • 31 3 7 • 31 PSL(2, 26) 26 • 32 • 5 • 7 • 13 3 7 • 13 M23 27 • 32 • 5 • 7 • 11 • 23 3 7 • 11 • 23 M24 210 • 33 • 5 • 7 • 11 • 23 3 7 • 11 • 23 J1 23 • 3 • 5 • 7•11 • 19 3 7 • 11 • 19 PSL(2,p) p(p+12(p-1) (p > 29) Proof. If T is a sporadic simple group, by [6, p. 135-136], T = M22, M23, M24, J1 or J2. If T = An is an alternating group, since 34 does not divide |T |, we have n < 8, it then easily exclude that T = A5, A6, A7 or A8. Hence no T exists for this case. Suppose now T = X (q) is a simple group of Lie type, where X is one type of Lie groups, and q = rd is a prime power. If r > 5, as |T| has at most three 3-factors, two 5-factors and one p-factor, it easily follows from [6, p. 135] that the only possibility is T = PSL(2,p) with p > 29 (note that PSL(2,p) with 5 < p < 23 does not satisfy the condition of the lemma) or PSL(2,25), where p is the largest prime factor of n. If r < 3, as 212 and 34 do not divide |T|, then we have T = Sz(32), PSU(3,4), PSp(4,4), PSL(2,26), PSL(2, 28) or PSL(5,2). □ For a graph r and a positive integer s, an s-arc of r is a sequence a0, a1;..., as of vertices such that a® are adjacent for 1 < i < s and ai-1 = ai+1 for 1 < i < s - 1. In particular, a 1-arc is just an arc. Then r is called (G, s)-arc-transitive with G < Aut r if G is transitive on the set of s-arcs of r. A (G, s)-arc-transitive graph is called (G, s)-transitive if it is not (G, s + 1)-arc-transitive. In particular, a graph r is simply called s-transitive if it is (Aut r, s)-transitive. Let F20 denote the Frobenius group of order 20. The following lemma determines the stabilizers of pentavalent symmetric graphs, refer to [7, 20]. Lemma 2.5. Let r be a pentavalent (G, s)-transitive graph, where G < Aut r and s > 1. Let a G Vr. Then one of the following holds. (a) If Ga is soluble, then s < 3 and |Ga| I 80. Further, the pair (s,Ga) lies in the B. Ling et al.: Pentavalent symmetric graphs of order four times an odd square-free integer 85 following table. s Ga 1 Z5, D10, D2o 2 F2o, F20 x Z2 3 F20 x Z4 (b) If Ga is insoluble, then 2 < s < 5, and \Ga\ | 29 • 32 • 5. Further, the pair (s, Ga) lies in the following table. s Ga | Ga | 2 3 4 5 A5, S5 A4 x A5, (A4 x A5) : Z2, S4 x S5 ASL(2,4), AGL(2,4), AEL(2,4), ArL(2,4) Z6 : rL(2,4) 60, 120 720, 1440, 2880 960, 1920, 2880, 5760 23040 A typical method for studying vertex-transitive graphs is taking normal quotients. Let r be a G-vertex-transitive graph, where G < Aut r. Suppose that G has a normal subgroup N which is intransitive on V r. Let V rN be the set of N-orbits on Vr. The normal quotient graph rN of r induced by N is defined as the graph with vertex set VrN, and B is adjacent to C in rN if and only if there exist vertices ft G B and 7 G C such that ft is adjacent to 7 in r. In particular, if val(r) = val(rN), then r is called a normal cover of rN. A graph r is called G-locally primitive if, for each a G Vr, the stabilizer Ga acts primitively on r(a). Obviously, a pentavalent symmetric graph is locally primitive. The following theorem gives a basic method for studying vertex-transitive locally primitive graphs, see [17, Theorem 4.1] and [12, Lemma 2.5]. Theorem 2.6. Let r be a G-vertex-transitive locally primitive graph, where G < Aut r, and let N < G have at least three orbits on Vr. Then the following statements hold. (i) N is semi-regular on Vr, G/N < Aut rN, and r is a normal cover of rN; (ii) Ga = (G/N)Y, where a G Vr and 7 G VrN; (iii) r is (G, s)-transitive if and only if rN is (G/N, s)-transitive, where 1 < s < 5 or s = 7. For reduction, we need some information of pentavalent symmetric graphs of order 4pq, stated in the following lemma, see [8, Theorem 4.1] and [16, Theorem 3.1]. Lemma 2.7. Let r be a pentavalent symmetric graph of order 4pq, where q > p > 3 are primes. Then the pair (Aut r, (Aut r )a) lies in the following Table 3, where a G V r. Remark 2.8 (Remarks on Lemma 2.7). (a) Suppose that r is one of the graphs in Lemma 2.7 and M is an arc-transitive subgroup of Aut r. Then M is insoluble (for convenience, we prove this conclusion in Lemma 4.4 and we remark that Lemma 4.4 is independent where it is used). (b) By Magma [1], the graphs C^ and Cf32 in [8, Theorem 4.1] are isomorphic, Aut(Cf32) = PGL(2,11) x Z2. 100 Ars Math. Contemp. 16 (2019) 97-109 Table 3: Pentavalent symmetric graphs of order 4pq. r C60 r1 C132 2 < i < 4 C132, 2 — ' — C132 (2) C574 C4IO8 (p, q) Aut r (Aut r) (3, 5) A5 X D10 D10 (3,11) PSL(2,11) x Z2 D10 (3,11) PGL(2,11) D10 (3,11) PGL(2,11) x Z2 D20 (7,41) PSL(2,41) x Z2 A5 (13, 79) PSL(2, 79) A5 a The final lemma of this section gives some information about the pentavalent symmetric graphs of square-free order, refer to [19, Theorem 1.1] and [11, Theorem 1.1]. Lemma 2.9. Let r be a pentavalent symmetric graph of order 2n, where n is an odd square-free integer and has at least three prime factors. Then one of the following statements holds. (1) Aut r is soluble and Aut r = D2n : Z5. (2) Aut r = PSL(2,p) or PGL(2,p), where p > 5 is a prime. (3) The triple (r, 2n, Aut r) lies in the following Table 4. Table 4: Two 'sporadic' pentavalent symmetric graphs. r 2n Aut r (Aut r)a ~C390 390 PSL(2,25) F20 C2926 2926 Ji A5 3 Some examples In this section, we give some examples of pentavalent symmetric graphs of order 4n with n an odd square-free integer. In order to construct our graphs we first introduce the definition of a coset graph. Let G be a finite group and let H be a core-free subgroup of G. Let t g G and t2 G H. Define the coset graph Cos(G, H, t) of G with respect to H as the graph with vertex set [G : H] such that Hx, Hy are adjacent if and only if yx-1 G HtH. The following lemma about coset graphs is well known and the proof of the lemma follows from the definition of coset graphs. Lemma 3.1. Using the notation as above, the coset graph r = Cos(G, H, t) is G-arc-transitive graph and (1) val r = |H : H n HT (2) r is connected if and only if (H, t) = G. B. Ling et al.: Pentavalent symmetric graphs of order four times an odd square-free integer 87 Conversely, each G-arc-transitive graph E is isomorphic to the coset graph Cos(G, Gv,t), where t g NG(Gvw) is a 2-element such that t2 G Gv, and v G VE, w G E(v). We next introduce the definition of the bipartite double cover of a graph. Let r be a graph with vertex set Vr. The standard double cover of r is defined as the undirected bipartite graph r with biparts V0 and V, where V = {(v, i) | v G Vr}, such that two vertices (x, 0) and (y, 1) are adjacent if and only if x, y are adjacent in r. It is easily shown that thestandard double cover can be represented as a direct product: r = fx K2. Furthermore, r is connected if and only if r is connected and non-bipartite. For a given small permutation group X, we may determine all graphs which admit X as an arc-transitive automorphism group by using Magma [1]. It is then easy to have the following result. Example 3.2. (1) There is a unique pentavalent symmetric graph of order 5852 which admits Ji x Z2 as an arc-transitive automorphism group; and its full automorphism group is J1 x Z2. This graph is denoted by C5832 which satisfies the conditions in Row 6 of Table 1. (2) There are five pentavalent symmetric graphs of order 17556 admitting J1 as an arc-transitive automorphism group; and their full automorphism group are all isomorphic to J1. These five graphs are denoted by Cj7556 which satisfy the conditions in Row 1 to Row 5 of Table 1, where 1 < i < 5. (3) There are three pentavalent symmetric graphs of order 780 which admit PSL(2, 25) x Z2 as an arc-transitive automorphism group; and their full automorphism group are all isomorphic to PSL(2,25) x Z2. These three graphs are denoted by C^80 which satisfy the conditions in Row 7 to Row 9 of Table 1, where 1 < j < 3. Remark 3.3 (Remarks on Example 3.2). (a) Let r be a pentavalent symmetric graph of order 4n with n an odd square-free integer and having at least three prime factors. Then the graphs appearing in Example 3.2 are the only sporadic graphs of such r. In fact, let A = Aut r. If A is insoluble and has no nontrivial soluble normal subgroup, then Lemma 4.2 shows that Cj7556 with 1 < i < 5 are the only sporadic graphs. If A is insoluble and has a soluble minimal normal subgroup N = Z2, then Lemma 4.3 shows that C5832 and Cj80 with 1 < j < 3 are the only sporadic graphs. If A is soluble or has a soluble minimal normal subgroup N = Zr with r > 2, then Lemma 4.1 and Lemma 4.6 show that no such exists. (b) Since both C2926 and C390 are non-bipartite, the bipartite double cover of both C2926 and C390 is connected pentavalent symmetric graph of order 4n. In fact, the graph C5832 is isomorphic to the bipartite double cover of C2926 and the graph C380 is isomorphic to the bipartite double cover of C390. Example 3.4. Let p be a prime such that p = 49, 79, 81,111 (mod 160) and let A = PSL(2,p). Then by Lemma 2.2, A has a subgroup H = A5. Let K < H with K = A4. Then NA(K) = K : (t} = S4, where t g A — H is an involution. Let r = Cos(A, H, HtH). Then r is a connected pentavalent symmetric graph. 100 Ars Math. Contemp. 16 (2019) 97-109 Example 3.5. Let p be a prime such that p = 9, 39, 41, 71 (mod 80) and let A = PGL(2,p). Then by Lemma 2.2 and Lemma 2.3, A has a subgroup H = A5. Let K < H with K = A4. Then NA(K) = K : (t} = S4 is a maximal subgroup of A, where t g A — H is an involution, and so (H, t} = A. Let r = Cos(A, H, HtH). Then r is a connected pentavalent symmetric graph. Example 3.6. Let p be a prime such that p = 9, 39, 41, 71 (mod 80) and let A = PSL(2,p) x Z2 = T x (z}, where T = PSL(2,p) and (z} = Z2. Then T has a subgroup H = A5. Let K < H with K = A4. Then Na(K) = K : (t} x (z} = S4 x Z2, where t g T — H is an involution. Let r = Cos(A, H, HtzH). Then r is a connected pentavalent symmetric graph. Example 3.7. Let p be a prime such that p = 11,19, 21, 29 (mod 40) and let A = PGL(2,p) x Z2 = T x (z}, where T = PGL(2,p) and (z} = Z2. Then T has a subgroup H = A5. Let K < H with K = A4. Then Na(K) = K : (t} x (z} = S4 x Z2, where t g T — H is an involution. Let r = Cos(A, H, HtzH). Then r is a connected pentavalent symmetric graph. 4 Proof of Theorem 1.1 Let n be an odd square-free integer and n has at least three prime factors. Let r be a pentavalent symmetric graph of order 4n. Set A = Aut r. By Lemma 2.5, | Aa | | 29 • 32 • 5, and hence |A| | 211 • 32 • 5 • n. Assume that n = p1p2 • • • ps, where s > 3 and pj's are distinct primes. Lemma 4.1. The group A is insoluble. Proof. Suppose to the contrary that A is soluble. Let F be the Fitting subgroup of A. By Lemma 2.1, F = 1 and CA(F) < F. Further, F = O2(A) x OPl (A) x OP2 (A) x • • • x Ops (A), where 02(A), Op1 (A), Op2 (A), ..., Ops (A) denote the largest normal 2-, pi-, p2-,..., ps-subgroups of A, respectively. For each p4 g {p1,p2,... ,ps}, OPi(A) has at least three orbits on Vr, by Theorem 2.6, OPi(A) is semi-regular on Vr. Therefore, F is semi-regular on Vr and so |F| divides |Vr| = 4n. Since n = p1p2 • • • ps, we have OPi (A) < ZPi. This argument also proves O2(A) < Z4 or Z2. If O2(A) = Z4 or Z2, then by Theorem 2.6, the normal quotient graph rO2(A) is a pentavalent symmetric graph of odd order, which is a contradiction. Thus, O2(A) < Z2, F = Zm, where m | 2n. It implies that CA(F) > F, and so Ca(F ) = F. If F has at least three orbits on Vr, then, by Theorem 2.6, rF is A/F-arc-transitive. Since A/F = A/Ca(f) < Aut(F) is abelian, we have (A/F)s = 1, where 6 g VrF, which is a contradiction. B. Ling et al.: Pentavalent symmetric graphs of order four times an odd square-free integer 89 Thus, F has at most two orbits on Vr. If F is transitive on Vr, then F is regular on Vr, a contradiction with F = Zm, where m | 2n. Hence F has two orbits on Vr and F = Z2n. Let K = OP3 (A) x OP4 (A) x • • • x OPs (A). Then K = ZP3P4...Ps. Since K< A has 4pip2 orbits on Vr, by Theorem 2.6(i), rK is an A/K-arc-transitive pentavalent graph of order 4p1p2, and hence rK satisfies the conditions in Table 3. Since A/K is soluble, by Remark 2.8, a contradiction occurs. Hence A is insoluble. This completes the proof of the Lemma. □ We now consider the case where A is insoluble and has no nontrivial soluble normal subgroup. Lemma 4.2. Assume that A is insoluble and has no nontrivial soluble normal subgroup. Then Aut r = J1, PSL(2,p) or PGL(2,p) with p > 29. Further, if Aut r = J1, then r = q7556 satisfies the conditions in Row 1 to Row 5 of Table 1 of Theorem 1.1, where 1 < i < 5. If Aut r = PSL(2,p) or PGL(2,p), then r satisfies the conditions in Table 5. Table 5: Aut r is almost simple. Aut r (Aut r )a r Remark PSL(2, p) A5 Example 3.4 p = 49, 79, 81,111 (mod 160) PGL(2,p) A5 Example 3.5 p = 9, 39,41, 71 (mod 80) PSL(2, p) Dio p = 9, 39,41, 71 (mod 80) PGL(2,p) Dio p = 11,19, 21, 29 (mod 40) PSL(2, p) D20 p = 49, 79, 81,111 (mod 160) PGL(2,p) D20 p = 9, 39,41, 71 (mod 80) Proof. Let N be the socle of A. Then N is insoluble and 4 divides |NIf N has more than three orbits on Vr, then by Theorem 2.6, rN is a pentavalent symmetric graph of odd order, a contradiction. Hence, N has at most two orbits on Vr, so 2n divides |N|. Assume that A has at least two minimal normal subgroups Ni and N2. Then by a similar argument as above, we have that 2n divides both | N11 and | N21. Hence 4n2 divides |A| = 211 • 32 • 5 • n, and so n divides 29 • 32 • 5. It implies that n = 3 • 5, a contradiction with n having at least three prime factors. So A has a unique minimal normal subgroup and we may write N = Sd, where S is a nonabelian simple group and d > 1. Sinceps > 5, ps divides |N| andp2s does not divide |N| as |A| | 211 • 32 • 5 • p1p2 • • • ps, we conclude that d = 1 and N = S is a nonabelian simple group. Hence A is almost simple with socle S. If Sa = 1, then S acts regularly on Vr. Hence S is a non-abelian simple group such that |S| = 4n. By checking the orders of nonabelian simple groups (see [6, pp. 135-136] for example), we have that S = PSL(2,p) and so A < Aut(S) = PGL(2,p), which is impossible as A is transitive on Ar, |A|< 2|S| and ^r| = 5|S|. Hence Sa = 1. Since r is connected and S < A, we have 1 = S^(a) < A^(a), it follows that 5 | |Sa|, we thus have 10 • p1p2 • • • ps divides |S|. 100 Ars Math. Contemp. 16 (2019) 97-109 Thus, soc(A) = S is a nonabelian simple group such that |S| | 211 • 32 • 5 • n and 10 • n | |S|. Hence the triple (S, |S|, n) lies in Table 2 of Lemma 2.4. We will analyse all the candidates one by one in the following. Assume (S,n) = (J1, 3 • 7 • 11 • 19). Then |Vr| = 17556 and A = J1 as Out(J1) = 1. It then follows from Example 3.2 that r = Ci7556 satisfies the conditions in Row 1 to Row 5 of Table 1 of Theorem 1.1, where 1 < i < 5. Assume (S,n) = (Sz(32), 5 • 31 • 41). Since Out(Sz(32)) = Z5 (see Atlas [3] for example), A = Sz(32) or Sz(32).Z5, so |Aa| = -n = 1280 or 6400, which is not possible by Lemma 2.5. Similarly, for the case (S, n) = (PSL(5,2), 3 • 7 • 31), then A = PSL(5,2) or PSL(5, 2).Z2 as Out(PSL(5, 2)) = Z2. Thus, |Aa| = JA = 3840 or 7680, which is impossible by Lemma 2.5. For the case where (S, n) = (PSL(2, 28), 3 • 17 • 257), since A = PSL(2,28).O, where O < Out(PSL(2, 28)) = Z8, we have |Aa| = Jn = 2k • 5, where 6 < k < 9, which is also impossible by Lemma 2.5. For the case where (S, n) = (PSU(3,4), 3 • 5 • 13), since A = PSU(3,4).O, where O < Out(PSU(3,4)) = Z4, we have |Aa| = jn = 2k • 5, where 4 < k < 6, which is impossible by Lemma 2.5. Assume (S, n) = (PSp(4,4), 3 • 5 • 17). Since S < A < Aut(S) = PSp(4,4).Z4, we have |Aa| = -Jn = 960, 1920 or 3840. If |Aa| = 960 or 1920, then by Lemma 2.5, Aa ^ ASL(2,4) or A£L(2,4). However, by Atlas [3], PSp(4,4) has no subgroup isomorphic to ASL(2,4) and PSp(4,4)Z has no subgroup isomorphic to A£L(2,4). If |Aa| = 3840, then also by Lemma 2.5, a contradiction occurs. Assume (S,n) = (PSL(2,26), 3 • 7 • 13). Recall that S has at most two orbits on Vr, |Sa| = 4J = 240 or 2n = 480. However, by Lemma 2.2, PSL(2, 26) has no maximal subgroup with order a multiple of 240, a contradiction occurs. Similarly, for the case (S, n) = (J2,3 • 5 • 7), then |Sa| = 4§ = 2880 or 2n = 5760. By Atlas [3], J2 has no maximal subgroup with order a multiple of 2880, a contradiction also occurs. Assume S = M23. Then n = 3 • 7 • 11 • 23 or 7 • 11 • 23, and as Out(M23) = 1, we have A = S and |Aa| = ^r = 480 or 1440. By Lemma 2.5, it is impossible for the case |Aa| = 480. For the latter case, by a direct computation using Magma [1], no graph r exists. If (S, n) = (M22,7 • 11 • 23), as Out(M22) = Z2, we have A = M22 or M22.Z2, so | Aa | = = 480 or 960, a computation by MAGMA [1] shows that no graph r exists. Similarly, we can exclude the case where (S, n) = (PSL(2, 25), 3 • 5 • 13) by MAGMA [1]. Assume (S, n) = (M24, 3 • 7 • 11 • 23) or (J1, 3 • 7 • 11 • 19). Since Out(M24) = Out(J1) = 1, we always have A = S. Hence |Aa | = -Jn = 11520 or 10. A computation by MAGMA [1] also shows that no graph r exists. Finally, assume S = PSL(2,p) with p > 29 a prime. Then A = PSL(2,p) or PGL(2,p). By Lemma 2.2, Lemma 2.3 and Lemma 2.5, we have Aa = Z5, D1o, D2o or A5. If Aa = Z5, then r is an arc-regular pentavalent graph of order four times an odd square-free integer. However, by [15, Theorem 1.1], no such r exists. Hence Aa = D10, D20 or A5. If Aa = A5, then by Lemma 2.2 and Lemma 2.3, we have p = ±1 (mod 5). Since |A : Aa| = 4n, we have |A| is divisible by 16, but not by 32. Since |A| = | PSL(2,p)| = p(p-12(p+1) or | PGL(2,p)| = p(p - 1)(p + 1), we havep = ±15 (mod 32) for A = PSL(2,p) or p = ±7 (mod 16) for A = PGL(2,p). Since p = ±1 (mod 5), we have p = 49,79,81,111 (mod 160) for A = PSL(2,p) or p = 9, 39,41,71 (mod 80) for A = PGL(2,p). These graphs are constructed in Example 3.4 and Example 3.5. Similarly, if Aa = D10 or D20, then p satisfies the condition in Table 5. This completes the proof of the Lemma. □ B. Ling et al.: Pentavalent symmetric graphs of order four times an odd square-free integer 91 We next assume that A has a nontrivial soluble normal subgroup. Let N be a soluble minimal normal subgroup of A. Then there exists a prime r | 4n such that N = ZJ?. Further, N has at least three orbits on Vr. It follows from Theorem 2.6 that N is semi-regular on Vr, and so |N| = |Zr |d | |Vr| = 4n. If d > 2, then (r, d) = (2,2). It follows that rN is an arc-transitive graph of odd order, a contradiction. Hence d = 1, N = Zr. The next lemma consider the case where r = 2. Lemma 4.3. Assume that A is insoluble and has a soluble minimal normal subgroup N = Z2. Then one of the following statements holds: (1) Aut r = PSL(2,p) x Z2 or PGL(2,p) x Z2, where p > 29 is a prime. Furthermore, r satisfies the conditions in Table 6. (2) Aut r = PSL(2,25) x Z2 and r is isomorphic to Qso in Table 1, where 1 < i < 3. (3) Aut r = Ji x Z2 and r is isomorphic to C5852 in Table 1. Table 6: Aut r has a normal subgroup isomorphic to Z2. Aut r (Aut r )a r Remark PSL(2,p) x Z2 A5 Example 3.6 p = 9, 39,41, 71 (mod 80) PGL(2, p) x Z2 A5 Example 3.7 p = 11,19, 21, 29 (mod 40) PSL(2,p) x Z2 D10 p = 11,19, 21, 29 (mod 40) PSL(2,p) x Z2 D20 p = 9, 39,41, 71 (mod 80) PGL(2, p) x Z2 D20 p = 11,19, 21, 29 (mod 40) Proof. Since N has more than three orbits on Vr, then by Theorem 2.6, rN is an A/N-arc-transitive pentavalent graph of order n = 2n. It follows that rN is isomorphic to one of the graphs in Lemma 2.9. Since A/N < Aut rN and A/N is insoluble, we have that Aut rN is insoluble and so Aut rN = PSL(2,p), PGL(2,p), PSL(2, 25) or J1. Let A := Aut r. Suppose that A = PSL(2,p) or PGL(2,p). Since A/N is insoluble, by Lemma 2.2 and Lemma 2.3, A/N is isomorphic to A5, PSL(2,p) or PGL(2,p). If A/N = A5, then since rN is an A/N-arc-transitive pentavalent graph of order n = 2n, we have 2n • 5 | | A51. It implies that n divides 6, a contradiction with n having at least three odd prime factors. Thus, A/N is isomorphic to PSL(2,p) or PGL(2,p). Therefore, A = N. PSL(2,p) or N. PGL(2,p), that is, A = PSL(2,p) x Z2, SL(2,p), PGL(2,p) x Z2 or SL(2,p).Z2. Assume first that A = SL(2,p). Note that SL(2,p) has a unique central involution. Then by Lemma 2.5, Aa = Z5. It follows that |Vr| = |A : Aa| is divisible by 8 as | SL(2,p)| is divisible by 8, a contradiction. Assume next that A = SL(2,p).Z2. Then A contains a normal subgroup H isomorphic to SL(2,p). Since 8 | |H|, we have Ha = 1. By Theorem 2.6, H has at most two orbits on Vr and so 144 1 2. If H is transitive on Vr, then H \Ha\ I is arc-transitive. A similar argument with the case A = SL(2,p), a contradiction occurs. Therefore, H has two orbits on Vr and so Ha = Aa. Since H has a unique central involution, by Lemma 2.5, Aa = Z5, it follows that |Vr| = |A : Aa| is divisible by 16, a contradiction. Therefore, A = PSL(2,p) x Z2 or PGL(2,p) x Z2 in this case. By a similar 100 Ars Math. Contemp. 16 (2019) 97-109 argumentas for the case A = PSL(2,p) (the last paragraph in the proof of Lemma 4.2), we have that r satisfies the condition in Table 6. Note that since 16 divides | PGL(2, p) x Z21 and |A : Aa| = 4n, we have (A, Aa) = (PGL(2,p) x Z2, D10). Suppose that A = PSL(2,25). Since rN is A/N-arc-transitive, we have that 5 • 390 I |A/N|. By checking the maximal subgroup of PSL(2, 25) (see Atlas [3] for example), we have that A/N = A = PSL(2,25). It follows that A = SL(2, 25) or PSL(2, 25) x Z2. If A = PSL(2,25) x Z2, then by Example 3.2, r = C780 in Table 1, where 1 < i < 3. If A = SL(2,25), then by Magma [1], no graph r exists. Suppose that A = J1. Similarly, since rN is A/N-arc-transitive, we have that 5 • 2926 | | A/N|. By checking the maximal subgroup of J1 (see Atlas [3] for example), we have that A/N = A = J1. Since the Schur multiplier of J1 is Z1, A = N.J1 = J1 x Z2. By Example 3.2, r = C5852 in Table 1. □ Finally, suppose that r > 2. We first prove the following lemma. Lemma 4.4. Let E be a graph. Assume that E is isomorphic to one of the graphs appearing in Lemma 2.7, in Lemma 4.2 or in Lemma 4.3. If M is an arc-transitive subgroup of Aut E, then M contains the derived subgroup of Aut E. Proof. Let E be a graph and isomorphic to one of the graphs appearing in Lemma 2.7, in Lemma 4.2 or in Lemma 4.3. Let M be an arc-transitive subgroup of B = Aut E. Then B = MBaß, where (a, ß) G AE. In particular, m := |B : M| divides |Baß|. Assume first that E is isomorphic to one of the graphs appearing in Lemma 2.7. Then, in the first three rows of Table 3, we have that M has index at most two, and for the fourth row M has index at most four, so in particular, M contains B'. For the last two rows, we have that m | 12. Since there is no faithful representation of B in degree m for 2 < m < 12, we have I < m < 2 and so M also contains B'. Now assume that E is isomorphic to one of the graphs appearing in Lemma 4.2 or in Lemma 4.3. Then B is isomorphic to one of the groups PSL(2,p), PGL(2,p), PSL(2,p) x Z2, PGL(2,p) x Z2, J1, J1 x Z2 or PSL(2, 25) x Z2 withp > 29. If B = J1, then M has index at most two. If B = J1 x Z2, then M has index at most 12. If B = PSL(2, 25) x Z2, then M has index at most four. For these three cases, by a similar argument as above, we also have M contains B'. If B = PSL(2,p), then sincep | n and 20n | |M|, by Lemma 2.2, M < Zp : Zp-i or M = B = PSL(2,p). If M < Zp : Zp-i, then M = Zp : Z; for some II ■. Thus, M has a normal subgroup, say S = Zp, which has more than three orbits on VE. It then follows from Theorem 2.6 that the normal quotient graph ES is M/S-arc-transitive, a contradiction occurs as M/S = Z; is cyclic. Hence, M < Zp : Zp-i and so M = B' = PSL(2,p). If B = PGL(2,p), then since 20n | |M|, by Lemma 2.3, M < Zp : Zp-1, M < PSL(2,p) or M = B = PGL(2,p). With a similar argument, we can conclude that M > B' = PSL(2,p). Similarly, we can further show that M > B' = PSL(2,p) for the case B = PSL(2,p) x Z2 or PGL(2,p) x Z2. □ Now assume that A has a soluble minimal normal subgroup N = Zr for r > 2. Lemma 4.5. Assume that A has a soluble minimal normal subgroup N = Zr for r > 2. Then the normal quotient rN is not isomorphic to any graph appearing in Lemma 2.7, Lemma 4.2 or Lemma 4.3. B. Ling et al.: Pentavalent symmetric graphs of order four times an odd square-free integer 93 Proof. Suppose to the contrary that rN is isomorphic to one of the graphs appearing in Lemma 2.7, Lemma 4.2 or Lemma 4.3. Let M/N = (Aut rN )',andlet Q := {PSL(2,p), Ji, PSL(2, 25), A5 }. By checking the graphs appearing in Lemma 2.7, in Lemma 4.2 or in Lemma 4.3, we have that Aut rN is isomorphic to one of the groups PSL(2, p), PGL(2, p), PSL(2,p) x Z2, PGL(2,p) x Z2, J1, J1 x Z2, PSL(2, 25) x Z2 or A5 x D10. Thus, m/n is isomorphic to one of the groups in Q. Since the order of the Schur multiplier of a group in Q is less than or equal to 2 (see [10, Theorem 7.1.1] for PSL(2,p) and Atlas [3] for the others) and r > 2, we have that M' G Q. By Theorem 2.6, A/N < Aut rN is transitive on ArN. It follows from Lemma 4.4 that A/N contains the derived subgroup of Aut Tn , that is, M/N < A/N. Since M/N < Aut rN, we have M/N < A/N. Therefore, M' char M < A, it implies that M' 2. Proof. Suppose that, on the contrary, A has a soluble minimal normal subgroup N = Zr with r > 2. We prove the lemma by induction on the order of r. Assume first that n = pqt has three prime factors. (Note that, by Table 3, the conclusion of Lemma 4.6 does not hold for n = pq.) Without loss of generality, we may assume that r = t. Then rN is a pentavalent symmetric graph of order 4pq. By Lemma 2.7, rN is isomorphic to one of the graphs in Table 3, which contradicts to Lemma 4.5. Assume next that n has at least four prime factors. Note that Aut rN is insoluble. If Aut rN has no nontrivial soluble normal subgroup, then rN is isomorphic to one of the 100 Ars Math. Contemp. 16 (2019) 97-109 graphs in Lemma 4.2, which contradicts to Lemma 4.5. If Aut rN has a soluble minimal normal subgroup NV, then we can also conclude that NV = Zf with f a prime. 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