A 2-arc Transitive Hexavalent Nonnormal Cayley Graph on A₁₁₉

Abstract

A Cayley graph $\Gamma =\mathsf{Cay}(G,S)$ is said to be normal if the base group G is normal in $\mathsf{Aut}\Gamma$. The concept of the normality of Cayley graphs was first proposed by M.Y. Xu in 1998 and it plays a vital role in determining the full automorphism groups of Cayley graphs. In this paper, we construct an example of a 2-arc transitive hexavalent nonnormal Cayley graph on the alternating group ${\mathsf{A}}_{119}$. Furthermore, we determine the full automorphism group of this graph and show that it is isomorphic to ${\mathsf{A}}_{120}$.

1. Introduction

Throughout this paper, all graphs are assumed to be finite and undirected.

For a graph $\Gamma$, we use $V\Gamma$, $E\Gamma$, $Arc\Gamma$ and $\mathsf{Aut}\Gamma$ to denote the vertex set, edge set, arc set and full automorphism group of the graph $\Gamma$, respectively. A graph $\Gamma$ is said to be arc-transitive if the full automorphism group $\mathsf{Aut}\Gamma$ acts transitively on $Arc\Gamma$. We use $\mathsf{val}\Gamma$ to denote the valency of the $\Gamma$, and we say $\Gamma$ is a cubic, tetravalent, pentavalent or hexavalent graph, meaning $\mathsf{val}\Gamma =3$, 4, 5 or 6.

Let G be a finite group with identity element 1 and S (say Cayley subset) a subset of G such that $1\notin S$ and $S=S^{-1}:=\{x^{-1}\mid x\in S\}$. Define the Cayley graph $\mathsf{Cay}(G,S)$, that is, the Cayley graph of G with respect to the Cayley subset S as the graph with vertex set G such that $g,h\in G$ are adjacent if and only if $h{g}^{-1}\in S$. It is easy to see that the valency of $\mathsf{Cay}(G,S)$ is $\left|S\right|$. As we all know, $\mathsf{Cay}(G,S)$ is connected if and only if $\langle S\rangle =G$. On the other hand, letting $R\left(G\right)$ be the right regular representation of G and letting $\mathsf{Aut}\mathsf{Cay}(G,S)$ be the full automorphism group of $\mathsf{Cay}(G,S)$, there are clearly $R\left(G\right)\le \mathsf{Aut}\mathsf{Cay}(G,S)$, and $R\left(G\right)$ acts transitively on the vertices of $\mathsf{Cay}(G,S)$. Then, the graph $\mathsf{Cay}(G,S)$ is vertex-transitive, and G (or $R\left(G\right)$) can be viewed as a regular subgroup of $\mathsf{Aut}\mathsf{Cay}(G,S)$. Conversely, a connected graph $\Gamma$ is isomorphic to a Cayley graph of a group G if and only if the full automorphism group $\mathsf{Aut}\Gamma$ contains a subgroup which acts regularly on $V\Gamma$ and the subgroup is isomorphic to G (see [1]). A Cayley graph $\Gamma =\mathsf{Cay}(G,S)$ is said to be a normal Cayley graph if the base group G is normal in $\mathsf{Aut}\Gamma$; otherwise, $\Gamma$ is said to be a nonnormal Cayley graph (see [2]).

The study about Cayley graphs on finite non-abelian simple groups has always attracted much attention because of Cayley graphs with high levels of symmetry; for example, vertex-transitivity, edge-transitivity and arc-transitivity are widely used in the design of interconnection networks. For more detailed applications, we recommend that readers refer to [3,4]. Let G be a finite non-abelian simple group, and let $\Gamma =\mathsf{Cay}(G,S)$ be a connected arc-transitive Cayley graph on G. The main motivation for classifying 2-arc-transitive nonnormal Cayley graphs comes from the fact that Fang, Ma and Wang [5] proved all but finitely that many locally primitive Cayley graphs of valency $d\le 20$ or a prime number of the finite non-abelian simple groups are normal. In [5] (Problem 1.2), they proposed the following problem: classify nonnormal locally primitive Cayley graphs (note that 2-arc-transitive graphs must be locally primitive) of finite simple groups with valency $d\le 20$ or a prime number. To solve this problem, we should study each valency $d\le 20$ or a prime number. In the case where $\Gamma$ is a cubic graph (3-valent), Li [6] proved that $\Gamma$ must be normal, except for seven exceptions. On the basis of Li’s result, Xu et al. [7,8] proved that $\Gamma$ must be normal, except for two exceptions on ${\mathsf{A}}_{47}$. In the case where $\Gamma$ is a tetravalent graph (4-valent), Fang et al. in [9] proved that most of such $\Gamma$ are normal, except for Cayley graphs on a list of G. Further, Fang et al. in [10] proved that $\Gamma$ are normal when $\Gamma$ is 2-transitive, except for two graphs on ${\mathrm{M}}_{11}$. In the case where $\Gamma$ is a pentavalent graph (5-valent), Zhou and Feng [11] proved that all 1-transitive Cayley $\Gamma$ of simple groups are normal. Ling and Lou in [12] gave an example of a 2-transitive pentavalent nonnormal Cayley graph on ${\mathsf{A}}_{39}$. Therefore, the next natural problem is to study the case of the 6-valent. However, there are no known nonnormal examples of hexavalent 2-arc-transitive Cayley graphs on finite simple groups.

The aim of this paper is to construct a nonnormal example of a connected 2-arc transitive hexavalent Cayley graph on a finite non-abelian simple group. Our main result is the following theorem.

Theorem 1.

There exists a nonnormal example of a connected 2-arc-transitive hexavalent Cayley graph on the alternating group ${\mathsf{A}}_{119}$, and the full automorphism group of this graph is isomorphic to the alternating group ${\mathsf{A}}_{120}$.

2. Preliminaries

In this section, we give some necessary preliminary results which are used in later discussions.

Let G be a finite group and let H be a subgroup of G. Then we have the following result (see [13] (Ch. I, 1.4)).

Lemma 1.

Let G be a group and let H be a subgroup of G. Let ${\mathrm{N}}_G\left(H\right)$ be the normalizer of H in G, and let ${\mathrm{C}}_G\left(H\right)$ be the centralizer of H in G. Then, ${\mathrm{N}}_G\left(H\right)/{\mathrm{C}}_G\left(H\right)$ is isomorphic to a subgroup of the automorphism group $\mathsf{Aut}\left(H\right)$ of H.

We next introduce the definition of a Sabidussi coset graph. Let G be a group, $g\in G\backslash H$ such that $g^2\in H$, and let H be a core-free subgroup of G. Define the Sabidussi coset graph $\mathsf{Cos}(G,H,g)$ of G with respect to the core-free subgroup H as the graph with vertex set $[G:H]$ (the set of cosets of H in G) such that $Hx$ and $Hy$ are adjacent if and only if $y{x}^{-1}\in HgH$. The following lemma follows from [14], and it can be easily proved by the definition of the coset graphs (see [15] (Theorem 3) for example).

Lemma 2.

Let G be a group, $g\in G\backslash H$ such that $g^2\in H$, and let H be a core-free subgroup of G. Let $\Gamma =\mathsf{Cos}(G,H,g)$ be a Sabidussi coset graph of G with respect to H. Then, Γ is G-arc-transitive and the following holds:

(1) 

The valency of the graph Γ is equal to $|H:H\cap H^g|$.

(2) 

Γ is a connected graph if and only if $\langle H,g\rangle =G$.

(3) 

If G contains a subgroup R is regular on $V\Gamma$, then $\Gamma \cong \mathsf{Cay}(R,S)$, where $S=R\cap HgH$.

Conversely, if Σ is an X-arc-transitive graph, then Σ is isomorphic to a Sabidussi coset graph $\mathsf{Cos}(X,X_v,g)$, where $g\in {\mathrm{N}}_X\left(X_{vw}\right)$ is a 2-element such that $g^2\in X_v$, and $v\in V\Sigma$, $w\in \Sigma \left(v\right)$.

Proof. 

Let $\Sigma$ be an X-arc-transitive graph. Let $v\in V\Sigma$ be a vertex of $\Sigma$ and $w\in \Sigma \left(v\right)$. Since $\Sigma$ is X-arc-transitive, there is g such that $v^g=w$. For each $x\in X$, define $\phi :Hx⟶v^x$. Then we can verify that $\phi$ is a graph isomorphic from $\Sigma$ to $\mathsf{Cos}(X,X_v,g)$. Since $\Sigma$ is undirected, we have $g^2\in X_v$. Hence, ${(H\cap Hg)}^g=H\cap Hg$. Thus, we can choose a 2-element g satisfying $g\in {\mathrm{N}}_X\left(X_{vw}\right)$.    □

Let $t_1\ge 0$ and $t_2\ge 0$ be two integers. We denote by the $\{2,3\}$-group the finite group of the order $2^{t_1}{3}^{t_2}$. Following the definition of relevant objects in [16] (Theorem 3.1), we have the following lemma, which is about the stabilizers of arc-transitive hexavalent graphs. For the structure of the received stabilizers, see the proof in [16] (Page 926).

Lemma 3.

Let s be a positive integer, and let Γ be a connected hexavalent $(G,s)$-transitive graph for some $G\le \mathsf{Aut}\Gamma$. Let $v\in V\Gamma$. Then $s\le 4$ and one of the following statements holds:

(1) 

For $s=1$, the stabilizer $G_v$ is a $\{2,3\}$-group.

(2) 

For $s=2$, the stabilizer $G_v\cong \mathrm{PSL}(2,5)$, $\mathrm{PGL}(2,5)$, ${\mathsf{A}}_6$ or ${\mathrm{S}}_6$.

(3) 

For $s=3$, the stabilizer $G_v\cong {\mathrm{D}}_{10}\times \mathrm{PSL}(2,5)$, $F_{20}\times \mathrm{PGL}(2,5)$, ${\mathsf{A}}_5\times {\mathsf{A}}_6$, ${\mathrm{S}}_5\times {\mathrm{S}}_6$. $({\mathrm{D}}_{10}\times \mathrm{PSL}(2,5))·{\mathbb{Z}}_2$ with ${\mathrm{D}}_{10}·{\mathbb{Z}}_2=F_{20}$ and $\mathrm{PSL}(2,5)·{\mathbb{Z}}_2=\mathrm{PGL}(2,5)$, or $({\mathsf{A}}_5\times {\mathsf{A}}_6)⋊{\mathbb{Z}}_2$ with ${\mathsf{A}}_5⋊{\mathbb{Z}}_2={\mathrm{S}}_5$ and ${\mathsf{A}}_6⋊{\mathbb{Z}}_2={\mathrm{S}}_6$.

(4) 

For $s=4$, the stabilizer $G_v\cong {\mathbb{Z}}_5^2⋊\mathrm{GL}(2,5)=\mathrm{AGL}(2,5)$.

3. A 2-arc Transitive Hexavalent Nonnormal Cayley Graph on ${\mathsf{A}}_{119}$

In this section, we construct a connected 2-arc transitive hexavalent nonnormal Cayley graph on ${\mathsf{A}}_{119}$ and determine its full automorphism group. In fact, if $\Gamma :=\mathsf{Cay}(G,S)$ is a Cayley graph of a non-abelian simple group G, then G is core free in X, where $G\le X\le \mathsf{Aut}\Gamma$. Let $v\in V\Gamma$ and $H=X_v$. Suppose that $\left|H\right|=n$. Then by Lemma 3, n may be 60, 120, etc. Consider the action of X on the set of $[X:G]$ by right multiplication; then, $X\lesssim {\mathrm{S}}_n$. So, we may construct the nonnormal Cayley graph in ${\mathrm{S}}_n$, where $n=60$, 120, etc. The following example is really the case where we construct $n=120$.

Construction 1.

Let G be the alternating group on the set $\{2,3,\dots ,120\}$. Then, $G\cong {\mathsf{A}}_{119}$. Let $H=\langle a,b\rangle <X:={\mathsf{A}}_{120}$ (the alternating group on $\{1,2,\dots ,120\}$), where the following holds:

$$\begin{array}{cc}a=\hfill & \left(1243\right)\left(5131217\right)\left(6141118\right)\left(7151019\right)\left(816920\right)\left(216110181\right)\hfill \\ & \left(226210282\right)\left(236310383\right)\left(246410484\right)\left(256510585\right)\left(266610686\right)\hfill \\ & \left(276710787\right)\left(286810888\right)\left(296910989\right)\left(307011090\right)\left(317111191\right)\hfill \\ & \left(327211292\right)\left(337311393\right)\left(347411494\right)\left(357511595\right)\left(367611696\right)\hfill \\ & \left(377711797\right)\left(387811898\right)\left(397911999\right)\left(4080120100\right)\left(41565946\right)\hfill \\ & \left(42555748\right)\left(43546045\right)\left(44535847\right)\left(49515250\right),\hfill \end{array}$$

$$\begin{array}{cc}b=\hfill & \left(12141\right)\left(22242\right)\left(32343\right)\left(42444\right)\left(52545\right)\left(62646\right)\left(72747\right)\hfill \\ & \left(82848\right)\left(92949\right)\left(103050\right)\left(113151\right)\left(123252\right)\left(133353\right)\left(143454\right)\hfill \\ & \left(153555\right)\left(163656\right)\left(173757\right)\left(183858\right)\left(193959\right)\left(204060\right)\left(6185110\right)\hfill \\ & \left(6286112\right)\left(6387109\right)\left(6488111\right)\left(6595106\right)\left(6693108\right)\left(6796105\right)\hfill \\ & \left(6894107\right)\left(6981113\right)\left(7082114\right)\left(7183115\right)\left(7284116\right)\left(73100119\right)\hfill \\ & \left(7499117\right)\left(7598120\right)\left(7697118\right)\left(7791101\right)\left(7889102\right)\left(7992103\right)\hfill \\ & \left(8090104\right).\hfill \end{array}$$

Take $x\in X$ as follows:

$$\begin{array}{cc}x=\hfill & \left(179\right)\left(280\right)\left(360\right)\left(458\right)\left(5113\right)\left(664\right)\left(7114\right)\left(863\right)\left(9112\right)\left(10111\right)\left(1112\right)\hfill \\ & \left(1347\right)\left(1573\right)\left(16106\right)\left(1943\right)\left(2041\right)\left(21118\right)\left(22120\right)\left(2324\right)\left(2550\right)\left(2649\right)\hfill \\ & \left(2768\right)\left(2866\right)\left(3062\right)\left(3261\right)\left(3342\right)\left(3444\right)\left(35103\right)\left(36101\right)\left(37107\right)\left(3845\right)\hfill \\ & \left(4075\right)\left(48108\right)\left(5354\right)\left(55115\right)\left(56116\right)\left(57119\right)\left(59117\right)\left(65109\right)\left(67110\right)\hfill \\ & \left(69102\right)\left(70104\right)\left(7172\right)\left(76105\right)\left(8189\right)\left(8293\right)\left(8499\right)\left(8598\right)\left(8791\right)\left(8894\right)\hfill \\ & \left(90100\right)\left(9697\right).\hfill \end{array}$$

Define $\Sigma =\mathsf{Cos}(X,H,x)$.

Lemma 4.

The graph $\Sigma =\mathsf{Cos}(X,H,x)$ in Construction 1 is a connected 2-arc-transitive graph and isomorphic to the nonnormal hexavalent Cayley graph $\mathsf{Cay}(G,S)$ of G, determined by $S=\{x_1,x_1^{-1},x_2,x_3,x_4,x_5\}$ with the following:

$$\begin{array}{cc}{x}_1=\hfill & \left(2338369510145608077\right)\left(437971034635961075878\right)\hfill \\ & \left(5901066651281610011374\right)\left(618647698109146585105\right)\hfill \\ & \left(79211473684952262715\right)\left(82031416375598311740\right)\hfill \\ & \left(98810411699312054218111523569111171\right)\hfill \\ & \left(108711624558911853228210212709411272\right)\hfill \\ & \left(133430176244476786110\right)\left(1932\right)\left(2942489911939578410833\right)\hfill \\ & \left(4361\right),\hfill \\ x_2=\hfill & \left(297\right)\left(499\right)\left(573\right)\left(674\right)\left(78\right)\left(941\right)\left(1088\right)\left(1178\right)\left(1358\right)\left(1496\right)\hfill \\ & \left(1560\right)\left(1695\right)\left(1766\right)\left(1865\right)\left(1952\right)\left(2050\right)\left(2240\right)\left(2327\right)\left(2432\right)\left(2529\right)\hfill \\ & \left(2633\right)\left(2839\right)\left(3136\right)\left(3538\right)\left(43116\right)\left(4477\right)\left(4562\right)\left(4648\right)\left(4761\right)\left(49107\right)\hfill \\ & \left(51105\right)\left(53110\right)\left(5475\right)\left(55109\right)\left(5676\right)\left(63120\right)\left(64119\right)\left(6791\right)\left(6892\right)\hfill \\ & \left(69103\right)\left(7094\right)\left(71104\right)\left(7293\right)\left(79113\right)\left(8086\right)\left(81112\right)\left(83111\right)\left(87114\right)\hfill \\ & \left(8990\right)\left(98117\right)\left(100118\right)\left(101102\right),\hfill \\ x_3=\hfill & \left(339\right)\left(437\right)\left(5106\right)\left(6105\right)\left(750\right)\left(849\right)\left(935\right)\left(1045\right)\left(1136\right)\hfill \\ & \left(1247\right)\left(13113\right)\left(14114\right)\left(1516\right)\left(18118\right)\left(1928\right)\left(2056\right)\left(2192\right)\left(2391\right)\hfill \\ & \left(26120\right)\left(2794\right)\left(2930\right)\left(31107\right)\left(32108\right)\left(33112\right)\left(34110\right)\left(3897\right)\left(4098\right)\hfill \\ & \left(4142\right)\left(43102\right)\left(44101\right)\left(5187\right)\left(5285\right)\left(53117\right)\left(5496\right)\left(5789\right)\left(5890\right)\hfill \\ & \left(59116\right)\left(60115\right)\left(6280\right)\left(6367\right)\left(6472\right)\left(6569\right)\left(6673\right)\left(6879\right)\left(7176\right)\hfill \\ & \left(7578\right)\left(8182\right)\left(83109\right)\left(84111\right)\left(93119\right)\left(99104\right)\left(100103\right),\hfill \\ x_4=\hfill & \left(220\right)\left(37\right)\left(412\right)\left(59\right)\left(613\right)\left(819\right)\left(1116\right)\left(1518\right)\left(2149\right)\left(2384\right)\hfill \\ & \left(24101\right)\left(25112\right)\left(2628\right)\left(27110\right)\left(2994\right)\left(3060\right)\left(3196\right)\left(3259\right)\left(3385\right)\hfill \\ & \left(34120\right)\left(3587\right)\left(36118\right)\left(3853\right)\left(4055\right)\left(4276\right)\left(4474\right)\left(45119\right)\left(46117\right)\hfill \\ & \left(4748\right)\left(5064\right)\left(51102\right)\left(5467\right)\left(5665\right)\left(57108\right)\left(58106\right)\left(62104\right)\left(6382\right)\hfill \\ & \left(66116\right)\left(68114\right)\left(6986\right)\left(7188\right)\left(7397\right)\left(7599\right)\left(77105\right)\left(7880\right)\left(79107\right)\hfill \\ & \left(81103\right)\left(8991\right)\left(90115\right)\left(92113\right)\left(98109\right)\left(100111\right),\hfill \\ x_5=\hfill & \left(2111\right)\left(389\right)\left(432\right)\left(521\right)\left(6118\right)\left(723\right)\left(8117\right)\left(995\right)\left(1193\right)\hfill \\ & \left(1399\right)\left(1427\right)\left(15100\right)\left(1628\right)\left(17106\right)\left(1820\right)\left(19105\right)\left(2273\right)\left(2475\right)\hfill \\ & \left(29112\right)\left(3069\right)\left(33108\right)\left(3480\right)\left(35107\right)\left(3679\right)\left(3738\right)\left(39101\right)\left(40103\right)\hfill \\ & \left(4149\right)\left(4253\right)\left(4459\right)\left(4558\right)\left(4751\right)\left(4854\right)\left(5060\right)\left(5657\right)\left(61116\right)\hfill \\ & \left(6264\right)\left(63114\right)\left(6594\right)\left(6696\right)\left(67102\right)\left(68104\right)\left(7190\right)\left(72110\right)\left(7784\right)\hfill \\ & \left(7882\right)\left(85120\right)\left(86119\right)\left(8788\right)\left(91109\right)\left(97113\right)\left(98115\right).\hfill \end{array}$$

Proof. 

Let $\Delta :=\{1,2,\dots ,120\}$. Then, X has a natural action on $\Delta$. By Magma [17], $\langle H,x\rangle =X$, and so the graph $\Sigma$ is connected by Lemma 2 (2). Furthermore, by Magma [17], we have that H is regular on $\Delta$. However, G is the stabilizer of point 1 in X. Hence, X has a factorization $X=GH=HG$ with $G\cap H=1$. Therefore, G is regular on $[X:H]$. By Lemma 2 (3), $\Sigma$ is isomorphic to a Cayley graph of $G={\mathsf{A}}_{119}$. Additionally, by the computation of Magma [17] (for the Magma code, see Appendix A), we have $\frac{\left|H\right|}{|H\cap H^x|}=6$. Hence, Lemma 2 (1) implies that $\Sigma$ is a hexavalent graph. Since $H\cong \mathrm{PGL}(2,5)$, Lemma 3 implies that $\Sigma$ is 2-arc transitive. Since X is a non-abelian simple group, G is not normal in $X\le \mathsf{Aut}\Sigma$. It follows that $\Sigma$ is nonnormal. Let $x_1$, $x_2$, $x_3$, $x_4$, $x_5$ and S be defined as in this lemma. By the computation of Magma [17] (for the Magma code, see Appendix B), we have $G\cap \left(HxH\right)=S$. Thus, by Lemma 2 (3), we have that $\Sigma$ is isomorphic to $\mathsf{Cay}(G,S)$. This completes the proof of the lemma.    □

In the next lemma, we show that the full automorphism group $\mathsf{Aut}\Sigma$ is isomorphic to alternating group ${\mathsf{A}}_{120}$.

Lemma 5.

The full automorphism group $\mathsf{Aut}\Sigma$ of the 2-arc-transitive hexavalent graph $\Sigma =\mathsf{Cos}(X,H,x)$ in Construction 1 is isomorphic to alternating group ${\mathsf{A}}_{120}$.

Proof. 

Let $\mathsf{A}=\mathsf{Aut}\Sigma$. Assume first that the full automorphism group $\mathsf{A}$ is quasiprimitive on $V\Sigma$. Let N be a minimal normal subgroup of $\mathsf{A}$. Then, N is transitive on $V\Sigma$. It implies that N is insoluble. Thus, N is isomorphic to $T_1\times T_2\times \cdots \times T_d=T^d$, where $T_i\cong T$ for each $1\le i\le d$, T is a non-abelian simple group, and $d\ge 1$. Let p be the largest prime factor of the order of ${\mathsf{A}}_{119}$. Then, $p>5$ and $p^2/|\left|{\mathsf{A}}_{119}\right|$. Since N is transitive on $V\Sigma$ and $|V\Sigma |=|{\mathsf{A}}_{119}|$, we have that p divides $\left|N\right|$. Assume that $d\ge 2$. Then, $p^d$ divides $\left|N\right|$. However, by Lemma 3, the order of the stabilizer ${\mathsf{A}}_v$ divides $2^7·3^3·5^3$, and so $\left|\mathsf{A}\right|$ divides $2^7·3^3·5^3·\left|{\mathsf{A}}_{119}\right|$ which is divisible by $p^d$, a contradiction. Hence, we have $d=1$ and $N=T⊴\mathsf{A}$. Let $C={\mathrm{C}}_{\mathsf{A}}\left(T\right)$ be the centralizer of T in $\mathsf{A}$. Then, $C⊴N_{\mathsf{A}}\left(T\right)=\mathsf{A}$ and $CT=C\times T$. If $C\ne 1$, since $\mathsf{A}$ is quasiprimitive on $V\Sigma$, this implies that C is transitive on $V\Sigma$. It implies that p divides $\left|C\right|$. Therefore, $p^2$ divides $\left|CT\right|$, which divides $\left|\mathsf{A}\right|$, and so we have that $p^2$ divides $\left|\mathsf{A}\right|$, a contradiction. Hence, $C=1$, and $\mathsf{A}\le \mathsf{Aut}\left(T\right)$ is almost simple.

Since $T\cap X⊴X\cong {\mathsf{A}}_{120}$, it follows that $T\cap X=1$ or X. If $T\cap X=1$, then since $\frac{\left|\mathsf{A}\right|}{\left|X\right|}|2^4·3^2·5^2$, we have $\left|T\right||2^4·3^2·5^2$; note that $p>5$, $p\left|\right|T|$, a contradiction. Thus, $T\cap X=X$, and so $X\le T$. It follows that $|T:X|$ divides $|\mathsf{A}:X|$, which divides $2^4·3^2·5^2$. By [18] (pp. 135–136), we can conclude that $T=X\cong {\mathsf{A}}_{120}$. Thus, $\mathsf{A}\le \mathsf{Aut}\left(T\right)\cong {\mathrm{S}}_{120}$. If $\mathsf{A}\cong {\mathrm{S}}_{120}$, then $|{\mathsf{A}}_v|=\frac{\left|\mathsf{A}\right|}{\left|G\right|}=240$, a contradiction to Lemma 3. Hence, $\mathsf{A}\cong {\mathsf{A}}_{120}$.

Now assume that the full automorphism group $\mathsf{A}$ is not quasiprimitive on $V\Sigma$. Then there is a minimal normal subgroup M of $\mathsf{A}$ that acts nontransitively on $V\Sigma$. Since $M\cap X⊴X$, we have $M\cap X=1$ or X. For the latter case $M\cap X=X$, we have $X\le M$, and so M is transitive on $V\Sigma$, a contradiction. For the former case, $M\cap X=1$, then we have that $\left|M\right|$ divides $\frac{\left|\mathsf{A}\right|}{\left|X\right|}$, which divides $2^4·3^2·5^2$.

Assume that M is insoluble. Since $\left|M\right|$ divides $2^4·3^2·5^2$, and the simple groups ${\mathsf{A}}_5$, ${\mathsf{A}}_6$, $\mathrm{PSp}(4,3)$ are the only $\{2,3,5\}$-factor non-abelian simple groups (see [19] (Table 1), and note that the definition of the $\{2,3,5\}$-group is similar to $\{2,3\}$-group); by checking the orders of these groups, it is easy to figure out $M\cong {\mathsf{A}}_5$ or ${\mathsf{A}}_5^2$ or ${\mathsf{A}}_6$. Then since $\left|M\right|·|{\mathsf{A}}_{120}|=|M|·|X|=|L|=|V\Sigma |·|L_v|=|{\mathsf{A}}_{119}|·|L_v|$, we have $|L_v|=2^5·3^2·5^2$ or $2^7·3^3·5^3$ or $2^6·3^3·5^2$, a contradiction to the description of the orders of the stabilizers in Lemma 3.

Assume that M is soluble. Then $M\cong {\mathbb{Z}}_2^r$ or ${\mathbb{Z}}_3^s$ or ${\mathbb{Z}}_5^l$, where $1\le r\le 4$, $1\le s\le 2$ and $1\le l\le 2$. Let $L=MX$. Then $L=M:X$, a split expansion of M by X. Further, we have $L/{\mathrm{C}}_L\left(M\right)\lesssim \mathsf{Aut}\left(M\right)\cong \mathrm{GL}(r,2)$ or $\mathrm{GL}(s,3)$ or $\mathrm{GL}(l,5)$. We note that M is a subgroup of ${\mathrm{C}}_L\left(M\right)$. If $M={\mathrm{C}}_L\left(M\right)$, then we have $L/{\mathrm{C}}_L\left(M\right)=L/M\cong X\cong {\mathsf{A}}_{120}\lesssim \mathrm{GL}(r,2)$ or $\mathrm{GL}(s,3)$ or $\mathrm{GL}(l,5)$. However, for each $1\le r\le 4$, $1\le s\le 2$ and $1\le l\le 2$, $\mathrm{GL}(r,2)$, $\mathrm{GL}(s,3)$ or $\mathrm{GL}(l,5)$ has no subgroup isomorphic to the alternating group ${\mathsf{A}}_{120}$. Hence, we have $M<{\mathrm{C}}_L\left(M\right)$ and $1\ne {\mathrm{C}}_L\left(M\right)/M⊴L/M\cong {\mathsf{A}}_{120}$. It implies that ${\mathsf{A}}_{120}\cong {\mathrm{C}}_L\left(M\right)/M$; then $|{\mathrm{C}}_L\left(M\right)|=|M|·|X|=|L|$ since ${\mathrm{C}}_L\left(M\right)⊴L$, we have ${\mathrm{C}}_L\left(M\right)=L=MX$, and X centralizes M. Hence, $L=M\times X$. Then $L_v/X_v=L_v/L_v\cap X\cong L_{v}X/X\cong L/X\cong M$. Thus, $L_v\cong X_v.M$. Note that with the order of the stabilizers given in Lemma 3, we conclude $M\cong {\mathbb{Z}}_3$ or ${\mathbb{Z}}_5$. In the case where $M\cong {\mathbb{Z}}_3$, we have $|L_v|=|X_v|·|M|=360$, then $L_v\cong {\mathsf{A}}_6$, ${\mathsf{A}}_6\cong \mathrm{PGL}(2,5).{\mathbb{Z}}_3$, but there is no normal subgroup which is isomorphic to $\mathrm{PGL}(2,5)$ in ${\mathsf{A}}_6$, a contradiction. In the case where $M\cong {\mathbb{Z}}_5$, we have $|L_v|=|X_v|·|M|=600$, then $L_v\cong {\mathrm{D}}_{10}\times \mathrm{PSL}(2,5)$, ${\mathrm{D}}_{10}\times \mathrm{PSL}(2,5)\cong \mathrm{PGL}(2,5).{\mathbb{Z}}_5$; by [17], there is no normal subgroup with order 120 in ${\mathrm{D}}_{10}\times \mathrm{PSL}(2,5)$, so clearly, $\mathrm{PGL}(2,5)⋬{\mathrm{D}}_{10}\times \mathrm{PSL}(2,5)$, which also leads to a contradiction. This completes the proof of the lemma.    □

Proof of Theorem 1.

Now we are ready to prove our main Theorem 1. Let $\Sigma =\mathsf{Cos}(X,H,x)$ be the graph as in Construction 1. Then, Lemma 4 shows that $\Sigma$ is a connected 2-arc-transitive graph and isomorphic to a nonnormal hexavalent Cayley graph $\mathsf{Cay}(G,S)$, with $G\cong {\mathsf{A}}_{119}$. This proves the statement of the former part of Theorem 1. The next Lemma 5 shows that the full automorphism group $\mathsf{Aut}\Sigma$ of the graph $\Sigma$ is isomorphic to alternating group ${\mathsf{A}}_{120}$. This proves the statement of the latter part of Theorem 1, and so completes the proof of Theorem 1.    □