Pentavalent symmetric graphs of order four times an odd square-free integer

TL;DR

This paper classifies all connected pentavalent symmetric graphs of order four times an odd square-free integer, identifying their automorphism groups or specific isomorphism types.

Contribution

It generalizes previous classifications by determining all such graphs for a broader class of orders, including explicit automorphism group structures.

Findings

Automorphism groups are isomorphic to PSL(2,p), PGL(2,p), or their products with Z_2.

Identifies 8 specific graphs with unique structures.

Provides a complete classification for the specified order class.

Abstract

A graph is said to be symmetric if its automorphism group is transitive on its arcs. Guo et al. (Electronic J. Combin. 18, \#P233, 2011) and Pan et al. (Electronic J. Combin. 20, \#P36, 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 of such graphs $\it\Gamma$, either the full automorphism group ${\sf Aut}\it\Gamma$ is isomorphic to ${\sf PSL}(2,p)$, ${\sf PGL}(2,p)$, ${\sf PSL}(2,p){\times}\mathbb{Z}_2$ or ${\sf PGL}(2,p){\times}\mathbb{Z}_2$, or $\it\Gamma$ is isomorphic to one of 8 graphs.

Tables

Table 1. Table 1 .

Table 2. Table 2 .

Table 3. Table 3 .

Table 4. Table 4 .

Equations

$$\begin{array}[]{|c|c|c|c|}\hline\cr s&1&2&3\\ \hline\cr G_{\alpha}&\mathbb{Z}_{5},~{}{\sf D}_{10},~{}{\sf D}_{20}&{\sf F}_{20},~{}{\sf F}_{20}\times\mathbb{Z}_{2}&{\sf F}_{20}\times\mathbb{Z}_{4}\\ \hline\cr\end{array}$$

$$\begin{array}[]{|c|c|c|c|c|}\hline\cr s&2&3&4&5\\ \hline\cr G_{\alpha}&{\sf A}_{5},{\sf S}_{5}&{\sf A}_{4}\times{\sf A}_{5},({\sf A}_{4}\times{\sf A}_{5}){:}\mathbb{Z}_{2},&{\sf ASL}(2,4),{\sf AGL}(2,4),&\mathbb{Z}_{2}^{6}{:}{\sf\Gamma L}(2,4)\\ &&{\sf S}_{4}\times{\sf S}_{5}&{\sf A\Sigma L}(2,4),{\sf A\Gamma L}(2,4)&\\ \hline\cr|G_{\alpha}|&60,120&720,1440,2880&960,1920,2880,5760&23040\\ \hline\cr\end{array}$$

Full text

Keywords. Arc-transitive graph; Normal quotient; Automorphism group.

2000 MR Subject Classification 05C25, 05E18, 20B25.

This work was partially supported by the National Natural Science Foundation of China (11301468,11231008,11461004)

∗Corresponding author. E-mails: bolinggxu@163.com (B. Ling).

1. Introduction

All graphs in this paper are assumed to be finite, simple, connected and undirected. Let ${\it\Gamma}$ be a graph and denote $V{\it\Gamma}$ and $A{\it\Gamma}$ the vertex set and arc set of ${\it\Gamma}$, respectively. Let $G$ be a subgroup of the full automorphism group ${\sf Aut}{\it\Gamma}$ of ${\it\Gamma}$. Then ${\it\Gamma}$ is called $G$-vertex-transitive and $G$-arc-transitive if $G$ is transitive on $V{\it\Gamma}$ and $A{\it\Gamma}$, respectively. An arc-transitive graph is also called symmetric. It is well known that ${\it\Gamma}$ is $G$-arc-transitive if and only if $G$ is transitive on $V{\it\Gamma}$ and the stabilizer $G_{\alpha}:=\{g\in G\mid\alpha^{g}=\alpha\}$ for some $\alpha\in V{\it\Gamma}$ is transitive on the neighbor set ${\it\Gamma}(\alpha)$ of $\alpha$ in ${\it\Gamma}$.

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, 18]. Also, for distinct primes $p$, $q$ and $r$, the classifications of pentavalent symmetric graphs of order $2pq$ and $2pqr$ are presented in [9, 17], 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 $2p^{2}$ are presented in [8, 14, 5]. The main purpose of this paper is to extend the results in [8, 14] 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 ${\it\Gamma}$ be a pentavalent symmetric graph of order $4n$. If $n$ has at least three prime factors, then one of the following statements holds.

• (1)

${\sf Aut}{\it\Gamma}\cong{\sf PSL}(2,p)$, ${\sf PGL}(2,p)$, ${\sf PSL}(2,p)\times\mathbb{Z}_{2}$ or ${\sf PGL}(2,p)\times\mathbb{Z}_{2}$, where $p\geq 29$ is a prime.

• (2)

The triple $({\it\Gamma},n,{\sf Aut}{\it\Gamma})$ lies in the following Table 1.

Remark 1.1.

• (a)

The graphs in Table 1 are introduced in Example 3.1.

• (b)

It seems not feasible to determine all the possible values of $p$ in part (1) for general odd square-free integer $n$. However, if the number of the prime divisors of $n$ is fixed, then it is not difficult to determine the possible values of $p$ and hence all corresponding graphs ${\it\Gamma}$.

2. Preliminaries

We now give some necessary preliminary results. The first one is a property of the Fitting subgroup, see [16, P. 30, Corollary].

Lemma 2.1**.**

Let $F$ be the Fitting subgroup of a group $G$. If $G$ is soluble, then $F\neq 1$ and the centralizer $C_{G}(F)\leq F$.

The maximal subgroups of ${\sf PSL}(2,q)$ are known, see [4, Section 239].

Lemma 2.2**.**

Let $T={\sf PSL}(2,q)$, where $q=p^{n}\geq 5$ with $p$ a prime. Then a maximal subgroup of $T$ is isomorphic to one of the following groups, where $d=(2,q-1)$.

1. (1)

${\sf D}_{2(q-1)/d}$, where $q\neq 5,7,9,11$; 2. (2)

${\sf D}_{2(q+1)/d}$, where $q\neq 7,9$; 3. (3)

$\mathbb{Z}_{p}^{n}{:}\mathbb{Z}_{{(q-1)/d}}$; 4. (4)

${\sf A}_{4}$, where $q=p=5$ or $q=p\equiv 3,13,27,37~{}({\sf mod~{}}40)$; 5. (5)

${\sf S}_{4}$, where $q=p\equiv\pm 1~{}({\sf mod~{}}8)$ 6. (6)

${\sf A}_{5}$, where $q=p\equiv\pm 1~{}({\sf mod~{}}5)$, or $q=p^{2}\equiv-1~{}({\sf mod~{}}5)$ with $p$ an odd prime; 7. (7)

${\sf PSL}(2,p^{m})$* with $n/m$ an odd integer;* 8. (8)

${\sf PGL}(2,p^{n/2})$* with $n$ an even integer.*

By [2, Theorem 2], we may easily derive the maximal subgroups of ${\sf PGL}(2,p)$.

Lemma 2.3**.**

Let $T={\sf PGL}(2,p)$ with $p\geq 5$ a prime. Then a maximal subgroup of $T$ is isomorphic to one of the following groups:

1. (1)

$\mathbb{Z}_{p}:\mathbb{Z}_{p-1}$; 2. (2)

${\sf D}_{2(p+1)}$; 3. (3)

${\sf D}_{2(p-1)}$, where $p\geq 7$; 4. (4)

${\sf S}_{4}$, where $p\equiv\pm 3~{}({\sf mod~{}}8)$; 5. (5)

${\sf PSL}(2,p)$.

For a graph ${\it\Gamma}$ and a positive integer $s$, an $s$-arc of ${\it\Gamma}$ is a sequence $\alpha_{0},\alpha_{1},\dots,\alpha_{s}$ of vertices such that $\alpha_{i-1},\alpha_{i}$ are adjacent for $1\leq i\leq s$ and $\alpha_{i-1}\not=\alpha_{i+1}$ for $1\leq i\leq s-1$. In particular, a $1$-arc is just an arc. Then ${\it\Gamma}$ is called $(G,s)$-arc-transitive with $G\leq{\sf Aut}{\it\Gamma}$ if $G$ is transitive on the set of $s$-arcs of ${\it\Gamma}$. A $(G,s)$-arc-transitive graph is called $(G,s)$-transitive if it is not $(G,s+1)$-arc-transitive. In particular, a graph ${\it\Gamma}$ is simply called $s$-transitive if it is $({\sf Aut}{\it\Gamma},s)$-transitive.

Let ${\sf F}_{20}$ denote the Frobenius group of order $20$. The following lemma determines the stabilizers of pentavalent symmetric graphs, refer to [7, 18].

Lemma 2.4**.**

Let ${\it\Gamma}$ be a pentavalent $(G,s)$-transitive graph, where $G\leq{\sf Aut}{\it\Gamma}$ and $s\geq 1$. Let $\alpha\in V{\it\Gamma}$. Then one of the following holds.

• (a)

If $G_{\alpha}$ is soluble, then $s\leq 3$ and |G_{\alpha}|\,\big{|}\,80. Further, the couple $(s,G_{\alpha})$ lies in the following table.

[TABLE]

• (b)

If $G_{\alpha}$ is insoluble, then $2\leq s\leq 5$, and |G_{\alpha}|\,\big{|}\,2^{9}\cdot 3^{2}\cdot 5. Further, the couple $(s,G_{\alpha})$ lies in the following table.

[TABLE]

From [6, pp. 134-136], we can obtain the following lemma by checking the orders of nonabelian simple groups.

Lemma 2.5**.**

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 $2^{i}\cdot 3^{j}\cdot 5\cdot n$, where $1\leq i\leq 11$ and $0\leq j\leq 2$. Let $p$ be the largest prime factor of $n$. Then $T$ is listed Table 2.

*Proof.  *If $T$ is a sporadic simple group, by [6, P. 135-136], $T={\sf M}_{22}$, ${\sf M}_{23}$, ${\sf M}_{24}$, ${\sf J}_{1}$ or ${\sf J}_{2}$. If $T={\sf A}_{n}$ is an alternating group, since $3^{4}$ does not divide $|T|$, we have $n\leq 8$, it then easily exclude that $T={\sf A}_{5}$, ${\sf A}_{6}$, ${\sf A}_{7}$ or ${\sf A}_{8}$. 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 $q={r}^{d}$ is a prime power. If $r\geq 5$, 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 $T={\sf PSL}(2,p)$ with $p\geq 29$ (note that ${\sf PSL}(2,p)$ with $5\leq p\leq 23$ does not satisfy the condition of the lemma) or ${\sf PSL}(2,25)$, where $p$ is the largest prime factor of $n$. If $r\leq 3$, as $2^{12}$ and $3^{4}$ do not divide $|T|$, then we have $T={\sf Sz}(32)$, ${\sf PSU}(3,4)$, ${\sf PSp}(4,4)$, ${\sf PSL}(2,2^{6})$, ${\sf PSL}(2,2^{8})$ or ${\sf PSL}(5,2)$. $\Box$

A typical method for studying vertex-transitive graphs is taking normal quotients. Let ${\it\Gamma}$ be a $G$-vertex-transitive graph, where $G\leq{\sf Aut}{\it\Gamma}$. Suppose that $G$ has a normal subgroup $N$ which is intransitive on $V{\it\Gamma}$. Let $V{\it\Gamma}_{N}$ be the set of $N$-orbits on $V{\it\Gamma}$. The normal quotient graph ${\it\Gamma}_{N}$ of ${\it\Gamma}$ induced by $N$ is defined as the graph with vertex set $V{\it\Gamma}_{N}$, and $B$ is adjacent to $C$ in ${\it\Gamma}_{N}$ if and only if there exist vertices $\beta\in B$ and $\gamma\in C$ such that $\beta$ is adjacent to $\gamma$ in ${\it\Gamma}$. In particular, if ${\sf val}({\it\Gamma})={\sf val}({\it\Gamma}_{N})$, then ${\it\Gamma}$ is called a *normal cover * of ${\it\Gamma}_{N}$.

A graph ${\it\Gamma}$ is called $G$-locally primitive if, for each $\alpha\in V{\it\Gamma}$, the stabilizer $G_{\alpha}$ acts primitively on ${\it\Gamma}(\alpha)$. Obviously, a pentavalent symmetric graph is locally primitive. The following theorem gives a basic method for studying vertex-transitive locally primitive graphs, see [15, Theorem 4.1] and [12, Lemma 2.5].

Theorem 2.6**.**

Let ${\it\Gamma}$ be a $G$-vertex-transitive locally primitive graph, where $G\leq{\sf Aut}{\it\Gamma}$, and let $N\lhd G$ have at least three orbits on $V{\it\Gamma}$. Then the following statements hold.

• (i)

$N$* is semi-regular on $V{\it\Gamma}$, $G/N\leq{\sf Aut}{\it\Gamma}_{N}$, and ${\it\Gamma}$ is a normal cover of ${\it\Gamma}_{N}$;*

• (ii)

$G_{\alpha}\cong(G/N)_{\gamma}$, where $\alpha\in V{\it\Gamma}$ and $\gamma\in V{\it\Gamma}_{N}$;

• (iii)

${\it\Gamma}$* is $(G,s)$-transitive if and only if ${\it\Gamma}_{N}$ is $(G/N,s)$-transitive, where $1\leq s\leq 5$ 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 [14, Theorem 3.1].

Lemma 2.7**.**

Let ${\it\Gamma}$ be a pentavalent symmetric graph of order $4pq$, where $q>p\geq 3$ are primes. Then the couple $({\sf Aut}{\it\Gamma},({\sf Aut}{\it\Gamma})_{\alpha})$ lies in the following Table $3$, where $\alpha\in V{\it\Gamma}$.

Remark 2.8.

• (a)

Suppose that ${\it\Gamma}$ is one of the graphs in Lemma 2.7 and $M$ is an arc-transitive subgroup of ${\sf Aut}{\it\Gamma}$. 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 $\mathcal{C}_{66}^{(2)}$ and $\mathcal{C}_{132}^{5}$ in [8, Theorem 4.1] are isomorphic, ${\sf Aut}(\mathcal{C}_{132}^{5})\cong{\sf PGL}(2,11){\times}\mathbb{Z}_{2}$.

The final lemma of this section gives some information about the pentavalent symmetric graphs of square-free order, refer to [17, Theorem 1.1] and [11, Theorem 1.1].

Lemma 2.8**.**

Let ${\it\Gamma}$ 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)

${\sf Aut}{\it\Gamma}$* is soluble and ${\sf Aut}{\it\Gamma}\cong{\sf D}_{2n}:\mathbb{Z}_{5}$.*

• (2)

${\sf Aut}{\it\Gamma}={\sf PSL}(2,p)$* or ${\sf PGL}(2,p)$, where $p\geq 5$ is a prime.*

• (3)

The triple $({\it\Gamma},2n,{\sf Aut}{\it\Gamma})$ lies in the following Table 4.

3. Some examples

In this section, we give some examples of pentavalent symmetric graphs of order $4n$ with $n$ an odd square-free integer.

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 3.1**.**

(1) There is a unique pentavalent symmetric graph of order $5852$ which admits ${\sf J}_{1}\times\mathbb{Z}_{2}$ as an arc-transitive automorphism group; and its full automorphism group is ${\sf J}_{1}\times\mathbb{Z}_{2}$. This graph is denoted by ${\mathcal{C}}_{5832}$ which satisfies the conditions in Row 6 of Table 1.

(2) There are five pentavalent symmetric graphs of order $17556$ admitting ${\sf J}_{1}$ as an arc-transitive automorphism group; and their full automorphism group are all isomorphic to ${\sf J}_{1}$. These five graphs are denoted by ${\mathcal{C}}_{17556}^{i}$ which satisfy the conditions in Row 1 to Row 5 of Table 1, where $1\leq i\leq 5$.

(3) There are two pentavalent symmetric graphs of order $780$ which admits ${\sf PSL}(2,25)\times\mathbb{Z}_{2}$ as an arc-transitive automorphism group; and their full automorphism group are all isomorphic to ${\sf PSL}(2,25)\times\mathbb{Z}_{2}$. These two graphs are denoted by ${\mathcal{C}}_{780}^{j}$ which satisfy the conditions in Row 7 to Row 8 of Table 1, where $1\leq j\leq 2$.

4. Proof of Theorem 1.1

Let $n$ be an odd square-free integer and $n$ has at least three prime factors. Let ${\it\Gamma}$ be a pentavalent symmetric graph of order $4n$. Set ${\sf A}={\sf Aut}{\it\Gamma}$. By Lemma 2.4, |{\sf A}_{\alpha}|\,\big{|}\,2^{9}\cdot 3^{2}\cdot 5, and hence |{\sf A}|\,\big{|}\,2^{11}\cdot 3^{2}\cdot 5\cdot n. Assume that $n=p_{1}p_{2}\cdots p_{s}$, where $s\geq 3$ and ${p_{i}}^{\prime}s$ are distinct primes.

We first consider the case where ${\sf A}$ is soluble.

Lemma 4.1**.**

Assume that ${\sf A}$ is soluble. Then no graph ${\it\Gamma}$ exists.

*Proof.  *Let $F$ be the Fitting subgroup of ${\sf A}$. By Lemma 2.1, $F\neq 1$ and ${\bf C}_{{\sf A}}(F)\leq F$. Further, $F={\bf O}_{2}({\sf A})\times{\bf O}_{p_{1}}({\sf A})\times{\bf O}_{p_{2}}({\sf A})\times\cdots\times{\bf O}_{p_{s}}({\sf A})$, where ${\bf O}_{2}({\sf A})$, ${\bf O}_{p_{1}}({\sf A})$, ${\bf O}_{p_{2}}({\sf A})$,$\dots$,${\bf O}_{p_{s}}({\sf A})$ denote the largest normal $2$-, $p_{1}$-, $p_{2}$-, $\dots$, $p_{s}$-subgroups of ${\sf A}$, respectively.

For each $p_{i}\in\{p_{1},p_{2},\dots,p_{s}\}$, ${\bf O}_{p_{i}}({\sf A})$ has at least three orbits on $V{\it\Gamma}$, by Theorem 2.6, ${\bf O}_{p_{i}}({\sf A})$ is semi-regular on $V{\it\Gamma}$. Therefore, $F$ is semi-regular on $V{\it\Gamma}$ and ${\bf O}_{p_{i}}({\sf A})\leq\mathbb{Z}_{p_{i}}$. This argument also proves ${\bf O}_{2}({\sf A})\leq\mathbb{Z}_{4}$ or $\mathbb{Z}_{2}^{2}$. If ${\bf O}_{2}({\sf A})=\mathbb{Z}_{4}$ or $\mathbb{Z}_{2}^{2}$, then by Theorem 2.6, the normal quotient graph ${\it\Gamma}_{{\bf O}_{2}({\sf A})}$ is a pentavalent symmetric graph of odd order, which is a contradiction. Thus, ${\bf O}_{2}({\sf A})\leq\mathbb{Z}_{2}$, $F\cong\mathbb{Z}_{m}$, where m\,\big{|}\,2n. It implies that ${\bf C}_{{\sf A}}(F)\geq F$, and so ${\bf C}_{{\sf A}}(F)=F$.

If $F$ has at least three orbits on $V{\it\Gamma}$, then, by Theorem 2.6, ${\it\Gamma}_{F}$ is ${\sf A}/F$-arc-transitive. Since ${\sf A}/F={\sf A}/{\bf C}_{{\sf A}}(F)\leq{\sf Aut}(F)$ is abelian, we have $({\sf A}/F)_{\delta}=1$, where $\delta\in V{\it\Gamma}_{F}$, which is a contradiction.

Thus, $F$ has at most two orbits on $V{\it\Gamma}$. If $F$ is transitive on $V{\it\Gamma}$, then $F$ is regular on $V{\it\Gamma}$. It follows that ${\it\Gamma}\cong{\sf Cay}(F,S)$ is a normal arc-transitive Cayley graph of $F$. Then we easily conclude that $S$ consists of the involutions of $F$. Since $F$ has at most one involution, it follows that $F=\langle S\rangle\leq\mathbb{Z}_{2}$, which is a contradiction.

Hence $F$ has two orbits on $V{\it\Gamma}$ and $F\cong\mathbb{Z}_{2n}$ and $K:={\bf O}_{p_{3}}({\sf A})\times{\bf O}_{p_{4}}({\sf A})\times{\bf O}_{p_{s}}({\sf A})\cong\mathbb{Z}_{p_{3}p_{4}\ldots p_{s}}$. Since $K\unlhd{\sf A}$ has $4p_{1}p_{2}$ orbits on $V{\it\Gamma}$, by Theorem 2.6(i), ${\it\Gamma}_{K}$ is an ${\sf A}/K$-arc-transitive pentavalent graph of order $4p_{1}p_{2}$, and hence ${\it\Gamma}_{K}$ satisfies Lemma 2.7. Since ${\sf A}/K$ is soluble, by Remark 2.8, a contradiction occurs. $\Box$

We next consider the case where ${\sf A}$ is insoluble.

Lemma 4.2**.**

Assume that ${\sf A}$ is insoluble and has no nontrivial soluble normal subgroup. Then ${\sf Aut}{\it\Gamma}\cong{\sf J}_{1}$, ${\sf PSL}(2,p)$ or ${\sf PGL}(2,p)$. Further, If ${\sf Aut}{\it\Gamma}\cong{\sf J}_{1}$, then ${\it\Gamma}\cong{\mathcal{C}}_{17556}^{i}$ which satisfy the conditions in Row 2 to Row 5 of Table 1 of Theorem 1.1, where $1\leq i\leq 5$.

*Proof.  *Let $N$ be a minimal normal subgroup of ${\sf A}$. Then $N=S^{d}$, where $S$ is a nonabelian simple group and $d\geq 1$.

If $N$ has more than three orbits on $V{\it\Gamma}$, then by Theorem 2.6, $N$ is semi-regular on $V{\it\Gamma}$ and so $|N|$ divides $4n$. Since $N$ is a direct product of nonabelian groups, it implies that $4$ divides $|N|$. Again by Theorem 2.6, ${\it\Gamma}_{N}$ is a pentavalent symmetric graph of odd order $n$, a contradiction. Hence, $N$ has at most two orbits on $V{\it\Gamma}$, so $2n$ divides $|N|$.

Moreover, since $p_{s}>5$, $p_{s}$ divides $|N|$ and $p_{s}^{2}$ does not divide $|N|$ as |{\sf A}|\,\big{|}\,2^{11}\cdot 3^{2}\cdot 5\cdot p_{1}p_{2}\cdots p_{s}, we conclude that $d=1$ and $N=S$ is a nonabelian simple group. Let $C={\bf C}_{\sf A}(S)$. Then $C\lhd{\sf A}$, $C\cap S=1$ and $\langle C,S\rangle=C\times S$. If $C\not=1$, then $C$ is insoluble as $C\lhd{\sf A}$ and ${\sf A}$ has no soluble normal subgroup. It follows that $4$ divides $|C|$. A similarly argument with the above paragraph, we have $2n$ divides $|C|$. Hence $4n^{2}$ divides $|{\sf A}|=2^{11}\cdot 3^{2}\cdot 5\cdot n$, and so $n$ divides $2^{9}\cdot 3^{2}\cdot 5$. It implies that $n=3\cdot 5$, a contradiction with $n$ having at least three prime factors. So $C=1$ and ${\sf A}={\sf A}/C\leq{\sf Aut}(S)$, that is, ${\sf A}$ is almost simple with socle $S$.

If $S_{\alpha}=1$, then $S$ acts regularly on $V{\it\Gamma}$. Hence $S$ is a non-abelian simple group such that $|S|=4n$. By checking the orders of nonabelian simple groups (see [6, P. 135-136] for example), we have that $S={\sf PSL}(2,p)$ and so ${\sf A}\leq{\sf Aut}(S)={\sf PGL}(2,p)$, which is impossible as ${\sf A}$ is transitive on $A{\it\Gamma}$, $|{\sf A}|\leq 2|S|$ and $|A{\it\Gamma}|=5|S|$. Hence $S_{\alpha}\not=1$. Since ${\it\Gamma}$ is connected and $S\lhd{\sf A}$, we have $1\neq S_{\alpha}^{{\it\Gamma}(\alpha)}\lhd{\sf A}_{\alpha}^{{\it\Gamma}(\alpha)}$, it follows that 5\,\big{|}\,|S_{\alpha}|, we thus have $10\cdot p_{1}p_{2}\cdots p_{s}$ divides $|S|$.

Thus, ${\sf soc}({\sf A})=S$ is a nonabelian simple group such that |S|\,\big{|}\,2^{11}\cdot 3^{2}\cdot 5\cdot n and 10\cdot n\,\big{|}\,|S|. Hence the triple $(S,|S|,n)$ lies in Table 2 of Lemma 2.5. We will analyse all the candidates one by one in the following.

If $\cong{\sf PSL}(2,p)$ with $p\geq 29$ a prime, then ${\sf A}\cong{\sf PSL}(2,p)$ or ${\sf PGL}(2,p)$, the Lemma holds. If $(S,n)=({\sf J}_{1},3{\cdot}7{\cdot}11{\cdot}19)$, then $|V{\it\Gamma}|=17556$ and ${\sf A}\cong{\sf J}_{1}$ as ${\sf Out}({\sf J}_{1})=1$. It then follows from Example 3.1 that ${\it\Gamma}\cong{\mathcal{C}}_{17556}^{i}$ which satisfy the conditions in Row 1 to Row 5 of Table 1 of Theorem 1.1, where $1\leq i\leq 5$.

Assume $(S,n)=({\sf Sz}(32),5{\cdot}31{\cdot}41)$. Since ${\sf Out}({\sf Sz}(32))\cong\mathbb{Z}_{5}$ (see Atlas [3] for example), ${\sf A}\cong{\sf Sz}(32)$ or ${\sf Sz}(32).\mathbb{Z}_{5}$, so $|{\sf A}_{\alpha}|=\frac{|{\sf A}|}{4n}=1280$ or 6400, which is not possible by Lemma 2.4. Similarly, for the case $(S,n)=({\sf PSL}(5,2),3{\cdot}7{\cdot}31)$, then ${\sf A}\cong{\sf PSL}(5,2)$ or ${\sf PSL}(5,2).\mathbb{Z}_{2}$ as ${\sf Out}({\sf PSL}(5,2))\cong\mathbb{Z}_{2}$. Thus, $|{\sf A}_{\alpha}|=\frac{|{\sf A}|}{4n}=3840$ or 7680, which is impossible by Lemma 2.4. For the case where $(S,n)=({\sf PSL}(2,2^{8}),3{\cdot}17{\cdot}257)$, since ${\sf A}\cong{\sf PSL}(2,2^{8}).O$, where $O\leq{\sf Out}({\sf PSL}(2,2^{8}))\cong\mathbb{Z}_{8}$, we have $|{\sf A}_{\alpha}|=\frac{|{\sf A}|}{4n}=2^{k}{\cdot}5$, where $6\leq k\leq 9$, which is also impossible by Lemma 2.4. For the case where $(S,n)=({\sf PSU}(3,4),3{\cdot}5{\cdot}13)$, since ${\sf A}\cong{\sf PSU}(3,4).O$, where $O\leq{\sf Out}({\sf PSU}(3,4))\cong\mathbb{Z}_{4}$, we have $|{\sf A}_{\alpha}|=\frac{|{\sf A}|}{4n}=2^{k}{\cdot}5$, where $4\leq k\leq 6$, which is impossible by Lemma 2.4.

Assume $(S,n)=({\sf PSp}(4,4),3{\cdot}5{\cdot}17)$. Since $S\leq{\sf A}\leq{\sf Aut}(S)\cong{\sf PSp}(4,4).\mathbb{Z}_{4}$, we have $|{\sf A}_{\alpha}|=\frac{|{\sf A}|}{4n}=960$, 1920 or 3840. If $|{\sf A}_{\alpha}|=960$ or $1920$, then by Lemma 2.4, ${\sf A}_{\alpha}\cong{\sf ASL}(2,4)$ or ${\sf A\Sigma L}(2,4)$. However, by Atlas [3], ${\sf PSp}(4,4)$ has no subgroup isomorphic to ${\sf ASL}(2,4)$ and ${\sf PSp}(4,4).\mathbb{Z}_{2}$ has no subgroup isomorphic to ${\sf A\Sigma L}(2,4)$. If $|{\sf A}_{\alpha}|=3840$, then also by Lemma 2.4, a contradiction occurs.

Assume $(S,n)=({\sf PSL}(2,2^{6}),3{\cdot}7{\cdot}13)$. Recall that $S$ has at most two orbits on $V{\it\Gamma}$, $|S_{\alpha}|=\frac{|S|}{4n}=240$ or $\frac{|S|}{2n}=480$. However, by Lemma 2.2, ${\sf PSL}(2,2^{6})$ has no maximal subgroup with order a multiple of 240, a contradiction occurs. Similarly, for the case $(S,n)=({\sf J}_{2},3{\cdot}5{\cdot}7)$. Then $|S_{\alpha}|=\frac{|S|}{4n}=2880$ or $\frac{|S|}{2n}=5760$. By Atlas [3], ${\sf J}_{2}$ has no maximal subgroup with order a multiple of 2880, a contradiction also occurs.

Assume $S\cong{\sf M}_{23}$. Then $n=3{\cdot}7{\cdot}11{\cdot}23$ or $7{\cdot}11{\cdot}23$, and as ${\sf Out}({\sf M}_{23})=1$, we have ${\sf A}=S$ and $|{\sf A}_{\alpha}|=\frac{|{\sf M}_{23}|}{4n}=480$ or $1440$. By Lemma 2.4, it is impossible for the case $|{\sf A}_{\alpha}|=480$. For the latter case, by a direct computation using Magma [1], no graph ${\it\Gamma}$ exists. If $(S,n)=({\sf M}_{22},7{\cdot}11{\cdot}23)$, as ${\sf Out}({\sf M}_{22})\cong\mathbb{Z}_{2}$, we have ${\sf A}\cong{\sf M}_{22}$ or ${\sf M}_{22}.\mathbb{Z}_{2}$, so $|{\sf A}_{\alpha}|=\frac{|{\sf A}|}{4n}=480$ or 960, a computation by Magma [1] shows that no graph ${\it\Gamma}$ exists. Similarly, we can exclude the case where $(S,n)=({\sf PSL}(2,25),3{\cdot}5{\cdot}13)$ by Magma [1].

Finally, assume $(S,n)=({\sf M}_{24},3{\cdot}7{\cdot}11{\cdot}23)$ or $({\sf J}_{1},3{\cdot}7{\cdot}11{\cdot}19)$. Since ${\sf Out}({\sf M}_{24})={\sf Out}({\sf J}_{1})=1$, we always have ${\sf A}=S$. Hence $|{\sf A}_{\alpha}|=\frac{|{\sf A}|}{4n}=11520$ or 10. A computation by Magma [1] also shows that no graph ${\it\Gamma}$ exists. $\Box$

We next assume that ${\sf A}$ has a nontrivial soluble normal subgroup. Let $N$ be a minimal soluble normal subgroup of ${\sf A}$. Then there exists a prime r\,\big{|}\,4n such that $N\cong\mathbb{Z}_{r}^{d}$. Further, $N$ has at least three orbits on $V{\it\Gamma}$. It follows from Theorem 2.6 that $N$ is semi-regular on $V{\it\Gamma}$, and so |N|=|\mathbb{Z}_{r}|^{d}\,\big{|}\,|V{\it\Gamma}|=4n. If $d\geq 2$, then $(r,d)=(2,2)$. It follows that ${\it\Gamma}_{N}$ is an arc-transitive graph of odd order, a contradiction. Hence $d=1$, $N=\mathbb{Z}_{r}$. The next lemma consider the case where $r=2$.

Lemma 4.3**.**

Assume that ${\sf A}$ is insoluble and has a minimal soluble normal subgroup $N=\mathbb{Z}_{2}$. Then one of the following statements holds:

• (1)

${\sf Aut}{\it\Gamma}\cong{\sf PSL}(2,p)\times\mathbb{Z}_{2}$* or ${\sf PGL}(2,p)\times\mathbb{Z}_{2}$, where $p\geq 29$ is a prime.*

• (2)

${\sf Aut}{\it\Gamma}\cong{\sf PSL}(2,25)\times\mathbb{Z}_{2}$* and ${\it\Gamma}$ is isomorphic to ${\mathcal{C}}_{780}^{i}$ in Table 1, where $1\leq i\leq 2$.*

• (3)

${\sf Aut}{\it\Gamma}\cong{\sf J}_{1}\times\mathbb{Z}_{2}$* and ${\it\Gamma}$ is isomorphic to ${\mathcal{C}}_{5852}$ in Table 1.*

*Proof.  *Since $N$ has more than three orbits on $V{\it\Gamma}$, then by Theorem 2.6, ${\it\Gamma}_{N}$ is an ${\sf A}/N$-arc-transitive pentavalent graph of order $\bar{n}=2n$. It follows that ${\it\Gamma}_{N}$ is isomorphic to one of the graphs in Lemma 2.8. Since ${\sf A}/N\leq{\sf Aut}{\it\Gamma}_{N}$ and ${\sf A}/N$ is insoluble, we have that ${\sf Aut}{\it\Gamma}_{N}$ is insoluble and so ${\sf Aut}{\it\Gamma}_{N}\cong{\sf PSL}(2,p)$, ${\sf PGL}(2,p)$, ${\sf PSL}(2,25)$ or ${\sf J}_{1}$. Let $\bar{\sf A}:={\sf Aut}\bar{\it\Gamma}$.

Suppose that $\bar{\sf A}\cong{\sf PSL}(2,p)$ or ${\sf PGL}(2,p)$. Since ${\sf A}/N$ is insoluble, by Lemma 2.2 and Lemma 2.3, ${\sf A}/N$ is isomorphic to ${\sf A}_{5}$, ${\sf PSL}(2,p)$ or ${\sf PGL}(2,p)$. Since ${\sf A}/N$ is transitive on $A{\it\Gamma}_{N}$, we can further conclude that ${\sf A}/N$ is isomorphic to ${\sf PSL}(2,p)$ or ${\sf PGL}(2,p)$. Therefore, ${\sf A}\cong N.{\sf PSL}(2,p)$ or $N.{\sf PGL}(2,p)$, that is, ${\sf A}\cong{\sf PSL}(2,p){\times}\mathbb{Z}_{2}$, ${\sf SL}(2,p)$, ${\sf PGL}(2,p){\times}\mathbb{Z}_{2}$ or ${\sf SL}(2,p).\mathbb{Z}_{2}$. Assume first that ${\sf A}\cong{\sf SL}(2,p)$. Note that ${\sf SL}(2,p)$ has a unique central involution. Then by Lemma 2.4, ${\sf A}_{\alpha}\cong\mathbb{Z}_{5}$. It follows that $|V{\it\Gamma}|=|{\sf A}:{\sf A}_{\alpha}|$ is divisible by 8 as $|{\sf SL}(2,p)|$ is divisible by 8, a contradiction. Assume next that ${\sf A}\cong{\sf SL}(2,p).\mathbb{Z}_{2}$. Then ${\sf A}$ contains a normal subgroup $H$ isomorphic to ${\sf SL}(2,p)$. Since 8\,\big{|}\,|H|, we have $H_{\alpha}\not=1$. By Theorem 2.6, $H$ has at most two orbits on $V{\it\Gamma}$ and so \frac{|{\sf A}_{\alpha}|}{|H_{\alpha}|}\,\big{|}\,2. If $H$ is transitive on $V{\it\Gamma}$, then $H$ is arc-transitive. A similar argument with the case ${\sf A}\cong{\sf SL}(2,p)$, a contradiction occurs. Therefore, $H$ has two orbits on $V{\it\Gamma}$ and so $H_{\alpha}={\sf A}_{\alpha}$. Since $H$ has a unique central involution, by Lemma 2.4, ${\sf A}_{\alpha}\cong\mathbb{Z}_{5}$, it follows that $|V{\it\Gamma}|=|{\sf A}:{\sf A}_{\alpha}|$ is divisible by 16, a contradiction. Therefore, ${\sf A}\cong{\sf PSL}(2,p){\times}\mathbb{Z}_{2}$ or ${\sf PGL}(2,p){\times}\mathbb{Z}_{2}$ in this case.

Suppose that $\bar{\sf A}\cong{\sf PSL}(2,25)$. Since ${\it\Gamma}_{N}$ is ${\sf A}/N$-arc-transitive, we have that 5{\cdot}390\,\big{|}\,|{\sf A}/N|. By checking the maximal subgroup of ${\sf PSL}(2,25)$ (see Atlas [3] for example), we have that ${\sf A}/N=\bar{\sf A}\cong{\sf PSL}(2,25)$. It follows that ${\sf A}\cong{\sf SL}(2,25)$ or ${\sf PSL}(2,25){\times}\mathbb{Z}_{2}$. If ${\sf A}\cong{\sf PSL}(2,25){\times}\mathbb{Z}_{2}$, then by Example 3.1, ${\it\Gamma}\cong{\mathcal{C}}_{780}^{i}$ in Table 1, where $1\leq i\leq 2$. If ${\sf A}\cong{\sf SL}(2,25)$, then by Magma [1], no graph exists.

Suppose that $\bar{\sf A}\cong{\sf J}_{1}$. Similarly, since ${\it\Gamma}_{N}$ is ${\sf A}/N$-arc-transitive, we have that 5{\cdot}2926\,\big{|}\,|{\sf A}/N|. By checking the maximal subgroup of ${\sf J}_{1}$ (see Atlas [3] for example), we have that ${\sf A}/N=\bar{\sf A}\cong{\sf J}_{1}$. Since the Schur multiplier of ${\sf J}_{1}$ is $\mathbb{Z}_{1}$, ${\sf A}\cong N.{\sf J}_{1}\cong{\sf J}_{1}{\times}\mathbb{Z}_{2}$. By Example 3.1, ${\it\Gamma}\cong{\mathcal{C}}_{5852}$ in Table 1. $\Box$

Finally, suppose that $r>2$. We first prove the following lemma.

Lemma 4.4**.**

Assume that $\Sigma$ is isomorphic to one of the graphs listed in Lemma 2.7, in Lemma 4.2 and in Lemma 4.3. If $M$ is an arc-transitive subgroup of ${\sf Aut}\Sigma$, then $M$ contains a subgroup isomorphic to ${\sf PSL}(2,p)$, ${\sf J}_{1}$, ${\sf PSL}(2,25)$ or ${\sf A}_{5}$.

*Proof.  *By checking the graphs in Lemma 2.7, in Lemma 4.2 and in Lemma 4.3, we have that ${\sf Aut}\Sigma$ is isomorphic to one of the groups ${\sf PSL}(2,p)$, ${\sf PGL}(2,p)$, ${\sf PSL}(2,p){\times}\mathbb{Z}_{2}$, ${\sf PGL}(2,p){\times}\mathbb{Z}_{2}$, ${\sf J}_{1}$, ${\sf J}_{1}{\times}\mathbb{Z}_{2}$, ${\sf PSL}(2,25){\times}\mathbb{Z}_{2}$ or ${\sf A}_{5}{\times}{\sf D}_{10}$. If ${\sf Aut}\Sigma\cong{\sf PSL}(2,p)$, then since p\,\big{|}\,n and 20n\,\big{|}\,|M|, by Lemma 2.2, $M\leq\mathbb{Z}_{p}:\mathbb{Z}_{\frac{p-1}{2}}$ or $M={\sf Aut}\Sigma\cong{\sf PSL}(2,p)$. If $M\leq\mathbb{Z}_{p}:\mathbb{Z}_{\frac{p-1}{2}}$, then $M\cong\mathbb{Z}_{p}:\mathbb{Z}_{l}$ for some l\,\big{|}\,\frac{p-1}{2}. Thus, $M$ has a normal subgroup, say $S\cong\mathbb{Z}_{p}$, which has more than three orbits on $V\Sigma$. It then follows from Theorem 2.6 that the normal quotient graph $\Sigma_{S}$ is $M/S$-arc-transitive, a contradiction occurs as $M/S\cong\mathbb{Z}_{l}$ is cyclic. Hence, $M\not\leq\mathbb{Z}_{p}:\mathbb{Z}_{\frac{p-1}{2}}$ and so $M={\sf Aut}\Sigma\cong{\sf PSL}(2,p)$. If ${\sf Aut}\Sigma\cong{\sf PGL}(2,p)$, then since 20n\,\big{|}\,|M|, by Lemma 2.3, ${\sf M}\leq\mathbb{Z}_{p}:\mathbb{Z}_{p-1}$, $M\leq{\sf PSL}(2,p)$ or $M={\sf Aut}\Sigma\cong{\sf PGL}(2,p)$. A similar argument, we can conclude that $M\geq{\sf PSL}(2,p)$. Similarly, we can further show that $M\geq{\sf PSL}(2,p)$ for the case ${\sf Aut}\Sigma\cong{\sf PSL}(2,p){\times}\mathbb{Z}_{2}$ or ${\sf PGL}(2,p){\times}\mathbb{Z}_{2}$.

If ${\sf Aut}\Sigma\cong{\sf J}_{1}$, then since $\Sigma$ is isomorphic to ${\mathcal{C}}_{2926}$, we have that 5{\cdot}2926\,\big{|}\,|M|. By checking the maximal subgroup of ${\sf J}_{1}$ (see Atlas [3]), we have that $M={\sf Aut}\Sigma\cong{\sf J}_{1}$. We can further show that $M\geq{\sf J}_{1}$ for the case ${\sf Aut}\Sigma\cong{\sf J}_{1}{\times}\mathbb{Z}_{2}$, $M\geq{\sf PSL}(2,25)$ for the case ${\sf Aut}\Sigma\cong{\sf PSL}(2,25){\times}\mathbb{Z}_{2}$ and $M\geq{\sf A}_{5}$ for the case ${\sf Aut}\Sigma\cong{\sf A}_{5}{\times}{\sf D}_{10}$. $\Box$

Now assume that ${\sf A}$ has a minimal soluble normal subgroup $N=\mathbb{Z}_{r}$ for $r>2$.

Lemma 4.5**.**

Assume that ${\sf A}$ has a minimal soluble normal subgroup $N=\mathbb{Z}_{r}$ for $r>2$. Then the normal quotient ${\it\Gamma}_{N}$ is not isomorphic to any of the graphs $\Sigma$ listed in Lemma 4.4.

*Proof.  *Suppose to the contrary that ${\it\Gamma}_{N}$ is isomorphic to one of the graphs in Lemma 4.4. Then by Theorem 2.6, ${\sf A}/N\leq{\sf Aut}{\it\Gamma}_{N}$ is transitive on $A{\it\Gamma}_{N}$. Let $\Omega:=\{{\sf PSL}(2,p),{\sf J}_{1},{\sf PSL}(2,25),{\sf A}_{5}\}$. It follows from Lemma 4.4 that there exists a subgroup $M/N$ of ${\sf A}/N$ isomorphic to one of the groups in $\Omega$. Since now $M/N\leq{\sf A}/N\leq{\sf Aut}{\it\Gamma}_{N}$, it follows from the structure of ${\sf Aut}{\it\Gamma}_{N}$ that $M/N\unlhd{\sf A}/N$. Therefore, $M^{\prime}{\sf char}M\unlhd{\sf A}$, it implies that $M^{\prime}\unlhd{\sf A}$. On the other hand, since the order of the Schur multiplier of a group in $\Omega$ is less than or equal to 2 (see [10, Theorem 7.1.1] for ${\sf PSL}(2,p)$ and Atlas [3] for the others) and $r>2$, we have that $M^{\prime}\in\Omega$ and 4\,\big{|}\,|M^{\prime}|. If $M^{\prime}$ has more than three orbits on $V{\it\Gamma}$, then by Theorem 2.6, ${\it\Gamma}_{M^{\prime}}$ is a pentavalent symmetric graph of odd order, a contradiction. Thus, $M^{\prime}$ has at most two orbits on $V{\it\Gamma}$ and so $2n$ divides $|M^{\prime}|$. Let $\bar{\sf A}:={\sf Aut}{\it\Gamma}_{N}$, $\bar{n}:=\frac{n}{r}$ and $\bar{M}:=M/N$. Then $M^{\prime}\cong\bar{M}$.

Assume first that $r>5$. Then since |\bar{M}|\,\big{|}\,|\bar{\sf A}|\,\big{|}\,2^{11}{\cdot}3^{2}{\cdot}5{\cdot}\frac{n}{r} and $n$ is an odd square-free integer, we have that $r$ does not divide $|\bar{M}|=|M^{\prime}|$. It implies that $M^{\prime}$ has at least $r$ orbits on $V{\it\Gamma}$, a contradiction.

Assume next that $r=3$. Since $\bar{M}\cong M^{\prime}$ has at most two orbits on $V{\it\Gamma}_{N}$ (if not $({\it\Gamma}_{N})_{\bar{M}}$ is a pentavalent symmetric graph of odd order, a contradiction), we have that $|\bar{M}:\bar{M}_{\delta}|=2\bar{n}$ or $4\bar{n}$, where $\delta\in V\bar{\it\Gamma}$. Now $2n$ divides $|\bar{M}|$ and $|\bar{M}:\bar{M}_{\delta}|=\frac{2n}{r}$ or $\frac{4n}{r}$. It implies that $r=3$ divides $\bar{M}_{\delta}$. Therefore 3\,\big{|}\,|\bar{\sf A}_{\delta}|. By Lemma 2.4, $\bar{\sf A}_{\delta}$ is nonsolvable, because $|\bar{\sf A}_{\delta}|$ does not divide $80$, forcing that $\bar{M}_{\delta}$ is nonsolvable. If $\bar{M}\cong{\sf PSL}(2,p)$, then by Lemma 2.2, $\bar{M}_{\delta}\cong{\sf A}_{5}$. Hence $M^{\prime}_{\alpha}\leq(M^{\prime}N)_{\alpha}\cong(M^{\prime}N/N)_{\delta}=\bar{M}_{\delta}\cong{\sf A}_{5}$ by Theorem 2.6 (ii). Note that $|M^{\prime}_{\alpha}|=20$, it contradicts that ${\sf A}_{5}$ has no subgroup of order $20$. If $\bar{M}\cong{\sf J}_{1}$, then ${\it\Gamma}_{N}\cong{\mathcal{C}}_{5852}$ or ${\mathcal{C}}_{17556}^{i}$ in Table 1, where $1\leq i\leq 5$. If ${\it\Gamma}_{N}\cong{\mathcal{C}}_{17556}^{i}$, then $\bar{\sf A}_{\delta}\cong{\sf D}_{10}$ is soluble, a contradiction. If ${\it\Gamma}_{N}\cong{\mathcal{C}}_{5852}$, then $\bar{M}_{\delta}=\bar{\sf A}_{\delta}\cong{\sf A}_{5}$. A similar argument with the case $\bar{M}\cong{\sf PSL}(2,p)$ leads to a contradiction. If $\bar{M}\cong{\sf A}_{5}$, then ${\it\Gamma}_{N}\cong{\mathcal{C}}_{60}$ in Table 3 and $\bar{\sf A}_{\delta}\cong{\sf D}_{10}$ is soluble, a contradiction. If $\bar{M}\cong{\sf PSL}(2,25)$, then ${\it\Gamma}_{N}\cong{\mathcal{C}}_{780}^{1}$ or ${\mathcal{C}}_{780}^{2}$ in Table 1 and $\bar{\sf A}_{\delta}\cong{\sf F}_{20}$ is soluble, also a contradiction.

Finally assume that $r=5$. Then $|M^{\prime}:M^{\prime}_{\alpha}|=2n$ or $4n$ as $M^{\prime}$ has at most two orbits on $V{\it\Gamma}$. Since ${\it\Gamma}$ is connected and $1\neq M^{\prime}_{\alpha}\lhd{\sf A}_{\alpha}$, we have $1\neq{M^{\prime}_{\alpha}}^{{\it\Gamma}(\alpha)}\lhd{\sf A}_{\alpha}^{{\it\Gamma}(\alpha)}$, it follows that 5\,\big{|}\,|M^{\prime}_{\alpha}|. On the other hand, since $\bar{M}$ has at most two orbits on $V{\it\Gamma}_{N}$, we have that $|\bar{M}:\bar{M}_{\delta}|=2\bar{n}$ or $4\bar{n}$. Note that $\bar{M}\cong M^{\prime}$ and $r=5$. Hence $\frac{|\bar{M}_{\delta}|}{|M^{\prime}_{\alpha}|}=5$, it follows that 5^{2}\,\big{|}\,|\bar{M}_{\delta}|\,\big{|}\,|\bar{\sf A}_{\delta}|, a contradiction with |\bar{\sf A}_{\delta}|\,\big{|}\,2^{9}\cdot 3^{2}\cdot 5 by Lemma 2.4, a contradiction. $\Box$

The final lemma completes the proof of Theorem 1.1.

Lemma 4.6**.**

Assume ${\sf A}$ is insoluble. Then ${\sf A}$ has no minimal soluble normal subgroup isomorphic to $\mathbb{Z}_{r}$ with $r>2$.

*Proof.  *Suppose that, on the contrary, ${\sf A}$ has a minimal soluble normal subgroup $N=\mathbb{Z}_{r}$ with $r>2$. We prove the lemma by induction on the order of ${\it\Gamma}$.

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 ${\it\Gamma}_{N}$ is a pentavalent symmetric graph of order $4pq$. By Lemma 2.7, ${\it\Gamma}_{N}$ 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 ${\sf Aut}{\it\Gamma}_{N}$ is insoluble. If ${\sf Aut}{\it\Gamma}_{N}$ has no nontrivial soluble normal subgroup, then ${\it\Gamma}_{N}$ is isomorphic to one of the graphs in Lemma 4.2, which contradicts to Lemma 4.5. If ${\sf Aut}{\it\Gamma}_{N}$ has a minimal soluble normal subgroup $\bar{N}$, then we can also conclude that $\bar{N}\cong\mathbb{Z}_{f}$ with $f$ a prime. If $f>2$, then by induction, no such ${\it\Gamma}_{N}$ exists, a contradiction. If $f=2$, then ${\it\Gamma}_{N}$ is isomorphic to one of the graphs in Lemma 4.3, which also contradicts to Lemma 4.5. This completes the proof of the lemma. $\Box$