Symmetric graphs of valency seven and their basic normal quotient graphs

TL;DR

This paper classifies basic normal quotient graphs of all connected 7-valent symmetric graphs with specific order structures, revealing finiteness results and generalizing previous theorems in graph symmetry.

Contribution

It completely determines the basic normal quotient graphs of certain 7-valent symmetric graphs of order 2pq^n, including an infinite family and specific small graphs, extending prior work.

Findings

Identifies an infinite family of dihedrants of order 2p with p ≡ 1 (mod 7)

Classifies 6 specific graphs with order up to 310

Shows finiteness of 2-arc-transitive 7-valent graphs of certain orders

Abstract

A graph $\Gamma$ is basic if Aut$\Gamma$ has no normal subgroup $N\ne1$ such that $\Gamma$ is a normal cover of the normal quotient graph $\Gamma_N$. In this paper, we completely determine the basic normal quotient graphs of all connected 7-valent symmetric graphs of order $2pq^n$ with $p < q$ odd primes, which consist of an infinite family of dihedrants of order $2p$ with $p\equiv1$(mod 7), and 6 specific graphs with order at most 310. As a consequence, it shows that, for any given positive integer n, there are only finitely many connected 2-arc-transitive 7-valent graphs of order $2pq^n$ with $7\ne p<q$ primes, partially generalizing Theorem 1 of Conder, Li and Potocnik [On the orders of arc-transitive graphs, J. Algebra 421 (2015), 167-186].

Tables

Table 1. Table 1. Basic normal quotient graphs of symmetric graphs of order 2 ​ p ​ q n 2 𝑝 superscript 𝑞 𝑛 2pq^{n}

Table 2. Table 2 . Some simple K 4 subscript 𝐾 4 K_{4} -groups

Table 3. Table 3. Some simple K 5 subscript 𝐾 5 K_{5} -groups

Table 4. Table 4 . Certain simple K 4 subscript 𝐾 4 K_{4} -groups

$T$	$\left|T\right|$	$T$	$\left|T\right|$	$T$	$\left|T\right|$
${𝖩}_2$	$2^7\cdot 3^3\cdot 5^2\cdot 7$	${𝖠}_{10}$	$2^7\cdot 3^4\cdot 5^2\cdot 7$	$\mathrm{𝖯𝖲𝖴}(3,5)$	$2^4\cdot 3^2\cdot 5^3\cdot 7$
$\mathrm{𝖯𝖲𝗉}(4,7)$	$2^8\cdot 3^2\cdot 5^2\cdot 7^4$	$\mathrm{𝖯𝖲𝖫}(2,49)$	$2^4\cdot 3\cdot 5^2\cdot 7^2$

Table 5. Table 5 . Certain simple K 5 subscript 𝐾 5 K_{5} -groups

$T$	$\left|T\right|$	$T$	$\left|T\right|$
${𝖠}_{11}$	$2^7\cdot 3^4\cdot 5^2\cdot 7\cdot 11$	${𝖠}_{12}$	$2^9\cdot 3^5\cdot 5^2\cdot 7\cdot 11$
${𝖬}_{22}$	$2^7\cdot 3^2\cdot 5\cdot 7\cdot 11$	$\mathrm{HS}$	$2^9\cdot 3^2\cdot 5^3\cdot 7\cdot 11$
$\mathrm{𝖯𝖲𝖫}(2,2^6)$	$2^6\cdot 3^2\cdot 5\cdot 7\cdot 13$	$\mathrm{𝖯𝖲𝖫}(2,5^3)$	$2^3\cdot 3^2\cdot 5^3\cdot 7\cdot 31$
$\mathrm{𝖯𝖲𝖫}(2,29)$	$2^2\cdot 3\cdot 5\cdot 7\cdot 29$	$\mathrm{𝖯𝖲𝖫}(2,41)$	$2^3\cdot 3\cdot 5\cdot 7\cdot 41$
$\mathrm{𝖯𝖲𝖫}(2,71)$	$2^3\cdot 3^2\cdot 5\cdot 7\cdot 71$	$\mathrm{𝖯𝖲𝖫}(2,251)$	$2^2\cdot 3^2\cdot 5^3\cdot 7\cdot 251$
$\mathrm{𝖯𝖲𝖫}(2,449)$	$2^6\cdot 3^2\cdot 5^2\cdot 7\cdot 449$

Equations

$$x^6+x^5+\cdots +x+1\equiv 0(\mathsf{mod}m).$$

$$\mathrm{CD}(2m,7)=\mathsf{Cay}(G,\{b,ab,a^{k+1}b,\dots ,a^{k^5+k^4+\cdots +1}b\}).$$

$$\begin{array}[]{l|l|l|l}\hline\cr s&1&2&3\\ \hline\cr G_{\alpha}&\mathbb{Z}_{7},~{}{\sf F}_{14},~{}{\sf F}_{21},~{}{\sf F}_{14}\times\mathbb{Z}_{2},~{}{\sf F}_{21}\times\mathbb{Z}_{3}&{\sf F}_{42},~{}{\sf F}_{42}\times\mathbb{Z}_{2},~{}{\sf F}_{42}\times\mathbb{Z}_{3}&{\sf F}_{42}\times\mathbb{Z}_{6}\\ \hline\cr\end{array}$$

$$\begin{array}[]{l|l|l}\hline\cr s&2&3\\ \hline\cr G_{\alpha}&{\sf PSL}(3,2),{\sf ASL}(3,2),&{\sf PSL}(3,2){\times}{\sf S}_{4},{\sf A}_{7}{\times}{\sf A}_{6},{\sf S}_{7}{\times}{\sf S}_{6},({\sf A}_{7}{\times}{\sf A}_{6}){:}\mathbb{Z}_{2},\\ &{\sf ASL}(3,2){\times}\mathbb{Z}_{2},{\sf A}_{7},{\sf S}_{7}&\mathbb{Z}_{2}^{6}{:}({\sf SL}(2,2){\times}{\sf SL}(3,2)),[2^{20}]{:}({\sf SL}(2,2){\times}{\sf SL}(3,2))\\ \hline\cr|G_{\alpha}|&2^{3}{\cdot}3{\cdot}7,2^{6}{\cdot}3{\cdot}7,&2^{6}{\cdot}3^{2}{\cdot}7,~{}2^{6}{\cdot}3^{4}{\cdot}5^{2}{\cdot}7,~{}2^{8}{\cdot}3^{4}{\cdot}5^{2}{\cdot}7,~{}2^{7}{\cdot}3^{4}{\cdot}5^{2}{\cdot}7,\\ &2^{7}{\cdot}3{\cdot}7,2^{3}{\cdot}3^{2}{\cdot}5{\cdot}7,2^{4}{\cdot}3^{2}{\cdot}5{\cdot}7&2^{10}{\cdot}3^{2}{\cdot}7,~{}~{}2^{24}{\cdot}3^{2}{\cdot}7\\ \hline\cr\end{array}$$

$$2^{26}\nmid |T|,3^3\nmid |T|,7^2\nmid |T|,r^2\nmid |T|.$$

$$|T|=|\mathsf{PSL}(2,q)|=\frac{1}{(2,q-1)}q(q-1)(q+1).$$

$$|T|=|\mathsf{PSU}(3,q)|=\frac{1}{(3,q+1)}{q}^3(q-1)(q+1{)}^2(q^2-q+1).$$

$$|T|=|\mathsf{PSL}(3,q)|=\frac{1}{(3,q-1)}{q}^3(q-1{)}^2(q+1)(q^2+q+1).$$

$$|T|=|O_5(q)|=\frac{1}{2}{q}^4(q^4-1)(q^3-1)(q^2-1).$$

$$2^{10}\nmid |T|,3^6\nmid |T|,5^4\nmid |T|,7^3\nmid |T|.$$

$$2^{10}\nmid |T|,3^5\nmid |T|,5^3\nmid |T|,7^2\nmid |T|,r^2\nmid |T|.$$

$$|T|=|\mathsf{PSU}(4,q)|=\frac{1}{(4,q+1)}{q}^6(q^2-1{)}^2(q^2+1)(q^3+1).$$

$$|T|=|\mathsf{PSL}(4,q)|=\frac{1}{(4,q-1)}{q}^6(q^2-1)(q^3-1)(q^4-1),$$

Full text

1991 MR Subject Classification 20B15, 20B30, 05C25.

This work was partially supported by National Natural Science Foundation of China (11461007, 11231008).

keywords. symmetric graph, basic normal quotient graph, normal cover

1. Introduction

In this paper, graphs are undirected and have no loops and multiple edges. For a graph ${\it\Gamma}$, we denote by $V{\it\Gamma}$ and $A{\it\Gamma}$ its vertex set and arc set respectively, and by ${\sf Aut}{\it\Gamma}$ its full automorphism group. The size $|V{\it\Gamma}|$ is called the order of ${\it\Gamma}$. If there is a group $G\leq{\sf Aut}{\it\Gamma}$ acting transitively on $V{\it\Gamma}$ or $A{\it\Gamma}$, then ${\it\Gamma}$ is called $G$-vertex-transitive or $G$-arc-transitive, respectively. An arc-transitive graph is also called a symmetric graph.

In the field of algebraic combinatorics, symmetric graphs with order a fixed number (generally small) times a prime power have received a lot of attention. Chao [2], Cheng and Oxley [3] and Wang and Xu [22] classified symmetric graphs of prime order $p$, of order $2p$ and of order $3p$, respectively. For valency $3,4$ and $5$ case, see [4, 8, 11, 18, 23, 24] and references therein for examples. Recently, a classification of 7-valent symmetric graphs of order $2pq$ with $p,q$ distinct primes was given by Hua, Chen and Xiang [13], and a characterization of 7-valent symmetric graphs of order $4p^{n}$ was obtained in [20].

A typical method for studying symmetric graphs is taking normal quotient graphs, stated as following. Let ${\it\Gamma}$ be a $G$-arc-transitive graph. For an intransitive normal subgroup $N$ of $G$, denote by $V{\it\Gamma}_{N}$ the set of $N$-orbits in $V{\it\Gamma}$. The normal quotient graph ${\it\Gamma}_{N}$ of ${\it\Gamma}$ induced by $N$ is defined with vertex set $V{\it\Gamma}_{N}$ and two vertices $B,C\in V{\it\Gamma}_{N}$ are adjacent if and only if some vertex in $B$ is adjacent in ${\it\Gamma}$ to some vertex in $C$. If ${\it\Gamma}$ and ${\it\Gamma}_{N}$ have the same valency, then ${\it\Gamma}$ is called a normal cover of ${\it\Gamma}_{N}$. In particular, ${\it\Gamma}$ is called basic if ${\sf Aut}{\it\Gamma}$ has no nontrivial normal subgroup $N$ such that ${\it\Gamma}$ and ${\it\Gamma}_{N}$ have the same valency. From the definition, there is a natural ‘two-steps strategy’ for studying symmetric graphs:

Step 1. Determine their normal quotient graphs.

Step 2. Reconstruct the original graphs by determining the normal covers of their normal quotient graphs.

The main purpose of this paper is to determine all the basic normal quotient graphs of the connected $7$-valent symmetric graphs of order $2pq^{n}$ with $p<q$ odd primes and $n\geq 2$.

The notation used in this paper are standard, see [5]. For example, for a positive integer $n$, we use $\mathbb{Z}_{n}$ and ${\sf D}_{2n}$ to denote the cyclic group of order $n$ and the dihedral group of order $2n$, respectively. For two groups $N$ and $H$, denote by $N\times H$ the direct product of $N$ and $H$, by $N.H$ an extension of $N$ by $H$, and if such an extension is split, then we write $N:H$ instead of $N.H$.

For a positive integer $s$, an $s$-arc of ${\it\Gamma}$ is a sequence $v_{0},v_{1},\dots,v_{s}$ of $s+1$ vertices of ${\it\Gamma}$ such that $v_{i-1},v_{i}$ are adjacent for $1\leq i\leq s$ and $v_{i-1}\neq v_{i+1}$ for $1\leq i\leq s-1$. If $G\leq{\sf Aut}{\it\Gamma}$ is transitive on the set of $s$-arcs of ${\it\Gamma}$, then ${\it\Gamma}$ is called $(G,s)$-arc-transitive; if ${\it\Gamma}$ is $(G,s)$-arc-transitive but not $(G,s+1)$-arc-transitive, then ${\it\Gamma}$ is called $(G,s)$-transitive. Particularly, a $({\sf Aut}{\it\Gamma},s)$-transitive graph is simply called $s$-transitive.

Our main result is as follows. For convenience, graphs appearing in Table 1 are introduced in Section 2.

Theorem 1.1**.**

Let ${\it\Gamma}$ be a connected symmetric graph of valency $7$ and order $2pq^{n}$, with $p<q$ odd primes and $n\geq 2$. Then ${\it\Gamma}$ is $s$-transitive with $1\leq s\leq 3$, and is a normal cover of one of the basic graphs ${\it\Sigma}$ listed in Table $1$.

For any given positive integer $k$, a result of Conder, Li and Potočnik [4] asserts that there are only finitely many connected $2$-arc-transitive $7$-valent graphs of order $kp$ or $kp^{2}$ with $p$ a prime. Theorem 1.1, together with [20, Theorem 1.1] (for case $p=2$) have the following corollary.

Corollary 1.2**.**

• (1)

For any given positive integer $n$, there are only finitely many connected $2$-arc-transitive $7$-valent graphs of order $2pq^{n}$ with $7\neq p<q$ primes.

• (2)

For any given positive integer $n$, there is no connected $2$-arc-transitive $7$-valent graphs of order $2pq^{n}$ with $7<p<q$ primes.

2. Preliminaries

2.1. Examples

As usual, for a positive integer $n$, denote by ${\sf K}_{n},{\sf K}_{n,n}$ and ${\sf K}_{n,n}-n{\sf K}_{2}$ the complete graph of order $n$, the complete bipartite graph of order $2n$, and the graph deleted a 1-match from ${\sf K}_{n,n}$, respectively. Also, the Hoffman-Singleton graph of order $50$ and valency 7 is denoted by ${\rm HS}(50)$.

For a group $G$ and a subset $S\subseteq G\setminus\{1\}$, with $S=S^{-1}:=\{g^{-1}\mid g\in S\}$, the Cayley graph of the group $G$ with respect to $S$ is with vertex set $G$ and two vertices $g$ and $h$ are adjacent if and only if $hg^{-1}\in S$. This Cayley graph is denoted by ${\sf Cay}(G,S)$.

Example 2.1**.**

Let $G=\langle a,b\mid a^{m}=b^{2}=1,a^{b}=a^{-1}\rangle\cong{\sf D}_{2m}$ be a dihedral group, with $m$ a positive integer. Let $k$ be a solution of the congruence equation

[TABLE]

Define a Cayley graph

[TABLE]

Then ${\rm CD}(2m,7)$ is a connected arc-transitive graph of valency $7$. In particular, if $m\geq 31$, then ${\rm CD}(2m,7)$ is arc-regular and ${\sf Aut}({\rm CD}(2m,7))={\sf D}_{2m}:\mathbb{Z}_{7}$, see [7, Theorem B; Proposition 4.1].

The following are several specific examples, see [13, Section 3].

Example 2.2**.**

• (i)

There is unique connected symmetric $7$-valent graph of order $30$, denoted by ${\mathcal{C}}_{30}$, and ${\sf Aut}({\mathcal{C}}_{30})={\sf S}_{8}$;

• (ii)

There are exactly two connected symmetric $7$-valent graphs of order $78$, denoted by ${\mathcal{C}}_{78}^{1}$ and ${\mathcal{C}}_{78}^{2}$, and ${\sf Aut}({\mathcal{C}}_{78}^{1})={\sf PSL}(2,23)$ and ${\sf Aut}({\mathcal{C}}_{78}^{2})={\sf PGL}(2,23)$;

• (iii)

There is unique connected symmetric $7$-valent graph of order $310$, denoted by ${\mathcal{C}}_{310}$, and ${\sf Aut}({\mathcal{C}}_{310})={\sf Aut}({\sf PSL}(5,2))$.

Lemma 2.3**.**

Let ${\it\Gamma}$ be a connected $7$-valent symmetric graph. Then the following statements hold, where $p<q$ are primes.

• (1)

([3, Table])* If $|V{\it\Gamma}|=2p$, then ${\it\Gamma}={\sf K}_{7,7}$ or ${\rm CD}(2p,7)$ with $7\mid p-1$.*

• (2)

([13, Section 4])* If $|V{\it\Gamma}|=2pq$, then ${\it\Gamma}={\mathcal{C}}_{30},{\mathcal{C}}_{78}^{1},{\mathcal{C}}_{78}^{2},{\mathcal{C}}_{310}$ or ${\rm CD}(2pq,7)$ with $p=7$ or $7\mid p-1$ and $7\mid q-1$. *

2.2. Background results

For a positive integer $m$ and a group $T$, denote by $\pi(m)$ the number of the primes which divide $m$, and by $\pi(T)$ the set of the primes dividing $|T|$. The group $T$ is called a $K_{n}$-group if $|\pi(T)|=n$. The simple $K_{n}$-groups with $3\leq n\leq 6$ are classified in [12] and [17].

Theorem 2.4**.**

([17, Theorem A])* Let $T$ be a simple $K_{5}$-group. Then one of the following holds:*

• (a)

$T={\sf PSL}(2,q)$* with $\pi(q^{2}-1)=4$;*

• (b)

$T={\sf PSU}(3,q)$* with $\pi((q^{2}-1)(q^{3}+1))=4$;*

• (c)

$T={\sf PSL}(3,q)$* with $\pi((q^{2}-1)(q^{3}-1))=4$;*

• (d)

$T=O_{5}(q)$* with $\pi(q^{4}-1)=4$;*

• (e)

$T={\sf Sz}(2^{2m+1})$* with $\pi((2^{2m+1}-1)(2^{4m+2}+1))=4$;*

• (f)

$T=R(3^{2m+1})$* with $\pi((3^{4m+2}-1))=3$ and $\pi(3^{4m+2}-3^{2m+1}+1)=1$;*

• (g)

$T={\sf A}_{11},{\sf A}_{12},{\sf M}_{22},J_{3},HS,He,McL,{\sf PSL}(4,4),{\sf PSL}(4,5),{\sf PSL}(4,7),{\sf PSL}(5,2),\\ {\sf PSL}(5,3),{\sf PSL}(6,2),O_{7}(3),O_{9}(2),{\sf PSp}(6,3),{\sf PSp}(8,2),{\sf PSU}(4,4),{\sf PSU}(4,5),\\ {\sf PSU}(4,7),{\sf PSU}(4,9),{\sf PSU}(5,3),{\sf PSU}(6,2),O^{+}(8,3),O^{-}(8,2),^{3}D_{4}(3),G_{2}(4),G_{2}(5),\\ G_{2}(7)$* or $G_{2}(9)$.*

The vertex stabilizers of connected 7-valent symmetric graphs were determined independently by [10, Theorem 1.1] and [14, Theorem 3.4], where ${\sf F}_{n}$ with $n$ a positive integer denotes the Frobenius group of order $n$.

Lemma 2.5**.**

Let ${\it\Gamma}$ be a connected $7$-valent $(G,s)$-transitive graph, where $G\leq{\sf Aut}{\it\Gamma}$ and $s\geq 1$. Then $s\leq 3$ and one of the following holds, where $\alpha\in V{\it\Gamma}$.

• (a)

If $G_{\alpha}$ is soluble, then $|G_{\alpha}|\mid 2^{2}\cdot 3^{2}\cdot 7$. Further, the couple $(s,G_{\alpha})$ is listed in the following table.

[TABLE]

• (b)

If $G_{\alpha}$ is insoluble, then $|G_{\alpha}|\mid 2^{24}\cdot 3^{4}\cdot 5^{2}\cdot 7$. Further, the couple $(s,G_{\alpha})$ is listed in the following table.

[TABLE]

In particular, if $5\mid|G_{\alpha}|$, then $|G_{\alpha}|\mid 2^{8}\cdot 3^{4}\cdot 5^{2}\cdot 7$, and $G_{\alpha}^{{\it\Gamma}(\alpha)}\cong{\sf A}_{7}$ or ${\sf S}_{7}$; if $5{\not|}~{}|G_{\alpha}|$, then $|G_{\alpha}|\mid 2^{24}\cdot 3^{2}\cdot 7$.

The following theorem is a special case of [15, Lemma 2.5] which slightly improves a nice result of Praeger [21, Theorem 4.1].

Theorem 2.6**.**

Let ${\it\Gamma}$ be a connected $G$-arc-transitive graph of odd prime valency, and let $N\lhd G$ have more than two orbits on $V{\it\Gamma}$, where $G\leq{\sf Aut}{\it\Gamma}$. Then the following statements hold.

• (1)

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

• (2)

${\it\Gamma}$* is $(G,s)$-arc-transitive if and only if ${\it\Gamma}_{N}$ is $(G/N,s)$-arc-transitive, where $1\leq s\leq 5$ or $s=7$;*

• (3)

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

A transitive permutation group $X\leq{\sf Sym}({\it\Omega})$ is called quasiprimitive if each minimal normal subgroup of $X$ is transitive on ${\it\Omega}$, while $X$ is called biquasiprimitive if each of its minimal normal subgroups has at most two orbits and there exists one which has exactly two orbits on ${\it\Omega}$.

We have a next generalization of [8, Lemma 5.1].

Lemma 2.7**.**

Let ${\it\Gamma}$ be a connected $G$-arc-transitive $r$-valent graph of order $2q^{n}$, where $G\leq{\sf Aut}{\it\Gamma}$, $n\geq 2$, and $r\geq 5$ and $q\geq 5$ are primes. Then either ${\it\Gamma}={\sf K}_{7,7}$ or ${\rm HS}(50)$, or $G$ has a minimal normal elementary abelian $q$-subgroup.

*Proof.  *If $G$ is quasiprimitive or biquasiprimitive on $V{\it\Gamma}$, by [19, Theorem 1.2], Lemma 2.7 is true. Suppose that $G$ is neither quasiprimitive nor biquasiprimitive on $V{\it\Gamma}$. Then $G$ has a minimal normal subgroup $N$ which has at least three orbits on $V{\it\Gamma}$, by Theorem 2.6, $N$ is semiregular on $V{\it\Gamma}$ and hence $|N|\mid 2q^{n}$. It follows that either $N=\mathbb{Z}_{2}$ or $N=\mathbb{Z}_{q}^{d}$ for some $d<n$. For the former case, the normal quotient graph ${\it\Gamma}_{N}$ is arc-transitive of odd order $q^{n}$ and odd valency $r$, a contradiction. Therefore $N=\mathbb{Z}_{q}^{d}$, as required. $\Box$

3. Technical lemmas

The two lemmas regarding simple groups in this section are based on the classifications of simple $K_{n}$-groups with $3\leq n\leq 6$, obtained in [12] and [17].

Lemma 3.1**.**

Let $r<s$ be odd primes, and let $T$ be a nonabelian simple group such that $|T|\mid 2^{25}\cdot 3^{2}\cdot 7rs^{l}$ and $7rs^{l}\mid|T|$ for $l\geq 1$. Then one of the following holds.

• (i)

$|\pi(T)|=4$, and $T$ is isomorphic to one of the groups listed in Table 2.

• (ii)

$|\pi(T)|=5$, and $T$ is isomorphic to one of the groups listed in Table 3. In particular, $l=1$.

*Proof.  *Clearly, $3\leq|\pi(T)|\leq 5$. If $|\pi(T)|=3$, by [12, Theorem I], there are exactly 8 specific simple $K_{3}$-groups listed in [12, Table 1], checking the orders, no group $T$ exists in this case.

(i). Suppose $|\pi(T)|=4$. By [12, Theorem I], either

• (a)

$T$ is isomorphic to one of the groups listed in [12, Table 2]; or

• (b)

$T={\sf PSL}(2,q)$ for some prime power $q$.

Assume (a) occurs. Suppose $5\in\pi(T)$. As $7\in\pi(T)$, by checking [12, Table 2], $T$ is a $\{2,3,5,7\}$-group, and so $r,s\in\{3,5,7\}$. If $s=7$, then $r=3$ or 5, and one easily checks that no group $T$ exists in the case. If $s=5$, then $r=3$, so $|T|\mid 2^{25}\cdot 3^{3}\cdot 5^{l}\cdot 7$ and $7\cdot 3\cdot 5^{l}\mid|T|$, and one may derive that $T={\sf J}_{2},{\sf A}_{7},{\sf A}_{8},{\sf PSL}(3,4)$ and ${\sf PSU}(3,5)$. Suppose now $5\notin\pi(T)$. By [12, Table 2], $r=3$ or 7 and $s>7$, hence $|T|\mid 2^{25}\cdot 3^{3}\cdot 7^{2}\cdot s^{l}$, by checking the orders, we obtain $T={\sf PSL}(3,8)$.

Now assume (b) occurs. If $q$ is a power of $2,3$ or $7$, by [12, TABLE 3], the only example is $T={\sf PSL}(2,27)$. For the other cases, by [12, Theorem 3.2], $q\geq 11$ is a prime, notice that $|{\sf PSL}(2,q)|$ is always divisible by $3$, we conclude that $T$ is a $\{2,3,7,q\}$-group, hence $s=q$, $l=1$, and $r\in\{3,7\}$. Further, since $|T|\mid 2^{25}\cdot 3^{2}\cdot 7rs^{l}$, we have ${q-1\over 2}\cdot{q+1\over 2}\mid 2^{24}\cdot 3^{3}\cdot 7^{2}$, and as $({q-1\over 2},{q+1\over 2})=1$, it follows that either $q+1\mid 2\cdot 3^{3}\cdot 7^{2}$ if $2\mid{q-1\over 2}$, or $q-1\mid 2\cdot 3^{3}\cdot 7^{2}$ if $2\mid{q+1\over 2}$. Then a computation by Magma [1] shows $q\in\{13,17,19,41,43,53,97,127,293,379,881,883\}$. Checking the orders, we obtain $T={\sf PSL}(2,13),{\sf PSL}(2,97)$ or ${\sf PSL}(2,127)$.

(ii). Suppose $|\pi(T)|=5$. Then $T$ is a $\{2,3,7,r,s\}$-group and satisfies part (a)-(g) of Theorem 2.4. We analyse these cases one by one in the following. Notice that $|T|\mid 2^{25}\cdot 3^{2}\cdot 7rs^{l}$, we obtain

[TABLE]

Assume part (a) of Theorem 2.4 occurs. Then

[TABLE]

Since $|T|\mid 2^{25}\cdot 3^{2}\cdot 7rs^{l}$ and $7rs^{l}\mid|T|$, we have $q\neq 3,3^{2},7$ or $r$, and then derive from Eq.(1) that $q=2^{i}$ with $1\leq i\leq 25$ or $q=s^{l}$. For the former case, since $\pi(q^{2}-1)=4$, we get $q=2^{6},2^{8},2^{9},2^{11}$ or $2^{23}$, and by checking the orders, we obtain $T={\sf PSL}(2,2^{6})$. For the latter case, we have ${q+1\over 2}\cdot{q-1\over 2}\mid 2^{24}\cdot 3^{2}\cdot 7r$, and as $({q+1\over 2},{q-1\over 2})=1$, it follows that either $q-1\mid 2^{25}\cdot 3^{2}\cdot 7$ if $r\mid{q+1\over 2}$, or $q+1\mid 2^{25}\cdot 3^{2}\cdot 7$ if $r\mid{q-1\over 2}$. Recall that $\pi(q^{2}-1)=4$, $q=s^{l}$ with $s>7$ and $T$ satisfies Eq.(1), a direct computation by Magma[1] shows that $q=29,41,43,71,83,113,167,223,503,673,2017,3583,64513,2752513$ or $16515073$, as in Table 3.

Assume part (b) occurs. Then

[TABLE]

By Eq.(1), $q$ is a 2-power or a $s$-power. If $q$ is a 2-power, then $q=2^{i}$ with $1\leq i\leq 8$, and as $\pi((q^{2}-1)(q^{3}+1))=4$, we get $q=2^{4},2^{5}$ or $2^{7}$; however, in these three cases, $|T|$ is always not divisible by $7$, a contradiction. If $q$ is an $s$-power, as $7^{2}{\not|}~{}|T|$ and $r^{2}{\not|}~{}|T|$, we obtain $(q+1)^{2}\mid 2^{25}\cdot 3^{2}$, or equivalently $q+1\mid 2^{12}\cdot 3$. Since $\pi((q^{2}-1)(q^{3}+1))=4$, computation in Magma [1] shows $q=11$ and 23; however, in both cases, $|T|$ is not divisible by $7$, also a contradiction.

Assume part (c) occurs. Then

[TABLE]

By Eq.(1), one derives $q$ is a 2-power or a $s$-power. Then with similar discussion as in part (b) above, one may draw a contradiction.

Assume part (d) occurs. Then

[TABLE]

By Eq.(1), $q$ is a 2-power or a $s$-power. If $q$ is a 2-power, then $q=2^{i}$ with $1\leq i\leq 6$. Since $\pi(q^{4}-1)=4$, we have $q=2^{3}$ or $2^{4}$, and $|T|=2^{12}\cdot 3^{4}\cdot 5\cdot 7^{2}\cdot 13$ or $2^{16}\cdot 3^{2}\cdot 5^{2}\cdot 17^{2}\cdot 257$ respectively, contradicting Eq.(1). If $q$ is a $s$-power, as $7^{2}{\not|}~{}|T|$ and $r^{2}{\not|}~{}|T|$, we have $(q^{2}-1)^{2}\mid 2^{25}\cdot 3^{2}$, and so ${q+1\over 2}\cdot{q-1\over 2}\mid 2^{11}\cdot 3$. Noting that $({q+1\over 2},{q-1\over 2})=1$, we conclude that either ${q+1\over 2}\mid 3$ or ${q-1\over 2}\mid 3$, implying $s\leq q\leq 7$, a contradiction.

Assume part (e) occurs. Then $|T|=|{\sf Sz}(2^{2m+1})|=2^{4m+2}(2^{4m+2}+1)(2^{2m+1}-1)$. Since $2^{26}{\not|}~{}|T|$ and $\pi((2^{2m+1}-1)(2^{4m+2}+1))=4$, we derive $m=3$; however, $|{\sf Sz}(2^{7})|=2^{14}\cdot 5\cdot 29\cdot 113\cdot 127$ is not divisible by 7, a contradiction.

Assume part (f) occurs. Then $|T|=|R(3^{2m+1})|=3^{6m+3}(3^{6m+3}+1)(3^{2m+1}-1)$, so $3^{9}\mid|T|$, contradicting $3^{3}{\not|}~{}|T|$.

Finally, assume part (g) occurs. Checking the orders of the 30 specific simple groups there, we obtain $T={\sf M}_{22}$ and ${\sf PSL}(5,2)$. $\Box$

Lemma 3.2**.**

Let $r<s$ be odd primes, and let $T$ be a nonabelian simple group such that $|T|\mid 2^{9}\cdot 3^{4}\cdot 5^{2}\cdot 7rs^{l}$ and $35rs^{l}\mid|T|$ with $l\geq 1$. Then one of the following holds.

• (i)

$|\pi(T)|=4$, and $T$ is isomorphic to one of the groups listed in Table 4.

• (ii)

$|\pi(T)|=5$, $l=1$ and $T$ is isomorphic to one of the groups listed in Table 5.

• (iii)

$|\pi(T)|=6$, $l=1$ and $T={\sf J}_{1},{\sf M}_{23}$, or ${\sf PSL}(2,q)$ with $q=139,181,211,239,281,\\ 349,379,421,601,631,701,769,811,839,1009,1049,1051,1399,1511,1889,2099,\\ 2239,2267,2269,2591,2689,2801,3779,4481,6481,6719,7559,10079,12601,15121,\\ 21601,26881,28351,30241,37799,53759,56701,69119,96769,172801,201599,\\ 453599,483839$ or $907199$.

*Proof.  *Obviously, $3\leq|\pi(T)|\leq 6$. If $|\pi(T)|=3$, by [12, Theorem I], $T$ is isomorphic to one of the 8 groups listed in [12, Table 1]. However, the order of each group there is not divisible by $35$, a contradiction.

(i). Assume $|\pi(T)|=4$. By [12, Theorem I], $T$ is isomorphic to one of the groups listed in [12, Table 3] or $T={\sf PSL}(2,q)$ for some prime power $q$. For the former case, since $35rs^{l}\mid|T|$ and $|T|\mid 2^{9}\cdot 3^{4}\cdot 5^{2}\cdot 7rs^{l}$, one easily derives that $T={\sf J}_{2},{\sf A}_{10},{\sf PSU}(3,5)$ or ${\sf PSp}(4,7)$. For the later case, notice that $3\mid|{\sf PSL}(2,q)|$ and $35\mid|T|$, $T$ is a $\{2,3,5,7\}$-group, then by [12, TABLE 3], the only example is $T={\sf PSL}(2,49)$.

(ii). Assume $|\pi(T)|=5$. Then $s>7$ and $T$ satisfies parts (a)-(g) of Theorem 2.4. Since $|T|\mid 2^{9}\cdot 3^{4}\cdot 5^{2}\cdot 7rs^{l}$, we have

[TABLE]

Suppose $T={\sf PSL}(2,q)$, as in part (a) of Theorem 2.4. Then $T$ is a $\{2,3,5,7,s\}$-group as $3\mid|{\sf PSL}(2,q)|$. If $q$ is a 2-power, then $q=2^{6},2^{8}$ or $2^{9}$ since $2^{10}{\not|}~{}|T|$ and $\pi(q^{2}-1)=4$, by checking the orders, we obtain $T={\sf PSL}(2,2^{6})$. If $q$ is a 3-power, then $q\in\{3,3^{2},3^{3},3^{4},3^{5}\}$ since $3^{6}{\not|}~{}|T|$, it follows that $\pi(q^{2}-1)\neq 4$, a contradiction. If $q$ is a 5-power, then $q=5^{3}$ because $5^{4}{\not|}~{}|T|$ and $\pi(q^{2}-1)=4$, which gives rise to an example $T={\sf PSL}(2,5^{3})$. If $q$ is a 7-power, then $q=7$ or $7^{2}$ since $7^{3}{\not|}~{}|T|$, contradicting $\pi(q^{2}-1)=4$. Now, assume that $q$ is a $s$-power. Then ${q+1\over 2}\cdot{q-1\over 2}\mid 2^{9}\cdot 3^{5}\cdot 5^{3}\cdot 7^{2}$. Since $({q+1\over 2},{q-1\over 2})=1$, we have ${q-1\over 2}\mid 3^{5}\cdot 5^{3}\cdot 7^{2}$ or ${q+1\over 2}\mid 3^{5}\cdot 5^{3}\cdot 7^{2}$. Recall that $\pi(q^{2}-1)=4$, computation in Magma[1] shows $q\in\{29,41,43,71,89,149,151,251,269,271,293,449,751,809,2251,2647,4051,7937,12149,\\ 20249,23813\}$. Checking the orders, we obtain the examples $T={\sf PSL}(2,29),{\sf PSL}(2,41)$ and ${\sf PSL}(2,449)$.

Suppose $T={\sf PSU}(3,q)$, as in part (b). Since $\pi((q^{2}-1)(q^{3}+1))=4$, by Eq.(3) and (6), we derive that $q$ is a $s$-power, and $(q+1)^{2}\mid 2^{10}\cdot 3^{5}\cdot 5^{3}\cdot 7^{2}$, so $q+1\mid 2^{5}\cdot 3^{2}\cdot 5\cdot 7$. Since $\pi((q^{2}-1)(q^{3}+1))=4$, a computation by Magma [1] shows that $q\in\{11,13,17,19,23\}$; however, by checking the orders, no group $T$ exists in the case. Similarly, one may exclude part (c), namely $T={\sf PSL}(3,q)$.

Suppose $T={\sf Sz}(2^{2m+1})$ or $R(3^{2m+1})$, as in part (d) or (e). Then $|T|=2^{4m+2}(2^{4m+2}\\ +1)(2^{2m+1}-1)$ or $3^{6m+3}(3^{6m+3}+1)(3^{2m+1}-1)$, respectively. Since $|\pi({\sf Sz}(2^{3}))|=4$, $T\neq{\sf Sz}(2^{3})$, and hence $2^{10}\mid|T|$ or $3^{9}\mid|T|$, contradicting Eq.(6).

Suppose $T=O_{5}(q)$, as in part (f). Since $\pi(q^{4}-1)=4$, by Eq.(5) and (6), we conclude that $q$ is a $s$-power, and $(q-1)^{3}\mid 2^{10}\cdot 3^{5}\cdot 5^{3}$, hence $q-1\mid 2^{3}\cdot 3\cdot 5$. It follows $q=11$ and $|T|=|O_{5}(11)|=2^{8}\cdot 3^{2}\cdot 5^{2}\cdot 11^{4}\cdot 61$ is not divisible by $7$, a contradiction.

Finally, suppose $T$ lies in the groups listed in part (g). Checking the orders, we obtain $T={\sf A}_{11},{\sf A}_{12},{\sf M}_{22},$ or ${\rm HS}$.

(iii). Assume that $|\pi(T)|=6$. Then $7<r<s$ and $s>11$. By [17, Theorem B], one of the following holds:

• (a)

$T={\sf PSL}(2,q)$ where $\pi(q^{2}-1)=5$;

• (b)

$T={\sf PSL}(3,q)$ where $\pi((q^{2}-1)(q^{3}-1))=5$;

• (c)

$T={\sf PSL}(4,q)$ where $\pi((q^{2}-1)(q^{3}-1)(q^{4}-1))=5$;

• (d)

$T={\sf PSU}(3,q)$ where $\pi((q^{2}-1)(q^{3}+1))=5$;

• (e)

$T={\sf PSU}(4,q)$ where $\pi((q^{2}-1)(q^{3}+1)(q^{4}-1))=5$;

• (f)

$T=O_{5}(q)$ where $\pi(q^{4}-1)=5$;

• (g)

$T=G_{2}(q)$ where $\pi(q^{6}-1)=5$;

• (h)

$T={\sf Sz}(2^{2m+1})$ where $\pi((2^{2m+1}-1)(2^{4m+2}+1))=5$;

• (i)

$T=R(3^{2m+1})$ where $\pi((3^{2m+1}-1)(3^{6m+3}+1))=5$;

• (j)

$T$ is one of the 38 groups listed in [17, Theorem B].

Recall that $|T|\mid 2^{9}\cdot 3^{4}\cdot 5^{2}\cdot 7rs^{l}$, then we have

[TABLE]

Since $|{\sf Sz}(2^{2m+1}|=2^{4m+2}(2^{4m+2}+1)(2^{2m+1}-1)$ and $2^{10}{\not|}~{}|T|$, we have $m=1$ and $|\pi(T)|=|\pi({\sf Sz}(8))|=4\neq 6$, this contradiction excludes case (h). Since $|R(3^{2m+1})|=3^{6m+3}(3^{6m+3}+1)(3^{2m+1}-1)$, $3^{9}\mid|T|$, contradicting Eq.(7), this excludes case (i). For case (j), by checking the orders, we have $T={\sf J}_{1}$ or ${\sf M}_{23}$.

Suppose case (a) occurs. Since $\pi(q^{2}-1)=5$, $q\neq 3,3^{2},3^{3},3^{4},5,5^{2},7$, and by Eq.(7), we have that either $q=2^{i}~{}(1\leq i\leq 9)$ or $s^{l}$. The former case does not give examples by checking the orders. For the latter case, by Eq.(2), we have ${q-1\over 2}\cdot{q+1\over 2}\mid 2^{8}\cdot 3^{4}\cdot 5^{2}\cdot 7r$, and as $({q-1\over 2},{q+1\over 2})=1$, it follows that either ${q+1\over 2}\mid 2^{9}\cdot 3^{4}\cdot 5^{2}\cdot 7$ or ${q-1\over 2}\mid 2^{9}\cdot 3^{4}\cdot 5^{2}\cdot 7$. Recall that $\pi(q^{2}-1)=5$ and $s>11$, computation in Magma [1] shows that $q$ lies in part (iii) of Lemma 3.2.

Suppose case (b) occurs. Since $\pi((q^{2}-1)(q^{3}-1))=5$, $q\neq 2,2^{2},2^{3}$ or $3$, and by Eq.(4) and (7), we derive that $q$ is a $s$-power, and $(q-1)^{2}\mid 2^{9}\cdot 3^{4}\cdot 5^{2}$, implying $q-1\mid 2^{4}\cdot 3^{2}\cdot 5$. Since $\pi((q^{2}-1)(q^{3}-1))=5$, a computation by Magma [1] shows that $q=37,41$ or 241, which does not give rise to examples by checking the orders. Similarly, one may exclude case (d).

Suppose case (e) occurs. Then

[TABLE]

By Eq.(7), one may conclude that $q$ is a $s$-power, and $(q^{2}-1)^{2}\mid 2^{9}\cdot 3^{4}\cdot 5^{2}$, hence ${q-1\over 2}\cdot{q+1\over 2}\mid 2^{2}\cdot 3^{2}\cdot 5$. As $({q-1\over 2},{q+1\over 2})=1$, we obtain ${q-1\over 2}\mid 2^{2}\cdot 5$ or ${q+1\over 2}\mid 2^{2}\cdot 5$. It follows that $q=19$ since $\pi((q^{2}-1)(q^{3}+1)(q^{4}-1))=5$, and $|T|=|{\sf PSU}(4,19)|=2^{7}\cdot 3^{4}\cdot 5^{3}\cdot 7^{3}\cdot 19^{6}\cdot 181$, contradiction Eq.(7). For case (c), then

[TABLE]

a similar argument as in case (e) may draw a contradiction.

Suppose case (f) occurs. Since $\pi(q^{4}-1)=5$, $q\neq 2,2^{2}$ and $3$, then by Eq.(5) and (7), we conclude that $q$ is a $s$-power, and $(q^{2}+1)(q^{3}-1)(q^{2}-1)^{2}\mid 2^{10}\cdot 3^{4}\cdot 5^{2}\cdot 7r$, hence $(q^{2}-1)^{2}\mid 2^{10}\cdot 3^{4}\cdot 5^{2}$, or equivalently $q^{2}-1\mid 2^{5}\cdot 3^{2}\cdot 5$. As discussed in the above paragraph, one may derive that $q=17,19,29,31$ or $89$, which does not give rise to example by checking the orders. Finally, for case (g), $|T|=|G_{2}(q)|=q^{6}(q^{6}-1)(q^{2}-1)$, a similar discussion may draw a contradiction. $\Box$

4. Vertex quasiprimitive and vertex biquasiprimitive cases

Let ${\it\Gamma}$ be a connected $G$-arc-transitive 7-valent graph of order $2pq^{n}$, where $G\leq{\sf Aut}{\it\Gamma}$, $p<q$ are odd primes and $n\geq 2$. Let $N$ be a minimal normal subgroup of $G$. Then $N=T^{d}$, with $T$ a simple group and $d\geq 1$. Let $\alpha\in V{\it\Gamma}$.

Lemma 4.1**.**

If $N$ is nonabelian, then $d=1$.

*Proof.  *Suppose for a contradiction that $N$ is nonabelian and $d\geq 2$. Then $|N|{\not|}~{}2pq^{n}$, $N_{\alpha}\neq 1$, and $N$ has at most two orbits on $V{\it\Gamma}$ by Theorem 2.6. Set $N=T_{1}\times T_{2}\times\cdots\times T_{d}$ with each $T_{i}\cong T$.

Assume first $N$ is transitive on $V{\it\Gamma}$. Since $1\neq N_{\alpha}\lhd G_{\alpha}$ and ${\it\Gamma}$ is connected, we have $1\neq N_{\alpha}^{{\it\Gamma}(\alpha)}\lhd G_{\alpha}^{{\it\Gamma}(\alpha)}$. It follows that $N_{\alpha}^{{\it\Gamma}(\alpha)}$ is transitive, and ${\it\Gamma}$ is $N$-arc-transitive. If $T_{1}$ is transitive on $V{\it\Gamma}$, then the centralizer $C_{N}(T_{1})$ is semiregular on $V{\it\Gamma}$ (see [6, Theorem 4.2A]), so is $T_{2}$, which is a contradiction as $|T_{2}|{\not|}$ does not divide $|V{\it\Gamma}|=2pq^{n}$; if $T_{1}$ has at least three orbits on $V{\it\Gamma}$, by Theorem 2.6, $T_{1}$ is semiregular, again a contradiction. Therefore, $T_{1}$ has exactly 2 orbits, say $U$ and $W$, on $V{\it\Gamma}$. Since $T_{1}\lhd N$, $U$ and $W$ form a $N$-block system on $V{\it\Gamma}$. It follows that the set stabilizer $N_{U}$ is of index 2 in $N$, which is a contradiction because $N=T^{d}$ has no subgroup with index 2.

Assume now $N$ has exactly two orbits, say ${\it\Delta}_{1}$ and ${\it\Delta}_{2}$, on $V{\it\Gamma}$. Then ${\it\Gamma}$ is a bipartite graph with bipartitions ${\it\Delta}_{1}$ and ${\it\Delta}_{2}$. Let $G^{+}=G_{{\it\Delta}_{1}}=G_{{\it\Delta}_{2}}$, the stabilizer on the bipartitions. If $G^{+}$ acts unfaithfully on ${\it\Delta}_{1}$, by [9, Lemma 5.2], ${\it\Gamma}$ is a complete bipartite graph, so ${\it\Gamma}={\sf K}_{7,7}$ as ${\sf val}({\it\Gamma})=7$ and hence $|V{\it\Gamma}|=14$, a contradiction. Suppose $G^{+}$ acts faithfully on ${\it\Delta}_{1}$. Then $N\leq G^{+}$ can be viewed as a transitive permutation group on ${\it\Delta}_{1}$. If $T_{1}$ is transitive on ${\it\Delta}_{1}$, then [6, Theorem 4.2A] implies $T_{2}$ is semiregular on ${\it\Delta}_{1}$, hence $|T_{2}|\mid pq^{n}$, a contradiction. Thus $T_{1}$ has at least two orbits on ${\it\Delta}_{1}$. It then follows from [18, Lemma 3.2] that $T_{1}$ is semiregular on ${\it\Delta}_{1}$, also a contradiction. $\Box$

The next two lemmas exclude the vertex quasiprimitive and vertex biquasiprimitive cases.

Lemma 4.2**.**

If $G$ is quasiprimitive on $V{\it\Gamma}$, then no graph ${\it\Gamma}$ exists.

*Proof.  *Since $G$ is quasiprimitive on $V{\it\Gamma}$, $N$ is transitive on $V{\it\Gamma}$. If $N$ is abelian, then $N$ is regular on $V{\it\Gamma}$ and so $|T|^{d}=|N|=2pq^{n}$, a contradiction. Thus $N$ is nonabelian, and then by Lemma 4.1, we have $d=1$ and $N=T$. Further, since $T_{\alpha}\neq 1$, we conclude that ${\it\Gamma}$ is $T$-arc-transitive, and hence $T_{\alpha}$ satisfies Lemma 2.5. We divided our proof into two cases depending on whether 5 divides $|T_{\alpha}|$ or not.

Case 1. Assume $5{\not|}~{}|T_{\alpha}|$.

By Lemma 2.5, $|T_{\alpha}|\mid 2^{24}\cdot 3^{2}\cdot 7$, and by the transitivity of $T$, we have $|T|=|V{\it\Gamma}||T_{\alpha}|$ divides $2^{25}\cdot 3^{2}\cdot 7pq^{n}$; on the other hand, since ${\it\Gamma}$ is $T$-arc-transitive, we have $7\mid|T_{\alpha}|$, and so $7pq^{n}\mid|T|$. Therefore, $T$ satisfies Lemma 3.1; in particular, $|\pi(T)|=4$ or 5. If $|\pi(T)|=5$, by Lemma 3.1(ii), $n=1$, a contradiction.

Suppose $|\pi(T)|=4$. Noting that $n\geq 2$, by Lemma 3.1(i), we easily conclude that the couple $(T,p,q^{n})=({\sf J}_{2},3,5^{2})$ or $({\sf PSU}(3,5),3,5^{3})$. For the former case, $|V{\it\Gamma}|=2\cdot 3\cdot 5^{2}$, so $|T_{\alpha}|={|T|\over|V{\it\Gamma}|}=2^{6}\cdot 3^{2}\cdot 7$; however, by Atlas [5], ${\sf J}_{2}$ has no subgroup with order $2^{6}\cdot 3^{2}\cdot 7$, a contradiction. For the latter case, $|V{\it\Gamma}|=2\cdot 3\cdot 5^{3}$, hence $|T_{\alpha}|={|T|\over|V{\it\Gamma}|}=2^{3}\cdot 3\cdot 7$, then by Lemma 2.5, we obtain $T_{\alpha}={\sf PSL}(3,2)$; however, computation in Magma [1] shows that no graph ${\it\Gamma}$ exists in this case.

Case 2. Assume $5\mid|T_{\alpha}|$.

By Lemma 2.5, $|T_{\alpha}|\mid 2^{8}\cdot 3^{4}\cdot 5^{2}\cdot 7$, and $T_{\alpha}^{{\it\Gamma}(\alpha)}\cong{\sf A}_{7}$ or ${\sf S}_{7}$. It follows that $|T|=|V{\it\Gamma}||T_{\alpha}|$ divides $2^{9}\cdot 3^{4}\cdot 5^{2}\cdot 7pq^{n}$; moreover, as $35\mid|T_{\alpha}|$, we have $35pq^{n}$ divides $|T|$. Therefore, $T$ satisfies Lemma 3.2; in particular $4\leq\pi(T)\leq 6$. If $|\pi(T)|=5$ or 6, by Lemma 3.2, we have $n=1$, a contradiction.

Suppose $|\pi(T)|=4$. Since $n\geq 2$, by Lemma 3.2(i), the only possibility is $T={\sf PSp}(4,7)$ and $(p,q)=(3,7^{3})$ or $(5,7^{3})$. Consequently, $|V{\it\Gamma}|=2\cdot 3\cdot 7^{3}$ or $2\cdot 5\cdot 7^{3}$, and $|T_{\alpha}|={|T|\over|V{\it\Gamma}|}=2^{7}\cdot 3\cdot 5^{2}\cdot 7$ or $2^{7}\cdot 3^{2}\cdot 5\cdot 7$, respectively. By Lemma 2.5, it is a contradiction. $\Box$

Lemma 4.3**.**

If $G$ is biquasiprimitive on $V{\it\Gamma}$, then no graph ${\it\Gamma}$ exists.

*Proof.  *Since $G$ is biquasiprimitive on $V{\it\Gamma}$, $G$ has a minimal normal subgroup $N=T^{d}$ which has exactly two orbits (say ${\it\Delta}_{1}$ and ${\it\Delta}_{2}$) on $V{\it\Gamma}$. Then ${\it\Gamma}$ is a bipartite graph with bipartition ${\it\Delta}_{1}$ and ${\it\Delta}_{2}$. Let $G^{+}=G_{{\it\Delta}_{1}}=G_{{\it\Delta}_{2}}$. Then $N\leq G^{+}$, $|G:G^{+}|=2$ and $G_{\alpha}=G^{+}_{\alpha}$. If $N$ is abelian, then $N$ is regular on ${\it\Delta}_{1}$ and so $|T|^{d}=|N|=pq^{n}$, a contradiction. Hence $N$ is nonabelian, and by Lemma 4.1, we further conclude that $N=T$ is a nonabelian simple group.

If $G^{+}$ acts unfaithfully on ${\it\Delta}_{1}$ or ${\it\Delta}_{2}$, by [9, Lemma 5.2], ${\it\Gamma}$ is a complete bipartite graph, so ${\it\Gamma}={\sf K}_{7,7}$ and $|V{\it\Gamma}|=14$, a contradiction.

Assume now $G^{+}$ acts faithfully on ${\it\Delta}_{1}$ and ${\it\Delta}_{2}$. Then by [16, Theorem 1.5]), either

• (1)

$G^{+}$ is quasiprimitive on ${\it\Delta}_{i}$; or

• (2)

$G^{+}$ has two normal subgroups $M_{1}$ and $M_{2}$ such that $M_{1}\cong M_{2}$ are semiregular on $V{\it\Gamma}$. Further, the group $M_{1}\times M_{2}$ is regular on ${\it\Delta}_{i}$.

For case (2), we have $|M_{1}|^{2}=|{\it\Delta}_{i}|=pq^{n}$, a contradiction.

Suppose case (1) occurs. Since $G^{+}$ is quasiprimitive on ${\it\Delta}_{i}$ and has a simple minimal normal subgroup $T$, by O’Nan-Scott-Praeger theorem ([21]), ${\sf soc}(G^{+})=T$ or $T^{2}$. For the latter case, $G^{+}$ is of holomorph type and $T$ is regular on ${\it\Delta}_{i}$, so $|T|=pq^{n}$, a contradiction. Therefore, ${\sf soc}(G^{+})=T$. Further, if $T$ is not the unique minimal normal subgroup of $G$, since $G=G^{+}.\mathbb{Z}_{2}$, one easily derives $G=G^{+}\times\mathbb{Z}_{2}$, hence the normal subgroup $\mathbb{Z}_{2}$ has $pq^{n}$ orbits on $V{\it\Gamma}$, contradicting the biquasiprimitivity of $G$. Thus $G$ is almost simple with socle $T$, and we may set $G=T.o$, and $G^{+}=T.{o^{\prime}}$ with $\mathbb{Z}_{2}\leq o\leq{\sf Out}(T)$ and $|o:o^{\prime}|=2$.

Case 1. Assume $5{\not|}~{}|T_{\alpha}|$.

Since $T_{\alpha}\leq G_{\alpha}$, by Lemma 2.5, $|T_{\alpha}|\mid 2^{24}\cdot 3^{2}\cdot 7$, and hence $|T|=|{\it\Delta}_{1}||T_{\alpha}|$ divides $2^{24}\cdot 3^{2}\cdot 7pq^{n}$; on the other hand, noting that $T_{\alpha}\neq 1$, we obtain $7\mid|T_{\alpha}|$, and so $7pq^{n}\mid|T|$. Therefore, $T$ satisfies Lemma 3.1, and $\pi(T)=4$ or $5$.

If $\pi(T)=5$, by Lemma 3.1(ii), we have $n=1$, a contradiction.

Suppose now $|\pi(T)|=4$. Then by Lemma 3.1(i) and notices that $n\geq 2$, we have $(T,p,q^{n})=({\sf J}_{2},3,5^{2})$ or $({\sf PSU}(3,5),3,5^{3})$. For the former case, as ${\sf Out}({\sf J}_{2})\cong\mathbb{Z}_{2}$, we have $o=\mathbb{Z}_{2}$, $o^{\prime}=1$ and $G^{+}=T$. It follows $|G_{\alpha}|=|G^{+}_{\alpha}|=|T_{\alpha}|={|T|\over|{\it\Delta}_{1}|}=2^{7}\cdot 3^{2}\cdot 7$, which is a contradiction by Lemma 2.5. For the latter case, $|T_{\alpha}|={|T|\over|{\it\Delta}_{1}|}=2^{4}\cdot 3\cdot 5\cdot 7$. Since ${\sf Out}({\sf PSU}(3,5))\cong{\sf S}_{3}$, we have that either $o={\sf S}_{3}$ and $o^{\prime}=\mathbb{Z}_{3}$, or $o=\mathbb{Z}_{2}$ and $o^{\prime}=1$. Thus, $|G_{\alpha}|=|G^{+}_{\alpha}|=|T_{\alpha}|\cdot|o^{\prime}|=2^{4}\cdot 3\cdot 5\cdot 7$ or $2^{4}\cdot 3^{2}\cdot 5\cdot 7$. By Lemma 2.5, the only possibility is $|G_{\alpha}|=2^{4}\cdot 3^{2}\cdot 5\cdot 7$, and $G_{\alpha}\cong S_{7}$; however, computation in Magma [1] shows that no graph ${\it\Gamma}$ exist in the case.

Case 2. Assume $5\mid|T_{\alpha}|$.

As $T_{\alpha}\lhd G_{\alpha}$, by Lemma 2.5, $|T_{\alpha}|\mid 2^{8}\cdot 3^{4}\cdot 5^{2}\cdot 7$, and $T_{\alpha}^{{\it\Gamma}(\alpha)}\cong{\sf A}_{7}$ or ${\sf S}_{7}$. It follows that $|T|=|V{\it\Gamma}||T_{\alpha}|$ divides $2^{8}\cdot 3^{4}\cdot 5^{2}\cdot 7pq^{n}$; moreover, as $35\mid|T_{\alpha}|$, we have $35pq^{n}$ divides $|T|$. Hence $T$ satisfies Lemma 3.2.

If $|\pi(T)|=5$ or 6, by Lemma 3.2, $n=1$, a contradiction.

Suppose $|\pi(T)|=4$. By Lemma 3.2(i), we have $T={\sf PSp}(4,7)$, and $(p,q^{n})=(3,7^{3})$ or $(5,7^{3})$, so $|T_{\alpha}|={|T|\over|{\it\Delta}_{1}|}=2^{8}\cdot 3\cdot 5^{2}\cdot 7$ or $2^{8}\cdot 3^{2}\cdot 5^{2}\cdot 7$, respectively. Further, as ${\sf Out}({\sf PSp}(4,7))\cong\mathbb{Z}_{2}$, we have $o=\mathbb{Z}_{2}$, $o^{\prime}=1$ and $G^{+}=T.o^{\prime}=T$. Thus $|G_{\alpha}|=|G^{+}_{\alpha}|=|T_{\alpha}|=2^{8}\cdot 3\cdot 5^{2}\cdot 7$ or $2^{8}\cdot 3^{2}\cdot 5^{2}\cdot 7$, by Lemma 2.5, which is a contradiction. $\Box$

5. Proofs of Theorems 1.1

We will complete the proof of Theorem 1.1 in this final section.

Lemma 5.1**.**

Let $p<q$ be odd primes, and let ${\it\Gamma}$ be a connected $G$-arc-transitive $7$-valent graph of order $2pq^{m}$, where $G\leq{\sf Aut}{\it\Gamma}$ and $m\geq 2$. Then either ${\it\Gamma}$ is a $\mathbb{Z}_{3}$-cover of ${\rm HS}(50)$, or $G$ has a normal elementary abelian $q$-subgroup.

*Proof.  *By Lemmas 4.2 and  4.3, $G$ is neither quasiprimitive nor biquasiprimitive on $V{\it\Gamma}$. Hence $G$ has a minimal normal subgroup, say $N$, which has at least three orbits on $V{\it\Gamma}$. By Theorem 2.6, $N$ is semiregular, and hence $|N|$ divides $|V{\it\Gamma}|=2pq$. It follows that $N$ is soluble and $N\cong\mathbb{Z}_{2},\mathbb{Z}_{p}$ or $\mathbb{Z}_{q}^{s}$ with $s\leq n$. For the last case, we are done. For the first case, by Theorem 2.6, ${\it\Gamma}_{N}$ is connected arc-transitive of odd order $pq^{n}$ and odd valency $7$, a contradiction. For the second case, again by Theorem 2.6, ${\it\Gamma}_{N}$ is $G/N$-arc-transitive of order $2q^{m}$ and valency 7. Clearly, ${\it\Gamma}_{N}\neq{\sf K}_{7,7}$. It then follows from Lemma 2.7 that either ${\it\Gamma}_{N}={\rm HS}(50)$, or $G/N$ has a minimal normal subgroup $M/N\cong\mathbb{Z}_{q}^{k}$ for some positive integer $k$. For the former case, $(p,q)=(3,5)$, and ${\it\Gamma}$ is a $\mathbb{Z}_{3}$-cover of ${\rm HS}(50)$. For the latter case, $M=\mathbb{Z}_{p}.\mathbb{Z}_{q}^{k}$, and as $p<q$, the Sylow $q$-subgroup $\mathbb{Z}_{q}^{k}$ of $M$ is characteristic in $M$, and hence normal in $G$. $\Box$

Now, we are ready to prove Theorems 1.1 Corollary 1.2.

Proof of Theorems 1.1. Suppose ${\it\Gamma}$ is $G$-arc-transitive with $G\leq{\sf Aut}{\it\Gamma}$. Let $M$ be a maximal normal $q$-subgroup of $G$. Clearly, $M$ has at least $2p\geq 6$ orbits on $V{\it\Gamma}$. Then by Theorem 1.1, ${\it\Gamma}$ is a normal cover of ${\it\Gamma}_{M}$, and ${\it\Gamma}_{M}$ is a connected $G/M$-arc-transitive graph of order $2pq^{m}$ with $0\leq m\leq n-1$.

If $m=0$, then ${\it\Gamma}_{M}$ is of order $2p$, by Lemma 2.3(1), ${\it\Gamma}={\sf K}_{7,7}$ or ${\rm CD}_{2p}$ with $7\mid p-1$.

If $m=1$, then ${\it\Gamma}_{M}$ is of order $2pq$, by Lemma 2.3(2), ${\it\Gamma}_{M}={\sf K}_{8},{\mathcal{C}}_{30},{\mathcal{C}}_{78}^{1},{\mathcal{C}}_{78}^{2},{\mathcal{C}}_{310}$, or ${\rm CD}(2pq,7)$ with $7\mid q-1$ For the last case, as $2pq>31$, by Example 2.1, ${\it\Gamma}_{M}={\rm CD}(2pq,7)$ is arc-regular, so $G/M={\sf Aut}({\it\Gamma}_{M})\cong{\sf D}_{2pq}:\mathbb{Z}_{7}$. Now, $G$ has a normal subgroup $H=M.\mathbb{Z}_{q}$ which has $2p$ orbits on $V{\it\Gamma}$, and ${\it\Gamma}$ is a normal cover of ${\it\Gamma}_{H}={\rm CD}(2p,7)$ by Theorem 1.1.

Now assume $m\geq 2$. By Lemma 5.1, either ${\it\Gamma}$ is a $\mathbb{Z}_{3}$-cover of ${\rm HS}(50)$, or $G$ has a normal elementary abelian $q$-subgroup, say $X/M$. For the former case, ${\it\Gamma}$ is a normal $M.\mathbb{Z}_{3}$-cover of ${\rm HS}(50)$. For the latter case, $X$ is a normal $q$-subgroup of $G$, and by the maximality of $M$, $X$ has at most two orbits on $V{\it\Gamma}$. It follows that $pq^{n}\mid|X|$, a contradiction.

Finally, one easily verifies that the graphs in Table 1 are basic graphs. This completes the proof of Theorems 1.1. $\Box$

Proof of Corollary 1.2. Let ${\it\Gamma}$ be a connected $2$-arc-transitive $7$-valent graphs of order $2pq^{n}$ with $q>p$ primes.

If $n=1$, notice that ${\rm CD}(2pq,7)$ is not 2-arc-transitive, Corollary 1.2 is true by Lemma 2.3(2).

Assume $n\geq 2$. If $p\neq 7$, by Theorems 1.1 and 2.6, ${\it\Gamma}$ is a normal cover of ${\mathcal{C}}_{30}$, ${\rm HS}(50)$ or ${\mathcal{C}}_{310}$, hence $(p,q)=(3,5)$ or $(5,31)$. Now Corollary 1.2 easily follows. $\Box$