Classifying Seven-Valent Symmetric Graphs of Order 8pq

Abstract

A graph is symmetric if its automorphism group is transitive on the arcs of the graph. Guo et al. determined all of the connected seven-valent symmetric graphs of order $8p$ for each prime p. We shall generalize this result by determining all of the connected seven-valent symmetric graphs of order $8pq$ with p and q to be distinct primes. As a result, we show that for each such graph of $\mathsf{\Gamma }$, it is isomorphic to one of seven graphs.

1. Introduction

We assume that the graphs in this paper are finite, simple, connected and undirected. For undefined terminologies of groups and graphs, we refer the reader to [1,2].

Let $\mathsf{\Gamma }$ be a graph. We denote $V\mathsf{\Gamma }$, $E\mathsf{\Gamma }$, $A\mathsf{\Gamma }$ and $\mathsf{Aut}\mathsf{\Gamma }$ as a vertex set, edge set, arc set and full automorphism group of the graph $\mathsf{\Gamma }$, respectively. We define that the graph $\mathsf{\Gamma }$ is vertex-transitive if $\mathsf{Aut}\mathsf{\Gamma }$ is transitive on the vertex set $V\mathsf{\Gamma }$ of $\mathsf{\Gamma }$, and $\mathsf{\Gamma }$ is an arc-transitive graph if $\mathsf{Aut}\mathsf{\Gamma }$ is transitive on the arc set $A\mathsf{\Gamma }$ of $\mathsf{\Gamma }$. An arc-transitive graph is also called a symmetric graph.

Let G be a group, and let S be a subset of G such that $S=S^{-1}:=\left\{s^{-1}\right|s\in S\}$. The Cayley graph $Cay(G,S)$ is defined to have a vertex set G and edge set $\left\{\right\{g,sg\}\parallel g\in G,s\in S\}$. Now, we denote the following Cayley graphs of dihedral groups by ${\mathcal{CD}}_{2pq}^k$.

Set ${\mathcal{CD}}_{2pq}^k=Cay(G,\{b,ab,a^{k+1}b,\cdots ,a^{k^5+k^4+\cdots +k+1}b\})$, where $G=\langle a,b|a^{pq}=b^2=1,a^b=a^{-1}\rangle \cong D_{2pq}$, and k is a solution of the equation $x^6+x^5+\cdots +x+1\equiv 0\left(\mathsf{mod}pq\right)$.

There are many graph parameters to characterize the reliability and vulnerability of an interconnection network, such as spectral characterization, main eigenvalues, distance characteristic polynomials, and arc-transitivity. Among these parameters, the spectral characterizations, main eigenvalues, and distance characteristic polynomials are the better ones to measure the stability of a network; see [3,4,5,6,7], for example. For arc-transitivity, see [8], as an example. In this paper, we study the arc-transitivity of graphs.

Let p and q be distinct primes. By [9,10,11], symmetric graphs of orders p, $2p$, and $3p$ have been classified. Furthermore, Praeger et al. determined symmetric graphs of order $pq$ in [12,13].

Recently, the classification of symmetric graphs with certain valency and with a restricted order has attracted much attention. For example, all cubic symmetric graphs of an order up to 768 have been determined by Conder and Dobcsa $\acute{\mathrm{n}}$ yi [14]. Tetravalent s-transitive graphs of order $6p$, $6p^2$, $8p$, $8p^2$, $10p$ or $10p^2$ were classified in [15,16,17]. More recently, a large number of papers on seven-valent symmetric graphs have been published. The classification of seven-valent symmetric graphs of order $8p$, $12p$, $16p$, $24p$ or $2pq$ were presented in [18,19,20,21,22]. We shall generalize these results by determining all connected seven-valent symmetric graphs of the order $8pq$.

In this paper, the main result we obtain is the following theorem.

Theorem 1. 

Let $p<q$ be primes and let Γ be a seven-valent symmetric graph of the order $8pq$. Then, Γ is isomorphic to one of the graphs in

Table 1. Seven-valent symmetric graphs of order $8pq$.

$\mathbf{\Gamma }$	$\mathsf{Aut}\mathbf{\Gamma }$	$(\mathit{p},\mathit{q})$
${\mathcal{C}}_{48}$	$\mathrm{PGL}(2,7)\times {\mathrm{D}}_8$	(2,3)
${\mathcal{C}}_{112}$	$({\mathbb{Z}}_2^3\times {\mathrm{D}}_{14}):{\mathrm{F}}_{21}$	(2,7)
${\mathcal{C}}_{120}$	${\mathrm{S}}_7$	(3,5)
${\mathcal{C}}_{312}^i$	$\mathrm{PGL}(2,13)\times {\mathbb{Z}}_2$	$(3,13),i=1,2,3,4$
${\mathcal{C}}_{312}^5$	$(\mathrm{PSL}(2,13)\times {\mathbb{Z}}_2):{\mathbb{Z}}_2$	(3,13)
${\mathcal{C}}_{312}^6$	$\mathrm{PSL}(2,13):{\mathrm{D}}_8$	(3,13)
${\mathcal{C}}_{(2^3,2q)}$	$({\mathbb{Z}}_2^3\times {\mathrm{D}}_{2q}):{\mathbb{Z}}_7$	$(2,7|q-1)$

.

Some of the properties in Table 1 are obtained with the help of the Magma system [23]. The method of proving Theorem 1 is to reduce the automorphism groups of the graphs to some nonabelian simple groups. To make this method effective, we need to know the classification result of stabilizers of symmetric graphs. If the valency is a prime p, the method may still work. However, we need information about the stabilizers of prime-valent symmetric graphs and a more detailed discussion. Additionally, the term symmetric graph that is used in this paper has been also used for a different type of symmetry in other research works; see [24], for example. It studied the symmetry of graphs through characteristic polynomials, which is more interesting and detailed.

2. Preliminary Results

In this section, we will provide some necessary preliminary results to be used in later discussions.

For a graph $\mathsf{\Gamma }$ and its full automorphism group $\mathsf{Aut}\mathsf{\Gamma }$, let G be a vertex-transitive subgroup of $\mathsf{Aut}\mathsf{\Gamma }$ and let N be an intransitive normal subgroup of G on $V\mathsf{\Gamma }$. We use $V_N$ to denote the set of N-orbits in $V\mathsf{\Gamma }$. The normal quotient graph ${\mathsf{\Gamma }}_N$ is a graph that satisfies the vertex set of $V_N$ and two N-orbits B, and $C\in V_N$ are adjacent in ${\mathsf{\Gamma }}_N$ if and only if some vertex of B is adjacent in $\mathsf{\Gamma }$ to some vertex of C. The following Lemma ([25] Theorem 9) provides a basic method for studying our seven-valent symmetric graphs.

Lemma 1. 

Let Γ be an G-arc-transitive graph of the prime valency p, where $p>2$ and $G\le \mathsf{Aut}\mathsf{\Gamma }$, and let N be a normal subgroup of G and have at least three orbits on $V\mathsf{\Gamma }$. Then, the following statements hold.

(i) 

N is semi-regular on $V\mathsf{\Gamma }$ and $G/N\le \mathsf{Aut}{\mathsf{\Gamma }}_N$, and Γ is a normal cover of ${\mathsf{\Gamma }}_N$;

(ii) 

Γ is $(G,s)$-transitive if and only if ${\mathsf{\Gamma }}_N$ is $(G/N,s)$-transitive, where $1\le s\le 5$ or $s=7$.

By ([26] Theorem 3.4) and ([27] Theorem 1.1), we have the following lemma, which describes the vertex stabilizers of symmetric seven-valent graphs.

Lemma 2. 

Let Γ be a seven-valent $(G,s)$-transitive graph, where $G\le \mathsf{Aut}\mathsf{\Gamma }$ and $s\ge 1$ are integers. Let $\alpha \in V\mathsf{\Gamma }$. Then, $s\le 3$ and one of the following holds, where ${\mathrm{F}}_{14}$, ${\mathrm{F}}_{21}$ and ${\mathrm{F}}_{42}$ denote the Frobenius group of order 14, 21 and 42, respectively.

(i) 

If $G_{\alpha }$ is soluble, then $|G_{\alpha }\left|\right|2^2·3^2·7$. Further, the couple $(s,G_{\alpha })$ lie in the following table.

s	1	2	3
${\mathrm{G}}_{\alpha }$	${\mathbb{Z}}_7,{\mathrm{F}}_{14},{\mathrm{F}}_{21},{\mathrm{F}}_{14}\times {\mathbb{Z}}_2,{\mathrm{F}}_{21}\times {\mathbb{Z}}_3$	${\mathrm{F}}_{42},{\mathrm{F}}_{42}\times {\mathbb{Z}}_2,{\mathrm{F}}_{42}\times {\mathbb{Z}}_3$	${\mathrm{F}}_{42}\times {\mathbb{Z}}_6$

(ii) 

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

s	2	3
${\mathrm{G}}_{\alpha }$	$\begin{array}{c}\hfill \mathrm{PSL}(3,2),\mathrm{ASL}(3,2),\end{array}$$\mathrm{ASL}(3,2)\times {\mathbb{Z}}_2,$ ${\mathrm{A}}_7,{\mathrm{S}}_7$	$\begin{array}{c}\hfill \mathrm{PSL}(3,2)\times {\mathrm{S}}_4,{\mathrm{A}}_7\times {\mathrm{A}}_6,\end{array}$ ${\mathrm{S}}_7\times {\mathrm{S}}_6,({\mathrm{A}}_7\times {\mathrm{A}}_6):{\mathbb{Z}}_2,$ ${\mathbb{Z}}_2^6:(\mathrm{SL}(2,2)\times \mathrm{SL}(3,2)),\left[2^{20}\right]:(\mathrm{SL}(2,2)\times \mathrm{SL}(3,2))$
$|G_{\alpha }|$	$\begin{array}{c}\hfill 2^3·3·7,2^6·3·7,2^7·3·7,\end{array}$ $\begin{array}{c}\hfill 2^3·3^2·5·7,2^4·3^2·5·7\end{array}$	$\begin{array}{c}\hfill 2^6·3^2·7,2^6·3^4·5^2·7,2^8·3^4·5^2·7,\end{array}$ $\begin{array}{c}\hfill 2^7·3^4·5^2·7,2^{10}·3^2·7,2^{24}·3^2·7\end{array}$

To construct seven-valent symmetric graphs, we need to introduce the Sabidussi coset graph. Let G be a finite group, and H is a core-free subgroup of G. Suppose D is a union of some double cosets of H in G, such that $D^{-1}=D$. The Sabidussi coset graph $\mathsf{Cos}(G,H,D)$ of G with respect to H and D is defined to have a vertex set $V\mathsf{\Gamma }=[G:H]$ (the set of right cosets of H in G), and the edge set $E\mathsf{\Gamma }=\left\{\right\{Hg,Hdg\left\}\right|g\in G,d\in D\}$ [28,29].

Proposition 1 

([30] Proposition 2.9). Let Γ be a graph and let G be a vertex-transitive subgroup of $\mathsf{Aut}\left(\mathsf{\Gamma }\right)$. Then, Γ is isomorphic to a Sabidussi coset graph $\mathsf{Cos}(G,H,D)$, where $H=G_{\alpha }$ is the stabilizer of $\alpha \in V\mathsf{\Gamma }$ in G and D consists of all elements of G with a map of α to one of its neighbors. Further,

(i) 

Γ is connected if and only if D generates the group G;

(ii) 

Γ is G-arc-transitive if and only if D is a single double coset. In particular, if $g\in G$ interchanges α and one of its neighbors, then $g^2\in H$ and $D=HgH$;

(iii) 

The valency of the graph Γ is equal to $\left|D\right|/\left|H\right|=|H:H\cap H^g|$.

In the following lemmas, we provide classification information of seven-valent symmetric graphs of order $8p$ and $2pq$, where p and q are two distinct primes. By [19], we obtain the classification of seven-valent symmetric graphs of order $8p$.

Lemma 3. 

Let Γ be a seven-valent symmetric graph of order $8p$. Then $\mathsf{\Gamma }\cong {\mathrm{K}}_{8,8}-8{\mathrm{K}}_2$ or ${\mathcal{C}}_{24}$.

By [22], we can describe seven-valent symmetric graphs of order $2pq$.

Lemma 4. 

Let $3\le p<q$ be primes and let Γ be a seven-valent symmetric graph of order $2pq$. Then, the following statements hold:

(i) 

$\mathsf{\Gamma }\cong {\mathcal{CD}}_{2pq}^k$, where k is a solution of the equation $x^6+x^5+\cdots +x+1\equiv 0\left(\mathsf{mod}pq\right)$, and $\mathsf{Aut}\mathsf{\Gamma }\cong D_{2pq}:{\mathbb{Z}}_7$, where $p|q-1$.

(ii) 

Γ lies in

Table 2. Seven-valent symmetric graphs of order $2pq$.

$\mathbf{\Gamma }$	$\mathsf{Aut}\mathbf{\Gamma }$	$(\mathit{p},\mathit{q})$
${\mathcal{C}}_{78}^1$	$\mathrm{PGL}(2,13)$	(3, 13)
${\mathcal{C}}_{78}^2$	$\mathrm{PSL}(2,13)$	(3, 13)
${\mathcal{C}}_{310}$	$\mathrm{PSL}(5,2).{\mathbb{Z}}_2$	(5, 31)
${\mathcal{C}}_{30}$	${\mathrm{S}}_8$	(3, 5)

.

Next, we need some information about nonabelian simple groups. The first one has information about maximal subgroups of $\mathrm{PSL}(2,t)$ and $\mathrm{PGL}(2,t)$, where t is an odd prime; refer to ([31] Section 239) and ([32] Theorem 2).

Lemma 5. 

Let $G=\mathrm{PSL}(2,t)$ or $\mathrm{PGL}(2,t)$, where $t\ge 5$ is a prime, and let M be a maximal subgroup of G.

(i) 

If $G=\mathrm{PSL}(2,t)$, then $M\in \{{\mathrm{D}}_{t-1},{\mathrm{D}}_{t+1},Z_2:Z_{(t-1)/2},{\mathrm{A}}_4,{\mathrm{S}}_4,{\mathrm{A}}_5\}$;

(ii) 

If $G=\mathrm{PGL}(2,t)$, then $M\in \{{\mathrm{D}}_{2(t-1)},{\mathrm{D}}_{2(t+1)},Z_2:Z_{t-1},{\mathrm{S}}_4,\mathrm{PSL}(2,t)\}$.

The next proposition is about nonabelian simple groups of order that are divisible by at most seven primes. By [2] (pp. 134–136), we have the following proposition.

Proposition 2. 

Let T be a nonabelian simple group, such that $28pq\left|\right|T|$ and $\left|T\right||2^{27}·3^4·5^2·7$$·p·q$, where $5\le p<q$ are primes. Then, T is one of the groups in

Table 3. Simple group T with order dividing $2^{27}·3^4·5^2·7·p·q.$

T	$\left|\mathit{T}\right|$	$(\mathit{p},\mathit{q})$	T	$\left|\mathit{T}\right|$	$(\mathit{p},\mathit{q})$
${\mathrm{M}}_{22}$	$2^7·3^2·5·7·11$	$(5,11)$	$\mathrm{PSL}(3,8)$	$2^9·3^2·7^2·73$	$(7,73)$
${\mathrm{M}}_{23}$	$2^7·3^2·5·7·11·23$	$(11,23)$	$\mathrm{PSL}(3,16)$	$2^{12}·3^2·5^2·7·13·17$	$(13,17)$
${\mathrm{M}}_{24}$	$2^{10}·3^3·5·7·11·23$	$(11,23)$	$\mathrm{PSL}(2,5^3)$	$2^2·3^2·5^3·7·31$	$(5,31)$
${\mathrm{J}}_1$	$2^3·3·5·7·11·19$	$(11,19)$	$\mathrm{PSL}(2,7^2)$	$2^4·3·5^2·7^2$	$(5,7)$
$\mathrm{HS}$	$2^9·3^2·5^3·7·11$	$(5,11)$	$\mathrm{PSL}(4,4)$	$2^{12}·3^4·5^2·7·17·17$	$(5,17)$
${\mathrm{A}}_{11}$	$2^7·3^4·5^2·7·11$	$(5,11)$	$\mathrm{PSL}(5,2)$	$2^{10}·3^2·5·7·31$	$(5,31)$
$\mathrm{Sz}\left(8\right)$	$2^6·5·7·13$	$(5,13)$	$\mathrm{PSL}(6,2)$	$2^{15}·3^4·5·7^2·31$	$(7,31)$
$\mathrm{PSp}(4,8)$	$2^{12}·3^4·5·7^2·13$	$(7,13)$	${}^{3}D_4\left(2\right)$	$2^{12}·3^4·7^2·13$	$(7,13)$
$\mathrm{PSL}(2,2^6)$	$2^6·3^2·5·7·13$	$(5,13)$	${}^{2}D_4\left(2\right)$	$2^{12}·3^4·5·7·17$	$(5,17)$
$\mathrm{PSL}(2,2^9)$	$2^9·3^2·7·19·73$	$(19,73)$	$G_2\left(4\right)$	$2^{12}·3^3·5^2·7·13$	$(5,13)$
$\mathrm{PSL}(2,q)$	$\frac{q(q+1)(q-1)}{2}$

.

Proof. 

Suppose T is a sporadic simple group, by [2] (pp.135–136), $T={\mathrm{M}}_{22}$, ${\mathrm{M}}_{23}$, ${\mathrm{M}}_{24}$, ${\mathrm{J}}_1$, or $\mathrm{HS}$. Suppose $T={\mathrm{A}}_n$ is an alternating group. Then, $T={\mathrm{A}}_{11}$ is the limitation of $\left|T\right|$.

Let X be one type of the Lie group, and let $t=r^f$ be a prime power. Now, suppose that $T=X\left(t\right)$ is a simple group of the Lie type, as T contains at most four 3-factors, three 5-factors, and two 7-factors [2] (p.135), and $T=\mathrm{PSL}(2,q)$, $\mathrm{PSL}(2,5^3)$ or $\mathrm{PSL}(2,7^2)$.

Similarly, if $r=2$, then $T=\mathrm{Sz}\left(8\right)$, $\mathrm{PSp}(4,8)$, $\mathrm{PSL}(2,2^6)$, $\mathrm{PSL}(2,2^9)$, $\mathrm{PSL}(3,8)$, $\mathrm{PSL}(3,16)$, $\mathrm{PSL}(4,4)$, $\mathrm{PSL}(5,2)$, $\mathrm{PSL}(6,2)$, ${}^{3}D_4\left(2\right)$, ${}^{2}D_4\left(2\right)$ or $G_4\left(2\right)$. □

3. The Proof of Theorem 1

We will prove Theorem 1 through a series of lemmas in this section. To prove Theorem 1, we need information on seven-valent symmetric graphs of order $4pq$. Therefore, we first prove the following lemma.

Lemma 6. 

Let $p<q$ be primes and let Γ be a seven-valent symmetric graph of order $4pq$. Then, $\mathsf{\Gamma }\cong {\mathcal{C}}_{24}$, ${\mathcal{C}}_{60}$, ${\mathcal{SG}}_{156}^i$ or ${\mathcal{CG}}_{156}^j$, where $i=1,2,3,4,5$ and $j=1,2,3,4$.

Proof. 

Let $\mathsf{\Gamma }$ be a seven-valent symmetric graph of the order $4pq$, where $p<q$ are primes. Let $\mathrm{A}=\mathsf{Aut}\mathsf{\Gamma }$. In Lemma 2, $\left|\mathrm{A}\right||2^{26}·3^4·5^2·7·p·q$ is $|{\mathrm{A}}_{\alpha }\left|\right|2^{24}·3^4·5^2·7$, where $\alpha \in V\mathsf{\Gamma }$. If $p=2$, then $\mathsf{\Gamma }$ has the order $8q$; in Lemma 3, we have $q=3$ and $\mathsf{\Gamma }\cong {\mathcal{C}}_{24}$. If $p=3$, then $\mathsf{\Gamma }$ has the order $12q$, and in [18,33], we have $q=5$ or 13 and $\mathsf{\Gamma }\cong {\mathcal{C}}_{60}$, ${\mathcal{SG}}_{156}^i$ or ${\mathcal{CG}}_{156}^j$, where $i=1,2,3,4,5$ and $j=1,2,3,4$. Therefore, we only need to prove that there is no seven-valent symmetric graph of order $4pq$ for $5\le p<q$, and the Lemma 6 is proved.

Now, we assume $5\le p<q$. By ([33] Theorem 1.1), we have $\mathrm{A}\cong \mathrm{PSL}(2,r)\times {\mathbb{Z}}_2$, $\mathrm{PGL}(2,r)\times {\mathbb{Z}}_2$, $\mathrm{PSL}(2,r)$ or $\mathrm{PGL}(2,r)$, where $r\equiv \pm 1\left(\mathsf{mod}7\right)$ is a prime. If $\mathrm{A}\cong \mathrm{PSL}(2,r)\times {\mathbb{Z}}_2$ or $\mathrm{PGL}(2,r)\times {\mathbb{Z}}_2$, then $\mathrm{A}$ has a normal subgroup $N\cong {\mathbb{Z}}_2$. It follows that ${\mathsf{\Gamma }}_N$ is a seven-valent symmetric graph of order $2pq$ and $\mathrm{A}/N\le \mathsf{Aut}{\mathsf{\Gamma }}_N$. Since $\mathrm{A}/N$ is isomorphic to $\mathrm{PSL}(2,r)$ or $\mathrm{PGL}(2,r)$ for $5\le p<q$, there exists no such graph in Lemma 4. Hence, $\mathrm{A}$ is not isomorphic to $\mathrm{PSL}(2,r)\times {\mathbb{Z}}_2$ or $\mathrm{PGL}(2,r)\times {\mathbb{Z}}_2$.

If $A\cong \mathrm{PSL}(2,r)$ or $\mathrm{PGL}(2,r)$, then A has a normal subgroup $N\cong \mathrm{PSL}(2,r)$. Assume that N has t orbits on the vertex set of $\mathsf{\Gamma }$, $t\ge 3$. Then, N is semi-regular on $V\mathsf{\Gamma }$ in Lemma 1 and thus $\left|N\right|$ divides $4pq$, contradicting with $N\cong \mathrm{PSL}(2,r)$ and $5\le p<q$. Hence, $N_{\alpha }\ne 1$, N has, at most, two orbits on $V\mathsf{\Gamma }$ and $2pq\left|\right|N:N_{\alpha }|$. Note that $\mathsf{\Gamma }$ is connected, $N⊴A$, and $N_{\alpha }\ne 1$. Then, we have $1\ne N_{\alpha }^{\mathsf{\Gamma }\left(\alpha \right)}⊴A_{\alpha }^{\mathsf{\Gamma }\left(\alpha \right)}$. This implies that 7 | $|N_{\alpha }|$; thus, we have that $14pq\left|\right|N|$. And, $\left|N\right||2^{26}·3^4·5^2·7·p·q$ is $\left|N\right|\left|\right|A|$. Since $|A:N|\le 2$, we have $|A_{\alpha }:N_{\alpha }|\le 2$. If $A_{\alpha }$ is insoluble, then $N_{\alpha }$ is also insoluble as $|A_{\alpha }:N_{\alpha }|\le 2$. In Lemma 5, $N_{\alpha }=A_5$ (the alternating group on {1, 2, 3, 4, 5}), which contradicts with $7\left|\right|N_{\alpha }|$. Therefore, $A_{\alpha }$ is soluble. It follows that $|A_{\alpha }|$ | 252 in Lemma 2; thus, $|N_{\alpha }|$ divides 252. This implies that $\left|N\right||1008·p·q$.

We claim that $r=q$, since $\left|V\mathsf{\Gamma }\right|=\left|A\right|/|A_{\alpha }|=4pq$ and $|A_{\alpha }\left|\right|252$. Then, we have $4pq={\displaystyle \frac{r(r-1)(r+1)}{2|A_{\alpha }|}}$ or ${\displaystyle \frac{r(r-1)(r+1)}{|A_{\alpha }|}}$. Since $r\equiv \pm 1\left(\mathsf{mod}7\right)$ is a prime and $|A_{\alpha }\left|\right|252$, we have $r=p$ or q. Assume that $r=p$. Then, $4q={\displaystyle \frac{(r-1)(r+1)}{2|A_{\alpha }|}}$ or ${\displaystyle \frac{(r-1)(r+1)}{|A_{\alpha }|}}$. This implies that $q=r+1$ as $q>p$, which is impossible because $r+1$ is not a prime. Thus, $r=q$ and $\left|N\right|={\displaystyle \frac{q(q-1)(q+1)}{2}}$. Note that $({\displaystyle \frac{q+1}{2}},{\displaystyle \frac{q-1}{2}})=1$. Assume that $p|{\displaystyle \frac{q-1}{2}}$. Then, $q+1|1008$. And then, we have $q=7$, 11, 13, 17, 23, 41, 47, 71, 83, 167, 251 or 503. Assume that $p|{\displaystyle \frac{q+1}{2}}$. Then, $q-1|1008$. And then, we have $q=7$, 13, 17, 19, 29, 37, 43, 73, 113, 127, 337 or 1009. Note that $14pq\left|\right|N|$, $\left|N\right||2^{26}·3^4·5^2·7·p·q$ and $5\le p<q$. Therefore, N is one of the groups in the following table:

N	Order	N	Order
$\mathrm{PSL}(2,29)$	$2^2·3·5·7·29$	$\mathrm{PSL}(2,41)$	$2^3·3·5·7·41$
$\mathrm{PSL}(2,43)$	$2^2·3·7·11·43$	$\mathrm{PSL}(2,71)$	$2^3·3^2·5·7·71$
$\mathrm{PSL}(2,83)$	$2^2·3·7·41·83$	$\mathrm{PSL}(2,113)$	$2^4·3·7·19·113$
$\mathrm{PSL}(2,167)$	$2^3·3·7·83·167$	$\mathrm{PSL}(2,251)$	$2^2·3^2·5^3·7·251$
$\mathrm{PSL}(2,337)$	$2^4·3·7·{13}^2·337$	$\mathrm{PSL}(2,503)$	$2^3·3^2·7·251·503$
$\mathrm{PSL}(2,1009)$	$2^4·3^2·5·7·101·1009$

Assume that $q=29$, 71, 113, 251 or 1009. Note that $|N:N_{\alpha }|=2pq$ or $4pq$. N has no subgroup of index $2pq$ or $4pq$ in Lemma 5, which is a contradiction.

Assume that $q=337$. Then, $N=\mathrm{PSL}(2,337)$, contradicting with $\left|N\right||2^{26}·3^4·5^2·7·p·q$.

Assume that $q=41$. Then, $N=\mathrm{PSL}(2,41)$ and $(p,q)=(5,41)$. Since N has no subgroup of index $2pq$ in Lemma 5, we have that N is transitive on $V\mathsf{\Gamma }$, and thus $|N_{\alpha }|=42$. Hence, $N_{\alpha }=F_{42}$ in Lemma 2. In Proposition 1, $\mathsf{\Gamma }=\mathsf{Cos}(N,N_{\alpha },N_{\alpha }g{N}_{\alpha })$, where g is a 2-element in N such that $g^2\in N_{\alpha }$ and $\langle N_{\alpha },g\rangle =N$. In Magma [23], there is no such $g\in N$, which is a contradiction.

Finally, assume that $q=43$. Then, $N=\mathrm{PSL}(2,43)$ and $(p,q)=(11,43)$. If N has two orbits on $V\mathsf{\Gamma }$, then $A=\mathrm{PGL}(2,43)$ and $A_{\alpha }=F_{42}$ in Lemma 2. This is impossible, as $\mathrm{PGL}(2,41)$ has no subgroup isomorphic to $F_{42}$. Therefore, N is transitive on $V\mathsf{\Gamma }$ and in Lemma 2, $N_{\alpha }=F_{21}$. In Lemma 5, $\mathrm{PSL}(2,41)$ has no subgroup isomorphic to $F_{21}$, which is a contradiction. Similarly, $q\ne 83$, 167 or 503. This completes the proof. □

Now, let $\mathsf{\Gamma }$ be a seven-valent symmetric graph of the order $8pq$, where $p<q$ are primes. Let $\mathrm{A}:=\mathsf{Aut}\mathsf{\Gamma }$. Take $\alpha \in V\mathsf{\Gamma }$. In Lemma 2, $|{\mathrm{A}}_{\alpha }\left|\right|2^{24}·3^4·5^2·7$, and hence $\left|\mathrm{A}\right||2^{27}·3^4·5^2·7·p·q$.

If $p=2$, then $\mathsf{\Gamma }$ has the order $16q$; by [20], we have $q=3$, 7 or $7|q-1$, and $\mathsf{\Gamma }$ is isomorphic to ${\mathcal{C}}_{48}$, ${\mathcal{C}}_{112}$ or ${\mathcal{C}}_{(2^3,2q)}$. If $p=3$, then $\mathsf{\Gamma }$ has the order $24q$; in [21], we have $q=5$ or 13, and $\mathsf{\Gamma }$ is isomorphic to ${\mathcal{C}}_{120}$, ${\mathcal{C}}_{312}^i$ with $i=1,2,3,4$, ${\mathcal{C}}_{312}^5$ or ${\mathcal{C}}_{312}^6$. Therefore, we only need to prove that there is no seven-valent symmetric graph of the order $8pq$ for $5\le p<q$, and the Theorem 1 is proved. For the remainder of this paper, we let $5\le p<q$.

In the next lemma, we deal with the case where there is a soluble minimal normal subgroup of A.

Lemma 7. 

Assume that A has a soluble minimal normal subgroup. Then, there exists no seven-valent symmetric graph of order $8pq$ for $5\le p<q$.

Proof. 

Assuming N is a soluble minimal normal subgroup of the full automorphism group A. Then, N is an elementary abelian group. Since $\left|V\mathsf{\Gamma }\right|=8pq$, we have $N\cong {\mathbb{Z}}_2$, ${\mathbb{Z}}_2^2$, ${\mathbb{Z}}_2^3$, ${\mathbb{Z}}_p$ or ${\mathbb{Z}}_q$. It is easy to prove that N has more than two orbits on $V\mathsf{\Gamma }$; if not, we have $4pq\left|\right|N|$, a contradiction. Therefore, in Lemma 1, $|N_{\alpha }|=1$, and the quotient graph ${\mathsf{\Gamma }}_N$ of $\mathsf{\Gamma }$ relative to N is a seven-valent symmetric graph, with $A/N$ as an arc-transitive subgroup of the automorphism of ${\mathsf{\Gamma }}_N$.

If $N\cong {\mathbb{Z}}_2^3$, then ${\mathsf{\Gamma }}_N$ is a seven-valent symmetric graph of the order $pq$ ($pq$ is an odd number), which is a contradiction, as symmetric graphs of the odd order odd valent do not exist. If $N\cong {\mathbb{Z}}_2$, then ${\mathsf{\Gamma }}_N$ is a seven-valent symmetric graph of the order $4pq$. In Lemma 6, we note that $5\le p<q$, ${\mathsf{\Gamma }}_N$ does not exist, which is a contradiction. If $N\cong {\mathbb{Z}}_p$, then ${\mathsf{\Gamma }}_N$ is a seven-valent symmetric graph of the order $8q$. ${\mathsf{\Gamma }}_N$ does not exist in Lemma 3, which is a contradiction. Similarly, we obtain that $N\ncong {\mathbb{Z}}_q$.

If $N\cong {\mathbb{Z}}_2^2$, then ${\mathsf{\Gamma }}_N$ is a seven-valent symmetric graph of the order $2pq$. In Lemma 4, ${\mathsf{\Gamma }}_N\cong {\mathcal{C}}_{310}$ or ${\mathcal{CD}}_{2pq}^k$, where k is a solution of the equation $x^6+x^5+\cdots +x+1\equiv 0\left(\mathsf{mod}pq\right)$ and $p|q-1$.

Let ${\mathsf{\Gamma }}_N\cong {\mathcal{C}}_{310}$. Then, $A/N\le \mathsf{Aut}{\mathcal{C}}_{310}=\mathrm{PSL}(5,2).{\mathbb{Z}}_2$. Furthermore, $A/N$ is arc-transitive on $V{\mathsf{\Gamma }}_N$. By Magma [23], $\mathsf{Aut}{\mathsf{\Gamma }}_N$ has a minimal arc-transitive subgroup, which is isomorphic to $\mathrm{PSL}(5,2)$. Thus, $\mathrm{PSL}(5,2)\le A/N\le \mathrm{PSL}(5,2).{\mathbb{Z}}_2$. Since the Schur Multiplier of $\mathrm{PSL}(5,2)$ is trivial, $A={\mathbb{Z}}_2^2\times \mathrm{PSL}(5,2)$ or $({\mathbb{Z}}_2^2\times \mathrm{PSL}(5,2)).{\mathbb{Z}}_2$. For the former case, in Proposition 1, $\mathsf{\Gamma }=\mathsf{Cos}(A,A_{\alpha },A_{\alpha }g{A}_{\alpha })$, where g is a 2-element in A such that $g^2\in A_{\alpha }$ and $\langle A_{\alpha },g\rangle =A$. By Magma [23], there is no such $g\in A$, which is a contradiction. For the latter case, $A/N$ has a normal subgroup, $M\cong \mathrm{PSL}(5,2)$. It is obvious that M has at most two orbits on $V\mathsf{\Gamma }$. Since M has no subgroup of order 16128, M is transitive on $V\mathsf{\Gamma }$, implying that $|M_{\alpha }|=8064$; this is impossible in Lemma 2.

Let ${\mathsf{\Gamma }}_N\cong {\mathcal{CD}}_{2pq}^k$, where k is a solution of the equation $x^6+x^5+\cdots +x+1\equiv 0\left(\mathsf{mod}pq\right)$. Note that $A/N$ is an arc-transitive subgroup of $\mathsf{Aut}\left({\mathsf{\Gamma }}_N\right)=D_{2pq}:{\mathbb{Z}}_7$. Hence, $2pq·7$ | |A/N|. This implies that $A/N=D_{2pq}:{\mathbb{Z}}_7$. Let H be a normal subgroup of the order $pq$ of $D_{2pq}$ and Q be a Sylow q-subgroup of H. Then, in the Sylow Theorem, Q char H and thus $Q⊴D_{2pq}$ is $H⊴D_{2pq}$. Note that Q is also a Sylow q-subgroup of $D_{2pq}$. Then, Q char $D_{2pq}$ and thus $Q⊴A/N$ is $D_{2pq}⊴A/N$. Then, $5\le p<q$ and $p|q-1$. Then, $q\ge 11$. Hence, Q is also a Sylow q-subgroup of $A/N$. Let $Q=G/N$. Then, $G/N\cong {\mathbb{Z}}_q$ and $\left|G\right|=2^2·q$. In the Sylow Theorem, the Sylow q-subgroup of G is normal, at say L. Then, $L\cong {\mathbb{Z}}_q$, and thus $G={\mathbb{Z}}_2^2\times {\mathbb{Z}}_q=N\times L$. Hence, $L⊴A$ is $G⊴A$. Then, the normal quotient graph ${\mathsf{\Gamma }}_L$ of $\mathsf{\Gamma }$ relative to L is a seven-valent symmetric graph of order $8p$. In Lemma 3, there exists no graph for this case, which is a contradiction.

Thus, we complete the proof of Lemma 7. □

Now we move on to the case where there is no soluble minimal normal subgroup of A. Then, we have the following lemma.

Lemma 8. 

Assume that A has no soluble minimal normal subgroup. Then, there exists no seven-valent symmetric graph of order $8pq$ for $5\le p<q$.

Proof. 

Let N be an insoluble minimal normal subgroup of A, and let $C=C_A\left(N\right)$ be the centralizer of N in A. Then, N is isomorphic to $T^d$, where $d\ge 1$ and T are non-abelian simple groups. Assume that N has t orbits on the vertex set of $\mathsf{\Gamma }$. If $t\ge 3$, then $N_{\alpha }=1$ by Lemma 1 and thus $\left|N\right|={\left|T\right|}^d|8pq$, since N is insoluble. Then, $\left|N\right|=4pq$ or $8pq$. Thus, N has two orbits or an orbit on $V\mathsf{\Gamma }$, which is a contradiction. Hence, N has at most two orbits on $V\mathsf{\Gamma }$, and it follows that $4pq\left|\right|N|$.

If $N_{\alpha }=1$, then $\left|N\right|=4pq$ or $8pq$, since $q\left|\right|N|$ and $q^2\nmid \left|N\right|$. Then, $N=T$. Note that $5\le p<q$ [34]; no such simple group exists, and this is a contradiction. Hence, $N_{\alpha }\ne 1$. Since $\mathsf{\Gamma }$ is connected to $N⊴A$ and $N_{\alpha }\ne 1$, we have $1\ne N_{\alpha }^{\mathsf{\Gamma }\left(\alpha \right)}⊴A_{\alpha }^{\mathsf{\Gamma }\left(\alpha \right)}$. It follows that 7 divides $|N_{\alpha }|$. Then, we have that $28pq\left|\right|N|$.

Now, we claim that $d=1$. Otherwise, $d\ge 2$, and thus $7^2|\left|N\right|$. We have $d=2$ as $\left|N\right||2^{27}·3^4·5^2·7·p·q$. So $p=7$ or $q=7$. If $p=7$, then $q>7$ and $q^2|{\left|T\right|}^2$, which contradicts with $\left|N\right||2^{27}·3^4·5^2·7·p·q$. If $q=7$, then $p=5$. This implies that $\left|T\right||2^{13}·3^2·5·7$. Note that $35\left|\right|T|$. By checking the nonabelian simple group of an order less than $2^{13}·3^2·5·7$, we have that $T=A_7$, $A_8$ or $\mathrm{PSL}(3,4)$, and $N={A_7}^2$, ${A_8}^2$ or $\mathrm{PSL}{(3,4)}^2$ as $d=2$. On the other side of the coin, $C⊴A$, $C\cap N=1$ and thus $\langle C,N\rangle =C\times N$. Because $|C\times N||2^{27}·3^4·5^2·7·p·q$ and $\left|N\right|={\left|T\right|}^2=2^6·3^4·5^2·7^2$ or $2^{12}·3^4·5^2·7^2$, C is a $\{2,p\}$-group, and hence soluble, where $p=5$. So, $C=1$ as A contains no soluble minimal normal subgroup. This implies $A=A/C\le \mathsf{Aut}\left(N\right)\cong \mathsf{Aut}\left(T\right)wr{\mathbb{Z}}_2$. By Magma [23], no such graph exists, which is a contradiction. Therefore, we have $d=1$, and $N=T⊴A$ is a nonabelian simple group.

We next prove that $C=1$. If $C\ne 1$, then C is insoluble, as $C⊴A$ and A contain no soluble minimal normal subgroup. In the same argument as for the case N, we have 7 divides $|C_{\alpha }|$. Because $\langle C,N\rangle =C\times N$ and C, $N⊴A$, we have $C_{\alpha }\times N_{\alpha }\le A_{\alpha }$. Note that 7 divides $|N_{\alpha }|$; this concludes that $7^2|\left|A_{\alpha }\right|$, which is a contradiction with Lemma 2. Therefore, we have $C=1$, and thus $A\le \mathsf{Aut}\left(T\right)$ is almost simple. It follows that $T=\mathsf{soc}\left(A\right)$ is a nonabelian simple group and satisfies the following condition.

Condition(*): $\left|T\right|$ lies in Table 3 such that $28pq\left|\right|T|$ and $\left|T\right||2^{27}·3^4·5^2·7·p·q$.

Assume first that $T\cong {\mathrm{M}}_{22}$, ${\mathrm{M}}_{23}$, ${\mathrm{J}}_1$, ${\mathrm{A}}_{11}$, $\mathrm{PSL}(2,2^9)$, $\mathrm{PSL}(3,16)$, $\mathrm{PSL}(2,5^3)$, $\mathrm{PSL}(2,7^2)$, $\mathrm{PSL}(4,4)$, $\mathrm{PSL}(6,2)$, $\mathrm{PSp}(4,8)$, $\mathrm{HS}$, ${}^{2}D_4\left(2\right)$, ${}^{3}D_4\left(2\right)$, or ${\mathrm{G}}_2\left(4\right)$. Note that $|T:T_{\alpha }|=4pq$ or $8pq$. T has no subgroup of index $4pq$ or $8pq$ by Atlas [35], which is a contradiction.

Assume that $T\cong {\mathrm{M}}_{24}$. Since T has no subgroup of index $4pq$, we show that T is transitive on $V\mathsf{\Gamma }$, and thus $|T_{\alpha }|=\mathrm{120,960}$. In Proposition 1, $\mathsf{\Gamma }=\mathsf{Cos}(T,T_{\alpha },T_{\alpha }g{T}_{\alpha })$, where g is a 2-element in T such that $g^2\in T_{\alpha }$ and $\langle T_{\alpha },g\rangle =T$. In Magma [23], there is no such $g\in T$, which is a contradiction. Similarly, T is not isomorphic to $\mathrm{Sz}\left(8\right)$, $\mathrm{PSL}(2,2^6)$ or $\mathrm{PSL}(5,2)$.

Assume that $T\cong \mathrm{PSL}(3,8)$. If T has two orbits on $V\mathsf{\Gamma }$, then $\mathsf{\Gamma }$ is bipartite and $|T_{\alpha }|=2^7·3^2·7$. Recall that A is almost simple. Thus, $\mathrm{A}\le \mathsf{Aut}\left(T\right)$. Since $\mathsf{Aut}\left(T\right)=\mathrm{PSL}(3,8).{\mathbb{Z}}_6$, we have $A\cong \mathrm{PSL}(3,8).{\mathbb{Z}}_2$, $\mathrm{PSL}(3,8).{\mathbb{Z}}_3$ or $\mathrm{PSL}(3,8).{\mathbb{Z}}_6$, and thus $|A_{\alpha }|=2^7·3^2·7$, $2^6·3^3·7$ or $2^7·3^3·7$, which is impossible according to Lemma 2. Thus, T is transitive on $V\mathsf{\Gamma }$. In Proposition 1, $\mathsf{\Gamma }=\mathsf{Cos}(T,T_{\alpha },T_{\alpha }g{T}_{\alpha })$, where g is a 2-element in T such that $g^2\in T_{\alpha }$ and $\langle T_{\alpha },g\rangle =T$. By Magma [23], there is no such $g\in T$, which is a contradiction.

Finally, assume that $T\cong \mathrm{PSL}(2,q)$. Then, $T\le A\le \mathsf{Aut}\left(T\right)=\mathrm{PGL}(2,q)$ ($\mathrm{PGL}(2,q)=\mathrm{PSL}(2,q).{\mathbb{Z}}_2$) and $|A:T|\le 2$. If $A_{\alpha }$ is insoluble, then $T_{\alpha }$ is also insoluble as $|A_{\alpha }:T_{\alpha }|\le 2$. $T_{\alpha }=A_5$ in Lemma 5, contradicting with 7, divides $|T_{\alpha }|$. Therefore, $A_{\alpha }$ is soluble, and $|A_{\alpha }|$ divides by 252 in Lemma 2, and so $|T_{\alpha }|$ divides 252. This implies that $\left|T\right||2016·p·q$. Note that $\left|T\right|={\displaystyle \frac{q(q-1)(q+1)}{2}}$ and $({\displaystyle \frac{q+1}{2}},{\displaystyle \frac{q-1}{2}})=1$. If $p|{\displaystyle \frac{q-1}{2}}$, then $q+1|2016$. It follows that $q=7$, 11, 13, 17, 23, 31, 41, 47, 71, 83, 167, 223, 251 or 503. If $p|{\displaystyle \frac{q+1}{2}}$, then $q-1|2016$. It follows that $q=7$, 13, 17, 19, 29, 37, 43, 73, 97, 113, 127, 337, 673, 1009 or 2017. Note that T meets the condition (*) and $5\le p<q$. Therefore, T is one of the groups in the following table:

T	Order	T	Order
$\mathrm{PSL}(2,29)$	$2^2·3·5·7·29$	$\mathrm{PSL}(2,41)$	$2^3·3·5·7·41$
$\mathrm{PSL}(2,43)$	$2^2·3·7·11·43$	$\mathrm{PSL}(2,71)$	$2^3·3^2·5·7·71$
$\mathrm{PSL}(2,83)$	$2^2·3·7·41·83$	$\mathrm{PSL}(2,97)$	$2^5·3·7^2·97$
$\mathrm{PSL}(2,113)$	$2^4·3·7·19·113$	$\mathrm{PSL}(2,167)$	$2^3·3·7·83·167$
$\mathrm{PSL}(2,223)$	$2^5·3·7·37·223$	$\mathrm{PSL}(2,251)$	$2^2·3^2·5^3·7·251$
$\mathrm{PSL}(2,337)$	$2^4·3·7·{13}^2·337$	$\mathrm{PSL}(2,503)$	$2^3·3^2·7·251·503$
$\mathrm{PSL}(2,673)$	$2^5·3·7·337·673$	$\mathrm{PSL}(2,1009)$	$2^4·3^2·5·7·101·1009$
$\mathrm{PSL}(2,2017)$	$2^5·3^2·7·1009·2017$

Assume that $q=29$, 71, 97, 113, 223, 251, 337 or 1009. Note that $|T:T_{\alpha }|=4pq$ or $8pq$. T has no subgroup of index $4pq$ or $8pq$ in Lemma 5, which is a contradiction.

Assume that $q=337$. Then, $T=\mathrm{PSL}(2,337)$, which contradicts with $\left|T\right||2^{27}·3^4·5^2·7·p·q$.

Assume that $q=43$. Then, $T=\mathrm{PSL}(2,43)$ and $(p,q)=(11,43)$, since T has no subgroup of index $8pq$. Then, T is not transitive to $V\mathsf{\Gamma }$. If T has two orbits on $V\mathsf{\Gamma }$, then $|T_{\alpha }|=21$. As A is almost simple, $A=\mathrm{PGL}(2,43)$, and $A_{\alpha }=F_{21}$ in Lemma 2. In Proposition 1, $\mathsf{\Gamma }=\mathsf{Cos}(A,A_{\alpha },A_{\alpha }g{A}_{\alpha })$, where g is a 2-element in A such that $g^2\in A_{\alpha }$ and $\langle A_{\alpha },g\rangle =A$. In Magma [23], there is no such $g\in A$, which is a contradiction.

Finally, assume that $q=41$. Then, $T=\mathrm{PSL}(2,41)$ and $(p,q)=(5,11)$. If T has two orbits on $V\mathsf{\Gamma }$, then $|T_{\alpha }|=42$. As A is almost simple, $A=\mathrm{PGL}(2,41)$, and $A_{\alpha }=F_{42}$ in Lemma 2. This is impossible, as $\mathrm{PGL}(2,41)$ has no subgroup isomorphic to $F_{42}$. Therefore, T is transitive to $V\mathsf{\Gamma }$ and in Lemma 2, $T_{\alpha }=F_{21}$. In Lemma 5, $\mathrm{PSL}(2,41)$ has no subgroup isomorphic to $F_{21}$, which is a contradiction. Similarly, $q\ne 167$, 503, 673 or 2017.

Thus, we complete the proof of Lemma 8. □

By combining Lemma 6, 7 and 8, we have completed the proof of Theorem 1.

4. Conclusions

Through the classification of seven-valent symmetric graphs of the order $8pq$, we obtain many highly symmetric graphs in Table 1. These graphs can be applied to the design of the interconnection network. With induction, we may further classify seven-valent symmetric graphs of the order $8n$, where n is an odd square-free integer. We can even classify p-valent symmetric graphs of the order $2^{k}n$, where k is a positive integer and n is an odd square-free integer.