Semisymmetric prime-valent graphs of order $2p^3$

Abstract

A simple undirected graph is said to be semisymmetric if it is regular and edge-transitive but not vertex-transitive. Let p be a prime and $\Gamma$ a semisymmetric prime-valent graph of order $2p^3$. Then, $\Gamma$ is bipartite. Denote by $\mathrm{Aut}(\Gamma )$ the full automorphism group of $\Gamma$. In Du and Wang (J Algebraic Combin 41:275–302, 2015), Wang and Du (Eur J Combin 36:393–405, 2014) and Wang et al. (Ars Math Contemp 7:40–53, 2014), the first author and Du proved that there is no prime-valent graph when $\mathrm{Aut}(\Gamma )$ acts unfaithfully on at least one bipart of $\Gamma$. The first author in Wang (Ars Combin 133:3–15, 2017) gave a complete classification when $\mathrm{Aut}(\Gamma )$ acts faithfully and primitively on at least one bipart of $\Gamma$, and as a result there is only one such graph, that is, cubic Gray graph. Due to these efforts, there is only one remaining case for classifying such graphs: $\mathrm{Aut}(\Gamma )$ acts faithfully and imprimitively on both biparts of $\Gamma$, which is dealt with in this paper. We prove that there are two infinite families of such graphs. Thus, combining these results, we can get the complete classification of semisymmetric prime-valent graphs of order $2p^3$.

1 Introduction

In this paper, we consider undirected finite connected graphs without loops or multiple edges. For a graph $\Gamma$, we use $V(\Gamma )$, $E(\Gamma )$ and $\mathrm{Aut}(\Gamma )$ to denote its vertex set, edge set and full automorphism group, respectively. For $u,v\in V(\Gamma )$, denote by $\{u,v\}$ the edge incident to u and v in $\Gamma$. A graph $\Gamma$ is said to be regular if all the vertices have the same valency.

For a graph $\Gamma$ and a subgroup G of $\mathrm{Aut}(\Gamma )$, $\Gamma$ is said to be G-vertex-transitive and G-edge-transitive if G is transitive on the sets of vertices and edges of $\Gamma$ respectively. Similarly, a graph is G-semisymmetric if it is regular and G-edge-transitive but not G-vertex-transitive. A graph $\Gamma$ is said to be vertex-transitive, edge-transitive and semisymmetric if it is $\mathrm{Aut}(\Gamma )$-vertex-transitive, $\mathrm{Aut}(\Gamma )$-edge-transitive and $\mathrm{Aut}(\Gamma )$-semisymmetric respectively. It can be shown that a semisymmetric graph is necessarily bipartite.

Let $\Gamma$ be a regular G-edge-transitive bipartite graph with the bipartition $V(\Gamma )=W\cup U$ and $G\le \mathrm{Aut}(\Gamma )$. Set $G^+$ be a maximal subgroup of G preserving both W and U. It is easy to see that either $|G:G^+|=2$ or $G=G^+$, depending on whether or not there exists an automorphism which interchanges W and U. For $A\le G^+,$ $\Gamma$ is said to be A-semitransitive if A acts transitively on both W and U, while a $G^+$-semitransitive graph is simply said to be semitransitive. Clearly, a semisymmetric graph is semitransitive with the bipartition having equal size.

The class of semisymmetric graphs was first introduced by Folkman [13], where several infinite families of such graphs were constructed and eight open problems were posed which spurred the interest in this topic (see for example [4, 5, 17, 20, 24]). By using group-theoretic methods, Iofinova and Ivanov [16] in 1985 classified cubic semisymmetric graphs whose automorphism group acts primitively on both biparts. This was the first classification theorem for such graphs. Let p, q be two distinct primes. Folkman [13] proved that there are no cubic semisymmetric graphs of order 2p or $2p^2$. In [12], Du and Xu classified semisymmetric graphs of order 2pq. Note that a semisymmetric graph cannot be a covering of the complete graph $K_4$ of order 4 because $K_4$ is not bipartite. A simple observation then shows that there are finitely many connected cubic semisymmetric graphs of order 4p, $4p^2$ or $4p^3$. In fact, one can show that there are no connected cubic semisymmetric graphs of order 4p, $4p^2$ or $4p^3$. Lu et al. [18] classified cubic semisymmetric graphs of order $6p^2$. For more results about semisymmetric graphs, the reader can refer to [9, 10]. Recently, Zhou [26] characterized the automorphisms of bi-Cayley graphs, this can be viewed as an useful method for constructing semisymmetric graphs. Malnič et al. [19] proved that the Gray graph of order 54 is the only cubic semisymmetric graph of order $2p^3$. As a natural process, we classified semisymmetric graphs of order $2p^3$ with prime valency.

Let $\Gamma$ be a semisymmetric graph of order $2p^3$ with prime valency r. Note that $\Gamma$ is bipartite. Let W and U be the bipartition. By the different action of $\mathrm{Aut}(\Gamma )$ on each bipartition, we can divide these graphs into the following two classes:

• Class (I): $\mathrm{Aut}(\Gamma )$ is unfaithful on at least one of W and U;

• Class (II): $\mathrm{Aut}(\Gamma )$ is faithful on both W and U.

For Class (I), the first author and Du in [11, 21, 22] gave a complete classification. As a result, there is no semisymmetric graph of order $2p^3$ with prime valency in this case. Thus, we only need to consider the Class (II), which can be divided into the following two types:

• Type (i): $\mathrm{Aut}(\Gamma )$ is primitive on at least one of W and U;

• Type (ii): $\mathrm{Aut}(\Gamma )$ is imprimitive on both W and U.

For the Type (i), the first author [23] proved that there is only one such graph, that is, cubic Gray Graph. In this paper, we consider the Type (ii) of Class (II), and the following is the main result:

Theorem 1.1

Let p be a prime and let $\Gamma$ be a semisymmetric graph of order $2p^3$ with prime valency r. Suppose that $\mathrm{Aut}(\Gamma )$ is faithful and imprimitive on both biparts. Then, $\Gamma \cong \Gamma _p$ or $\Gamma _r(m,l)$ as defined in Construction 2.1. Moreover, $\mathrm{Aut}(\Gamma _p)\cong {\mathbb {Z}}_p^3\rtimes ({{\mathbb {Z}}}_p\rtimes {{\mathbb {Z}}}_{p-1}^2)$ and $\mathrm{Aut}(\Gamma _r(m,l))\cong P\rtimes {{\mathbb {Z}}}_r,$ where P is a nonabelian group of order $p^3$.

Combining the results of [11, 21,22,23], we can get the complete classification of semisymmetric graphs of order $2p^3$ with prime valency.

Theorem 1.2

Let p and r be two odd primes. Then, any semisymmetric graph of order $2p^3$ with prime valency r is isomorphic to one of the following graphs listed in Table 1:

Table 1 Semisymmetric prime-valent graphs of order $2p^3$

The layout of the paper is as follows: In Sect. 2, two infinite families of semisymmetric prime-valent graphs of order $2p^3$ will be constructed and some preliminary results will be quoted. In Sect. 3, we shall give a reduction of Theorem 1.1 and prove that the valency r of such graphs must be $r\le p$. In Sect. 4, we will deal with the case $r=p$ and have the graph $\Gamma _p$. In Sect. 5, we deal with the case $r<p$. We prove that $r\bigm |(p-1)$ or $r\bigm |(p+1)$. For $r\bigm |(p-1)$, we have the graph $\Gamma _r(m,l)$, and for $r\bigm |(p+1)$, we prove that there is no such graph.

2 Graph constructions and preliminaries

For group-theoretic concepts and notations, the reader can refer to [1, 6, 15]. Moreover, for a permutation group G on $\Omega$, a subset $\Delta \subset \Omega$ and a subgroup $N\le G$ preserving $\Delta$, by $G_{\Delta }$ and $G_{(\Delta )}$ we denote the stabilizer of G relative to $\Delta$ setwise and pointwise, respectively, and by $\Delta _N$ the set of N-orbits on $\Delta$. For a prime p, by $p^i{\bigm |\bigm |} n$ we mean that $p^i{\bigm |} n$ but $p^{i+1}{\not \mid } n$. For a prime p, by $F_p$ we denote a finite field about integers modulo p with addition and multiplication. For a prime power $q=p^n$, by $F_q$ we denote the extension field of $F_p$. Let $F_q^*$ be the multiplicative group of all the nonzero elements in $F_q$. For a group G and a subgroup H of G, by [G : H] we denote the set of right cosets of H in G, where the action of G on [G : H] is always assumed to be the right multiplication action. For any $\alpha$ in the n-dimensional row vector space $\mathbf{V=}{} \mathbf{V}(n,p)$ over $F_p$, we denote by $t_\alpha$ the translation corresponding to $\alpha$ in the affine group $\mathrm{AGL}(n,p)$ and by N the translation subgroup. Then, $\mathrm{AGL}(n, p)\cong N\rtimes \mathrm{GL}(n, p)$. We adopt matrix notation for $\mathrm{GL}(n, p)$ and so we have $g^{-1}t_\alpha g=(t_\alpha )^g=t_{\alpha g}$, for any $t_\alpha \in N\le \mathrm{AGL}(n,p)$ and any $g\in \mathrm{GL}(n, p)$. By ||a, b, c||, we denote a diagonal matrix of order 3 with a, b, c on the diagonal.

Construction 2.1

Suppose that p is an odd prime. Define two graphs with bipartition $V=W\cup U$, where

$$\begin{aligned} W=\{(i,j,k)\bigm |i,j,k\in {{\mathbb {Z}}}_p\}, \quad U=\{[x,y,z]\bigm |x,y,z\in {{\mathbb {Z}}}_p\}, \end{aligned}$$

and with the respective edge set:

• (1) Graph $\Gamma _p(p\ge 5)$:

  $$\begin{aligned} \begin{array}{lll} E(\Gamma _p)=&{}\{ \{(i,j,k),[x,y,z]\}\bigm |x=j-2li,y=2il^2-2lj+k,z=l,\\ &{}\quad i,j,k,x,y,z,l\in {{\mathbb {Z}}}_p\}.\end{array} \end{aligned}$$

• (2) Graph $\Gamma _r(m,l)$: Let B be a subgroup of order r in $F_p^*$. For some given $m\in B,$

  $$\begin{aligned} \begin{array}{lll} E(\Gamma _r(m,l))&{}=&{}\{ \{(i,j,k),[x,y,z]\}\bigm |x=b+i,\\ &{}&{}\quad y=b^{m}+j,\,z=k+ib^{m}+lb^{(m+1)},\\ &{}&{}\quad i,j,k,x,y,z\in {{\mathbb {Z}}}_p,b\in B\},\end{array} \end{aligned}$$

  where $m=2,\cdots ,\frac{r-1}{2}$ and $l=1,\cdots ,\frac{p-1}{2}$.

Remark 2.2

For the above two graphs $\Gamma _p$ and $\Gamma _r(m,l)$, we may get the following properties from Sects. 4 and 5:

• (1) Graph $\Gamma _p$: $\mathrm{Aut}(\Gamma _p)={{\mathbb {Z}}}_p^3\rtimes ({{\mathbb {Z}}}_p\rtimes {{\mathbb {Z}}}_{p-1}^2)=N\rtimes (\langle e\rangle \rtimes R),$ where N is the translation subgroup of $\mathrm{AGL}(3,p),$ and

  $$\begin{aligned} e=\left( \begin{array}{ccc} 1 &{}\quad 2 &{}\quad 2 \\ 0 &{}\quad 1 &{}\quad 2 \\ 0 &{}\quad 0 &{}\quad 1 \end{array}\right) ,\, R=\langle \left( \begin{array}{ccc} s^2t &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad s &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad t \end{array}\right) |s,t\in F_p^*\rangle . \end{aligned}$$

  Particularly, the valency of $\Gamma _p$ is p with $p\ge 5$.

• (2) Graph $\Gamma _r(m,l)$: $\mathrm{Aut}(\Gamma _r(m,l))=P\rtimes {{\mathbb {Z}}}_r,$ where P is a nonabelian group of order $p^3,\mathrm{exp}(P)=p$ and $r|p-1$. Particularly, the valency of $\Gamma _r(m,l)$ is r with $r\ge 5$.

Let G be a group, L and R two subgroups of G and let D be a union of double cosets of R and L in G, namely, $D=\bigcup _i Rd_iL$. Define a bipartite graph $\Gamma =\mathbf{B}(G,L,R;D)$ with bipartition $V(\Gamma )=[G:L]\cup [G:R]$ and edge set $E(\Gamma )=\{ \{Lg,Rdg\}\bigm |g\in G, d\in D\}$. This graph is called the bi-coset graph of G with respect to L, R and D. The following two propositions give some properties for bi-coset graphs, see [12].

Proposition 2.3

The graph $\Gamma =\mathbf{B}(G,L,R;D)$ is a well-defined bipartite graph. Under the right multiplication action of G on $V(\Gamma )$, the graph $\Gamma$ is G-semitransitive. The kernel of the action of G on $V(\Gamma )$ is $\mathrm{Core}_G(L)\cap \mathrm{Core}_G(R)$, the intersection of the cores of the subgroups L and R in G. Furthermore, we have

• (i) $\Gamma$ is G-edge-transitive if and only if $D=RdL$ for some $d\in G;$

• (ii) the valency of any vertex in [G : L] (resp. [G : R]) is equal to the number of right cosets of R (resp. L) in D (resp. $D^{-1})$, so $\Gamma$ is regular if and only if $|L|=|R|;$

• (iii) $\Gamma$ is connected if and only if G is generated by elements of $D^{-1}D;$

• (iv) $\Gamma \cong \mathbf{B}(G,L^a,R^b;D')$ where $D'=\bigcup _iR^b$ $(b^{-1}d_ia)L^a$, for any $a, b\in G,d_i\in D;$

• (v) $\Gamma \cong \mathbf{B}({\hat{G}},L^\sigma ,R^ \sigma ;D^\sigma )$ where $\sigma$ is an isomorphism from G to ${\hat{G}}$ (it does not appear in [12] but it is easy to prove.)

• (vi) Suppose that $\Gamma$ is a G-semitransitive graph with bipartition $V(\Gamma )=U(\Gamma )$ $\cup \, W(\Gamma )$. Take $u\in U(\Gamma )$ and $w\in W(\Gamma )$. Set $D=\{ g\in G\bigm |w^g \in Y_1(u)\}$. Then, D is a union of double cosets of $G_w$ and $G_u$ in G, and $\Gamma \cong \mathbf{B}(G,G_u,$ $G_w;D)$.

Proposition 2.4

Let $\Gamma =\mathbf{B}(G,L,R;D)$. If there exists an involutory automorphism $\sigma$ of G such that $L^\sigma =R$ and $D^\sigma =D^{-1},$ then $\Gamma$ is vertex-transitive. In particular,

• (1) If G is abelian and acts regularly on both biparts of $\Gamma ,$ then $\Gamma$ is vertex-transitive. In other words, bi-Cayley graphs of abelian groups are vertex-transitive.

• (2) If there exists an involutory automorphism $\sigma$ of G such that $L^\sigma =R,$ and the lengths of the orbits of L on [G : R] (or of the orbit of R on [G : L]) are all distinct, then $\Gamma$ is vertex-transitive.

• (3) If the representations of G on the two biparts of $\Gamma$ are equivalent and all suborbits of G relative to L are self-paired, then $\Gamma$ is vertex-transitive.

Next, we give some group-theoretic results, which will be used later. The first one is from [8, Theorems 3.4 and 3.5].

Proposition 2.5

For an odd prime p, let H be a maximal subgroup of $G=\mathrm{GL}(2,p)$ and $H\ne \mathrm{SL}(2,p)$. Then, up to conjugacy, H is isomorphic to one of the following subgroups:

• (i) $D\rtimes \langle b\rangle ,$ where D is the subgroup of diagonal matrices and $b={\left( \begin{array}{ll} 0 &{}\quad 1\\ 1 &{}\quad 0\end{array}\right) };$

• (ii) $\langle a\rangle \rtimes \langle b\rangle$, where $b={\left( \begin{array}{ll} 1&{}\quad 0\\ 0 &{}\quad -1\end{array}\right) }$ and $\langle a\rangle$ is the Singer subgroup of G, defined by a=$\left( \begin{array}{ll} \gamma &{}\quad \delta \theta \\ \delta &{}\quad \gamma \end{array}\right)$$\in G,$ where $F_p^*=\langle \theta \rangle$, $F_{p^2}=F_p(\mathbf{t})$ for $\mathbf{t} ^2=\theta ,$ and $F_{p^2}^*=\langle \gamma +\delta \mathbf{t}\rangle ;$

• (iii) $\langle a\rangle \rtimes D$, where $a={\left( \begin{array}{cc} 1 &{}\quad 1 \\ 0&{}\quad 1 \end{array} \right) };$

• (iv) $H/\langle z\rangle$ is isomorphic to $\mathrm{A}_4\times {\mathbb {Z}}_{\frac{p-1}{2}}$, for $p\equiv \pm 1(\mathrm{mod}\,8);$ $\mathrm{S}_4\times {\mathbb {Z}}_{\frac{p-1}{2}}$ for $p\equiv \pm 1(\mathrm{mod}\,8);$ or $\mathrm{A}_5\times {\mathbb {Z}}_{\frac{p-1}{2}}$ for $p\equiv \pm 1(\mathrm{mod}\,10),$ where $z={\left( \begin{array}{ll} -1 &{}\quad 0\\ 0 &{}\quad -1\end{array}\right) }$, ${\mathbb {Z}}_{\frac{p-1}{2}}=Z(G)/\langle z\rangle ;$

• (v) $H/\langle z\rangle =\mathrm{A}_4\rtimes \langle s\rangle$, $\langle s^2\rangle \le Z(G)/\langle z\rangle$, if $p\equiv 1(\mathrm{mod}\,4)$.

The next proposition is about the subgroups of $\mathrm{PSL}(3,p)$, which is extracted from [2].

Proposition 2.6

For an odd prime p, let ${\overline{G}}=\mathrm{PSL}(3,p)$ and ${\overline{H}}$ a proper subgroup of ${\overline{G}}$. Then, one of the following holds:

• (I) If ${\overline{H}}$ has no nontrivial normal elementary abelian subgroup, then ${\overline{H}}$ is conjugate in $\mathrm{GL}(3,p)/Z(\mathrm{SL}(3,p))$ to one of the following groups:

  • (i) $\mathrm{PSL}(2,7)$, with $p^3\equiv 1 (\mathrm{mod}\,7);$

  • (ii) $\mathrm{A}_6$, with $p\equiv 1, 19(\mathrm{mod}\,30);$

  • (iii) $\mathrm{PSL}(2,5)$, with $p\equiv \pm 1 (\mathrm{mod}\,10);$

  • (iv) $\mathrm{PSL}(2, p)$ or $\mathrm{PGL}(2,p)$ with $p\ge 5$.

• (II) If ${\overline{H}}$ has a nontrivial normal elementary abelian subgroup, then ${\overline{H}}$ is conjugate to a subgroup of one of the following subgroups:

  • (i) ${\mathbb {Z}}_{(p^2+p+1)/(3, p-1)}\rtimes {\mathbb {Z}}_3;$

  • (ii) the subgroup ${\overline{F}}$ of all matrices with only one nonzero entry in each row and column, and ${\overline{F}}$ contains the subgroup ${\overline{D}}$ of all diagonal matrices as a normal subgroup such that ${\overline{F}}/{\overline{D}}\cong \mathrm{S}_3;$

  • (iii) the point or line stabilizer of a given point $\langle (1,0,0)^T\rangle$ or the line $\langle (0,$ $\alpha , \beta )^T\bigm |\alpha , \beta \in F_p\rangle$;

  • (iv) the group ${\overline{M}}$ such that ${\overline{M}}$ contains a normal subgroup ${\overline{N}}\cong {\mathbb {Z}}_3^2$ and ${\overline{M}}/{\overline{N}}$ is isomorphic to $\mathrm{SL}(2,3)$ if $p\equiv 1(\mathrm{mod}\,9)$ or to $Q_8$ if $p\equiv 4$ or $7(\mathrm{mod}\,9)$.

From [14] we can get the following proposition, which is about the finite nonabelian simple groups with subgroups of prime power index.

Proposition 2.7

Let T be a nonabelian simple group with a subgroup $H<T$ satisfying $|T:H|=p^a,$ for p a prime. Then, one of the following holds:

• (i) $T=A_{n}$ and $H=A_{n-1}$ with $n=p^a;$

• (ii) $T=\mathrm{PSL}(n,q)$ and H is the stabilizer of a projective point or a hyperplane in $\mathrm{PG}(n-1,q),$ and $|T:H|=(q^n-1)/(q-1)=p^a;$

• (iii) $T=\mathrm{PSL}(2,11)$ and $H=A_5,$ where T has two subgroups isomorphic to $A_5$, which are not conjugate in T;

• (iv) $T=M_{11}$ and $H=M_{10};$

• (v) $T=M_{23}$ and $H=M_{22}$.

• (vi) $T=\mathrm{PSU}(4,2)$ and H is a subgroup of index 27.

The following results can be quoted from [11], except for Proposition 2.8(6) and (7), which are needed some simple proofs.

Proposition 2.8

For any odd prime p, let

$$\begin{aligned} P=\langle a, b, c\bigm |a^p=b^p=c^p=1, [b, a]=c, [c,a]=[c,b]=1\rangle . \end{aligned}$$

• (1) $\mathrm{Aut}(P)$ consists of the following automorphisms

  $$\begin{aligned} \pi : a\rightarrow a^ib^jc^k, \quad b\rightarrow a^mb^nc^l, \, \mathrm{\,where\, }\, in\ne jm. \end{aligned}$$

  Hence, $|\mathrm{Aut}(P)|=p^3(p-1)^2(p+1)$.

• (2) Let

  $$\begin{aligned} \pi _1: a\rightarrow ac, b\rightarrow b; \quad \pi _2: a\rightarrow a, b\rightarrow bc. \end{aligned}$$

  Then, $Q=\langle \pi _1, \pi _2\rangle \cong {\mathbb {Z}}_p^2,$ $Q\lhd \mathrm{Aut}(P)$ and $\mathrm{Aut}(P)/Q\cong \mathrm{GL}(2, p)$.

• (3) For any $s, t\in F_p^*$, let

  $$\begin{aligned} \phi (s, t): a\rightarrow a^s, b\rightarrow b^t. \end{aligned}$$

  Then, ${\hat{R}}=\langle \phi (s, t)\bigm |s, t\in F_p^*\rangle \cong {\mathbb {Z}}_{p-1}\times {\mathbb {Z}}_{p-1}$. For every subgroup $M\le \mathrm{Aut}(P)$ fixing a subgroup of P of order $p^2$ such that $|M|\bigm |(p-1)^2$, M is contained in a conjugacy of ${\hat{R}}$.

• (4) Let $\phi (s,t)\in {\hat{R}}$, where $(s, t)\ne (1, 1)$. Then,

  $$\begin{aligned} C_{\mathrm{Aut}(P)}(\phi (s,t))={\left\{ \begin{array}{lll} &{}\langle \pi : a\rightarrow a^ i, b\rightarrow b^nc^l\bigm |i, n\in F_p^*, l\in F_p\rangle , &{}\mathrm{\, if\, } s=1, t\ne 1; \\ &{}\langle \pi : a\rightarrow a^ ic^k, b\rightarrow b^n\bigm |i, n\in F_p^*, k\in F_p^*\rangle , &{}\mathrm{\, if\, } s\ne 1, t=1;\\ &{}{\hat{R}} &{}\mathrm{\, if\, } s\ne t ; s, t\ne 1;\\ &{}\langle \pi : a\rightarrow a^ ib^jc^{\frac{1}{2}ij}, b\rightarrow a^ mb^nc^{\frac{1}{2}mn}\bigm |&{}\mathrm{\, if\, } s=t\ge 2.\\ &{}i, j, m, n\in F_p^*, in\ne mj\rangle , &{}\\ \end{array} \right. } \end{aligned}$$

• (5) For $e, s\ne 1$ and $t^{|s|}\le \langle e\rangle$, $C_{\mathrm{Aut}(P)}(\langle \phi (1, e), \phi (s, t)\rangle ) ={\hat{R}}$.

• (6) Let

  $$\begin{aligned} \rho : a\rightarrow a^ib^{j\theta }, b\rightarrow a^jb^i, \end{aligned}$$

  where $F_p^*=\langle \theta \rangle ,F_{p^2}=F_p(\mathbf{t})$ and $\mathbf{t}^2=\theta , \langle i+j\mathbf{t}\rangle =F_{p^2}^*$. Then, $S=\langle \rho \rangle \cong {\mathbb {Z}}_{p^2-1}$. For every subgroup $L\le \mathrm{Aut}(P)$ and L is isomorphic to an irreducible subgroup of $\mathrm{GL}(2,p)$ such that $|L|\bigm |(p+1),$ L is contained in a conjugacy of S.

• (7) Let

  $$\begin{aligned} \mu : a\rightarrow a^ub^{v\theta }c^{l_1},\,b\rightarrow a^vb^uc^{l_2}, \end{aligned}$$

  where $u,v,l_1,l_2\in Z_p,u^2-v^2\theta =1,$ then

  $$\begin{aligned} \mu ^{-1}: a\rightarrow a^ub^{-v\theta }c^{l_3},\, b\rightarrow a^{-v}b^uc^{l_4}, \end{aligned}$$

  where $l_3=\frac{1}{2}uv\theta (u-1-v)+l_2v\theta +l_1u,l_4=\frac{1}{2}uv(u-v\theta -1)-l_2u-l_1v$.

Proof

From [11], it is easy to see that (1)–(5) hold. Thus, we only need to prove (6) and (7).

(6). First, we claim that $|S\cap Q|=1$. If $g\in S\cap Q,$ then $[g,s]=1,$ for $s\in S$. Set $g: a\rightarrow ac^{l_1},\,b\rightarrow bc^{l_2},$ Then,

$$\begin{aligned} \begin{array}{lllll}a^{gs}&{}=&{}a^ib^{j\theta }c^{l_1(i^2-j^2\theta )}&{}=&{}a^{sg}=a^ib^{j\theta }c^{l_1i+l_2j\theta },\\ b^{gs}&{}=&{}a^jb^ic^{l_2(i^2-j^2\theta )}&{}=&{}b^{sg}=a^jb^ic^{l_1j+l_2i}.\end{array} \end{aligned}$$

From these equations, we get $l_1(i^2-j^2\theta )=l_1i+l_2j\theta$ and $l_2(i^2-j^2\theta )=l_1j+l_2i$. If $l_1\ne 0,$ then $(i^2-j^2\theta -i)^2=j^2\theta$. Since $\langle \theta \rangle =F_p^*,$ $\theta$ is not a square element in $F_p^*$. However, $(i^2-j^2\theta -i)^2=j^2\theta$ forces that $\theta$ must be square, a contradiction. Then, $l_1=l_2=0$, that is, $g=1$. Thus, $|S\cap Q|=1$.

Since SQ/Q is isomorphic to a Singer subgroup of $\mathrm{GL}(2,p),$ we have that $p^2-1=|SQ/Q|=|S||Q|/|Q|=|S|$. Since L is isomorphic to an irreducible subgroup of $\mathrm{GL}(2,p)$ and $|L|\bigm |p+1,$ there exists $g\in \mathrm{Aut}(P)$ such that $(LQ/Q)^g\le SQ/Q$ from Proposition 2.5. Then, $L^g\le SQ$. Since $L^g$ must be contained in a $p-$complement of Q,  we have that there exists $h\in SQ$ such that $L^{gh}\le S$.

(7). Let

$$\begin{aligned} \mu ^{-1}: a\rightarrow a^{x_1}b^{y_1}c^{l_3},\,b\rightarrow a^{x_2}b^{y_2}c^{l_4}. \end{aligned}$$

Then,

$$\begin{aligned} \begin{array}{lll}a^{\mu \mu ^{-1}}&{}=&{}(a^ub^{v\theta }c^{l_1})^{\mu ^{-1}} =(a^{x_1}b^{y_1}c^{l_3})^u(a^{x_2}b^{y_2}c^{l_4})^{v\theta }c^{(x_1y_2-y_1x_2)l_1}\\ &{}=&{}a^{x_1u+x_2v\theta }b^{y_1u+y_2v\theta }c^{l_3u+l_4v\theta +x_2v\theta y_1u+(x_1y_2-y_1x_2)l_1}=a\\ b^{\mu \mu ^{-1}}&{}=&{}(a^vb^uc^{l_2})^{\mu ^{-1}} =(a^{x_1}b^{y_1}c^{l_3})^v(a^{x_2}b^{y_2}c^{l_4})^uc^{(x_1y_2-y_1x_2)l_2}\\ &{}=&{}a^{x_1v+x_2u}b^{y_1v+y_2u}c^{l_3v+l_4u+x_2uy_1v+(x_1y_2-y_1x_2)l_2}=b\end{array}. \end{aligned}$$

From these equations, we get,

$$\begin{aligned} \left\{ \begin{array}{lll}x_1u+x_2v\theta =1\\ y_1u+y_2v\theta =0\\ x_1v+x_2u=0\\ y_1v+y_2u=1\\ l_3u+l_4v\theta +x_2v\theta y_1u+(x_1y_2-y_1x_2)l_1=0\\ l_3v+l_4u+x_2uy_1v+(x_1y_2-y_1x_2)l_2=0.\end{array}\right. \end{aligned}$$

Solve it, we can get $x_1=y_2=u,\,y_1=-v\theta ,\,x_2=-v,\,l_3=\frac{1}{2}uv\theta (u-1-v)+l_2v\theta +l_1u,\, l_4=\frac{1}{2}uv(u-v\theta -1)-l_2u-l_1v$. $\square$

For convenience, we use the same notations in Proposition 2.8. Let R be a subgroup of ${\hat{R}}$ such that

$$\begin{aligned} R\cap \langle \phi (1, \theta )\rangle =\langle \phi (1, e)\rangle \cong {\mathbb {Z}}_{r_1}, \quad R/\langle \phi (1, e)\rangle =\langle \overline{\phi (s, t)}\rangle \cong {\mathbb {Z}}_{r_2}, \end{aligned}$$

where $e, s, t\in F_p^*$ and $r_1, r_2$ are divisors of $p-1$. Then,

$$\begin{aligned} R=\langle \phi (1, e), \phi (s, t)\rangle \quad \mathrm{and} \quad \phi (s, t)^{r_2}\in \langle \phi (1, e)\rangle . \end{aligned}$$

In other words, in $F_p^*$,

$$\begin{aligned} |e|=r_1, \quad |s|=r_2, \quad t^{r_2}\in \langle e\rangle , \end{aligned}$$

where if $e=1$, then $R=\langle \phi (s, t)\rangle \cong {\mathbb {Z}}_{r_2}$ and $\langle t\rangle \le \langle s\rangle \cong {\mathbb {Z}}_{r_2}$. Clearly, we may set

$$\begin{aligned} e=\theta ^{\frac{p-1}{r_1}} \quad \mathrm{and} \quad s=\theta ^{\frac{p-1}{r_2}}. \end{aligned}$$

Let $T=\{ t\in F_p^*\bigm |t^{r_2}\le \langle e\rangle \}$. Then, T is a subgroup of order $r_1{\cdot }(r_2, \frac{p-1}{r_1})$. Therefore, we set

$$\begin{aligned} t=(\theta ^ { \frac{p-1}{r_1{\cdot }(r_2, \frac{p-1}{r_1})}})^{r_3}, \end{aligned}$$

where $0\le r_3\le (r_2, \frac{p-1}{r_1})-1$. Clearly, $|R|=r_1{\cdot }r_2$.

Proposition 2.9

For $l=1, 2$, let $R_l=\langle \phi (1, e), \phi (s, t_l)\rangle$ be as above such that $R_1\ne R_2$ and $s\ne 1$. Set $A_l=P\rtimes R_l,$ where P is defined as Proposition 2.8. Suppose that $\gamma$ is an isomorphism from $A_1$ to $A_2$ such that $\gamma (R_1)=R_2$. Then, $\gamma$ satisfies

$$\begin{aligned} \gamma (a)=b^{j_1}, \,\, \gamma (b)=a^{i},\, \, \gamma (\phi (1, e))=\phi (e, 1),\,\, \gamma (\phi (s, t_1))=\phi (t_1, s), \end{aligned}$$

for some $i, j_1\in F_p^*,$ and $R_2=\langle \phi (e, 1), \phi (t_1, s)\rangle$.

Proposition 2.10

For any odd prime p, let

$$\begin{aligned} P=\langle a,b \bigm |a^{p^2}=b^p=1,[b,a]=c=a^p,[c,a]=1,[c,b]=1\rangle . \end{aligned}$$

Set $F^*_{p^2}=\langle \theta \rangle$ and $s=\theta ^p$.

• (1) $\mathrm{Aut}(P)$ consists of the following automorphisms

  $$\begin{aligned} \pi : a\rightarrow a^ib^j, \quad b\rightarrow a^{pk}b, \, \mathrm{\,where\, }\, i\in F^*_{p^2},j,k\in {{\mathbb {Z}}}_p. \end{aligned}$$

  Hence, $|\mathrm{Aut}(P)|=p^3(p-1)$.

• (2) Let

  $$\begin{aligned} \phi (\theta ): a\rightarrow a^\theta , b\rightarrow b; \quad \pi : a\rightarrow ab, b\rightarrow b. \end{aligned}$$

  Then, $\mathrm{Aut}(P)/\mathrm{Inn}(P)\cong \langle \pi \rangle \rtimes \langle \phi (s)\rangle \cong {\mathbb {Z}}_p\rtimes {\mathbb {Z}}_{p-1}$.

3 A reduction

From now on, let p be a prime (in fact, p is odd) and let $\Gamma$ be a semisymmetric graph of order $2p^3$ with prime valency r. Set $A=\mathrm{Aut}(\Gamma )$. Recall that we only need to determine that A acts faithfully and imprimitively on $W(\Gamma )$ and $U(\Gamma )$. Let

$$\begin{aligned} {\mathcal {B}}_W=\{B_1',\cdots ,B_{p^3/m}'\},\quad {\mathcal {B}}_U=\{B_1,\cdots ,B_{p^3/n}\}, \quad m,n\in \{p, p^2\} \end{aligned}$$

be the maximal complete imprimitive block systems of the action of A on $W(\Gamma )$ and $U(\Gamma ),$ respectively.

The main goal in this section is to give a boundary for the valency r.

Lemma 3.1

The valency $r\le p$. Moreover, for any Sylow p-subgroup P of A, $|P|=p^4$ if $r=p$; and $|P|=p^3$ if $r<p$.

Proof

First note that each vertex $w\in W$ (resp $u\in U$) is adjacent to at most one vertex in $B_i$ (resp $B_i'$) for any i; otherwise, r is not a prime.

With the above property, if one of n and m is $p^2$, say $n=p^2$, then $r=d(w)\le |{{\mathcal {B}}_U}|\le p$. Therefore, we only consider the case $m=n=p$.

Let $K_1$ and $K_2$ be the kernel of A on ${\mathcal {B}}_W$ and ${\mathcal {B}}_U,$ respectively. Suppose $K_1\not \le K_2$. Then, $K_1$ is transitive on each $B_i'$ and $(K_1 K_2)/K_2$ is transitive on ${\mathcal {B}}_U$. Noting $A/K_2$ is primitive on ${\mathcal {B}}_U$. This forces $r=p^2$, a contradiction. Therefore, $K_1\le K_2$. Similarly, $K_2\le K_1$. Set $K=K_1=K_2$. Then, ${\overline{A}}=A/K$ is faithful and primitive on both ${\mathcal {B}}_W$ and ${\mathcal {B}}_U$.

Suppose that $K=1$. Then, A can be viewed as a primitive group of degree $p^2$ and an imprimitive group of degree $p^3$. From the proof of [21, Lemma 4.1], we can get $\mathrm{soc}(A)=\mathrm{PSL}(2,8)$ or $\mathrm{soc}(A)={{\mathbb {Z}}}_p^2,$ where $p=3,5,7$ or 11. However, in both cases, $\mathrm{soc}(A)$ induces blocks of length $p^2$ and we exclude them.

Now suppose $K\ne 1$. Then, the subgraph $B_i\cup B_j$ is either a matching or an empty graph. By the well-known O’Nan–Scott Theorem, every primitive group G of degree $p^2$ is almost simple type, product type or affine type. Let $T=\mathrm{soc}({\overline{A}})$.

Case 1: ${\overline{A}}$ is almost simple type.

In this case, T is either $A_{p^2}$ or $\mathrm{PSL}(d,q)$, where $\frac{q^d-1}{q-1}=p^2$, by checking Proposition 2.7. Suppose that $T=A_{p^2}$. Since two representations of A/K on ${\mathcal {B}}_U$ and ${\mathcal {B}}_W$ are equivalent, it follows that the quotient graph ${\mathcal {B}}_U\cup {\mathcal {B}}_W$ is of valency $p^2-1$ and we exclude them.

Suppose $T=\mathrm{PSL}(d,q)$. With the same reason as in last paragraph, we only consider that the two representations of A/K on ${\mathcal {B}}_U$ and ${\mathcal {B}}_W$ are inequivalent. Then, the valency of the quotient graph (=r) is either $q^{d-1}$ or $\frac{q^{d-1}-1}{q-1}$. If $d=2,$ then we have that $p=3$ and $q=2^3$. Obviously, the valency is not a prime, a contradiction. Then, $d\ge 3$. Since $q^{d-1}$ is not a prime, we know $r=\frac{q^{d-1}-1}{q-1}$. Since $\frac{q^d-1}{q-1}=p^2,$ we have

$$\begin{aligned} r=p^2-q^{d-1}=q^{d-2}+q^{d-3}+\cdots +1=\frac{p^2-1}{q}=\frac{(p-1)(p+1)}{q}, \end{aligned}$$

It is known that $(p-1)(p+1)=4\cdot (p-1)/2\cdot (p+1)/2$. If $2\bigm |q$ and $\frac{(p-1)(p+1)}{q}$ is a prime, then $p=3$ and $d=2,$ a contradiction. If $2\not \mid q,$ then it must be $4\mid \frac{(p-1)(p+1)}{q}$. Obviously, $\frac{(p-1)(p+1)}{q}$ is not a prime and we exclude them.

Case 2: ${\overline{A}}$ is either product type or affine type.

Suppose that either ${\overline{A}}\cong (M\times M)\rtimes {{\mathbb {Z}}}_2$ or ${\overline{A}}\cong ({{\mathbb {Z}}}_p\times {{\mathbb {Z}}}_p)\rtimes H,$ where M is a nonabelian simple subgroup of $\mathrm{S}_p$ (so is 2-transitive) and H is an irreducible subgroup of $\mathrm{GL}(2,p)$. In this case, r is a prime. In both cases, the maximal prime divisor of |A| is p,  as desired.

From the above argument, we can get $r\le p$.

To prove the second part, take $w\in W$. Suppose that there exists a of order p in $A_w,$ which fixes a vertex $u\in \Gamma _1(w)$. Since $r\le p$, a fixes $\Gamma _1(w)$ pointwise. Since $a\in A_u$ fixes $w\in \Gamma _1(u)$, a fixes $\Gamma _1(u)$ pointwise. By the connectivity of $\Gamma$, $a=1$. From these properties, we get $p^2\not \mid |A_w|$. Therefore, for any Sylow p-subgroup P of A, $|P|=p^4$ if $r=p$; and $|P|=p^3$ if $r<p$. $\square$

4 The valency $r=p$

In this section, we always assume that p is an odd prime and $\Gamma$ is a semisymmetric graph of order $2p^3$ with valency p. Set $A=\mathrm{Aut}(\Gamma )$. Let P be a Sylow $p-$subgroup of A. First, we prove the following lemma which is about the structure of the Sylow $p-$subgroup of A.

Lemma 4.1

The center Z(P) of P is of order p.

Proof

By Lemma 3.1, $|P|=p^4$, which is transitive on both biparts. Suppose that $|Z(P)|\ge p^3$. Then, P is abelian, contradicting with the faithfulness of P. In what follows, we exclude the case $|Z(P)|=p^2$, resulting in $|Z(P)|=p$.

Suppose $|Z(P)|=p^2$. Noting $Z(P)_w=Z(P)_u=1$ for any w, u, we know that Z(P) induces P-blocks of length $p^2$ in both biparts, denotes by ${\mathcal {B}}_U$ and ${\mathcal {B}}_W$ again. Then, the block graph ${\mathcal {B}}_U\cup {\mathcal {B}}_W$ is isomorphic to $K_{p,p}$. Hence, $P/Z(P)\lesssim \mathrm{Aut}(K_{p,p})\cong (\mathrm{S}_p\times \mathrm{S}_p)\rtimes {{\mathbb {Z}}}_2$ and $P/Z(P)\cong {{\mathbb {Z}}}_p\times {{\mathbb {Z}}}_p$. Then, there exists $\overline{\langle x \rangle }\in P/Z(P)$ such that $\overline{\langle x \rangle }$ acts transitively on ${\mathcal {B}}_U$ and ${\mathcal {B}}_W$. Then, the abelian group $Z(P)\langle x\rangle$ acts transitively on both $W(\Gamma )$ and $U(\Gamma )$. By Proposition 2.4, the graph is symmetric, a contradiction. $\square$

In [25], we can get the group of order $p^4$ with $Z(P)\cong {{\mathbb {Z}}}_p$, which are listed as follows:

$P_1=\langle a,b \bigm |a^{p^2}=b^p=c^p=1,[a,b]=1,[c,a]=1,[c,b]=a^{ip}\rangle ,$ where $i=1$ or v,  v is square nonresidue of fixed mode p.

$$\begin{aligned} P_2= & {} \langle a,b \bigm |a^{p^2}=b^p=c^p=1,[a,b]=c,[c,a]=a^p,[c,b]=1\rangle ,\\ P_3= & {} \langle a,b \bigm |a^9=c^3=1,a^3=b^3,[a,b]=c,[c,a]=1,[c,b]=a^{-3}\rangle \,(p=3),\\ P_4= & {} \langle a,b \bigm |a^p=b^p=c^p=d^p=1,[a,b]=c,[c,a]=1,[c,b]=d\rangle \,(p\ge 5). \end{aligned}$$

Now we are ready to determine the graph $\Gamma$ by the structure of the Sylow $p-$subgroup P of A.

Lemma 4.2

The graph $\Gamma \cong \Gamma _p$ and $N\rtimes (\langle e\rangle R)\le \mathrm{Aut}(\Gamma ),$ where N is identified with the translation subgroup of $\mathrm{AGL}(3,p),$

$$\begin{aligned} e={\left( \begin{array}{lll} 1 &{}\quad 2&{}\quad 2\\ 0&{}\quad 1&{}\quad 2\\ 0&{}\quad 0&{}\quad 1\end{array}\right) }, R=\langle {\left( \begin{array}{lll} s^2t^{-1} &{}\quad 0&{}\quad 0\\ 0&{}\quad s&{}\quad 0\\ 0&{}\quad 0&{}\quad t\end{array}\right) }\rangle \cong {\mathbb {Z}}_{p-1}\times {\mathbb {Z}}_{p-1}. \end{aligned}$$

Proof

The proof is divided into the following three steps:

Step (1): Determination of P, $P_w$ and $P_u$

From Proposition 2.3, the graph $\Gamma \cong {\mathbf {B}}(P, P_w, P_u; P_uP_w),$ where $(u,w)\in E(\Gamma )$. According to Lemma 3.1, we can get $|P_u|=|P_w|=p$. For the above groups $P_1,P_2,$ and $P_3,$ we have $a\notin \langle x,y\rangle ,$ where $x,y\in P$ and $o(x)=o(y)=p$. Then, the bi-coset graph is not connected by Proposition 2.3. Thus, $P\cong P_4$, that is,

$$\begin{aligned} \begin{array}{lll}P&{}=&{}\langle a,b \bigm |a^p=b^p=c^p=d^p=1,[a,b]=c,[c,a]=1,[c,b]=d\rangle \\ &{}=&{}(\langle a\rangle \times \langle c\rangle \times \langle d\rangle )\rtimes \langle b\rangle =(\langle t_{v_1}\rangle \times \langle t_{v_2}\rangle \times \langle t_{v_3})\rangle \rtimes \langle e\rangle ,\end{array} \end{aligned}$$

where $t_{v_1}=t_{(1,0,0)},\,t_{v_2}=t_{(0,1,0)},\,t_{v_3}=t_{(0,0,1)},\,e=\left( \begin{array}{ccc} 1&{}\quad 2&{}\quad 2\\ 0&{}\quad 1&{}\quad 2\\ 0&{}\quad 0&{}\quad 1\\ \end{array} \right)$ and $p\ge 5$. Meanwhile,

$$\begin{aligned} t_{v_1}^e=t_{v_1}t_{v_2}^2t_{v_3}^2,\, t_{v_2}^e=t_{v_2}t_{v_3}^2,\, t_{v_3}^e=t_{v_3}. \end{aligned}$$

Let $N=\langle t_{v_1}\rangle \times \langle t_{v_2}\rangle \times \langle t_{v_3}\rangle$. In the following, we will determine $P_w$ and $P_u$ by discussing the centralizers of $P_w$ and $P_u$. Suppose that $|C_P(P_w)|\ne p^3$ and $|C_P(P_u)|\ne p^3$. Then, $P_w\cap N=P_u\cap N=1$. Since P is transitive on both biparts, then N acts transitively on $U(\Gamma )$ and $W(\Gamma )$. In the case, the graph is vertex-transitive from Proposition 2.4.

Without loss of generality, suppose that $|C_P(P_u)|=p^3$ and $|C_P(P_w)|=p^2$. According to the structure of the group P,  we can get $|C_P(t_{v}e^i)|=p^2,$ where $t_v\in N$ and $i\in F_p^*$. It is easy to verify that $t_{v}e\notin \langle t_{l}e \rangle$ for $t_v, t_l\in N$ and $t_v\notin t_l$. Let $\Delta =\{ \langle t_ve\rangle |t_v\in N\}$. Then, $\Delta$ is set of subgroup whose centralizers is order $p^2$ in group P. Choose a bijection $\pi (n)$ between P and P, such that $\pi (n): t_{v_1}\rightarrow t_{v_1}, t_{v_2}\rightarrow t_{v_2}, t_{v_3}\rightarrow t_{v_3}, e\rightarrow ne,$ where $n\in N$. Since $t_{v_1}^{ne}=t_{v_1}^e, \,t_{v_2}^{ne}=t_{v_2}^e,\, t_{v_3}^{ne}=t_{v_3}^{e},$ we have that $\pi (n)\in \mathrm{Aut}(P)$. So, we can set $P_w=\langle e\rangle$. By Proposition 2.3,

$$\begin{aligned} \Gamma \cong {\mathbf {B}}(P, \langle e\rangle , P_u; P_u\langle e\rangle ). \end{aligned}$$

It is easy to verify that ([8]) $N_{\mathrm{GL}(3,p)}(\langle e\rangle )=\langle e,e' \rangle \rtimes H,$ where

Consider the action of $N_{\mathrm{GL}(3,p)}(\langle e\rangle )$ on all the element of order p in N. Then,

By Proposition 2.3, in the isomorphism sense, let $P_u=\langle t_{v_1}\rangle , \langle t_{v_2}\rangle$ or $\langle t_{v_3}\rangle$. Then, the graph $\Gamma$ is isomorphic to one of the following bi-coset graphs.

$$\begin{aligned} \begin{array}{lll}X_1={\mathbf {B}}(P, \langle e\rangle ,\langle t_{v_1}\rangle ; D_1), where\, D_1=\langle t_{v_1}\rangle t_{v_2}^it_{v_3}^je^k\langle e\rangle \\ X_2={\mathbf {B}}(P,\langle e\rangle , \langle t_{v_2}\rangle ; D_2), where\, D_2=\langle t_{v_2}\rangle t_{v_1}^it_{v_3}^je^k\langle e\rangle \\ X_3={\mathbf {B}}(P, \langle e\rangle , \langle t_{v_3}\rangle ; D_3), where\, D_3=\langle t_{v_3}\rangle t_{v_1}^it_{v_2}^je^k\langle e\rangle . \end{array} \end{aligned}$$

Since $t_{v_2},t_{v_3}\in P'$ and $P'\le \Phi (P),$ we have that $\langle D_2^{-1} D_2\rangle < P$ and $\langle D_3^{-1}D_3\rangle < P$. From Proposition 2.3, the bi-coset graphs $X_2$ and $X_3$ are not connected. We only need to discuss $X_1={\mathbf {B}}(P, \langle e\rangle , \langle t_{v_1}\rangle ; \langle t_{v_1}\rangle t_{v_2}^it_{v_3}^je^k\langle e\rangle )$.

Step (2): The possible bi-coset graph isomorphic to $\Gamma$.

Let

$$\begin{aligned} W(\Gamma ):=[P:P_w]= & {} \{P_wt_{v_1}^it_{v_2}^jt_{v_3}^k\bigm |i,j,k\in {{\mathbb {Z}}}_p\},\\ U(\Gamma ):=[P: P_u]= & {} \{ P_ut_{v_2}^xt_{v_3}^ye^z\bigm |x,y,z\in {{\mathbb {Z}}}_p\}. \end{aligned}$$

For any $l\in Z_p,$ we have

$$\begin{aligned} D=D_1(i.j)=\langle t_{v_1}\rangle t_{v_2}^it_{v_3}^je^k\langle e\rangle =\langle t_{v_1}\rangle t_{v_2}^it_{v_3}^j\langle e\rangle . \end{aligned}$$

Choose $h=||i^2j^{-1},i,j||\in H$. Under the conjugacy action, h fixes $P_w$ and $P_u,$ and maps $D_1(1,1)$ to $D_1(i,j)$. Therefore, h induces an isomorphism between these two bi-coset graphs. So, we just need to consider two graphs $X_1(0,0)={\mathbf {B}}(P, \langle e\rangle , \langle t_{v_1}\rangle ; D_1(0,0))$ and $X_1(1,1)={\mathbf {B}}(P, \langle e\rangle , \langle t_{v_1}\rangle ; D_1(1,1))$.

Then, for any $i,j,k\in Z_p,$ the edge set is the following two cases

$$\begin{aligned} \begin{array}{lll}\langle t_{v_1}\rangle \langle e\rangle t_{v_1}^it_{v_2}^jt_{v_3}^k=\{\langle t_{v_1}\rangle e^lt_{v_1}^it_{v_2}^jt_{v_3}^k \bigm |l\in {{\mathbb {Z}}}_p\} =\{\langle t_{v_1}\rangle t_{v_2}^{j-2il}t_{v_3}^{2il^2-2jl+k}e^l\bigm |l\in {{\mathbb {Z}}}_p\}.\\ \langle t_{v_1}\rangle t_{v_2}t_{v_3}\langle e\rangle t_{v_1}^it_{v_2}^jt_{v_3}^k=\{\langle t_{v_1}\rangle e^lt_{v_1}^it_{v_2}^jt_{v_3}^k \bigm |l\in {{\mathbb {Z}}}_p\} =\{\langle t_{v_1}\rangle t_{v_2}^{j-2il+1}t_{v_3}^{2il^2-2jl+k+1}e^l\bigm |l\in {{\mathbb {Z}}}_p\}.\end{array} \end{aligned}$$

Therefore, the neighborhood of $\langle e\rangle t_{v_1}^it_{v_2}^jt_{v_3}^k$ in the bi-coset graph is

$$\begin{aligned} \{\langle t_{v_1}\rangle t_{v_2}^{j-2il}t_{v_3}^{2l^2i-2jl+k}e^l\bigm |l\in {{\mathbb {Z}}}_p\}. \end{aligned}$$

There is a mapping $\lambda$ from $X_1(0,0)$ to $X_1(1,1)$ such that the restriction of $\lambda$ on $W(X_1(0,0))$ to $W(X_1(1,1))$ is identity map, and on $U(X_1(0,0))$ to $U(X_1(1,1))$ is $t_{v_2}t_{v_3}$ multiplying each element in $U(X_1(0,0))$. It is easy to verify that $\lambda$ is an isomorphism between these two graphs. Thus, we may just consider the graph $X_1(0,0)={\mathbf {B}}(P, \langle e\rangle ,\langle t_{v_1}\rangle ; D_1(0,0))$.

Therefore, the neighborhood of $\langle e\rangle t_{v_1}^it_{v_2}^jt_{v_3}^k$ in the bi-coset graph is

$$\begin{aligned} \{\langle t_{v_1}\rangle t_{v_2}^{j-2il}t_{v_3}^{2l^2i-2jl+k}e^l\bigm |l\in {{\mathbb {Z}}}_p\}. \end{aligned}$$

Step (3): the bi-coset graph $X_1(0,0)$ is semisymmetric.

Let G be a subgroup of A preserving both biparts and Q the Sylow $p-$subgroup of G. First, we claim that Q is of order $p^4,$ that is, Q is conjugate to P. Suppose that $P\lneq Q\in \mathrm{Syl}_p(G)$. Then, $|Q|\ge p^5$ and $|Q_u|\ge p^2$. By the same argument as Lemma 3.1, Q acts unfaithfully on $V(\Gamma ),$ a contradiction.

In the following argument, we can get the graph $X_1(0,0)$ is semisymmetric. Suppose that $\Gamma$ is symmetric. Then, there exists $\sigma \in \mathrm{Aut}(G)$ such that $\langle t_{v_1}\rangle ^\sigma =\langle e\rangle$. Hence, $C_G( t_{v_1})\cong C_G( e)$. Since $C_G( t_{v_1})\ge C_P( t_{v_1})={{\mathbb {Z}}}_p^3\cong N,$ it follows $C_G( e)\ge {{\mathbb {Z}}}_p^3\cong N_1,$ where $N_1 \cong {{\mathbb {Z}}}_p^3=\langle e,t_{v_3},t\rangle$ and $t\not \in P$. Then, $\langle P,t\rangle \le C_G( t_{v_3})$. Furthermore, $\langle t_{v_3}\rangle \le Z(C_G(t_{v_3})),$ and $\langle t_{v_3}\rangle$ induces a $p-$block system ${\mathcal {B}}_U$ on $U(\Gamma )$. Considering the action of $C_G( t_{v_3})$ on ${\mathcal {B}}_U$ with the kernel K. The kernel K induces a $p-$block system ${\mathcal {B}}_W$ on $W(\Gamma )$. Then, $C_G(t_{v_3})/K$ acts faithfully on quotient graph ${\mathcal {B}}_U\cup {\mathcal {B}}_W$. The block graph ${\mathcal {B}}_U\cup {\mathcal {B}}_W$ is an edge-transitive connected graph of valency p. Then, $p\parallel |K|$. Therefore, $\langle t_{v_3}\rangle$ is a Sylow $p-$subgroup of K,  and $\langle t_{v_3}\rangle$ char K.

According to $|PK/K|=p^3,$ it follows that $C_G(t_{v_3})/K$ is either doubly transitive or $PK/K\unlhd C_A(t_{v_3})/K$ by [7, Theorem 3]. If $C_G(t_{v_3})/K$ is doubly transitive, the valency of block graph ${\mathcal {B}}_U\cup {\mathcal {B}}_W$ is not p. Then, $PK/K\unlhd C_A(t_{v_3})/K$. For any $Kg \in C_G( t_{v_3})/K$ and $Kx_1\in PK/K,$ there exists $Kx_2\in PK/K$ such that $Kg^{-1}Kx_1Kg=Kg^{-1}x_1g=Kx_2$. Then, $g^{-1}x_1gx_2^{-1}\in \langle t_{v_3}\rangle$. So, $g^{-1}x_1g\in P,$ and $P^g=P,$ where $g\in G$. In other words, $P\unlhd C_G(t_{v_3})$ and $P^t=P$. Therefore, $|P\langle t\rangle |=p^5,$ a contradiction. Thus, the graph $X_1(0,0)$ is semisymmetric.

By labeling $P_wt_{v_1}^it_{v_2}^jt_{v_3}^k,$ by (i, j, k) and $P_ut_{v_2}^xt_{v_3}^ye^z$ by [x, y, z],  that is

$$\begin{aligned} W(\Gamma )=\{(i,j,k)\bigm |i,j,k\in {{\mathbb {Z}}}_p\}, \, U(\Gamma )=\{[x, y, z]\bigm |x,y, z\in {{\mathbb {Z}}}_p\}, \end{aligned}$$

we get the respective edge set of the graph

$$\begin{aligned}&E=\{ \{(i,j,k),[x,y,z]\}\bigm |x=j-2il,y=2l^2i-2jl+k,z\\&=l\bigm |i,j,k,x,y,z,l\in {{\mathbb {Z}}}_p\}, \end{aligned}$$

where $p\ge 5$. Therefore, $\Gamma \cong \Gamma _p$. $\square$

Lemma 4.3

The full automorphism group $\mathrm{Aut}(\Gamma _p)$ of $\Gamma _p$ is isomorphic to ${\mathbb {Z}}_p^3\rtimes ({\mathbb {Z}}_p\rtimes {\mathbb {Z}}_{p-1}^2)$.

Proof

Note that A is imprimitive on both $W(\Gamma )$ and $U(\Gamma ),$ we may assume that ${\mathcal {B}}_U=\{B_0,B_1,\cdots ,B_{m-1}\}$ is a maximal imprimitive complete block system of A on $U(\Gamma ),$ where $m|p^3$ and $|B_i|=\frac{p^3}{m},i=0,1,\cdots , m-1$. Let K be the kernel of A acting on ${\mathcal {B}}_U$.

Suppose that $|{\mathcal {B}}_U|=p^2$. In this case, A/K acting primitively on ${\mathcal {B}}_U$ as ${\mathcal {B}}_U$ is maximal. Then, either $K=1$ or $p\Vert |K|$. If $K=1,$ A is a primitive group of degree $p^2$ and faithful transitive group of degree $p^3$. By checking [21, Lemma 4.1], $p^4\not \mid |A|,$ a contradiction. If $p\Vert |K|,$ the graph is $K-$regular covering of a $A/K-$graph so that $K=Z_p$. If $\mathrm{soc}(A/K)$ is nonabelian, then the valency of the quotient graph is not an odd prime by the above argument of Lemma 3.1. If $\mathrm{soc}(A/K)\cong M\times M,$ where M is a nonabelian simple group of the degree p,  $p^3\not \mid |A/K|$ Then, $p^4\not \mid |A|$. If $A/K\cong ({{\mathbb {Z}}}_p\times {{\mathbb {Z}}}_p)\rtimes H,$ where H is an irreducible subgroup of $\mathrm{GL}(2,p)$. But H is not permutation group of degree p. This is impossible because the valency of the quotient graph is p.

Suppose that $|{\mathcal {B}}_U|=p$. Then, $p\Vert |A/K|$. If K is not transitive on $W(\Gamma ),$ then the graph is $K-$regular covering of a $A/K-$graph and $|K|=p^2$. Then, $p^3\Vert |A|$. This is contrary to $p^4\Vert A|$. Then, K is transitive on $W(\Gamma )$. In this case, $K\cap P=N=\langle t_{v_1}\rangle \times \langle t_{v_2}\rangle \times \langle t_{v_3}\rangle$. It follows that N induces a $p^2-$block on $U(\Gamma )$ and acts transitively on $W(\Gamma )$. There exists $B_i\in {\mathcal {B}}_U$ such that $\langle t_{v_2}\rangle \times \langle t_{v_3}\rangle \cong N^{B_i}\le K^{B_i}$ and $\langle t_{v_1} \rangle \le K_{(B_i)}$. We claim that $N\unlhd A$.

First, we prove that $\langle t_{v_1}\rangle \unlhd K$. Suppose that $K_{(B_i)}=\langle t_{v_1} \rangle R,$ where $p\not \mid |R|$. Then, $R\le K_u,$ where $u\in B_i$. Since $|\Gamma _1(u)|=p,$ for any $u\in B_i,$ there exists $w\in \Gamma _1(u)$ such that $R\le A_w$ where $(w,u)\in E(\Gamma )$. Since $R\le K$ and $\Gamma _1(w)\cap B_j=1,$ where $j=1,\cdots ,p,$ we get R fixes points in $\Gamma _1(w)$. Since $|B_i|=p^2$ and $|\Gamma _1(w)|=p,$ we have R fixes each vertex in $W(\Gamma )$. Then, $R=1$ and $K_{(B_i)}=\langle t_{v_1} \rangle$. Thus, $\langle t_{v_1}\rangle \unlhd K$.

Second, we prove that $|K|\bigm |p^3(p-1)$. We claim that $K_u$ is faithful on $\Gamma _1(u),$ where $u\in B_i$. If not faithful, let $K_1$ be the kernel of this action, that is, $K_1\le A_w,$ where $(u,w)\in E(\Gamma )$. By the above result, $\langle t_{v_1}\rangle \unlhd K_u$ and $K_1\unlhd K_u$, we have $\langle t_{v_1} \rangle \le C_{K_u}(K_1)$. Since $u\in \Gamma _1(w)$ and $\Gamma _1(w)\cap B_j=1,$ where $j=1,\cdots ,p,$ we get $K_1$ fixes every vertex in $\Gamma _1(w)$. Then, $K_1$ is the kernel of $A_w$ on $\Gamma _1(w)$. Hence, $K_1\unlhd A_w$. Since $\langle e\rangle \le A_w$ as the above graph, $\langle e\rangle \le N_A(K_1)$. Since $P=\langle t_{v_1},e \rangle$ and P is transitive on $U(\Gamma ),$ we get $K_1$ fixes $W(\Gamma )$ pointwise. This is impossible because A is faithful on $W(\Gamma )$. Then, $K_u\lesssim {{\mathbb {Z}}}_p\rtimes {{\mathbb {Z}}}_{p-1}$. It forces that $|K|\bigm |p^3(p-1)$.

By Sylow theorem, we have $N\unlhd K$. Moreover, $N~\mathrm{char}~K$. Then, $N\unlhd A$. Since N is regular and normal in A,  we have that $A=N\rtimes H$ with H a reducible subgroup of $\mathrm{GL}(3,p)$. Moreover, by the same argument as Step(2) of proof in [11, Theorem 3.1], if $\langle e\rangle \unlhd H$, then

and N is the translation subgroup of $\mathrm{AGL}(3,p)$.

If $\langle e\rangle \ntrianglelefteq H$, then H contains a subgroup, say Q, such that $Q\cong \mathrm{SL}(2,p)$. Since $p\Vert |H|,$ we have that H is transitive on $\Gamma (w)$. By [14, Theorem 1], $p=5,7$ or 11,  and by Magma [3], $A=N\rtimes (\langle e\rangle \rtimes R),$ where N, e and R are described above. It implies that $\langle e\rangle \unlhd H$, a contradiction. $\square$

5 The valency $r < p$

In this section, we deal with the case $r<p$ for each odd prime $r\ge 3$ and prime p. The following lemma implies that $r\bigm |p-1$ or $r\bigm |p+1$.

Lemma 5.1

Let P be a Sylow $p-$subgroup of A. Then, $|P|=p^3$ and $A=P\rtimes R$ with $|R|=r$. Particularly, if $r \bigm |p-1,$ then $\exp (P)=p$ or $p^2;$ if $r\bigm |(p+1)$, then $\exp (P)=p$.

Proof

Let P be a Sylow $p-$subgroup of A. By Lemma 3.1, $|P|=p^3$ and P is transitive on both biparts. Moreover, P is nonabelian; otherwise, $\Gamma$ is vertex-transitive by Proposition 2.4.

Since A acts imprimitively on both $U(\Gamma )$ and $W(\Gamma ),$ we may let ${\mathcal {B}}_W$ be a maximal imprimitive complete block system of A on $W(\Gamma ),$ where $|{\mathcal {B}}_W|=p$ or $p^2$. Let K be the kernel of A on ${\mathcal {B}}_W$. In the following, we divide the proof into two cases:

Case 1: $|{\mathcal {B}}_W|=p$.

In the case, $p^2\bigm |\bigm ||K|$. Then, K induces a complete $p^2-$block system on both biparts. Again we know that the subgraph $B_i\cup B_j$ is either a matching or an empty graph, which implies that $\Gamma$ is $K-$regular covering of the quotient graph, that is, $|K|=p^2$. Now the quotient graph ${\mathcal {B}}_W\cup {\mathcal {B}}_U$ is a symmetric graph of order 2p and the valency is a prime $r<p$. By Burnside Theorem, A/K is either double transitive or affine. The first case implies $r=p-1$, a contradiction. Therefore, $A/K\cong {{\mathbb {Z}}}_p\rtimes {{\mathbb {Z}}}_r,$ where $r\bigm |p-1$ and r is an odd prime. In this case, $A=P\rtimes A_w$, where $A_w\cong {{\mathbb {Z}}}_r$.

Case 2: $|{\mathcal {B}}_W|=p^2$.

In this case, A/K acting primitively on ${\mathcal {B}}_W$ as ${\mathcal {B}}_W$ is maximal. Then, either $K=1$ or p||K|. With the completely same arguments in the proof of Lemma 3.1, we know $K\ne 1$. Then, K induces a complete $p-$block system and the graph is $K-$regular covering of the quotient graph so that $K\cong {{\mathbb {Z}}}_p$. Now, the quotient graph ${\mathcal {B}}_W\cup {\mathcal {B}}_U$ is symmetric graph of order $2p^2$ with prime valency r, and A/K is a primitive group of degree $p^2$, one of whose Sylow $p-$subgroup is $P/K\cong {{\mathbb {Z}}}_p^2$.

Checking O’Nan–Scott Theorem again, we get that A/K can be neither almost simple nor product type. The only possibility is $A/K\cong ({\mathbb {Z}}_p\times {\mathbb {Z}}_p)\rtimes H,$ where H is an irreducible subgroup of $\mathrm{GL}(2,p)$ and $p\not \mid |H|$. Clearly P/K is the unique Sylow p-subgroup of ${\overline{A}}$. Now $A=P\rtimes H$ and $H\cong {{\mathbb {Z}}}_r$. By Proposition 2.5, up to conjugacy, H is isomorphic to a subgroup of order r of Singer subgroup of $\mathrm{GL}(2,p)$ with $r\bigm |(p+1)$. $\square$.

5.1 The valency $r\bigm |(p-1)$

Be Lemma 5.1, there are two cases for $r<p$. In this subsection, we will treat the case $r\bigm |(p-1)$.

Lemma 5.2

Suppose that $r\bigm |(p-1)$. Then, $\Gamma \cong \Gamma _r(m,l)$ and $\mathrm{Aut}(\Gamma _r(m,l))=P\rtimes R,$ where P is a nonabelian subgroup of order $p^3,\exp (P)=p$ and $R={\mathbb {Z}}_r$.

Proof

Let P be a Sylow $p-$subgroup of A as above. Then, by Lemma 5.1, we have that $\exp (P)=p$ or $p^2$.

Suppose that $\exp (P)=p^2$. Then,

$$\begin{aligned} P=\langle a,b \bigm |a^{p^2}=b^p=1,[b,a]=c=a^p,[c,a]=1,[c,b]=1\rangle . \end{aligned}$$

By Proposition 2.8, there exists $a\in \mathrm{Aut}(P)$ such that $R<{\hat{R}}^a,$ where ${\hat{R}}=\langle \theta \rangle$ has the form in Proposition 2.8(3) and ${\mathbb {Z}}_{p^2}^*=\langle \theta \rangle$. Without loss of generality, suppose that $R<{\hat{R}}$. Then, $R=\langle \phi (t)\rangle ,$ where $t=\theta ^{p\frac{p-1}{r}}$.

According to $A=P\rtimes R$ and $(|P|,|A/P|)=1,$ it follows that $A_w$ and $A_u$ are conjugate in A, where $w\in W(\Gamma )$ and $u\in U(\Gamma )$. Without loss of generality, let $A_w=A_u=R$. By Proposition 2.8 and Lemma 5.1, $R=\langle \phi (t)\rangle ,$ where $t=\theta ^{p\frac{p-1}{r}}$ and R is the subgroup of order r in ${\hat{R}}$. Clearly, $\Gamma$ is isomorphic to $Z={\mathbf {B}}(A, R, R; D),$ where $D=RdR$ and $d\in P$. Set

$$\begin{aligned} W(\Gamma )=[A:A_u]=\{Ra^ib^jc^k\},\,U(\Gamma )=[A:A_w]=\{Ra^xb^yc^z\}, \end{aligned}$$

where $i,j,k,x,y,z\in {{\mathbb {Z}}}_p$.

According to ${\hat{R}}<N_A(\langle b,c\rangle )\cap N_A(\langle a,c\rangle ),$ for any $i,j,k\in F_p,$ $\langle R,a^ic^k\rangle <A$ and $\langle R,b^jc^k\rangle <A$. By Proposition 2.3, the bi-coset graph Z is not connected.

Assume that $d=a^{m_1}b^{m_2}c^{m_3},$ where $m_1,m_2,m_3\in {{\mathbb {Z}}}_p, m_1m_2\ne 0$. Since $[R,{\hat{R}}]=1,$ there exists $\phi (m_1+pm_3)\in {\hat{R}}$ such that

$$\begin{aligned} (Rab^{m_2}R)^{\phi (m_1+pm_3)}=Ra^{m_1}b^{m_2}c^{m_3}R. \end{aligned}$$

By Proposition 2.3, $\phi (m_1+pm_3)$ can induce an isomorphism from $X_1={\mathbf {B}}(A, R, R;D(1m))$ to $X_2={\mathbf {B}}(A, R, R; D(2m)),$ where $D(1m)=Rab^{m_2}R$ and $D(2m)=Ra^{m_1}b^{m_2}c^{m_3}R$. But $b\notin \langle R(ab^m)^{-1}RRab^mR\rangle ,$ that is, $\langle R(ab^m)^{-1}RRab^mR\rangle <A$. Then, the graph is not connected, a contradiction.

Suppose that $\exp (P)=p$. Take

$$\begin{aligned} P=\langle a,b\bigm |a^p=b^p=1,[b,a]=c,[c,a]=[c,b]=1\rangle . \end{aligned}$$

By Proposition 2.8, there exists $a\in \mathrm{Aut}(P)$ such that $R<{\hat{R}}^a,$ where ${\hat{R}}=\langle \phi (1, \theta ), \phi (\theta , 1)\rangle$ where $|\theta |=p-1$ as defined in Proposition 2.8. The following elements are the same as the form in Proposition 2.8. Without loss of generality, we may assume that $R<{\hat{R}}$. There are $r+1$ subgroups of order r in ${\hat{R}}$. Moreover, $\langle \phi (t, 1)\rangle$ and $\langle \phi (t^m,t)\rangle$ are all of subgroups of order r in ${\hat{R}},$ where $t=\theta ^{\frac{p-1}{r}},m\in {{\mathbb {Z}}}_r$. Choose $\pi \in \mathrm{Aut}(P),$ where $\pi : a\rightarrow b, b\rightarrow a$. Since $\phi (t, 1)^{\pi }=\phi (1,t),$ we only need to discuss $\phi (t^m,t)$ in the sense of isomorphism. Let $A_1=P\rtimes \langle \phi (t^m,t)\rangle$ and $A_2=P\rtimes \langle \phi (t^{(-m)},t)\rangle$. By Proposition 2.9, $A_1\cong A_2$. Thus, we only need to discuss $\phi (t^m,t),$ where $m=0,1,\cdots , \frac{r-1}{2}$.

The proof is divided into the following three steps:

Step (1): The possible edge-transitive bi-coset graph isomorphic to $\Gamma$

According to $A=P\rtimes R$ and $(|P|,|A/P|)=1,$ it follows that $A_w$ and $A_u$ are conjugate in A, where $w\in W(\Gamma )$ and $u\in U(\Gamma )$. Without loss of generality, let $A_w=A_u=R$. By the above argument, $R_m=\langle \varphi (t^m,t)\rangle ,$ where $m\in {{\mathbb {Z}}}_r$ and $R_m$ is a subgroup of order r in ${\hat{R}}$. Then, we can get the corresponding bi-coset graph $Z_m={\mathbf {B}}(A, R_m, R_m; D),$ where $D=R_mdR_m$ and $d\in P$. Set

$$\begin{aligned} W(\Gamma )=[A:A_u]=\{R_ma^ib^jc^k\},\quad U(\Gamma )=[A:A_w]=\{R_ma^xb^yc^z\}, \end{aligned}$$

where $i,j,k,x,y,z\in {{\mathbb {Z}}}_p$.

Since ${\hat{R}}\le N_A(\langle c\rangle )\cap N_A(\langle a,c\rangle )\cap N_A(\langle b,c\rangle ),$ it follows that $\langle R_m,a^ic^k\rangle <A$ and $\langle R_m,b^jc^k\rangle <A$ with $i,j,k\in F_p$. if $d=a^ic^k$ or $b^jc^k,$ and $i,j,k\in F_p,$ $\langle DD^{-1}\rangle \ne A$. By Proposition 2.3, the bi-coset graph Z is not connected. Thus, $d=a^ib^jc^k$ with $i,j,k\in {{\mathbb {Z}}}_p$ and $i,j\ne 0$.

Suppose that $m=0$. For any $i,j,k\in {{\mathbb {Z}}}_p,$ there exists $\pi (i,j,k)\in N_{\mathrm{Aut}(P)}(\varphi (1,t))$ such that $(R_0abR_0)^{\pi (i,j,k)}=R_0a^ib^jc^kR_0$ with $\pi (i,j,k): a\rightarrow a^i, b\rightarrow b^jc^k$ by Proposition 2.8(5). By Proposition 2.3, we only need to discuss the bi-coset graph $X_{D_0(0)}=\mathbf{B }(A, R_0, R_0; R_0abR_0)$. Since $b\notin \langle DD^{-1}\rangle ,$ we have that$\langle DD^{-1}\rangle \ne A$. It forces that $X_{D_0(0)}$ is not connected, a contradiction.

Suppose that $m=1$. For any $i,j,k\in F_p,$ there exists $\pi (i,j,k)\in N_{\mathrm{Aut}(P)}(\varphi (t,t))$ such that $(R_1abR_1)^{\pi (i,j,k)}=R_1a^ib^jc^kR_1,$ with $\pi (i,j_,k): a\rightarrow b^{2i^{-1}k-j}, b\rightarrow a^{i}b^{2j-2i^{-1}k}c^{ij-k}$ by Proposition 2.8(5). Similarly, we only need to discuss the following bi-coset graph $X_{D_1(0)}=\mathbf{B }(A, R_1, R_1; R_1abR_1)$. However, $a^ib^j\notin \langle DD^{-1}\rangle$ with $i\ne j$ implies that $\langle DD^{-1}\rangle \ne A$. This forces that $X_{D_1(0)}$ is not connected, a contradiction.

Suppose that $m\ne 0,1$. Since $[R_m,{\hat{R}}]=1,$ there exists $\varphi (j,i)\in {\hat{R}}$ such that

$$\begin{aligned} (R_mabc^{ki^{-1}j^{-1}}R_m)^{\varphi (j,i)}=R_ma^ib^jc^kR_m. \end{aligned}$$

Then, we only need to discuss

$$\begin{aligned} X_{D_m(l)}=\mathbf{B }(A, R_m, R_m; R_mabc^lR_m), \end{aligned}$$

where $l\in F_p,$ and $m\ne 0,1$.

For any $l\in {{\mathbb {Z}}}_p,$

$$\begin{aligned} D_m(l)= R_mabc^lR_m=\{R_ma^{t^s}b^{t^{ms}}c^{lt^{s(m+1)}}|s\in {{\mathbb {Z}}}_r\}. \end{aligned}$$

Furthermore, for any $i,j,k\in {{\mathbb {Z}}}_p,$ we can get

$$\begin{aligned} \begin{array}{lll}R_mabc^lR_ma^ib^jc^k&{}=\{R_ma^{t^s}b^{t^{ms}}c^{lt^{s(m+1)}}a^ib^jc^k|s\in {{\mathbb {Z}}}_r\}\\ &{}=\{R_ma^{t^s+i}b^{t^{ms}+j}c^{k+it^{ms}+lt^{s(m+1)}}|s\in {{\mathbb {Z}}}_r\}.\end{array} \end{aligned}$$

Then, for any $R_ma^ib^jc^k$ in $W(\Gamma ),$ the neighborhood is

$$\begin{aligned} \{R_ma^{t^s+i}b^{t^{ms}+j}c^{k+it^{ms}+lt^{s(m+1)}}|s\in {{\mathbb {Z}}}_r\}. \end{aligned}$$

Step (2): Isomorphic classes of the above bi-coset graphs.

From Proposition 2.9, if $m_1+m_2\ne r,$ the groups $PR_{m_1}$ and $PR_{m_2}$ are not isomorphic. Then, the corresponding bi-coset graphs are not isomorphic. Hence, we only need to consider whether $X_{D_{m}(l_i)}$ and $X_{D_{m}(l_j)}$ are isomorphic for different m with $m=2,\cdots ,\frac{r-1}{2}$

If $X_{D_m(l_i)}\cong X_{D_m(l_j)},$ where $l_i\ne l_j$, then there exists $\rho$ such that $P^\rho =P$ and $\langle \varphi (t^m,t)\rangle ^\rho =\langle \varphi (t^m,t)\rangle$. It follows that $\rho \in P\rtimes \mathrm{Aut}(P)$. Define two bijections $\rho ,\lambda$ of $V(\Gamma )$ to $V(\Gamma )$ as follows:

$$\begin{aligned} \begin{array}{lll} (1)&{}(A_wx)^\rho =A_wx^\rho , (A_uy)^\rho =A_uy^\rho \\ (2)&{}(A_wx)^\lambda =A_ux, (A_uy)^\lambda =A_uy \end{array}. \end{aligned}$$

where $x,y\in P$. If $X_{D_m(l_i)}\cong X_{D_m(l_j)}$ and $\rho ,\lambda$ preserve $E(\Gamma )$. Given $d\in D$. Since $A_w=A_u=R,$ we have $D=\langle d^r|r\in R\rangle$. So $\rho ,\lambda$ preserving $E(\Gamma )$ infers the following results: For $(A_wx,A_ud^rx)\in E(Z_{D_m(l_i)})$,

$$\begin{aligned} \begin{array}{lll} (A_wx,A_ud^rx)^\rho =(A_wx^\rho ,A_u(d^rx)^\rho )\in E(X_{D_m(l_j)}) &{}\quad \mathrm{(I)}\\ (A_wx,A_ud^rx)^\lambda =(A_wx^\lambda ,A_u(d^rx)^\lambda )=(A_ux,A_wd^rx)\in E(X_{D_m(l_j)}) &{}\quad \mathrm{(II)} \end{array}. \end{aligned}$$

In what follows, we discuss the bijections $\rho$ and $\lambda$ separately.

Case 1 For the bijection $\rho ,$ we shall show that $(Rabc^{l_i}Rx)^\rho \ne Rabc^{l_j}Rx^\rho ,$ that is, $\rho$ is not an isomorphic mapping between $X_{D_m(l_i)}$ and $X_{D_m(l_j)}$ with $l_i\ne l_j$.

If $(Rabc^{l_i}R)^\rho = Rabc^{l_j}R,$ then there exists $v\in {{\mathbb {Z}}}_r$ such that $(abc^{l_i})^{\rho \varphi (t^m,t)^v}=abc^{l_j}$.

Suppose that $[\rho ,P]= 1$. If $(abc^{l_i})^{\rho \varphi (t^m,t)^v}=abc^{l_j}$ with $v\in {{\mathbb {Z}}}_r,$ then $l_i=l_j,$ a contradiction.

Suppose that $[\rho ,P]\ne 1,$ $\rho \in \mathrm{Aut}(P)$. Since $\langle \varphi (t^m,t)\rangle ^\rho =\langle \varphi (t^m,t)\rangle$, we have $\rho \in N_{\mathrm{Aut}(P)}(\varphi (t^m,t))$. Suppose that $m\ne r-1$. Then, the form of $\rho$ is as follows:

$$\begin{aligned} \rho : a\rightarrow a^u, b\rightarrow b^w, \end{aligned}$$

where $u,w\in {{\mathbb {Z}}}_{p-1}$. In this case, $(abc^{l_i})^{\rho \varphi (t^m,t)^v}=a^{ut^v}b^{wt^{mv}}c^{uwt^{mv+v}l_i}=abc^{l_j}$. We can get $l_i=l_j$, also a contradiction.

Case 2 For $\lambda ,$ we prove that: $X_{D_m(l_i)}\cong X_{D_m(1-l_i)}$.

For equation (II),

$$\begin{aligned}&(A_wx,A_ud^rx)^\lambda =(A_wx^\lambda ,A_u(d^rx)^\lambda )\\&\quad =(A_ux,A_wd^rx)=(A_wx,A_ud^{(-r)}x)\in E(X_{D_m(l_j)}) \end{aligned}$$

Since $d^{(-r)}=(abc^{l_i})^{-r}=(a^{-1}b^{-1}c^{1-l_i})^{-r},$ we have $R_m(a^{-1}b^{-1}c^{1-l_i})R_m$. Choose $\phi (-1,-1)\in {\hat{R}}$. It is easy to see that $(R_m(abc^{1-l_i})R_m)^{\phi (-1,-1)}=R_m(a^{-1}b^{-1}c^{1-l_i})R_m$. Therefore, $X_{D_{m}(l_i)}\cong X_{D_{m}(1-l_i)}$.

Thus, we only need to consider that $X_{D_{m}(l)},$ where $m=2,\cdots , \frac{r-1}{2}$ and $l=1,\cdots ,\frac{p+1}{2}$

Step (3): Prove the graph $\Gamma$ is semisymmetric

Suppose that the graph X is vertex-transitive, then $|\mathrm{Aut}(X):A|=2$. Since $(|A|,2)=1$ and $A\unlhd \mathrm{Aut}(X),$ there exists an involution $\sigma$ in $\mathrm{Aut}(X)$ such that $\mathrm{Aut}(X)=A\rtimes \langle \sigma \rangle$. According to P char A,  it follows that $P^{\sigma }=P$. Since $P\unlhd \mathrm{Aut}(X)$ and $(|\mathrm{Aut}(X)/P|,|P|)=1,$ there exists a $p-$supplemented subgroup of P in $\mathrm{Aut}(X)$. Then, all subgroups of order 2r are conjugate. Since the subgroup of order r is normal in the group of order 2r,  we can get $R^{\sigma }=R$. Since X is vertex-transitive, $D=D^{-1}$. Note that D is the neighborhood of the vertex R in [A : R] and $R^{\sigma }=R$. Then, $D^{\sigma }=D^{-1}$. Therefore, there exists an involution $\sigma$ such that $P^{\sigma }=P,R^{\sigma }=R$ and $D^{\sigma }=D^{-1}$. In other words, if the graph is semisymmetric, there is no the above involution in $\mathrm{Aut}(X)$. In the following, we prove this conclusion in the method of reductio ad absurdum.

Suppose that there exists this involution, denoted by $\sigma$. If $[\sigma ,P]=1,$ then

$$\begin{aligned} (R_mabc^lR_m)^{\sigma }\ne R_ma^{-1}b^{-1}c^{1-l}R_m. \end{aligned}$$

This implies that $\sigma \in \mathrm{Aut}(P)$.

Next, every $R_m$ will be discussed with $m\in {{\mathbb {Z}}}_r$.

Note that $m\ne 0,1$. By Proposition 2.8,

$$\begin{aligned} N_{\mathrm{Aut}(P)}(\varphi (t^m,t))=\left\{ \begin{array}{lll} {\hat{R}},&{} \quad \mathrm{if}\; m\ne 0,1,p-1\\ \langle {\hat{R}},\pi : a\rightarrow b^j, b\rightarrow a^k\bigm |j, k\in F_p^* \rangle ,&{} \quad \mathrm{if}\; m=p-1 \end{array}\right. \end{aligned}$$

It is well known the involution in ${\hat{R}}$ has the following form:

$$\begin{aligned} \pi _1: a\rightarrow a^{-1}, b \rightarrow b^{-1}, \pi _2: a\rightarrow a^{-1}, b \rightarrow b, \pi _3: a\rightarrow a, b \rightarrow b^{-1}. \end{aligned}$$

If $R_mabR_m^{\pi _i}=R_m(ab)^{-1}R_m,$ then there exists $v\in {{\mathbb {Z}}}_r$ such that $(ab)^{{\pi _i}\varphi (t^m,t)^v}=a^{-1}b^{-1}c$. Note that r is a prime. Thus, $-1\not \in \langle t\rangle$.

• 1. $\pi _1: (abc^l)^{{\pi _l}\varphi (t^m,t)^v}=a^{-t^v}b^{-t^{mv}}c^{lt^{mv+v}}=a^{-1}b^{-1}c^{1-l},$ where $l=\frac{p+1}{2}$.

• 2. $\pi _2: (abc^l)^{{\pi _2}\varphi (t^m,t)^v}=a^{-t^v}b^{t^{mv}}c^{-lt^{mv+v}}\ne a^{-1}b^{-1}c^{1-l},$

• 3. $\pi _3: (abc^l)^{{\pi _3}\varphi (t^m,t)^v}=a^{t^v}b^{-t^{mv}}c^{-lt^{mv+v}}\ne a^{-1}b^{-1}c^{1-l}$.

In conclusion, there is no $\sigma$ such that $D^{\sigma }=D^{-1}$ except $l=\frac{p+1}{2}$. Then, the graph $X_{D_m(l)}$ is semisymmetric, when $m=2,\cdots , \frac{r-1}{2}$ and $l=1,\cdots , \frac{p-1}{2}$.

For given $m\in {\mathbb {Z}}_r,$ by labeling $R_ma^ib^jc^k$ by (i, j, k) and $R_ma^xb^yc^z$ by [x, y, z],  as the graph $\Gamma _p,$ that is,

$$\begin{aligned} W(\Gamma )=\{(i,j,k)\bigm |i,j,k\in {{\mathbb {Z}}}_p\}, \, U(\Gamma )=\{[x, y, z]\bigm |x,y, z\in {{\mathbb {Z}}}_p\}, \end{aligned}$$

we get the respective edge set of two graphs

$$\begin{aligned} \begin{array}{lll} E_m(l)&{}=&{}\{ ((i,j,k),[x,y,z])\bigm |x=b+i,\,y=b^{m}+j,\,z=k+ib^{m}+lb^{(m+1)},\\ &{}&{}\quad i,j,k,x,y,z\in {{\mathbb {Z}}}_p,b\in B\}.\end{array} \end{aligned}$$

where B is a subgroup of order r in $F_p^*$. Moreover, $m=2,\cdots , \frac{r-1}{2}$ and $l=1,\cdots , \frac{p-1}{2}$.

Therefore, $\Gamma \cong \Gamma _r(m,l)$. $\square$

Remark 5.3

Regarding graph $\Gamma _r(m,l)$, we can get $r\ne 3$. If $r=3,$ then $m\le \frac{3-1}{2}=1$. Since $m\ne 0,1,$ this case is impossible. Hence, $r\ne 3$.

5.2 The valency $r\bigm |(p+1)$

Finally, we treat the case $r\bigm |(p+1)$.

Lemma 5.4

Suppose that $r\bigm |(p+1)$. Then, there is no semisymmetric graph of order $2p^3$ with valency r.

Proof

By Lemma 5.1, $|P|=p^3$, $\exp (P)=p$, and

$$\begin{aligned} P=\langle a,b \bigm |a^p=b^p=1,[b,a]=c,[c,a]=[c,b]=1\rangle . \end{aligned}$$

Suppose S is the subgroup of order $p+1$ in $\mathrm{Aut}(P)$ defined as Proposition 2.8 and $S=\langle s\rangle$. Let H be an irreducible subgroup of order r in $\mathrm{GL}(2,p)$ with $r\bigm |p+1$. By proposition 2.8(6), there exists $a\in \mathrm{Aut}(P)$ such that $H<S^a$.

Without loss of generality, let $H<S$ and the form of s is as follows:

$$\begin{aligned} s: a\rightarrow a^{i_1}b^{j_1\theta },\, b\rightarrow a^{j_1}b^{i_1}, \end{aligned}$$

where $\langle \theta \rangle =F_p^*$ and $i_1^2-j_1^2\theta =1$. Then, $R=\langle s^{\frac{p+1}{r}}\rangle$. Meanwhile, there exist $i_2,j_2,k_1$ and $k_2$ such that

$$\begin{aligned} s^{\frac{p+1}{r}}: a\rightarrow a^{i_2}b^{j_2\theta }c^{k_1},\, b\rightarrow a^{j_2}b^{i_2}c^{k_2}. \end{aligned}$$

Moreover, there exist $i_{2e},j_{2e},k_{1e},k_{2e}$ such that

$$\begin{aligned} s^{e\frac{p+1}{r}} : a\rightarrow a^{i_{2e}}b^{j_{2e}\theta }c^{k_{1e}},\, b\rightarrow a^{j_{2e}}b^{i_{2e}}c^{k_{2e}}, \end{aligned}$$

where $e\in {{\mathbb {Z}}}_r,\,i_{2e},j_{2e},k_{1e},k_{2e}\in {{\mathbb {Z}}}_p$ and $i_{2e}^2-j_{2e}^2\theta =1$.

The proof is divided into the following two steps:

Step (1): The possible bi-coset graph isomorphic to $\Gamma$

Since $A=P\rtimes R$ and $(|P|,|A/P|)=1,$ we have that $A_w$ and $A_u$ are conjugate in A with $w\in W(\Gamma )$ and $u\in U(\Gamma )$. Without loss of generality, let $A_w=A_u=R$. Then, the graph $\Gamma$ is isomorphic to the bi-coset graph $Z={\mathbf {B}}(A, R, R; D)$ with $D=RdR$ and $d\in P$. Let

$$\begin{aligned} W(\Gamma )=[A:A_u]=\{Ra^ib^jc^k\},\,\,U(\Gamma )=[A:A_w]=\{Ra^xb^yc^z\}, \end{aligned}$$

where $i,j,k,x,y,z\in {{\mathbb {Z}}}_p$.

According to $\langle c\rangle =Z(P),$ it follows $\langle R,c\rangle < A$. If $d=c^k$ with $k\in F_p,$ then $\langle DD^{-1}\rangle \ne A$. By Proposition 2.3, the bi-coset graph Z is not connected. Then, $d=a^ib^jc^k$ with $i,j,k\in {{\mathbb {Z}}}_p$ and $(i,j)\ne (0,0)$.

We claim for any $i,j,k\in F_p$ and $(i,j)\ne (0,0)$, there exist $\tau$ and $l\in F_p$ in $\mathrm{Aut}(A)$ such that $(Rabc^l R)^{\tau }=Ra^ib^jc^kR$ with $\tau \in S$. Let $\tau :a\rightarrow a^{i_3}b^{j_3\theta }c^{k_3},\,b\rightarrow a^{j_3}b^{i_3}c^{k_4},$ where $i_3^2-j_3^2\theta \ne 0$ and $\tau \in S$. Then, $[\tau ,R]=1$. Suppose that $R(abc^l)^{\tau } R=Ra^ib^jc^kR$. Then,

$$\begin{aligned}&(abc^l)^\tau =a^{i_3}b^{j_3\theta }c^{k_3}a^{j_3}b^{i_3}c^{k_4}c^{l(i_3^2-j_3^2\theta )}\\&\quad =a^{i_3+j_3}b^{j_3\theta +i_3}c^{l(i_3^2-j_3^2\theta )+j_3^2\theta +k_3+k_4}=a^ib^jc^k. \end{aligned}$$

From this equation, we can get $i_3+j_3=i,\,j_3\theta +i_3=j,\,l(i_3^2-j_3^2\theta )+j_3^2\theta +k_3+k_4=k$. Solve it, we can get

$$\begin{aligned} i_3=\frac{j-i\theta }{1-\theta },\,j_3=\frac{i-j}{1-\theta },\,l=\frac{k-k_3-k_4-j_3^2\theta }{i_3^2-j_3^2\theta }. \end{aligned}$$

Then, there exist $\tau$ and l such that $(Rabc^l R)^{\tau }=Ra^ib^jc^kR$. By Proposition 2.3, $\tau$ induces an isomorphism between $X_1={\mathbf {B}}(A, R, R; D_1)$ and $X_2={\mathbf {B}}(A, R, R; D_2)$, where $D_1=Rabc^lR$ and $D_2=Ra^ib^jc^kR$.

For any $l\in {{\mathbb {Z}}}_p,$

$$\begin{aligned} \begin{array}{lll}D(l)&{}= Rabc^lR\\ &{}= \{Ra^{i_{2e}+j_{2e}}b^{j_{2e}\theta +i_{2e}}c^{k_{1e}+k_{2e}+j_{2e}^2\theta +l}|e\in {{\mathbb {Z}}}_r,\,i_{2e},j_{2e},k_{1e},k_{2e}\\ &{}\quad \in {{\mathbb {Z}}}_p,\,i_{2e}^2-j_{2e}^2\theta =1\}.\end{array} \end{aligned}$$

Then, for any $i,j,k\in F_p,$ we can get

$$\begin{aligned} \begin{array}{lll} Rabc^lRa^ib^jc^k&{}=&{} Ra^{i_{2e}+j_{2e}}b^{j_{2e}\theta +i_{2e}}c^{k_{1e}+k_{2e}+j_{2e}^2\theta +l}a^ib^jc^k\\ &{}=&{}Ra^{i_{2e}+j_{2e}+i}b^{j_{2e}\theta +i_{2e}+j}c^{k_{1e}+k_{2e}+j_{2e}^2\theta +l+k+i(j_{2e}\theta +i_{2e})}\end{array} \end{aligned}$$

Therefore, for any $Ra^ib^jc^k$ in $W(\Gamma )$, its neighborhood is

$$\begin{aligned} \begin{array}{lll}\{Ra^{i_{2e}+j_{2e}+i}b^{j_{2e}\theta +i_{2e}+j}c^{k_{1e}+k_{2e}+j_{2e}^2\theta +l+k+i(j_{2e}\theta +i_{2e})} \bigm |e\in {{\mathbb {Z}}}_r,\,\\ \qquad i_{2e},j_{2e},k_{1e},k_{2e}\in {{\mathbb {Z}}}_p,\,i_{2e}^2-j_{2e}^2\theta =1\}.\end{array} \end{aligned}$$

Next, we prove that $X_{D(0)}={\mathbf {B}}(A, R, R; D(0))$ and $X_{D(l)}={\mathbf {B}}(A, R, R;D(l))$ are isomorphic. Let

$$\begin{aligned} \rho :Ra^ib^jc^k\rightarrow Ra^ib^jc^k,\, Ra^xb^yc^z\rightarrow Ra^xb^yc^{z+l}. \end{aligned}$$

It is easy to verify $\rho$ is an isomorphism between $X_{D(0)}$ to $X_{D(l)}$. Thus, we only need to discuss the graph $X_{D(0)}$.

Step (2): The graph $X_{D(0)}$ is vertex-transitive.

By Proposition 2.4, if there exists an involution $\sigma$ in $\mathrm{Aut}(A)$ such that $R^\sigma =R$ and $D^\sigma =D^{-1},$ then $\Gamma$ is vertex-transitive. We claim the graph $X_{D(0)}$ is vertex-transitive. Then, we need to find this involution.

Suppose that there exists an involution $\sigma$ such that $\sigma \in A,\,R^\sigma =R$ and $D^\sigma =D^{-1}$. According to $A^\sigma =A,$ and P char A, it follows that $P^\sigma =P$.

By the structure of the subgroup of $\mathrm{GL}(2,p)$, there exists an involution $\mu$ such that $\langle s,\mu \rangle$ is a dihedral group. Then, for any $m\in {{\mathbb {Z}}}_{p+1},$ we can get $o(\mu s^m)=2$ and $s^{\mu s^m}=s^{-1}$. So, for any $l\in {{\mathbb {Z}}}_{p+1},$ we can get ${s^l}^{\mu s^m}=s^{-l}$ and $R^{\mu s^m}=R$. If there exists $m\in {{\mathbb {Z}}}_{p+1}$ such that $(RabR)^{\mu s^m}=Rb^{-1}a^{-1}R=Ra^{-1}b^{-1}cR,$ then the graph is vertex-transitive, where $m\in {{\mathbb {Z}}}_{p+1}$. Moreover, $(RabR)^{\mu s^m}=Ra^{-1}b^{-1}cR$ if and only if there exists e such that $(ab)^{\mu s^{m+e\frac{p+1}{r}}}=a^{-1}b^{-1}c,$ where $e\in {{\mathbb {Z}}}_r$. Then, if there exists $m_1\in {{\mathbb {Z}}}_p$ such that $m_1=m+e\frac{p+1}{r}$ and $(ab)^{\mu s^{m_1}}=a^{-1}b^{-1}c,$ it follows that the graph is vertex-transitive.

First, let

$$\begin{aligned} s^{m_1}: a\rightarrow a^ub^{v\theta }c^{l_5},\, b\rightarrow a^{v}b^{u}c^{l_6}, \end{aligned}$$

and

$$\begin{aligned} \mu :a\rightarrow ac^{l_1},\, b\rightarrow b^{-1}c^{l_2}, \end{aligned}$$

where $u^2-v^2\theta =1$ and $l_1,l_2\in {{\mathbb {Z}}}_p$. Since $o(\mu )=2$ and $l_2=0,$ by Proposition 2.8(7),

$$\begin{aligned} a^{s^{-1}}=a^{i_1}b^{-j_1\theta }c^{\frac{1}{2}i_1j_1\theta (i_1-j_1-1)},\, b^{s^{-1}}=a^{-j_1}b^{i_1}c^{\frac{1}{2}i_1j_1(i_1-j_1\theta -1)}, \end{aligned}$$

and

$$\begin{aligned} a^{s^{-m_1}}=a^ub^{-v\theta }c^{\frac{1}{2}uv\theta (u-1-v)+l_6v\theta +l_5u},\, b^{s^{-m_1}}=a^{-v}b^uc^{\frac{1}{2}uv(u-v\theta -1)-l_6u-l_5v}. \end{aligned}$$

Since $\langle s,\mu \rangle$ is a dihedral group, $s^\mu =s^{-1}$ and $(s^{m_1})^{\mu }=s^{-m_1},$ where $m_1=m+e\frac{p+1}{r}$. Then,

$$\begin{aligned} \begin{array}{lllllll}a^{\mu s\mu }&{}=&{}a^{i_1}b^{-j_1\theta }c^{l_1i_1-l_1}&{}=&{}a^{s^{-1}}&{}=&{}a^{i_1}b^{-j_1\theta }c^{\frac{1}{2}i_1j_1\theta (i_1-j_1-1)},\\ b^{\mu s\mu }&{}=&{}a^{-j_1}b^{i_1}c^{-(l_1+i_1)j_1}&{}=&{}b^{s^{-1}}&{}=&{}a^{-j_1}b^{i_1}c^{\frac{1}{2}i_1j_1(i_1-j_1\theta -1)},\\ a^{\mu s^{m_1}\mu }&{}=&{}a^ub^{-v\theta }c^{l_1u-l_5-l_1}&{}=&{}a^{s^{-m_1}}&{}=&{}a^ub^{-v\theta }c^{\frac{1}{2}uv\theta (u-1-v)+l_6v\theta +l_5u},\\ b^{\mu s^{m_1}\mu }&{}=&{}a^{-v}b^uc^{l_6-uv-l_1v}&{}=&{}b^{s^{-m_1}}&{}=&{}a^{-v}b^uc^{\frac{1}{2}uv(u-v\theta -1)-l_6u-l_5v}.\end{array} \end{aligned}$$

By solving it, we can get $l_1=\frac{1}{2}i_1(j_1\theta -1-i_1),\,l_5=0,\,l_6=\frac{v(2l_1+u+u^2-uv\theta )}{2(u+1)}$.

Next, we verify that $(ab)^{\mu s^{m_1}}=a^{-1}b^{-1}c,$

that is,

$$\begin{aligned} (ab)^{\mu s^{m_1}}=a^{u-v}b^{v\theta -u}c^{v^2\theta +l_1+uv-l_6}=a^{-1}b^{-1}c. \end{aligned}$$

Then,

$$\begin{aligned} \left\{ \begin{array}{lll}u-v=-1\\ v\theta -u=-1\\ v^2\theta +l_1+l_5+uv-l_6=1\end{array}\right. \end{aligned}$$

(I)

It is easy to get

$$\begin{aligned} u= & {} \frac{\theta +1}{1-\theta },\,v=\frac{2}{1-\theta },\, l_1=\frac{1}{2}i_1(j_1\theta -1-i_1),\,l_5=0,\,\\ l_6= & {} \frac{v(2l_1+u+u^2-uv\theta )}{2(u+1)} \end{aligned}$$

is solution of Eq. (I).

This shows that there exists an involution $\mu s^{m_1}\in \mathrm{Aut}(A)$ such that $R^{\mu s^{m_1}}=R$ and $(RabR)^{\mu s^{m_1}}=R(ab)^{-1}R,$ where $a^{s^{m_1}}=a^{\frac{\theta +1}{1-\theta }}b^{\frac{2\theta }{1-\theta }},\, b^{s^{m_1}}=a^{\frac{2}{1-\theta }}b^{\frac{\theta +1}{1-\theta }}c^{\frac{1}{2}(j\theta -1-i)}$. By Proposition 2.4, we can get the graph $X={\mathbf {B}}(A, R, R; RabR)$ is vertex-transitive, a contradiction. Thus, there is no semisymmetric graph of order $2p^3$ with valency $r\bigm |(p+1)$. $\square$

Proof of Theorem 1.1

Combining Sects. 3–5, we can easily deduce that any semisymmetric graph of order $2p^3$ with prime valency is isomorphic to $\Gamma _p$ or $\Gamma _r(m,l)$, whose automorphism group is faithful and imprimitive on both biparts. $\square$