Semisymmetric graphs of order twice prime powers with the same prime valency

Abstract

A simple undirected graph is said to be semisymmetric if it is regular and edge-transitive but not vertex-transitive. Folkman (J Combin Theory Ser B 3, 215–232, 1967) proved that a graph of order 2p or $2p^2$ is not semisymmetric for any prime p. Wang and Guo (J Algebra Comb 54, 49–73, 2021) proved that there is only one semisymmetric graph of order $2p^3$ with valency p. In this paper, we give a necessary condition for semisymmetric graphs of order $2p^n$ with valency p,  where p is an odd prime, and construct an infinite family of such graphs. As an application, a classification of semisymmetric graphs of order $2p^4$ with valency p is given.

1 Introduction

Throughout this paper, all graphs are finite, simple, connected and undirected. Let $X=(V(X),E(X))$ be a graph with vertex set V(X) and edge set E(X). For any vertex $v\in V(X)$, denote by $X_1(v)$ the set of vertices which are adjacent to v. The valency of v is the size of the set $X_1(v)$. A graph X is said to be regular if all the vertices have the same valency. A graph X is a bipartite graph if V(X) can be partitioned into two subsets U(X) and W(X), called partite sets, such that every edge of X joins a vertex of U(X) and a vertex of W(X). Denote, by $K_{m,n}$, the complete bipartite graph with partite sets of size m and n,  respectively.

Let $X_1$ and $X_2$ be graphs with vertex sets $V(X_1)$ and $V(X_2)$, respectively. An isomorphism $\sigma$ from $X_1$ to $X_2$ is a bijection from $V(X_1)$ to $V(X_2)$ such that for any $u,v\in V(X_1)$, u and v are adjacent in $X_1$ if and only if $u^\sigma$ and $v^\sigma$ are adjacent in $X_2$. Two graphs $X_1$ and $X_2$ are said to be isomorphic, denoted by $X_1\cong X_2$, if there is an isomorphism from $X_1$ to $X_2$. Let X be a graph, an automorphism of X is an isomorphism from X to itself. All automorphisms of X form a group under the composition of maps. This group is denoted by $\textrm{Aut}(X)$ and is called the full automorphism group of X. The graph X is said to be vertex-transitive or edge-transitive if $\textrm{Aut}(X)$ acts transitively on V(X) or on E(X), respectively. It is well known that a connected graph, which is edge-transitive but not vertex-transitive, is bipartite (see [9]). The complete bipartite graph $K_{m,n}$, where $m\ne n$, is a simple example of such graphs. But it is interesting to find such regular graphs. A graph X is said to be semisymmetric if it is regular and edge-transitive but not vertex-transitive.

Let X be a bipartite graph with the bipartition $V(X)=U(X)\cup W(X)$ and $A=\textrm{Aut}(X)$. Suppose that $A^+$ is the subgroup of A preserving both U(X) and W(X). The connectedness of the graph X implies that either $|A:A^+|=2$ or $A=A^+$, depending on whether or not there exists an automorphism, which interchanges U(X) and W(X). For $G\le A^+,$ X is said to be G-semitransitive if G acts transitively on both U(X) and W(X), while an $A^+$-semitransitive graph is simply called semitransitive. It can be checked easily that every semisymmetric graph is a semitransitive bipartite graph with two partite sets having the same size. Thus, the order of a semisymmetric graph must be even.

Semisymmetric graphs were first investigated by Folkman [9] in 1967. He gave some characterizations of semisymmetric graphs and constructed several infinite families of such graphs. Meanwhile, Folkman put forward 8 open problems, which spurred the interest in this topic. Whereafter, many semisymmetric graphs were constructed which nearly solved all Folkman’s open problems (see [2, 3, 8, 14, 15]). More recently, some new results on semisymmetric graphs have appeared by some group-theoretical methods, graph coverings and computer searching (see [4, 6, 10, 12, 18, 19, 25]).

One of Folkman’s problems is “for which pairs of integers v and d is there a connected semisymmetric graph with v vertices and d valency?” In response to this problem, Parker [20] studied semisymmetric cubic graphs of twice odd order. Li and Lu [16] classified pentavalent semisymmetric graphs of square-free order. But there are a few known semisymmetric graphs with twice prime powers vertices. Let p be a prime, Folkman [9] proved that there is no semisymmetric graph of order 2p or $2p^2$. Malnič, et al. [17] showed that cubic semisymmetric graph of order $2p^3$ is the Gray graph. Du and Wang et al. [7, 21,22,23,24] gave a partial classification of semisymmetric graphs of order $2p^3$. Especially, semisymmetric graphs of order $2p^3$ with prime valency have been completely classified [24]. As a natural process, for any pair of positive integer n and prime p,  whether there are semisymmetric graphs of order $2p^n$ with valency p is an interesting problem. In this paper, we give a necessary condition for semisymmetric graphs of order $2p^n$ with valency p. Applying this condition, an infinite family of such graphs is constructed and semisymmetric graphs of order $2p^4$ with valency p are classified. Note that when $p=2,$ the graph of order $2p^n$ with valency p is the cycle graph and vertex-transitive. It is also known that there is no semisymmetric graph of order 2p or $2p^2.$ Throughout this paper, let p be an odd prime and $n\ge 3.$ The main results in this paper are listed in the following three theorems.

Theorem 1.1

Let p be an odd prime, $n\ge 3$ and X be a semisymmetric graph of order $2p^n$ with valency p. Let P be a Sylow p-subgroup of $\textrm{Aut}(X).$ If $\textrm{Aut}(X)$ acts primitively on one partite set at least, then $p\bigm |n$ or P contains an elementary abelian group as a maximal subgroup.

Theorem 1.2

Let p be an odd prime, $n\ge 3$ and X be a semisymmetric graph of order $2p^n$ with valency p. Let P be a Sylow p-subgroup of $\textrm{Aut}(X).$ If P contains an elementary abelian group as a maximal subgroup, then $n \le p$ and X is isomorphic to the graph $X_{p,n}$, which is described in Construction 2.6.

It was shown that there is only one semisymmetric graph of order $2p^3$ with valency p. The graph $X_{p,3}$ is just such the graph [24]. Moreover, $X_{3,3}$ is the Gray graph and $\textrm{Aut}(X_{3,3})$ acts primitively on one partite set and imprimitively on the other. If $p\ge 5,$ $\textrm{Aut}(X_{p,3})$ acts imprimitively on both partite sets.

Theorem 1.3

Let $p\ge 5$ be a prime and X be a semisymmetric graph of order $2p^4$ with valency p. Then, X is isomorphic to the graph $X_{p,4}$ or $\widetilde{X}_{p,4}$, which are described in Constructions 2.6 and 2.7, respectively.

2 Preliminaries

Denote by $\mathbb {Z}_n$ the cyclic group of order n, and by $A_n$ and $S_n$ the alternating group and the symmetric group of degree n, respectively. An elementary abelian p-group is a direct product of several cyclic groups of order p, where p is a prime. For a transitive group G on $\Omega$ and a subset $\Omega _1$ of $\Omega$, denote by $G_{\Omega _1}$ and $G_{(\Omega _1)}$ the setwise stabilizer and the pointwise stabilizer of G relative to $\Omega _1$, respectively. For a group G and a subgroup H of G, we use Z(G),  $C_G(H)$ and $N_G(H)$ to denote the center of G, the centralizer and the normalizer of H in G,  respectively. Denote by [G : H] and |G : H| the set of right cosets and the index of H in G, respectively. The action of G on [G : H] is always assumed to be the right multiplication action. For a group G,  denote by $G'$ the derived group of G. A semidirect product of a group N by a group H is denoted by $N\rtimes H,$ where N is normal in $N\rtimes H.$ For any two sets A and B,  denote by $A{\setminus } B=\{a\mid a\in A, a\not \in B \}$ the difference set of A and B. For group-theoretical concepts and notations not defined here, the reader is refereed to [5, 13].

Let G be a group with subgroups L and R and $D=\bigcup _i Rg_iL$ be a union of double cosets of R and L in G, where $g_i\in G$. Define the bipartite graph $X=\textbf{B}(G,L,R;D)$ with bipartition $V(X)=[G:L]\cup [G:R]$ and edge set $E(X)=\{ \{ Lg_1,Rg_2\}\bigm |g_1,g_2\in G, g_2g_1^{-1}\in D\}$. This graph is called the bi-coset graph of G with respect to L, R and D. Clearly, the graph $X=\textbf{B}(G,L,R;D)$ is well-defined, i.e., the adjacency relation is independent of the choice of representatives of right cosets. Under the right multiplication action of G on V(X), the graph X is G-semitransitive.

Proposition 2.1

[8] Let $X=\textbf{B}(G,L,R;D)$ be the bi-coset graph of G with respect to L, R and D. Then,

• (i) X is G-edge-transitive if and only if D is a single double coset of R and L in G,  i.e., $D=RgL$ for some $g\in G$;

• (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 X is regular if and only if $|L|=|R|;$

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

• (iv) $X\cong \textbf{B}(G,L^a,R^b;D'),$ where $D'=\bigcup _{i}R^{b}(b^{-1}d_{i}a)L^{a}$, for any $a,b\in G,$ and $d_{i}\in D$;

• (v) $X\cong \textbf{B}(\widetilde{G},L^{\sigma },R^{\sigma };D^{\sigma }),$ where $\widetilde{G}$ is a group and $\sigma$ is an isomorphism from G to $\widetilde{G}.$

Proposition 2.2

[8] Suppose that X is a G-semitransitive graph with $V (X) = U(X)\cup W(X)$. Take u in U(X), w in W(X) and set $D = \{g \in G \mid w^{g}\in X_{1}(u)\}.$ Then D is a union of double cosets of $G_{u}$ and $G_{w}$ in G, and $X\cong \textbf{B}(G, G_{u}, G_{w}; D)$. Especially, any semisymmetric graph is a bi-coset graph.

In the following, a sufficient condition for judging the vertex transitivity of an edge-transitive bi-coset graph is given.

Lemma 2.3

Let $X=\textbf{B}(G,L,R;D)$ be a bi-coset graph, where $D=RgL$ for some $g\in G.$ If there exists $\sigma \in \textrm{Aut}(G)$ such that $L^{\sigma }=R,R^{\sigma }=L$ and $(D^{-1})^{\sigma }=D,$ then X is vertex-transitive.

Proof

Set $[G:L]=\{Lg_l\mid g_l\in G\}$ and $[G:R]=\{Rg_r\mid g_r\in G\}.$ By definition of a bi-coset graph, $V(X)=[G:L]\cup [G:R].$ Let $\overline{\sigma }$ induced by $\sigma$ be a mapping on V(X) as follows:

$$\begin{aligned} (Lg_l)^{\overline{\sigma }}=Rg_l^{\sigma }~~\textrm{and}~ ~(Rg_r)^{\overline{\sigma }}=Lg_r^{\sigma }, \end{aligned}$$

where $g_l,g_r\in G.$ We claim that $\overline{\sigma }\in \textrm{Aut}(X).$

Firstly, we prove that $\overline{\sigma }$ is a permutation on V(X). It is clear that $\overline{\sigma }$ is a surjection. For any $g_l,g_r\in G,$

$$\begin{aligned} Rg_{l_1}^{\sigma }=Rg_{l_2}^{\sigma }\Leftrightarrow (g_{l_1})^{\sigma }(g_{l_2}^{\sigma })^{-1}\in R \Leftrightarrow (g_{l_1}g_{l_2}^{-1})^{\sigma }\in R\Leftrightarrow g_{l_1}g_{l_2}^{-1}\in R^{\sigma ^{-1}}=L \Leftrightarrow Lg_{l_1}=Lg_{l_2}. \end{aligned}$$

Secondly, we prove that $\overline{\sigma }$ preserves edge set. Since $(D^{-1})^{\sigma }=D,$ it follows that $g_l^{\sigma }(g_r^{\sigma })^{-1}=(g_lg_r^{-1})^{\sigma }=((g_rg_l^{-1})^{-1})^{\sigma }\in D,$ for any $g_rg_l^{-1}\in D.$ Therefore, for any $\{Lg_l,Rg_r\}\in E(X),$ ${\{Lg_l,Rg_r\}}^{\overline{\sigma }}=\{Rg_l^{\sigma },Lg_r^{\sigma }\}=\{Lg_r^{\sigma },Rg_l^{\sigma }\}\in E(X).$ Thus $\overline{\sigma }\in \textrm{Aut}(X).$

Since G acts transitively on both [G : L] and [G : R],  respectively, $\langle G,\overline{\sigma }\rangle$ acts transitively on V(X). Thus, X is vertex-transitive. $\square$

The following lemma gives some sufficient conditions for a regular bipartite edge-transitive graph to be semisymmetric.

Lemma 2.4

Let X be a regular edge-transitive bipartite graph with $V(X)=U(X)\cup W(X),$ and P be a Sylow subgroup of $\textrm{Aut}(X).$ Then, the graph X is semisymmetric if P satisfies one of the following conditions:

• (i) $P_u\not \cong P_w,$ for any $u\in U(X)$ and $w\in W(X);$

• (ii) $N_P(P_u)\not \cong N_P(P_w),$ especially, $|N_P(P_u)|\ne |N_P(P_w)|,$ for any $u\in U(X)$ and $w\in W(X);$

• (iii) $C_P(P_u)\not \cong C_P(P_w),$ especially, $|C_P(P_u)|\ne |C_P(P_w)|,$ for any $u\in U(X)$ and $w\in W(X).$

Proof

In the following, we prove the result by a contradiction. Suppose that the graph is vertex-transitive. Then, there exists $\sigma \in \textrm{Aut}(X)$ such that $A_u^\sigma =A_w,$ where $u\in U(X)$ and $w\in W(X).$ Now $P_u=P\cap A_u$ and $P_w=P\cap A_w$ are Sylow subgroups of $A_u$ and $A_w$, respectively. Then, $P_u^\sigma \le A_w$ and so $P_u^\sigma =P_w^\eta$ for some $\eta \in A_w.$ Therefore, $\tau =\sigma \eta ^{-1}\in \textrm{Aut}(X)$ satisfies $P_u^\tau =P_w.$ Thus, $N_A(P_u)^\tau =N_A(P_w)$ and $C_A(P_u)^\tau =C_A(P_w).$ Note that $P\cap N_A(P_u)=N_P(P_u)$ and $P\cap N_A(P_w)=N_P(P_w)$ are Sylow subgroups of $N_A(P_u)$ and $N_A(P_w)$, respectively. Then, $N_P(P_u)$ and $N_P(P_w)$ are conjugate in $\textrm{Aut}(X).$ Similarly, $C_P(P_u)$ and $C_P(P_w)$ are conjugate in $\textrm{Aut}(X).$ Especially, $|N_P(P_u)|=|N_P(P_w)|$ and $|C_P(P_u)|=|C_P(P_w)|.$ Therefore, if P satisfies one of the conditions (i), (ii) or (iii), the graph X is not vertex-transitive. This implies that X is a semisymmetric graph. $\square$

Since any semisymmetric graph is a bi-coset graph, in order to study semisymmetric graphs of twice prime powers order, groups with subgroups of prime powers index are needed. The following proposition is the classification of nonabelian simple groups with a subgroup of prime powers index.

Proposition 2.5

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

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

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

• (iii) $S=\textrm{PSL}(2,11)$ and $H=A_5;$

• (iv) $S=M_{11}$ and $H=M_{10};$

• (v) $S=M_{23}$ and $H=M_{22};$

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

At the end of this section, we construct two families of bi-coset graphs.

Construction 2.6

Let $p\ge 5$ be a prime and n be a positive integer with $n\le p.$ Define the group P by

$$\begin{aligned} P=\langle x_1,x_2,\cdots ,x_{n},y\bigm |x_1^p=\cdots =x_{n}^p=y^p=1, [x_i,y]=x_{i+1},[x_{n},y]=[x_j,x_i]=1\rangle , \end{aligned}$$

$i=1,\cdots ,n-1,j=1,2,\cdots ,n.$ Then, the order of P is $p^{n+1}.$ Define the bi-coset Graph $X_{p,n}=\,\,\,\textbf{B}(P,\langle y\rangle ,\langle x_1\rangle ;\langle x_1\rangle \langle y\rangle ).$

Construction 2.7

Let $p\ge 5$ be a prime. Define the group Q by

$$\begin{aligned} \begin{array}{lll} Q&{}=&{}\langle a,b,c,d,e|a^p=b^p=c^p=d^p=e^p=1,[b,a]=c,[c,a]=d,[d,a]=[c,b]=e,\\ &{}&{}[d,b]=[d,c]=[e,a]=[e,b]=[e,c]=[e,d]=1\rangle . \end{array} \end{aligned}$$

Define the bi-coset Graph $\widetilde{X}_{p,4}= \textbf{B}(Q,\langle a\rangle ,\langle b\rangle ;\langle b\rangle \langle a\rangle ).$

3 Some results on p-groups

Lemma 3.1

Let p be a prime, n be a positive integer, and let P be a nonabelian group of order $p^{n+1}$, which is generated by two elements of order p. If P contains a maximal subgroup H which is an elementary abelian p-group, then

• (i) $n\le p$, and

  $$\begin{aligned} P=\langle x_1,x_2,\cdots ,x_{n},y\bigm |x_1^p=\cdots =x_{n}^p=y^p=1,[x_i,y]=x_{i+1},[x_{n},y]=[x_j,x_i]=1 \rangle \end{aligned}$$

  where $i=1,\cdots ,n-1$ and $j=1,\cdots ,n;$

• (ii) $C_P(y)=Z(P)\times \langle y\rangle \cong \mathbb {Z}_p^2$ and $C_P(h)=H\cong \mathbb {Z}_p^n,$ for any $h\in H{\setminus } Z(P);$

• (iii) For any $g=x_1^{i_1}\cdots x_n^{i_n}y^{i_{n+1}}\in P$ with $0\le i_1,\cdots ,i_n,i_{n+1}\le p-1,$ $o(g)=p^2$ if and only if $n=p$ and $i_1i_{n+1}\ne 0.$

Proof

(i) Let $P=\langle x,y\rangle$ with $o(x)=o(y)=p.$ Since H is a maximal subgroup of P,  it follows that $H\unlhd P,$ and at least one of x, y is not in H. Without loss of generality, we assume that $y\notin H.$ Then, $P=H\rtimes \langle y\rangle .$ Since $x\in P$ and $x\ne y,$ there exists $h\in H$ such that $x=hy^i$ for some $i\in \{0,1\cdots ,p-1\}.$ Since $P=\langle hy^i, y\rangle =\langle h,y\rangle$ and $o(h)=p,$ without loss of generality, let $x=h\in H.$ Note that P is a nonabelian group, then $[x,y]\ne 1,$ that is $x^y\ne x.$ Thus, $\langle x^{H\langle y\rangle }\rangle =\langle x^{\langle y\rangle }\rangle \le H.$

For any $g\in P=\langle x,y\rangle ,$ we have $g=x^{i_1}y^{j_1}x^{i_2}y^{j_2}\cdots$ $x^{i_s}y^{j_s},$ where $i_1,i_2,\cdots ,i_s,j_1,j_2,\cdots ,j_s\in \{0,1,\cdots ,p-1\}.$ Then there exist integers $k_1,k_2,\cdots , k_{s+1}$ such that

$$\begin{aligned} g=y^{k_1}(x^{i_1})^{y^{k_2}}\cdots (x^{i_s})^{y^{k_{s+1}}}\in \langle y\rangle \langle x^{\langle y\rangle }\rangle . \end{aligned}$$

Therefore, $P=\langle y\rangle \langle x^{\langle y\rangle }\rangle =\langle y\rangle H.$ It is clear that $\langle y\rangle \cap H=1$ and $\langle y\rangle \cap \langle x^{\langle y\rangle }\rangle =1.$ Thus, $|\langle x^{\langle y\rangle }\rangle |=|H|.$ Since $\langle x^{\langle y\rangle }\rangle \le H,$ it follows that $H=\langle x^{\langle y\rangle }\rangle .$ Since $|\langle y\rangle |=p$, we have $|\langle x^{\langle y\rangle }\rangle |\le p^p$. But $|\langle x^{\langle y\rangle }\rangle |=|H|=p^n,$ which implies that $n\le p.$

For any $g\in P,$ we have $g=x^{i_0}(x^y)^{i_1}\cdots (x^{y^{p-1}})^{i_{p-1}}y^{i_p}.$ From

$$\begin{aligned} g^y=({x^{i_0}(x^y)^{i_1}\cdots (x^{y^{p-1}})^{i_{p-1}}y^{i_p}})^y =x^{i_{p-1}}(x^y)^{i_0}(x^{y^2})^{i_1}\cdots (x^{y^{p-1}})^{i_{p-2}}y^{i_p}, \end{aligned}$$

we know that $g\in C_P(y)$ if and only if $i_0=i_1=\cdots =i_{p-1}.$ Hence, $C_P(y)=\langle xx^y\cdots x^{y^{p-1}}\rangle \times \langle y\rangle \cong \mathbb {Z}_p\times \mathbb {Z}_p$ and $Z(P)=\langle xx^y\cdots x^{y^{p-1}}\rangle \cong \mathbb {Z}_p.$ It is clear that $C_P(C_P(y))=C_P(y).$ Thus, P is a p-group of maximal class. Since $|P|=p^{n+1},$ the nilpotent class of P is n.

Suppose that $x_1=x, x_{i+1}=[x_{i},y]$ with $i=1,\cdots ,n-1.$ Since the nilpotent class of the group P is n,  one has $[x_{n},y]=1.$ Clearly, $x_i\in H$, for any $i\in \{1,2,\cdots ,n\}.$ Then, $\langle x_1, x_2,\cdots ,x_{n},y\rangle \le P.$ It is clear that $|\langle x_1, x_2,\cdots ,x_{n},y\rangle |= p^{n+1}=|P|.$ Thus,

$$\begin{aligned} P=\langle x_1,\cdots ,x_{n},y\bigm |x_1^p=\cdots =x_{n}^p=y^p=1,[x_{i},y]=x_{i+1},[x_{n},y]=[x_j,x_i]=1\rangle , \end{aligned}$$

where $i=1,\cdots ,n-1,j=1,\cdots ,n.$

(ii) We have already shown $C_P(y)=\mathbb {Z}_p^2$ in (i). Since H is an elementary abelian p-group, it is clear that $H\subseteq C_P(h)$ for any $h\in H\setminus Z(P).$ From $[h,y]\ne 1$ and H is a maximal subgroup of P,  we have $C_P(h)=H.$

(iii) For $j\in \{1,2,\cdots ,p-1\}$, we show first that $x_1^{y^j}=\prod _{k=1}^{j+1}x_k^{C_j^{k-1}},$ where $C_j^{k-1}$ is the binomial coefficient, by induction on j. It is clear that $x_1^y=x_1x_2,$ so the statement is true when $j=1.$ Suppose that it is true for $j=m.$ Now let $j=m+1,$ we have

$$\begin{aligned} \begin{array}{lll} x_1^{y^{m+1}}&{}=&{}(x_1^{y^m})^y=\left( \displaystyle \prod _{k=1}^{m+1}x_k^{C_m^{k-1}}\right) ^y =\displaystyle \prod _{k=1}^{m+1}(x_k^y)^{C_m^{k-1}}=\displaystyle \prod _{k=1}^{m+1}(x_kx_{k+1})^{C_m^{k-1}}\\ &{}=&{}x_1x_{m+2}\displaystyle \prod _{k=2}^{m+1}x_k^{C_m^{k-2}+C_m^{k-1}} =x_1x_{m+2}\prod _{k=1}^{m+1}x_k^{C_{m+1}^{k-1}}\\ &{}=&{}\displaystyle \prod _{k=1}^{m+2}x_k^{C_{m+1}^{k-1}}, \end{array} \end{aligned}$$

so the statement is true for $j=m+1.$ By induction principle, we have $x_1^{y^j}=\prod _{k=1}^{j+1}x_k^{C_j^{k-1}}$ for $j\in \{1,2,\cdots ,p-1\}.$

It can be checked easily that the derived group $P'$ of P is $\langle x_2\rangle \times \cdots \times \langle x_{n}\rangle$. If $n\le p-1,$ then $|P|\le p^p.$ Consequently, P is a regular p-group. For any $g=x_1^{i_1}x_2^{i_2}\cdots x_n^{i_n}y^{i_{n+1}}\in P,$ one get that

$$\begin{aligned} g^p=(x_1^{i_1}x_2^{i_2}\cdots x_n^{i_n}y^{i_{n+1}})^p=x_1^{pi_1}x_2^{pi_2}\cdots x_n^{pi_n}y^{pi_{n+1}}d_1^pd_2^p\cdots d_s^p, \end{aligned}$$

for some $d_i\in P',i=1,2,\cdots ,s.$ Since $P'$ is an elementary abelian p-subgroup, we have $g^p=1$ which follows $o(g)=p$ for any $1\ne g\in P.$

Now suppose that $n=p.$ Note that $P'\unlhd P,$ it follows that $P'\langle y\rangle \le P$ and $|P'\langle y\rangle |=p^n=p^p.$ Thus, $P'\langle y\rangle$ is a regular p-group. Similarly, we have $g_1^p=1$ for any $g_1\in P'\langle y\rangle .$ For any $g=x_1^{j_1}x_2^{j_2}\cdots x_p^{j_p}y^{j_{p+1}}\in P,$ where $j_1j_{p+1}\ne 0$, let $g_1=x_2^{j_2}\cdots x_p^{j_p}y^{j_{p+1}}$, then $g_1\in P'\langle y\rangle$. Therefore,

$$\begin{aligned} 1= & {} g_1^p=(x_2^{j_2}\cdots x_p^{j_p}y^{j_{p+1}})^p=(x_2^{j_2}\cdots x_p^{j_p})^{1+y^{j_{p+1}}+y^{2j_{p+1}}+\cdots +y^{(p-1)j_{p+1}}}\\= & {} (x_2^{j_2}\cdots x_p^{j_p})^{\sum _{l=0}^{p-1}(y^{j_{p+1}})^l}. \end{aligned}$$

Thus,

$$\begin{aligned} \begin{array}{lll} g^p&{}=&{}(x_1^{j_1}x_2^{j_2}\cdots x_p^{j_p}y^{j_{p+1}})^p=(x_1^{j_1}x_2^{j_2}\cdots x_p^{j_p})^{\sum _{l=0}^{p-1}(y^{j_{p+1}})^l}\\ &{}=&{}(x_1^{j_1})^{\sum _{l=0}^{p-1}(y^{j_{p+1}})^l}(x_2^{j_2}\cdots x_p^{j_p})^{\sum _{l=0}^{p-1}(y^{j_{p+1}})^l}\\ &{}=&{}(x_1^{j_1})^{\sum _{l=0}^{p-1}(y^{j_{p+1}})^l}\\ &{}=&{}\displaystyle \prod _{k=0}^{p-1} \left( \displaystyle \prod _{i=1}^{k+1} x_i^{C_k^{i-1}}\right) =\displaystyle \prod _{i=1}^{p}x_i^{i_1\sum _{j=1}^{p-i+1}C_j^{i-1}}\\ &{}=&{}x_p^{j_1}\displaystyle \prod _{i=1}^{p-1} x_i^{j_1\sum _{k=1}^{p-i+1} C_k^{i-1}}\end{array}. \end{aligned}$$

Note that $o\left( x_p\displaystyle \prod _{i=1}^{p-1} x_i^{\sum _{j=1}^{p-i+1} C_k^{i-1}}\right) =p$ and $j_1\ne 0,$ it follows that $o(g^p)=p.$ Therefore, $o(g)=p^2.$ $\square$

Lemma 3.2

Let $3\le n\le p$ and

$$\begin{aligned} P=\langle x_1,x_2,\cdots ,x_{n},y\bigm |x_1^p=\cdots =x_{n}^p=y^p=1,[x_i,y]=x_{i+1},[x_{n},y]=[x_j,x_i]=1 \rangle , \end{aligned}$$

• (i) If $n=p,$ then every automorphism of P is of the form

  $$\begin{aligned} \varphi :~ x_1\mapsto x_1^{i_1}x_2^{i_2}\cdots x_p^{i_{p}},\,\,\,\,\, y\mapsto x_2^{j_2}\cdots x_p^{j_{p}}y^{j_{p+1}}, \end{aligned}$$

  where $0\le i_l,j_k\le p-1,\, 1\le l\le p,2\le k\le p+1$, and $i_1,j_{p+1}\ne 0.$ Hence, $|\textrm{Aut}(P)|=p^{2p-2}(p-1)^2.$

• (ii) If $n< p,$ then every automorphism of P is of the form

  $$\begin{aligned} \phi :~ x_1\mapsto x_1^{i_1}x_2^{i_2}\cdots x_{n}^{i_{n}},\,\,\,\,\, y\mapsto x_1^{j_1}x_2^{j_2}\cdots x_{n}^{j_{n}}y^{j_{n+1}}, \end{aligned}$$

  where $0\le i_l,j_k\le p-1,\, 1\le l\le n,1\le k\le n+1$, and $i_1,j_{n+1}\ne 0.$ Hence, $|\textrm{Aut}(P)|=p^{2n-3}(p-1)^2.$

Proof

It is clear that P can be generated by $x_1$ and y,  so it is only need to give the images of $x_1$ and y for any automorphism of P. Define the following maps on P via the generators $x_1$ and y and preserving multiplications by

$$\begin{aligned} \begin{array}{llll} \sigma _i: &{}x_1\mapsto x_1x_i,~~ y\mapsto y;&{}\,\,\,\,\, \rho _i: &{}x_1\mapsto x_1,~~y\mapsto x_iy;\\ \mu :&{} x_1\mapsto x_1^2,~~ y\mapsto y;&{}\,\,\,\,\, \tau :&{} x_1\mapsto x_1, ~~y\mapsto y^2;\\ \end{array} \end{aligned}$$

where $i=2,\cdots ,n.$ It can be checked easily that all of them are automorphisms of the group P.

(i) Suppose that $n=p.$ In the follows, we will show that $\textrm{Aut}(P)=\langle \sigma _i,\rho _i,\mu ,\tau \rangle .$

If $i_1i_{p+1}\ne 0$, then from Lemma 3.1 we know that $o(x_1^{i_1}x_2^{i_2}\cdots x_{p}^{i_{p}}y^{i_{p+1}})=p^2$. Thus, there is no $\varphi$ in $\textrm{Aut}(P)$ such that $x_1^{\varphi }=x_1^{i_1}x_2^{i_2}\cdots x_{p}^{i_{p}}y^{i_{p+1}}$ or $y^{\varphi }=x_1^{i_1}x_2^{i_2}\cdots x_{p}^{i_{p}}y^{i_{p+1}},$ for any $x_1^{i_1}x_2^{i_2}\cdots x_{p}^{i_{p}}y^{i_{p+1}}\in P$ and $i_1i_{p+1}\ne 0.$ Since $C_P(h)=H$ for any $h\in H{\setminus } Z(P),$ and $C_P(y)=\mathbb {Z}_p^2$, there is no $\varphi$ in $\textrm{Aut}(P)$ such that $y^{\varphi }\in H.$ Thus, under the automorphisms of P, y might map to an element of the form $x_2^{j_2}\cdots x_{p}^{j_{p}}y^{j_{p+1}}.$ By $\rho _i,\tau \in \textrm{Aut}(G),$ it follows

$$\begin{aligned} y^{\langle \rho _i,\tau \rangle }=\{x_2^{j_2}\cdots x_{p}^{j_{p}}y^{j_{p+1}}\bigm |0\le j_k\le p-1, k=2,\cdots ,p+1, j_{p+1}\ne 0\}=y^\mathrm{Aut(P)}. \end{aligned}$$

Note that $P'=\langle x_2,x_3,\cdots , x_p\rangle$ and $P'\langle y\rangle <P,$ we have $x_1^{\varphi }\ne x_2^{i_2}\cdots x_{p}^{i_{p}}$ for any $\varphi \in \textrm{Aut}(P),$ since otherwise, $\langle x_1^{\varphi },y^{\varphi }\rangle < P.$ There is no $\varphi$ in $\textrm{Aut}(P)$ such that $x_1^{\varphi }=y,$ because of $|C_P(x_1)|\ne |C_P(y)|$ from Lemma 3.1. Under the automorphism of P, $x_1$ might map to an element of the form $x_1^{i_1}x_2^{i_2}\cdots x_{p}^{i_{p}}$ with $i_1\ne 0.$ Note that $\sigma _i,\mu \in \textrm{Aut}(P)$, it follows

$$\begin{aligned} x_1^{\langle \sigma _i,\mu \rangle }=\{x_1^{i_1}x_2^{i_2}\cdots x_{p}^{i_{p}}\bigm |0\le i_l\le p-1,\,l=1,2,\cdots ,p,i_1\ne 0\}=x_1^{\textrm{Aut}(P)}. \end{aligned}$$

Therefore, $\textrm{Aut}(P)=\langle \sigma _i,\rho _i,\mu ,\tau \rangle ,$ and hence $|\textrm{Aut}(P)|=p^{2p-2}(p-1)^2.$

(ii) Suppose that $n < p.$ Similarly, the map $\pi$ on P defined via the generators $x_1$ and y and preserving multiplications by

$$\begin{aligned} \pi : x_1\mapsto x_1,~~ y\mapsto x_1y, \end{aligned}$$

is an automorphism of the group P. In the follows, we will show $\textrm{Aut}(P)=\langle \sigma _i,\rho _i,\mu ,\tau ,\pi \rangle .$ Note that $C_P(y)=\mathbb {Z}_p\times \mathbb {Z}_p$ and $C_P(h)=H,$ for any $h\in H{\setminus } Z(P).$ Since $n\ge 3,$ it follows that $|H|\ge p^3$ and $p^2=|C_P(y)|\ne |H|.$ Thus, $y^\phi \notin H$ for any $\phi \in \textrm{Aut}(P).$ Therefore, under the automorphism of P, y might map to an element of the form $x_1^{i_1}x_2^{i_2}\cdots x_{n}^{i_{n}}y^{j_{n+1}}$ with $j_{n+1}\ne 0.$ By $\rho _j,\tau ,\pi \in \textrm{Aut}(P),$ we have

$$\begin{aligned} y^{\langle \rho _j, \tau , \pi \rangle } =\{x_1^{j_1}x_2^{j_2}\cdots x_{n}^{j_{n}}y^{j_{n+1}}\bigm |0\le j_k\le p-1,k=1,\cdots ,n+1,j_{n+1}\ne 0\}=y^{\textrm{Aut}(P)}. \end{aligned}$$

Note that $y^\phi \notin H,$ for any $\phi \in \textrm{Aut}(P).$ Since there exists $\phi$ in $\textrm{Aut}(P)$ such that $y^{\phi }=x_1^{i_1}x_2^{i_2}\cdots x_{n}^{i_{n}}y^{i_{n+1}}$ where $0\le i_k\le p-1$, $k=1,2,\cdots , n+1$ and $i_{n+1}\ne 0,$ it follows that $x_1^{\phi }\ne x_1^{i_1}x_2^{i_2}\cdots x_{n}^{i_{n}}y^{i_{n+1}}$ for any $\phi \in \textrm{Aut}(P).$ Since $P'$ is the characteristic subgroup of P and $x_1\not \in P',$ it follows that $x_1^{\phi }\not \in P'$ for any $\phi \in \textrm{Aut}(P).$ Thus, under the automorphisms of P, $x_1$ might map to an element of the form $x_1^{i_1}x_2^{i_2}\cdots x_{n}^{i_{n}}$ with $i_1\ne 0.$ Note that $\sigma _i,\mu \in \textrm{Aut}(P),$ then

$$\begin{aligned} x_1^{\langle \sigma _i,\mu \rangle }=\{x_1^{i_1}x_2^{i_2}\cdots x_{n}^{i_{n}}\bigm |i_k=0,1,\cdots ,p-1,\,k=1,2,\cdots n,i_1\ne 0\}=x_1^{\textrm{Aut}(P)}. \end{aligned}$$

Therefore, $\textrm{Aut}(P)=\langle \sigma _i,\rho _i,\mu ,\tau ,\pi \rangle ,$ and hence $|\textrm{Aut}(P)|=p^{2n-3}(p-1)^2.$ $\square$

4 Semisymmetric graphs of order $2p^n$ with valency p

In this section, let X be a p-valent semisymmetric graph with bipartitions $V(X)=U(X)\cup W(X)$ of order $2p^n$, $A=\textrm{Aut}(X)$, and P be a Sylow p-subgroup of $\textrm{Aut}(X).$

Lemma 4.1

Let X be a semisymmetric graph with prime valency. Then $\textrm{Aut}(X)$ is faithful on both partite sets of X.

Proof

We will prove the result by contradiction. Assume that $\textrm{Aut}(X)$ is unfaithful on at least one partite set of X. Without loss of generality, suppose that $\textrm{Aut}(X)$ is unfaithful on U(X). Let K be the kernel of A on U(X). In this case, $K\ne 1$ and K is not transitive on W(X),  since otherwise the graph is the complete bipartite graph. Now let $\mathcal{B_W}=\{B_1,B_2,\cdots ,B_m\}$ be the complete imprimitive block system induced by K,  where $m\bigm ||W(X)|.$ For any $u\in U(X),$ if $w\in X_1(u),$ then $w^K\subseteq X_1(u).$ Since $|X_1(u)\cap B_j|\bigm ||X_1(u)|$ and $|X_1(u)|$ is a prime, we have $|X_1(u)\cap B_j|=|X_1(u)|$ or 1. If $|X_1(u)\cap B_j|=|X_1(u)|$, then $X_1(u)\subseteq B_j$ and the graph is unconnected. Thus, $|X_1(u)\cap B_j|=1$. Since $B_j$ is an orbit of K on W(X),  it follows that $w^K=w,$ where $w\in X_1(u).$ According to the arbitrariness of u,  this implies that K fixes any vertex in W(X). Thus, K fixes any vertex in V(X). This contradicts with the faithfulness of A on V(X). The result then follows. $\square$

Lemma 4.2

Let X be a semisymmetric graph of order $2p^n$ with valency p and P be a Sylow p-subgroup of A,  where $n\ge 3,$ and p is an odd prime. Then, $|P|=p^{n+1}.$

Proof

Since A is transitive on both U(X) and W(X),  $A=A_uP=A_wP$ for any $u\in U(X)$ and $w\in W(X).$ By Frattini’s argument, P is transitive on both U(X) and W(X),  and it needs only to show that $|P_u|=p$ for some $u\in U(X).$ Since the bipartite graph X is edge-transitive, one has that $P_u$ is transitive on $X_1(u).$ In other words, we need only to prove that $P_u$ is regular on $X_1(u)$. For $w\in X_1(u),$ if $g\in P_u\cap P_w$, then g fixes every vertex in $X_1(u)$ because of $|X_1(u)|=p$. Since $|X_1(w)|=p$ and $(u,w)\in E(X),$ g fixes every vertex in $X_1(u)\cup X_1(w).$ By the connectivity of X, we can get that g fixes every vertex in V(X). This forces $g=1.$ Consequently, $|P_u|=p$ and $|P|=p^{n+1}.$ $\square$

Proof of Theorem 1.1

Let X be a semisymmetric graph of order $2p^n$ with valency p. Set $V(X)=U(X)\cup W(X).$ In [21], the conclusion has been proved to be correct when $n=3.$ Next, let $n\ge 4.$ Without loss of generality, suppose that A acts primitively on U(X). From O’Nan-Scott Theorem [5], A is of almost simple type, product type or affine group type.

(1) We claim that A cannot be of almost simple type. Conversely, suppose that A is of almost simple type, i.e., the socal S of A is a nonabelian simple group. By Proposition 2.5, S is $A_{p^n}$ or $\textrm{PSL}(m,q),$ where $\frac{q^m-1}{q-1}=p^n$ and $q=p_1^l$ for some prime $p_1$ and positive integer l. We distinguish the following two cases.

(a) A is primitive on W(X).

Assume that two representations of S on both partite sets are equivalent. Consider the action of $S_u$ on $[S:S_w],$ where $u\in U(X)$ and $w\in W(X).$ Then, the lengths of the orbits are 1 and $p^n-1,$ respectively. It is impossible as $p^n-1\ne p.$ Now Assume that two representations of S on the partite sets are not equivalent. Then, $S=\textrm{PSL}(m,q)$ and $\frac{q^m-1}{q-1}=p^n.$ Consider the action of $S_u$ on $[S:S_w],$ there are two orbits with different lengths $\frac{q^{m-1}-1}{q-1}$ and $q^{m-1},$ respectively. If $\frac{q^{m-1}-1}{q-1}=p,$ then

$$\begin{aligned} q^{m-1}=\frac{q^m-1}{q-1}-\frac{q^{m-1}-1}{q-1}=p^n-p=p(p^{n-1}-1). \end{aligned}$$

And so $p\bigm |q.$ This forces $p_1=p$ and $q=p^{l}.$ Since $p\not \mid p^{n-1}-1,$ it follows that $q^{m-1}=p^{l(m-1)}\ne p^n-p,$ which is a contradiction. If $q^{m-1}=p,$ then $q=p$ and $m=2.$ In this case, $\frac{q^m-1}{q-1}=p+1\ne p^n.$ This contradicts the conditions.

(b) A is imprimitive on W(X).

Let $\mathcal{B_W}$ be a maximal imprimitive block system of A on W(X) and $|\mathcal{B_W}|=p^k,$ where $k<n.$ Let K be the kernel of A on $\mathcal{B_W}$. Since $S\unlhd A,$ it follows that S is transitive on $\mathcal{B_W}$ or $S \le K.$ Note that S is a simple group, it is impossible since S has no faithful permutation representation of degree $p^t,$ where $t<n.$

(2) A is of product type.

In this case, there is a nonabelian simple group R of degree $p^s$ such that $A=R^k\rtimes M,$ where M is a transitive permutation group of degree k and $sk=n.$ Let P and Q be Sylow p-subgroups of A and R,  respectively. Since $Q^k$ is a Sylow p-subgroup of $R^k$ and $R^k$ is transitive on U(X),  we have $Q^k$ is transitive on U(X), and thus, $p^n\bigm ||Q^k|.$ By the action of $R^k$ on U(X) and $|P|=p^{n+1},$ we get $|Q^k|=p^n$ and $|Q|=p^s.$ Thus, every Sylow p-subgroup of M is $\mathbb {Z}_p.$ If $s=1,$ then $Q\cong \mathbb {Z}_p$ and $P=\mathbb {Z}_p^k\rtimes \mathbb {Z}_p.$ So P contains an elementary abelian group as a maximal subgroup. If $s>1,$ then $|Z(Q)|>p.$ Let $P_1$ be a subgroup of order p of Z(Q). By the structure of A,  $P_1^k\rtimes \mathbb {Z}_p=\mathbb {Z}_p^k\rtimes \mathbb {Z}_p$ is a subgroup of A. From Lemma 3.2 (1), we can get that $k\le p.$ Since $M\le S_k,$ it follows that $p\le k.$ This forces $p=k$ and then $p\bigm |n.$

(3) A is of affine group type.

In this case, the socal of A is $\mathbb {Z}_p^n$ and is regular on U(X). Note that $|P|=p^{n+1}.$ Thus, $\mathbb {Z}_p^n\le P$ and $\mathbb {Z}_p^n$ is a maximal subgroup of P, i.e., P contains an elementary abelian group as a maximal subgroup. $\square$

Next, we will classify semisymmetric graphs X whose one Sylow p-subgroup of $\textrm{Aut}(X)$ contains an elementary abelian p-subgroup as a maximal subgroup.

Proof of Theorem 1.2

Every semisymmetric graph is a semitransitive bipartite graph, and every semitransitive bipartite graph is a bi-coset graph. In the follows, we study semisymmetric graphs by means of bi-coset graphs. Since P is transitive on both partite sets U(X) and W(X), from Proposition 2.2, let $X=\textbf{B}(P,P_u,P_w;D),$ where $u\in U(X)$, $w\in W(X)$ and $D=P_wdP_u$ for some $d\in P.$ From Proposition 2.1(iv), $\textbf{B}(P,P_u,P_w;P_wdP_u)\cong \textbf{B}(P,P_u,P_w^d;P_w^dP_u).$ So we need only to consider the graph $X=\textbf{B}(P,P_u,P_w;P_wP_u)$ and its vertex transitivity.

Since the graph X is connected, we can get $P=\langle P_u,P_w\rangle$, which implies that P is generated by two elements of order p. Thus,

$$\begin{aligned} P=\langle x_1,x_2,\cdots ,x_{n},y\bigm |x_1^p=\cdots =x_{n}^p=y^p=1,[x_i,y]=x_{i+1},[x_{n},y]=[x_j,x_i]=1 \rangle , \end{aligned}$$

where $i=1,\cdots ,n-1$ and $j=1,\cdots ,n,$ from Lemma 3.1(i). Let $P_u=\langle x_1^{j_1}\cdots x_n^{j_n}y^{j_{n+1}}\rangle$ and $P_w=\langle x_1^{i_1}\cdots x_n^{i_n}y^{i_{n+1}}\rangle ,$ where $0\le i_k,j_l\le p-1,1\le k,l\le n+1.$ Now, $(i_{n+1},j_{n+1})\ne (0,0),$ since otherwise $\langle P_u,P_w\rangle <P.$ Similarly, $(i_{1},j_{1})\ne (0,0)$. Since $x_2^{i_2}\cdots x_n^{i_n}\in P'$ and $P'\le \Phi (P),$ where $\Phi (P)$ is the Frattini subgroup of P, $i_{1}$ and $i_{n+1}$ cannot be equal to 0. Similarly, $j_{1}=0$ and $j_{n+1}=0$ cannot simultaneously occur.

Case 1: $n=p.$

Note that $|P_w|=|P_u|=p,$ then $i_1i_{p+1}=0$ and $j_1j_{p+1}= 0,$ by Lemma 3.1(iii). Without loss of generality, let $i_1\ne 0,j_{p+1}\ne 0$ and $j_1=i_{p+1}=0.$ For any $x_1^{i_1}x_2^{i_2}\cdots x_{p}^{i_{p}}\in P,$ there exists $\varphi \in \textrm{Aut}(P)$ such that $x_1^{\varphi }=x_1^{i_1}x_2^{i_2}\cdots x_{p}^{i_p}$ and $y^{\varphi _1}=x_2^{j_2}\cdots x_p^{j_p}y^{j_{p+1}}.$ Thus, under the automorphism of P, one can set $P_u=\langle y\rangle$ and $P_w=\langle x_1\rangle .$ From Proposition 2.1, we can get

$$\begin{aligned} X=\textbf{B}(P,P_u,P_w;P_wP_u)\cong \textbf{B}(P,\langle y\rangle ,\langle x_1\rangle ;\langle x_1\rangle \langle y\rangle ). \end{aligned}$$

From Lemma 3.1(ii), $C_P(P_u)=\mathbb {Z}_p^{2}$ and $C_P(P_w)=\mathbb {Z}_p^p.$ Note that p is an odd prime, the graph X is not vertex transitive by Lemma 2.4 and so X is semisymmetric.

Case 2: $n < p.$

Without loss of generality, let $j_{n+1}\ne 0.$ Then, there exists $\phi$ in $\textrm{Aut}(P)$ such that $y^{\phi }=x_1^{j_1}\cdots x_n^{j_n}y^{j_{n+1}}$ and $(x_1^{i_1}\cdots x_n^{i_n}y^{i_{n+1}})^{\phi }=x_1^{i_1}x_2^{l_2}\cdots x_n^{l_n}y^{l_{n+1}},$ where $l_{n+1}\ne 0.$ In this case, $i_1\ne 0,$ since otherwise X is disconnected. If $i_{n+1}=0,$ there exists $\phi _1$ in $\langle \sigma _i \rangle ,$ where $i=2,\cdots ,n,$ such that $x_1^{\phi _1}=x_1^{i_1}x_2^{l_2}\cdots x_n^{l_n}$ and $y^{\phi _1}=y.$ Without loss of generality, let $P_u=\langle y\rangle$ and $P_w=\langle x_1\rangle$. If $i_{n+1}\ne 0,$ there exists $\phi _2$ in $\langle \sigma _i,\tau \rangle$ such that $(x_1y)^{\phi _2}=x_1^{i_1}x_2^{l_2}\cdots x_n^{l_n}y^{l_{n+1}}$ and $\langle y\rangle ^{\phi _2}=\langle y\rangle .$ In this case, we can set $P_u=\langle y\rangle$ and $P_w=\langle x_1y\rangle .$ By Proposition 2.1, the connected edge-transitive regular graph $\textbf{B}(P,P_u,P_w;P_wP_u)$ is isomorphic to one of the following two bi-coset graphs

$$\begin{aligned} X=\textbf{B}(P,\langle y\rangle ,\langle x_1\rangle ;\langle x_1\rangle \langle y\rangle )~~\textrm{and}~~ X_1=\textbf{B}(P,\langle y\rangle ,\langle x_1y\rangle ;\langle x_1y\rangle y). \end{aligned}$$

Similar to the case 1, we have $C_P(P_u)=\mathbb {Z}_p^{2}$ and $C_P(P_w)=\mathbb {Z}_p^n,$ where $n\ge 3.$ Then, the graph X is semisymmetric by Lemma 2.4.

For the graph $X_1$, take $\sigma \in \textrm{Aut}(P)$ such that

$$\begin{aligned} \sigma :~ y\mapsto x_1y,~~ x_1\mapsto x_1^{p-1}. \end{aligned}$$

Then $P_w^{\sigma }=P_u, P_u^{\sigma }=P_w$ and $D^{\sigma }=(P_wP_u)^{\sigma }=P_uP_w=D^{-1}.$ By Proposition 2.3, the graph $X_1$ is vertex transitive, and thus, $X_1$ is not a semisymmetric graph. $\square$

5 A classification of semisymmetric graphs of order $2p^4$ with valency p

In this section, we will classify semisymmetric graphs of order $2p^4$ with valency p. For $p=3,$ there is no cubic semisymmetric graph of order 162 (see [4]). In the follows, we assume that $p\ge 5.$ Let P be a Sylow p-subgroup of $\textrm{Aut}(X).$ By Lemma 4.2 and the proof of Theorem 1.2, we need only to consider the graph $\textbf{B}(P,P_u,P_w;P_wP_u)$ where $P=\langle P_u,P_w\rangle$ and $|P|=p^5.$ In [1, 26], groups of order $p^5$ which generated by two elements of order p are as follows:

$P_1=\langle a,b,c,d,e|a^p=b^p=c^p=d^p=e^p=1,[b,a]=c,[c,a]=d,[d,a]=e, [c,b]=[d,b]=[d,c]=[e,a]=[e,b]=[e,c]=[e,d]=1\rangle ;$

$P_2=\langle a,b,c,d,e|a^p=b^p=c^p=d^p=e^p=1,[b,a]=c,[c,a]=d,[c,b]=e, [d,a]=[d,b]=[d,c]=[e,a]=[e,b]=[e,c]=[e,d]=1\rangle ;$

$P_3=\langle a,b,c,d,e|a^p=b^p=c^p=d^p=e^p=1,[b,a]=c,[c,a]=d,[d,a]=[c,b]=e, [d,b]=[d,c]=[e,a]=[e,b]=[e,c]=[e,d]=1\rangle .$

It is clear that the group $P_1$ is just the group in Lemma 3.1 for $n=4$ and its automorphism group has already been given in Lemma 3.2. In the follows, we determine the automorphic groups of $P_2$ and $P_3.$

Lemma 5.1

Every element in $\textrm{Aut}(P_2)$ is of the form

$$\begin{aligned} \varphi :~ a\mapsto a^{i_1}b^{i_2}c^{i_3}d^{i_4}e^{i_5},\,\,\,\,\, b\mapsto a^{j_1}b^{j_2}c^{j_3}d^{j_4}e^{j_5}, \end{aligned}$$

where $0\le i_l,j_k\le p-1,1\le l,k\le 5$ and $i_1j_2-i_2j_1\not \equiv 0(\textrm{mod }p).$ Especially, $|\textrm{Aut}(P_2)|=p^7(p-1)^2(p+1).$

Proof

It is clear that $P_2$ can be generated by a and b, and the derived group of $P_2$ is $P_2'=\langle c\rangle \times \langle d\rangle \times \langle e\rangle .$ Note that $|P_2|=p^5\le p^p$ because of $p\ge 5.$ So $P_2$ is a regular p-group. For any $a^{i_1}b^{i_2}c^{i_3}d^{i_4}e^{i_5}\in P_2,$ $(a^{i_1}b^{i_2}c^{i_3}d^{i_4}e^{i_5})^p=a^{pi_1}b^{pi_2}d_1^{p}d_2^{p}\cdots d_s^{p}=1,$ where $d_1,d_2,\cdots ,d_s\in P_2'.$ It follows that exp$(P_2)=p.$ Let $\sigma _1,\sigma _2,\sigma _3,\sigma _4,$ $\sigma _5,\sigma _6,\sigma _{i_1i_2j_1j_2}$ be the following maps on $P_2$ defined via generators a and b and preserving multiplications by

$$\begin{aligned} \begin{array}{llll} \sigma _1: &{} a\mapsto ac, b\mapsto b;&{}\,\,\,\,\, \sigma _2: a\mapsto ad,b\mapsto b;&{}\,\,\,\,\, \sigma _3: a\mapsto ae,b\mapsto b;\\ \sigma _4:&{} a\mapsto a, b\mapsto bc;&{}\,\,\,\,\, \sigma _5: a\mapsto a, b\mapsto bd;&{}\,\,\,\,\, \sigma _6: a\mapsto a,b\mapsto be;\\ \sigma _{i_1i_2j_1j_2}:&{} a\mapsto a^{i_1}b^{j_1}, b\mapsto a^{i_2}b^{j_2}.&{}&{} \end{array} \end{aligned}$$

It is easy to prove that $\sigma _1,\sigma _2,\sigma _3,\sigma _4,\sigma _5,\sigma _6\in \textrm{Aut}(P_2).$ Then, every element $\sigma$ in $\langle \sigma _1,\sigma _2,\sigma _3,\sigma _4,$ $\sigma _5,\sigma _6\rangle$ is of the form:

$$\begin{aligned} \sigma :~ a\mapsto ac^{i_3}d^{i_4}e^{i_5},~~ b\mapsto bc^{j_3}d^{j_4}e^{j_5}, \end{aligned}$$

where $1\le i_l,j_k\le p-1, 3\le l,k\le 5.$ Consider the induced action of $\textrm{Aut}(P_2)$ on $P_2/P_2'\cong \mathbb {Z}_p\times \mathbb {Z}_p,$ the kernel of this action is $K=\langle \sigma _1,\sigma _2,\sigma _3,\sigma _4,\sigma _5,\sigma _6\rangle .$ Hence $\textrm{Aut}(P_2)/K \lesssim \textrm{GL}(2,p).$

In the follows, we consider the maps $\sigma _{i_1i_2j_1j_2}.$ If $\sigma _{i_1i_2j_1j_2}\in \textrm{Aut}(P_2),$ then

$$\begin{aligned} \begin{array}{lll} c^{\sigma _{i_1i_2j_1j_2}}&{}=&{}[b^{\sigma _{i_1i_2j_1j_2}},a^{\sigma _{i_1i_2j_1j_2}}] =[a^{i_2}b^{j_2},a^{i_1}b^{j_1}]\\ &{}=&{}c^{i_1j_2-i_2j_1}d^{j_2C_{i_1}^2-j_1C_{i_2}^2}e^{i_1C_{j_2+1}^2+i_2C_{p-j_1+1}^2+j_1(i_1j_2-j_1i_2-i_2j_2)},\\ d^{\sigma _{i_1i_2j_1j_2}}&{}=&{}[c^{\sigma _{i_1i_2j_1j_2}},a^{\sigma _{i_1i_2j_1j_2}}]=[c^{i_1j_2-i_2j_1},a^{i_1}b^{j_1}]=d^{i_1(i_1j_2-i_2j_1)}e^{j_1(i_1j_2-i_2j_1)},\\ \end{array} \end{aligned}$$

and

$$\begin{aligned} e^{\sigma _{i_1i_2j_1j_2}}=[c^{\sigma _{i_1i_2j_1j_2}},b^{\sigma _{i_1i_2j_1j_2}}]=[c^{i_1j_2-i_2j_1},a^{i_2}b^{j_2}]=d^{i_2(i_1j_2-i_2j_1)}e^{j_2(i_1j_2-i_2j_1)}. \end{aligned}$$

Since $Z(P_2)=\langle d,e\rangle$ and $c\not \in Z(P_2),$ it follows that $c^{\sigma _{i_1i_2j_1j_2}}\not \in Z(P_2).$ This forces that $i_1j_2-i_2j_1\not \equiv 0(\textrm{mod }p).$ It is clear that $\sigma _{i_1i_2j_1j_2}\not \in K,$ so $K\cap \langle \sigma _{i_1i_2j_1j_2}\rangle =1.$ Thus

$$\begin{aligned} |\langle \sigma _1,\sigma _2,\sigma _3,\sigma _4,\sigma _5,\sigma _6,\sigma _{i_1i_2j_1j_2}\rangle |=|K||\textrm{GL}(2,p)|. \end{aligned}$$

Note that $|\textrm{Aut}(P_2)|\le |K||\textrm{GL}(2,p)|.$ Therefore, $\textrm{Aut}(P_2)=\langle \sigma _1,\sigma _2,\sigma _3,\sigma _4,\sigma _5,\sigma _6,\sigma _{i_1i_2j_1j_2}\rangle ,$ where $i_1j_2-i_2j_1\not \equiv 0(\textrm{mod }p).$ Hence $|\textrm{Aut}(P_2)|=p^7(p-1)^2(p+1).$ $\square$

Lemma 5.2

Every element in $\textrm{Aut}(P_3)$ is of the form

$$\begin{aligned} \phi : a\mapsto a^{i_1}b^{i_2}c^{i_3}d^{i_4}e^{i_5},\,\,\,\,\, b\mapsto b^{i_1^2}c^{j_3}d^{j_4}e^{j_5}, \end{aligned}$$

where $0\le i_l,j_k\le p-1,1\le l\le 5,3\le k\le 5$ and $i_1\ne 0.$ Especially, $|\textrm{Aut}(P_3)|=p^7(p-1).$

Proof

By the same argument as in Lemma 5.1, $P_3$ is a regular p-group and exp$(P_3)=p.$ Let $\sigma _1,\sigma _2,\sigma _3,\sigma _4,$ $\sigma _5,\sigma _6,\sigma _7,\sigma _{ij}$ be the follows maps on $P_3$ defined via generators a and b and preserving multiplications by

$$\begin{aligned} \begin{array}{lll} \sigma _1: a\mapsto ac, b\mapsto b;&{}\,\,\,\,\, \sigma _2: a\mapsto ad,b\mapsto b;&{}\,\,\,\,\, \sigma _3: a\mapsto ae,b\mapsto b;\\ \sigma _4: a\mapsto a, b\mapsto bc;&{}\,\,\,\,\, \sigma _5: a\mapsto a, b\mapsto bd;&{}\,\,\,\,\, \sigma _6: a\mapsto a,b\mapsto be\\ \sigma _7:a\mapsto ab, b\mapsto b;&{}\,\,\,\,\, \sigma _{ij}: a\mapsto a^i, b\mapsto b^j;&{} \end{array} \end{aligned}$$

where $1\le i,j\le p-1.$ It can be checked easily that $\sigma _1,\sigma _2,\sigma _3,\sigma _4,\sigma _5,\sigma _6,\sigma _7\in \textrm{Aut}(P_3).$ Note that $P_3'=\langle c,d,e\rangle \unlhd P_3.$ Consider the induced action of $\textrm{Aut}(P_3)$ on $P_3/P_3'\cong \mathbb {Z}_p\times \mathbb {Z}_p.$ The kernel of this action is $K=\langle \sigma _1,\sigma _2,\sigma _3,\sigma _4,\sigma _5,\sigma _6\rangle .$ Thus, $\textrm{Aut}(P_3)/K \lesssim \textrm{GL}(2,p).$

Now consider the maps $\sigma _{ij}$, we have

$$\begin{aligned} \begin{array}{lll} c^{\sigma _{ij}}&{}=&{}[b^{\sigma _{ij}},a^{\sigma _{ij}}]=[b^j,a^i]=c^{ij}d^{jC_i^2}e^{iC_j^2+j\Sigma _{k=1}^{j-1}C_{k}^2},\\ d^{\sigma _{ij}}&{}=&{}[c^{\sigma _{ij}},a^{\sigma _{ij}}]=[c^{ij}d^{jC_i^2}e^{iC_j^2+j\Sigma _{k=1}^{j-1}C_{k}^2},a^i]=[c^{ij}d^{jC_i^2},a^i]=d^{i^2j}e^{2ijC_i^2},\\ e^{\sigma _{ij}}&{}=&{}[c^{\sigma _{ij}},b^{\sigma _{ij}}]=[c^{ij}d^{jC_i^2}e^{iC_j^2+j\Sigma _{k=1}^{j-1}C_{k}^2},b^j]=[c^{ij},b^j]=e^{ij^2}, \end{array} \end{aligned}$$

and

$$\begin{aligned} e^{\sigma _{ij}}=[d^{\sigma _{ij}},a^{\sigma _{ij}}]=[d^{i^2j}e^{2ijC_i^2},a^i]=[d^{i^2j},a^i]=e^{i^3j}. \end{aligned}$$

Thus if $\sigma _{ij}\in \textrm{Aut}(P_3),$ then $e^{ij^2}=e^{i^3j}.$ It follows that $ij(j-i^2)\equiv 0(\mathrm{\textrm{mod }}p).$ Since $i,j\ne 0,$ one has $j\equiv i^2(\mathrm{\textrm{mod }}p).$ That is, $\sigma _{ii^2}\in \textrm{Aut}(P_3).$

Note that $C_{P_3}(a)=\langle a\rangle \times \langle e\rangle$ and $C_{P_3}(b)=\langle b\rangle \times \langle d\rangle \times \langle e\rangle .$ Thus, a and b are not conjugate in $\textrm{Aut}(P_3).$ By the above argument, for any $a^{i_1}b^{i_2}c^{i_3}d^{i_4}e^{i_5}\in P_3,$ there exists $\sigma \in \textrm{Aut}(P_3)$ such that $a^{\sigma }=a^{i_1}b^{i_2}c^{i_3}d^{i_4}e^{i_5},$ where $0\le i_k\le p-1,k=1,\cdots ,5,$ and $i_1\ne 0.$ Thus, there is no $\phi$ in $\textrm{Aut}(P_3)$ such that $b^{\phi }=a^{i_1}b^{i_2}c^{i_3}d^{i_4}e^{i_5}$, where $i_1\ne 0.$ Since $b\not \in P_3',$ it follows that $b^\phi \ne c^{i_3}d^{i_4}e^{i_5},$ for any $\phi \in \textrm{Aut}(P_3).$ Therefore, under the automorphisms of $P_3,$ b might map to an element of the form $b^{ j_2}c^{j_3}d^{j_4}e^{j_5},$ where $j_2\ne 0.$ From the above discussions on $\sigma _{ij},$ we have

$$\begin{aligned} b^{\langle \sigma _4,\sigma _5,\sigma _6,\sigma _{i_1i_1^2}\rangle }=\{b^{i_1^2}c^{j_3}d^{j_4}e^{j_5}\mid 0\le i_1,j_3,j_4,j_5\le p-1, i_1\ne 0\}=b^{\textrm{Aut}(P_3)}. \end{aligned}$$

Therefore $\textrm{Aut}(P_3)=\langle \sigma _1,\sigma _2,\sigma _3,\sigma _4,\sigma _5,\sigma _6,\sigma _7,\sigma _{ii^2}\rangle .$ Hence, $|\textrm{Aut}(P_3)|=p^7(p-1).$ $\square$

Proof of Theorem 1.3

It is clear that $P_1$ has a subgroup $\langle b\rangle \times \langle c\rangle \times \langle d\rangle \times \langle e\rangle$ which is an elementary abelian p-group of order $p^4.$ It has already been proved that the semisymmetric graph of order $2p^4$ with valency p which is a bi-coset graph of $P_1$ is $X_{p,4}$ in Theorem 1.2. Now we consider two groups $P_2$ and $P_3.$ By the proof of Theorem 1.2, we need only to consider the bi-coset graph $\textbf{B}(P,P_u,P_w;P_wP_u),$ where $P=P_2$ or $P_3.$

Case 1: $P=P_2.$

Firstly, we determine the structures of $P_u$ and $P_w.$ From $|P_u|=|P_w|=p,$ let $P_u=\langle a^{i_1}b^{i_2}c^{i_3}d^{i_4}e^{i_5}\rangle$ and $P_w=\langle a^{j_1}b^{j_2}c^{j_3}d^{j_4}e^{j_5}\rangle ,$ where $0\le i_k,j_l\le p-1$, $k,l=1,\cdots ,5.$ Note that $P'=\langle c,d,e\rangle \unlhd P,$ then $(i_1,j_1)\ne (0,0)$, since otherwise $\langle P_u,P_w\rangle <P.$ Without loss of generality, we assume that $i_1\ne 0.$ By Lemma 5.1, for any $a^{i_1}b^{i_2}c^{i_3}d^{i_4}e^{i_5}\in P_2$ with $i_1\ne 0,$ there exists $\phi$ in $\textrm{Aut}(P_2)$ such that $a^{\phi }= a^{i_1}b^{i_2}c^{i_3}d^{i_4}e^{i_5}.$ By Proposition 2.1, set $P_u=\langle a \rangle .$ Since $P=\langle P_u,P_w \rangle =\langle a,a^{j_1}b^{j_2}c^{j_3}d^{j_4}e^{j_5}\rangle ,$ it follows that $j_2\ne 0.$ Note that there exists $\varphi$ in $\langle \sigma _4,\sigma _5,\sigma _6,\sigma _{11j_1j_2}\rangle$ such that $a^{\varphi }= a$ and $b^{\varphi }=a^{j_1}b^{j_2}c^{j_3}d^{j_4}e^{j_5},$ where $j_2\ne 0.$ From Proposition 2.1, we have

$$\begin{aligned} \textbf{B}(P,P_u,P_w;P_wP_u)\cong \textbf{B}(P,\langle a\rangle ,\langle b\rangle ;\langle b\rangle \langle a\rangle ). \end{aligned}$$

Secondly, we can show that the graph $\textbf{B}(P,\langle a\rangle ,\langle b\rangle ;\langle b\rangle \langle a\rangle )$ is vertex-transitive. Take $\sigma \in \textrm{Aut}(P_2)$ such that

$$\begin{aligned} \sigma :~ a\mapsto b,~ b\mapsto a. \end{aligned}$$

Then it is clear that $P_u^{\sigma }=\langle a\rangle ^{\sigma }=\langle b\rangle =P_w, P_w^{\sigma }=\langle b\rangle ^{\sigma }=\langle a\rangle =P_u$ and $(D^{-1})^{\sigma }=(P_wP_u)^{-1})^{\sigma }=(P_uP_w)^{\sigma }=P_wP_u=D.$ From Lemma  2.3, the graph $\textbf{B}(P_2,\langle a\rangle ,\langle b\rangle ;\langle b\rangle \langle a\rangle )$ is vertex-transitive and thus the graph $\textbf{B}(P_2,\langle a\rangle ,\langle b\rangle ;\langle b\rangle \langle a\rangle )$ is not semisymmetric.

Case 2: $P=P_3.$

Similarly, we first determine the structures of $P_u$ and $P_w.$ Let $P_u=\langle a^{i_1}b^{i_2}c^{i_3}d^{i_4}e^{i_5}\rangle$ and $P_w=\langle a^{j_1}b^{j_2}c^{j_3}d^{j_4}e^{j_5}\rangle ,$ where $0\le i_k,j_l\le p-1$ with $k,l=1,\cdots ,5.$ By the same argument as in Case 1, let $i_1\ne 0.$ By Lemma 5.2, there exists $\psi$ in $\textrm{Aut}(P_3)$ such that $a^{\psi }= a^{i_1}b^{i_2}c^{i_3}d^{i_4}e^{i_5},$ where $i_1\ne 0.$ Without loss of generality, let $P_u=\langle a\rangle .$ Since $P=\langle P_w,P_u\rangle ,$ one has $j_2\ne 0.$ If $j_1=0,$ for any $b^{j_2}c^{j_3}d^{j_4}e^{j_5}\in P_3,$ there exists $\omega \in \langle \sigma _4,\sigma _5,\sigma _6\rangle$ such that $a^{\omega }=a$ and $(b^{j_2})^{\omega }=b^{j_2}c^{j_3}d^{j_4}e^{j_5}\in P_3.$ Thus $P=\langle a,b\rangle .$ If $j_1\ne 0,$ for any $a^{j_1}b^{j_2}c^{j_3}d^{j_4}e^{j_5}\in P_3,$ there exists $\nu$ in $\langle \sigma _4,\sigma _5,\sigma _6,\sigma _{ii^2}\rangle$ such that $\langle a \rangle ^{\nu }=\langle a\rangle$ and $(ab^{k})^{\nu }=a^{j_1}b^{j_2}c^{j_3}d^{j_4}e^{j_5},$ where $kj_1^2\equiv j_2(\mathrm{\textrm{mod }}p).$ Therefore $P=\langle a,ab^k\rangle .$ By Proposition 2.1, the connected edge-transitive regular graph $\textbf{B}(P_3,P_u,P_w;P_wP_u)$ is isomorphic to one of the following two bi-coset graphs

$$\begin{aligned} X=\textbf{B}(P_3,\langle a\rangle ,\langle b\rangle ;\langle b\rangle \langle a\rangle ),\quad \text{ and }\quad X_k= \textbf{B}(P_3,\langle a\rangle ,\langle ab^k\rangle ;\langle ab^k\rangle \langle a\rangle ). \end{aligned}$$

We claim that the graph $X_k$ is vertex-transitive. Take $\sigma _k\in \textrm{Aut}(P_3)$ such that

$$\begin{aligned} \sigma _k: a\mapsto (ab^k)^{-1},~ b\mapsto bc^{-1}de^{j}, \end{aligned}$$

where $j\equiv (k-1)(\mathrm{\textrm{mod }}p).$ Then,

$$\begin{aligned} \begin{array}{lll} (ab^k)^{\sigma _k}&{}=&{}b^{-k}a^{-1}(bc^{-1}d^{-1}e^j)^k\\ &{}=&{}b^{-k}a^{-1}b^kc^{-k}d^{k}e^{-C_k^2+jk}\\ &{}=&{}a^{-1}c^kd^{-k}e^{-C_k^2}c^{-k}d^{k}e^{-C_k^2+jk}\\ &{}=&{}a^{-1}e^{jk-k(k-1)}\\ &{}=&{}a^{-1}. \end{array} \end{aligned}$$

Hence, we have

$$\begin{aligned} \begin{array}{lll} P_u^{\sigma _k}&{}=&{}\langle a\rangle ^{\sigma _k}=\langle (ab^k)^{-1}\rangle =\langle ab^k\rangle =P_w,\\ P_w^{\sigma _k}&{}=&{}\langle ab^k\rangle ^{\sigma _k}=\langle a^{-1}\rangle =\langle a\rangle =P_u. \end{array} \end{aligned}$$

Thus, $(D^{-1})^{\sigma _k}=((P_wP_u)^{-1})^{\sigma _k}=(P_uP_w)^{\sigma _k}=P_wP_u=D.$ By Lemma 2.3, the graph $X_k$ is vertex-transitive.

Now consider the graph X. Note that $C_{P_3}(a)=\mathbb {Z}_p\times \mathbb {Z}_p$ and $C_{P_3}(b)=\mathbb {Z}_p\times \mathbb {Z}_p\times \mathbb {Z}_p.$ By Lemma 2.4, the graph X is not vertex-transitive and so it is semisymmetric.

Therefore, from Theorem 1.2 and the above argument, the semisymmetric graph of order $2p^4$ with valency p is isomorphic to $X_{p,4}$ or $\widetilde{X}_{p,4}$, which are described in Constructions 2.6 and 2.7. Furthermore, $\textrm{Aut}(\widetilde{X}_{p,4})$ is imprimitive on both partite sets by Theorem 1.1. $\square$