On edge-primitive and 2-arc-transitive graphs

TL;DR

This paper investigates the structure of edge-primitive, 2-arc-transitive graphs, proving they have almost simple automorphism groups unless they are cycles or complete bipartite graphs, and provides examples with specific symmetry properties.

Contribution

It establishes a classification result for finite 2-arc-transitive edge-primitive graphs and presents explicit examples with high symmetry.

Findings

Finite 2-arc-transitive edge-primitive graphs have almost simple automorphism groups.

Examples of 3-arc-transitive graphs with faithful vertex-stabilizers are provided.

The classification excludes cycles and complete bipartite graphs.

Abstract

A graph is edge-primitive if its automorphism group acts primitively on the edge set. In this short paper, we prove that a finite 2-arc-transitive edge-primitive graph has almost simple automorphism group if it is neither a cycle nor a complete bipartite graph. We also present two examples of such graphs, which are 3-arc-transitive and have faithful vertex-stabilizers.

Equations

$${\mathbf{N}}_N(N_{\alpha })=N\cap {\mathbf{N}}_G(N_{\alpha })=N\cap G_{\alpha }=N_{\alpha }.$$

$$((T_i{)}_{\alpha }{)}^g=(T_i\cap G_{\alpha }{)}^g=T_i^g\cap G_{\alpha }^g=(T_i{)}_{\alpha }^g=(T_j{)}_{\alpha }\text{mbox}forsomej.$$

$$G_v^{[1]}=G_{uv}^{[1]}.(G_v^{[1]}{)}^{\Gamma (u)},G_v=(G_{uv}^{[1]}.(G_v^{[1]}{)}^{\Gamma (u)}).G_v^{\Gamma (v)},G_{uv}=G_v^{[1]}.(G_v^{\Gamma (v)}{)}_u.$$

$$\begin{array}[]{lcl}|G|&=&{|N||G_{\{u,v\}}|\over|N\cap G_{\{u,v\}}|}={|N||G_{\{u,v\}}|\over|N_{\{u,v\}}|}={2|N||G_{uv}|\over|N_{\{u,v\}}|}={2|N||G_{v}|\over d|N_{\{u,v\}}|}\\ &=&{|N||G_{v}|\over|N_{v}|}\cdot{2|N_{v}|\over d|N_{\{u,v\}}|}=|NG_{v}|{2|N_{v}|\over d|N_{\{u,v\}}|}.\end{array}$$

$$Q_2\times \cdots \times Q_k\cong M_{\{u,v\}}/Q_1=M_{uv}{Q}_1/Q_1\cong M_{uv}/(M_{uv}\cap Q_1).$$

$$G_{\{u,v\}}/M_{\{u,v\}}=G_{\{u,v\}}/(M\cap G_{\{u,v\}})\cong M{G}_{\{u,v\}}/M=G/M\le (\mathsf{Out}(T_1)\times \mathsf{Out}(T_1)):2.$$

$$((T_i{)}_v{)}^h=(T_i\cap G_v{)}^h=T_i^h\cap (G_v{)}^h=T_i\cap G_v=(T_i{)}_v,i=1,2.$$

$$(X_{uv}{)}_{u_1}=(X_{\{u,v\}}{)}_{u_1}=(X_{\{u,v\}}{)}_{v_1^x}=((X_{\{u,v\}}{)}_{v_1}{)}^x=((X_{uv}{)}_{v_1}{)}^x.$$

Full text

Keywords**. Primitive group, almost simple group, edge-primitive graph, $2$-arc-transitive graph.**

2010 Mathematics Subject Classification. 05C25, 20B25

Supported by the National Natural Science Foundation of China (11731002) and the Fundamental Research Funds for the Central Universities.

1. Introduction

All graphs and groups considered in this paper are assumed to be finite.

A graph in this paper is a pair ${\it\Gamma}=(V,E)$ of a nonempty set $V$ and a set $E$ of $2$-subsets of $V$. The elements in $V$ and $E$ are called the vertices and edges of ${\it\Gamma}$, respectively. The number $|V|$ of vertices is called the order of ${\it\Gamma}$. For $v\in V$, the set ${\it\Gamma}(v)=\{u\in V\mid\{u,v\}\in E\}$ is called the neighborhood of $v$ in ${\it\Gamma}$, while $|{\it\Gamma}(v)|$ is the valency of $v$. We say that ${\it\Gamma}$ has valency $d$ or ${\it\Gamma}$ is $d$-regular if its vertices all have equal valency $d$. For an integer $s\geq 1$, an $s$-arc in ${\it\Gamma}$ is an $(s+1)$-tuple $(v_{0},v_{1},\ldots,v_{s})$ of vertices such that $\{v_{i},v_{i+1}\}\in E$ and $v_{i}\neq v_{i+2}$ for all possible $i$. A $1$-arc is also called an arc.

Let ${\it\Gamma}=(V,E)$ be a graph. A permutation $g$ on $V$ is called an automorphism of ${\it\Gamma}$ if $\{u^{g},v^{g}\}\in E$ for all $\{u,v\}\in E$. Let ${\sf Aut}{\it\Gamma}$ denote the set of all automorphisms of ${\it\Gamma}$. Then ${\sf Aut}{\it\Gamma}$ is a subgroup of the symmetric group ${\rm Sym}(V)$, and called the automorphism group of ${\it\Gamma}$. Note that the group ${\sf Aut}{\it\Gamma}$ has a natural action on the edge set $E$ (and also on the set of $s$-arcs). The graph ${\it\Gamma}$ is called edge-transitive if $E\neq\emptyset$ and for each pair of edges there exists some $g\in{\sf Aut}{\it\Gamma}$ mapping one of these two edges to the other one. (Similarly, we may define vertex-transitive, arc-transitive or $s$-arc-transitive** graphs.) An edge-transitive graph is called edge-primitive if some (and hence every) edge-stabilizer, the subgroup of its automorphism group which fixes a given edge, is a maximal subgroup of the automorphism group.**

It is well-known that edge-transitive graphs and hence edge-primitive graphs are either bipartite or vertex-transitive. As a subclass of the edge-transitive graphs, edge-primitive graphs posses more restrictions on their symmetries and automorphism groups. For example, a connected edge-primitive graph is necessarily arc-transitive provided that it is not a star graph. In ****[9]****, appealing to the O’Nan-Scott Theorem for (quasi)primitive groups ****[22]****, Giudici and Li investigated the structural properties of edge-primitive graphs, particularly, on their automorphism groups. Let ${\it\Gamma}=(V,E)$ be an arc-transitive and edge-primitive graph which is neither a cycle nor a complete bipartite graph. If ${\it\Gamma}$ is bipartite then let ${\sf Aut}^{+}{\it\Gamma}$ be the subgroup of ${\sf Aut}{\it\Gamma}$ preserving the bipartition. By ****[9]****, as a primitive group on $E$, only $4$ of the eight O’Nan-Scott types for (quasi)primitive groups may occur for ${\sf Aut}{\it\Gamma}$, say SD, CD, PA and AS. For the first two types, ${\it\Gamma}$ is bipartite and ${\sf Aut}^{+}{\it\Gamma}$ is quasiprimitive of type CD on each bipartite half. For the last two types, with one exception case, ${\sf Aut}{\it\Gamma}$ or ${\sf Aut}^{+}{\it\Gamma}$ is quasiprimitive on $V$ or on each bipartite half respectively of the same type for ${\sf Aut}{\it\Gamma}$ on $E$. In this paper, we will work on the types of ${\sf Aut}{\it\Gamma}$ on $E$ and on $V$ under the further assumption that ${\it\Gamma}$ is $2$-arc-transitive.

The interests for edge-primitive graphs arises partially from the fact that many (almost) simple groups may be represented as the automorphism groups of edge-primitive graphs. Consulting the Atlas ****[3]****, one may get first-hand such examples. For example, the sporadic Higman-Sims group ${\rm HS}$ is the automorphism group of a rank $3$ graph with order $100$ and valency $22$, which is in fact a $2$-arc-transitive and edge-primitive graph; the sporadic Rudvalis group ${\rm Ru}$ is the automorphism group of a rank $3$ graph with order $4060$ and valency $2304$, which is edge-primitive but not $2$-arc-transitive. Besides, the almost groups ${\rm PSU}(3,5).2$, ${\rm M}_{22}.2$, ${\rm J}_{2}.2$ and ${\rm McL}.2$ all have representations on edge-primitive graphs. The reader may refer to ****[11, 12, 18, 21, 26]**** for more examples of edge-primitive graphs which have almost simple automorphism groups. Of course, using the constructions given in ****[9]****, one can easily construct examples of edge-primitive graphs with automorphism groups not almost simple.

We have a strong impression from the known examples for edge-primitive graphs in the literature that a $2$-arc-transitive and edge-primitive graph has almost simple automorphism group unless it is a cycle or a complete bipartite graph. Yet could it be so? Yes, it is true! We shall prove the following result in Section 3.

Theorem 1.1**.**

Let ${\it\Gamma}=(V,E)$ be an edge-primitive $d$-regular graph for some $d\geq 3$. If ${\it\Gamma}$ is $2$-arc-transitive, then either ${\it\Gamma}$ is a complete bipartite graph, or ${\it\Gamma}$ has almost simple automorphism group.

Remarks on Theorem 1.1.

(1) Li and Zhang ****[18]**** proved that $4$-arc-transitive and edge-primitive graphs have almost automorphism groups. Further, as a sequence of their classification on almost simple primitive groups with soluble point-stabilizers, they give a complete list for $4$-arc-transitive and edge-primitive graphs.

(2) By Theorem 1.1, appealing to the classification of almost simple groups with soluble maximal subgroups, it might be feasible to classify $2$-arc-transitive and edge-primitive graphs with soluble edge-stabilizers.

2. Preliminaries

For the subgroups of (almost) simple groups, we sometimes follow the notation used in the Atlas ****[3]****, while we also use $\mathbb{Z}_{l}$ and $\mathbb{Z}_{p}^{k}$ to denote respectively the cyclic group of order $l$ and the elementary abelian group of order $p^{k}$.

2.1. Primitive groups

In this subsection, $\Omega$ is nonempty finite set, and $G$ is a transitive subgroup of the symmetric group ${\rm Sym}(\Omega)$. Let ${\rm soc}(G)$ be the socle of $G$, that is, ${\rm soc}(G)$ is generated by all minimal normal subgroups of $G$.

Consider the point-stabilizer $G_{\alpha}:=\{g\in G\mid\alpha^{g}=\alpha\}$, where $\alpha\in\Omega$. Then

1. (1)

$G$** is primitive if $G_{\alpha}$ is a maximal subgroup of $G$;** 2. (2)

$G$** is ${3\over 2}$-transitive if $G_{\alpha}$ is ${1\over 2}$-transitive on $\Omega\setminus\{\alpha\}$, that is, all $G_{\alpha}$-orbits on $\Omega\setminus\{\alpha\}$ have equal length $>1$;** 3. (3)

$G$** is a Frobenius group if $G_{\alpha}$ is semiregular on $\Omega\setminus\{\alpha\}$;** 4. (4)

$G$** is $2$-transitive if $G_{\alpha}$ is transitive on $\Omega\setminus\{\alpha\}$.**

Note that (4) implies (1) and (2), and (2) implies (1) or (3) (refer to ****[29, Theorem 10.4]****).

Let $1\neq N\unlhd G$, a normal subgroup of $G$. Then $N$ is ${1\over 2}$-transitive, and $N_{\alpha}=N\cap G_{\alpha}\unlhd G_{\alpha}$, and so $G_{\alpha}$ is contained in the normalizer ${\bf N}_{G}(N_{\alpha})$ of $N_{\alpha}$ in $G$. Thus, if $G_{\alpha}$ is maximal then either $N_{\alpha}\unlhd G$ or ${\bf N}_{G}(N_{\alpha})=G_{\alpha}$. The former case yields $N_{\alpha}=1$, while the latter case gives

[TABLE]

Then we have following simple fact for primitive groups.

Lemma 2.1**.**

If $G$ is primitive and $N\neq 1$ then either $N$ is regular on $\Omega$ or $N_{\alpha}$ is self-normalized; if $G$ is $2$-transitive and $N\neq 1$ then $N$ is either regular or ${3\over 2}$-transitive on $\Omega$.

For an almost simple $2$-transitive group $G$, each non-trivial normal subgroup $N$ of $G$ is primitive, and in fact $2$-transitive except for the case where $N={\rm soc}(G)={\rm PSL}(2,8)$ acting on $28$ pionts, refer to ****[1, page 197, Table 7.4]****. Next we consider the normal subgroups of affine $2$-transitive groups. Refer to ****[1, page 195, Table 7.3]**** for a complete list of affine $2$-transitive groups. We consider the affine $2$-transitive groups in their natural actions.

Lemma 2.2**.**

Let $G$ be an affine $2$-transitive group and $1\neq N\unlhd G$. If $N$ is imprimitive on $\Omega$, then $N$ is a soluble Frobenius group, $N_{0}$ is cyclic, and either $G_{0}\leq{\rm\Gamma L}(1,q)$ or $N_{0}\leq{\bf Z}(G_{0})$, where $q$ is not a prime.

Proof.  Assume that $N$ is imprimitive. Then $N\neq G$, and so $N_{0}\neq G_{0}$. Further, by Lemma 2.1 and ****[29**, Theorem 10.4], $N$ is a Frobenius group. Let $|\Omega|=p^{k}$ for a prime $p$. We may write $G_{0}\leq{\rm GL}(k,p)$, $G=\mathbb{Z}_{p}^{k}{:}G_{0}$ and $N=\mathbb{Z}_{p}^{k}{:}N_{0}$. Since $N$ is imprimitive, $N_{0}$ is not maximal in $N$, and thus $N_{0}$ is a normal reducible subgroup of $G_{0}$. Then, by ****[13**, Lemma 5.1]****, $N_{0}$ is cyclic and $|N_{0}|$ is a divisor of $p^{l}-1$, where $l<k$ and l\,\big{|}\,k. Finally, the lemma follows from checking all affine $2$-transitive groups one by one. **$\square$

If every minimal normal subgroup of $G$ is transitive on $\Omega$, then $G$ called a quasiprimitive group. Praeger ****[22, 24]**** generalized the O’Nan-Scott Theorem for primitive groups to quasiprimitive groups, which says that a quasiprimitive group has one of the following eight types: HA, HS, HC, TW, AS, SD, CD and PA. In particular, if $G$ is quasiprimitive then $G$ has at most two minimal normal subgroups, and if two (for HS and HC) then they are isomorphic and regular.

Suppose that $G$ has a transitive insoluble minimal normal subgroup $N$. Then $G=NG_{\alpha}$ for $\alpha\in\Omega$. Write $N=T_{1}\times\cdots\times T_{k}$ for isomorphic nonabelian simple groups $T_{i}$ and integer $k\geq 1$. Then $G_{\alpha}$ acts transitively on $\{T_{i}\mid 1\leq i\leq k\}$ by conjugation. Note that, for $g\in G_{\alpha}$ and $1\leq i\leq k$,

[TABLE]

Then $G_{\alpha}$ acts transitively on $\{(T_{i})_{\alpha}\mid 1\leq i\leq k\}$ by conjugation. Clearly, $(T_{1})_{\alpha}\times\cdots\times(T_{k})_{\alpha}\leq N_{\alpha}$; however, the equality is not necessarily holds even if $G$ is quasiprimitive. A sufficient condition for this equality is that $G$ is primitive and of type AS or PA, refer to ****[4, Theorem 4.6]**** and its proof. In survey, we have the simple fact as follows.

Lemma 2.3**.**

Assume that $G$ has a transitive minimal normal subgroup $N=T_{1}\times\cdots\times T_{k}$, where $T_{i}$ are isomorphic nonabelian simple groups. Let $\alpha\in\Omega$. Then $G_{\alpha}$ acts transitively on $\{(T_{i})_{\alpha}\mid 1\leq i\leq k\}$ by conjugation. If further $G$ is primitive and of type AS or PA, then $N_{\alpha}=(T_{1})_{\alpha}\times\cdots\times(T_{k})_{\alpha}$.

2.2. Locally-primitive graphs

In this subsection, ${\it\Gamma}$ is a connected $d$-regular graph for some $d\geq 3$, and $G\leq{\sf Aut}{\it\Gamma}$. Assume further that the graph ${\it\Gamma}$ is $G$-locally primitive, that is, $G_{v}$ acts primitively on ${\it\Gamma}(v)$ for all $v\in V$.

Fix an edge $\{u,v\}\in E$. Note that $G_{v}$ induces a primitive permutation group $G_{v}^{{\it\Gamma}(v)}$ (on ${\it\Gamma}(v)$). Let $G_{v}^{[1]}$ be the kernel of $G_{v}$ acting on ${\it\Gamma}(v)$. Then $G_{v}^{{\it\Gamma}(v)}\cong G_{v}/G_{v}^{[1]}$. Set $G_{uv}^{[1]}=G_{u}^{[1]}\cap G_{v}^{[1]}$. Then $G_{v}^{[1]}$ induces a normal subgroup of $(G_{u}^{{\it\Gamma}(u)})_{v}$ with the kernel $G_{uv}^{[1]}$. Writing $G_{v}^{[1]}$ and $G_{v}$ in group extensions,

[TABLE]

Assume that $G$ is transitive on $V$. Then $G_{uv}^{[1]}$ is a $p$-group for some prime $p$, refer to ****[6]****. Note that $G$ is transitive on the arcs of ${\it\Gamma}$. There is some element in $G$ interchanging $u$ and $v$. This implies that $(G_{v}^{[1]})^{{\it\Gamma}(u)}\unlhd(G_{v}^{{\it\Gamma}(v)})_{u}\cong(G_{u}^{{\it\Gamma}(u)})_{v}$. Thus we have the following lemma.

Lemma 2.4**.**

Assume that $G$ is transitive on $V$, and $\{u,v\}\in E$. Then $G_{uv}^{[1]}$ is a $p$-group, and $(G_{v}^{[1]})^{{\it\Gamma}(u)}$ is isomorphic to a normal subgroup of a point-stabilizer in $G_{v}^{{\it\Gamma}(v)}$. In particular, $G_{v}$ is soluble if and only if $G_{v}^{{\it\Gamma}(v)}$ is soluble.

The graph ${\it\Gamma}=(V,E)$ is said to be $(G,s)$-arc-transitive if ${\it\Gamma}$ has an $s$-arc and $G$ acts transitively on the set of $s$-arcs of ${\it\Gamma}$, where $s\geq 1$. Note that ${\it\Gamma}$ is $(G,2)$-arc-transitive if and only if $G$ is transitive on $V$, and $G_{v}^{{\it\Gamma}(v)}$ is a $2$-transitive group for some (and hence every) $v\in V$. By ****[7, 27, 28]****, we have the following result.

Theorem 2.5**.**

Assume that ${\it\Gamma}=(V,E)$ is $(G,2)$-arc-transitive. Then ${\it\Gamma}$ is not $(G,8)$-arc-transitive. Further,

1. (1)

if $G_{uv}^{[1]}=1$ then ${\it\Gamma}$ is not $(G,4)$-arc-transitive. 2. (2)

if $G_{uv}^{[1]}\neq 1$ then $G_{uv}^{[1]}$ is a nontrivial $p$-group, ${\bf O}_{p}(G_{uv}^{{\it\Gamma}(v)})\not=1$, ${\rm PSL}(n,q)\unlhd G_{v}^{{\it\Gamma}(v)}$, and $|{\it\Gamma}(v)|={q^{n}-1\over q-1}$, where $n\geq 2$ and $q$ is a power of $p$; in this case, ${\it\Gamma}$ is $(G,4)$-arc-transitive if and only if $n=2$.

3. The proof of Theorem 1.1

In this section, we let ${\it\Gamma}=(V,E)$ be a connected graph of valency $d\geq 3$, and $G\leq{\sf Aut}{\it\Gamma}$. Assume that ${\it\Gamma}$ is $G$-edge-primitive, that is, $G$ act primitively on $E$. Then, by ****[9, Lemma 3.4]****, $G$ acts transitively on the arc set of ${\it\Gamma}$. Thus, for an edge $\{u,v\}\in E$, $d=|G_{v}:G_{uv}|$ and $|G_{\{u,v\}}:G_{uv}|=2$.

Let $1\neq N\unlhd G$. Then $N$ is transitive on $E$, and so either $N$ is transitive on $V$ or $N$ has two orbits on $V$; for the latter case, $N_{v}$ is transitive on ${\it\Gamma}(v)$. This implies that either $G=NG_{v}$, or $|G:(NG_{v})|=2$ and $N_{uv}=N_{\{u,v\}}$. Note that $G=NG_{\{u,v\}}$ by the maximality of $G_{\{u,v\}}$ or the transitivity of $N$ on $E$. We have

[TABLE]

Then the next lemma follows.

Lemma 3.1**.**

Let $1\neq N\unlhd G$. If $N$ is transitive on $V$ then $2|N_{v}|=d|N_{\{u,v\}}|$; if $N$ is intransitive on $V$ then $|N_{v}|=d|N_{\{u,v\}}|=d|N_{uv}|$. In particular, $N_{v}\neq 1$ and $N_{v}\neq N_{\{u,v\}}$.

Let ${\sf K}_{d,d}$ and ${\sf K}_{d+1}$ be the complete bipartite graph and complete graph of valency $d$, respectively.

Corollary 3.2**.**

Let $1\neq N\unlhd G$. Then either ${\it\Gamma}\cong{\sf K}_{d,d}$, or $N_{uv}\neq 1$ and $N_{\{u,v\}}$ is self-normalized in $N$, where $\{u,v\}\in E$.

Proof.  Assume that ${\it\Gamma}\not\cong{\sf K}_{d,d}$. Then, by the O’Nan-Scott Theorem and ****[9**, Lemmas 6.1, 6.2 and Propersition 6.13], $G$ has no normal subgroup acting regularly on $E$. Thus $N_{\{u,v\}}\neq 1$, and so $N_{\{u,v\}}$ is self-normalized in $N$ by Lemma 2.1.

Suppose that $N_{uv}=1$. Then $N_{\{u,v\}}$ has order $2$, and so $N_{\{u,v\}}\leq{\bf C}_{N}(N_{\{u,v\}})\leq{\bf N}_{N}(N_{\{u,v\}})=N_{\{u,v\}}$. This implies that ${\bf C}_{N}(N_{\{u,v\}})={\bf N}_{N}(N_{\{u,v\}})$, and then $N_{\{u,v\}}$ is a Sylow $2$-subgroup of $N$. By Burnside’s transfer theorem (refer ****[14**, IV.2.6]****), $N$ has normal $2^{\prime}$-Hall subgroup, say $M$. Then this $M$ is normal in $G$ and regular on $E$, a contradiction. **$\square$

By ****[9]****, if ${\it\Gamma}\not\cong{\sf K}_{d,d}$ then $G$ has type SD, CD, AS or PA on $E$; in particular, $G$ has a unique (of course, insoluble) minimal normal subgroup. Thus, if ${\it\Gamma}\not\cong{\sf K}_{d,d}$ then $G$ is insoluble, and so $G_{\{u,v\}}$ is not abelian by ****[14, IV.7.4]****. If $G_{uv}$ is abelian the following result says that ${\it\Gamma}\cong{\sf K}_{d,d}$ or ${\sf K}_{d+1}$.

Theorem 3.3**.**

Assume that ${\it\Gamma}\not\cong{\sf K}_{d,d}$. Let $1\neq N\unlhd G$.

1. (1)

If $N_{\{u,v\}}$ has a normal Sylow subgroup $P\neq 1$ then $P$ is also a Sylow subgroup of $N$; in particular, $N_{\{u,v\}}$ is not abelain. 2. (2)

If $N_{uv}$ is abelian then $N$ is transitive on the arc set of ${\it\Gamma}$. 3. (3)

If $N_{uv}$ is an abelian $2$-group then ${\rm soc}(G)={\rm PSL}(2,q)$ and ${\it\Gamma}\cong{\sf K}_{q+1}$, where $q$ is a power of some prime with $q-1$ a power of $2$ greater than $8$. 4. (4)

If $G_{uv}$ is an abelian group then $d=q$ and either ${\rm soc}(G)\cong{\rm PSL}(2,q)$ and ${\it\Gamma}\cong{\sf K}_{q+1}$, or ${\rm soc}(G)={\rm Sz}(q)$, ${\sf Aut}{\it\Gamma}={\sf Aut}({\rm Sz}(q))$ and ${\it\Gamma}$ is $({\rm Sz}(q),2)$-arc-transitive, where $q$ is a power of some prime.

*Proof.  *(1) Assume that $P\neq 1$ is a normal Sylow $p$-subgroup of $N_{\{u,v\}}$. Then $P$ is a characteristic subgroup of $N_{\{u,v\}}$, and so $P\unlhd G_{\{u,v\}}$ as $N_{\{u,v\}}\unlhd G_{\{u,v\}}$. Thus ${\bf N}_{G}(P)\geq G_{\{u,v\}}$, and then ${\bf N}_{G}(P)=G_{\{u,v\}}$ by the maximality of $G_{\{u,v\}}$. This gives ${\bf N}_{N}(P)=N\cap{\bf N}_{G}(P)=N\cap G_{\{u,v\}}=N_{\{u,v\}}$. Choose a Sylow $p$-subgroup $Q$ of $N$ with $P\leq Q$. Then ${\bf N}_{Q}(P)\leq Q\cap{\bf N}_{G}(P)=Q\cap N_{\{u,v\}}=P$. This yields $P=Q$, so $P$ is a Sylow $p$-subgroup of $N$.

Suppose that $N_{\{u,v\}}$ is abelian. Then $N_{\{u,v\}}\leq{\bf C}_{N}(P)\leq{\bf N}_{N}(P)=N_{\{u,v\}}$, yielding ${\bf C}_{N}(P)={\bf N}_{N}(P)$. By Burnside’s transfer theorem, $P$ has a normal complement $H$ in $N$, that is $N=PH$ with $P\cap H=1$ and $H\unlhd N$. Note that $H$ is a Hall subgroup of $N$. It follows that $H$ is characteristic in $N$, and hence $H\unlhd G$. Let $P$ runs over the Sylow subgroup of $N_{\{u,v\}}$. Then the resulting normal complements intersect at a normal complement of $N_{\{u,v\}}$ in $N$, which is normal in $G$ and regular on $E$. This contradicts Corollary 3.2. Therefore, $N_{\{u,v\}}$ is nonabelian, and (1) of this theorem follows.

(2) Assume that $N_{uv}$ is abelian. Then $N_{uv}\neq N_{\{u,v\}}$ by (1), and thus $(u,v)=(v,u)^{x}$ for some $X\in N_{\{u,v\}}$. Since ${\it\Gamma}$ is $N$-edge-transitive, ${\it\Gamma}$ is $N$-arc-transitive.

(3) Assume that $N_{uv}$ is an abelian $2$-group. Recall that $G$ has a unique minimal normal subgroup, say $M$. Then $M\leq N$, and (1) and (2) hold for $M$. Then, since $M_{uv}$ is an abelian $2$-group, $M_{\{u,v\}}$ is a Sylow $2$-subgroup of $M$, and $M_{\{u,v\}}$ is not abelian.

Write $M=T_{1}\times\cdots\times T_{k}$, where $T_{i}$ are isomorphic nonabelian simple groups. Recall that $M_{\{u,v\}}$ is a Sylow $2$-subgroup of $M$. For each $i$, choose a Sylow $2$-subgroup of $T_{i}$ with $Q_{i}\leq M_{\{u,v\}}$. Then $M_{\{u,v\}}=Q_{1}\times\cdots\times Q_{k}$. Noting that $Q_{i}$ are all isomorphic, every $Q_{i}$ is nonabelian; otherwise, $M_{\{u,v\}}$ is abelian, a contradiction. In particular, $Q_{1}\not\leq M_{uv}$. Then $M_{\{u,v\}}=M_{uv}Q_{1}$, and so

[TABLE]

Since ${\rm M}_{uv}$ is abelian, the only possibility is $k=1$. Thus $M={\rm soc}(G)$ is simple.

By ****[10, Corollary 5]****, $M_{\{u,v\}}$ has cyclic commutator subgroup. Since $M_{\{u,v\}}$ is nonabelian, by ****[2]****, $M$ is isomorphic to one of the Mathieu group ${\rm M}_{11}$, ${\rm PSL}(2,q)$ (with $q^{2}-1$ divisible by $16$), ${\rm PSL}(3,q)$ (with $q$ odd) and ${\rm PSU}(3,q)$ (with $q$ odd). If $M\cong{\rm M}_{11}$, then $G=M$, and so $M_{\{u,v\}}$ is maximal in $M$; however, by the Atlas ****[3]****, a Sylow $2$-subgroup of ${\rm M}_{11}$ is not a maximal subgroup, a contradiction. Thus we next let $M\cong{\rm PSL}(2,q)$, ${\rm PSL}(3,q)$ or ${\rm PSU}(3,q)$.

Since $M$ is transitive on $E$, we know that $|E|=|M:M_{\{u,v\}}|$ is odd. Thus $G$ is an almost simple primitive group (on $E$) of odd degree. Noting that $M_{\{u,v\}}=M\cap G_{\{u,v\}}$, by ****[20]****, $M_{\{u,v\}}$ is known. Notice that the isomorphisms among simple groups (refer to ****[15, Proposition 2.9.1 and Theorem 5.1.1]****). Since $M_{\{u,v\}}$ is a Sylow $2$-subgroup of $M$, the only possibility is that $M\cong{\rm PSL}(2,q)$, and $M_{\{u,v\}}$ is the stabilizer of some orthogonal decomposition of a natural projective module associated with $M$ into $1$-dimensional subspaces. It follows that $M_{\{u,v\}}\cong{\rm D}_{q-1}$ or ${\rm D}_{q+1}$, and so $M_{uv}\cong\mathbb{Z}_{q-1\over 2}$ or $\mathbb{Z}_{q+1\over 2}$, respectively. Since $M$ is transitive on the arcs of ${\it\Gamma}$, we have $|M_{v}:M_{uv}|=d\geq 3$. Checking the subgroups of ${\rm PSL}(2,q)$ (refer to ****[14, II.8.27]****), we conclude that $M_{uv}\cong\mathbb{Z}_{q-1\over 2}$, $d=q$, $V=|M:M_{v}|=q+1$ and $M$ is $2$-transitive on $V$. Thus ${\it\Gamma}\cong{\sf K}_{q+1}$.

(4) Assume that $G_{uv}$ is abelian. Let $M$ be the unique minimal normal subgroup of $G$. If $M_{uv}$ is a $2$-group, then (4) of this theorem follows from (3).

We next assume that $|M_{uv}|$ has an odd prime divisor $p$. By (1), the unique Sylow $p$-subgroup of $M_{uv}$ is also a Sylow $p$-subgroup of $M$. Write $M=T_{1}\times\cdots\times T_{k}$ , where $T_{i}$ are isomorphic nonabelian simple groups. By (1) of this theorem, ${\rm M}_{\{u,v\}}$ is not abelian, so $M_{\{u,v\}}\not\leq G_{uv}$, and then $G_{\{u,v\}}=M_{\{u,v\}}G_{uv}$. Thus $G=MG_{uv}$, and hence $G_{uv}$ acts transitively on $\{T_{1},\ldots,T_{k}\}$ by conjugation. Choose, for each $i$, a Sylow $p$-subgroup $P_{i}$ of $T_{i}$ such that $P_{1}\times\cdots\times P_{k}$ is the unique Sylow subgroup $M_{uv}$. Since $G_{uv}$ is abelian, we have $P_{1}=P_{1}^{x}\leq T_{1}^{x}$ for $x\in G_{uv}$. It follows that $P_{1}\leq T_{i}$ for all $i$. The only possibility is that $k=1$, and so $M$ is simple.

Note that $G$ is an almost simple group with a soluble maximal subgroup $G_{\{u,v\}}$. Then, by ****[18]****, both $M={\rm soc}(G)$ and $M_{\{u,v\}}=M\cap G_{\{u,v\}}$ are known. Since $M_{\{u,v\}}$ has an abelian subgroup of index $2$, it follows that either $M\cong{\rm PSL}(2,q)$ and $M_{\{u,v\}}\cong{\rm D}_{2(q\pm 1)\over(2,q-1)}$, or $M={\rm Sz}(q)$ and $M_{\{u,v\}}\cong{\rm D}_{2(q-1)}$. Check the subgroups of $M$, refer to ****[25]**** for ${\rm Sz}(q)$. The former case yields that $M_{v}\cong[q]{:}\mathbb{Z}_{q-1\over(2,q-1)}$ and ${\it\Gamma}\cong{\sf K}_{q+1}$. Assume that $M={\rm Sz}(q)$ and $M_{\{u,v\}}\cong{\rm D}_{2(q-1)}$. Then $M_{v}\cong[q]{:}\mathbb{Z}_{q-1}$ and $d=q$; in this case, ${\it\Gamma}$ is $(M,2)$-arc-transitive. By ****[5**]****, we have that ${\sf Aut}{\it\Gamma}={\sf Aut}({\rm Sz}(q))$ and ${\it\Gamma}$ is unique up to isomorphism. Thus (4) of this theorem follows. **$\square$

Lemma 3.4**.**

Assume that $G$ has type PA on $E$. Let ${\rm soc}(G)=T_{1}\times\cdots\times T_{k}$. Then $(T_{i})_{uv}\neq 1$ for each $i$ and $\{u,v\}\in E$; in particular, every $T_{i}$ is not semiregular.

*Proof.  ***Let $M={\rm soc}(G)$. By Lemma 2.3, $M_{\{u,v\}}=(T_{1})_{\{u,v\}}\times\cdots\times(T_{k})_{\{u,v\}}$, and $(T_{i})_{\{u,v\}}$ all have equal order. By Theorem 3.3, $M_{\{u,v\}}$ is nonabelian. Thus $(T_{i})_{\{u,v\}}$ is nonabelian for all $i$. Then the lemma follows. **$\square$

For the case where ${\it\Gamma}$ is a bipartite graph, we let $G^{+}$ be the subgroup of $G$ preserving the bipartition of ${\it\Gamma}$. Then $|G:G^{+}|=2$, and each bipartite half of ${\it\Gamma}$ is a $G^{+}$-orbit on $V$.

Lemma 3.5**.**

Assume that the graph ${\it\Gamma}=(V,E)$ is $(G,2)$-arc-transitive, and $G$ has type PA on $E$. Then either ${\it\Gamma}\cong{\sf K}_{d,d}$, or one of the following holds:

1. (1)

$G$* is quasiprimitive on $V$;* 2. (2)

${\it\Gamma}$* is bipartite, and $G^{+}$ is faithful and quasiprimitive on each bipartite half of ${\it\Gamma}$.*

*Proof.  *Since $G$ is primitive on $E$, every minimal normal subgroup of $G$ is transitive on $E$, and so has at most two orbits on $V$. If ${\it\Gamma}$ is not bipartite then quasiprimitive on $V$.

Now let ${\it\Gamma}$ be bipartite with bipartition say $V=V_{1}\cup V_{2}$. Note that $G_{v}\leq G^{+}$ for each $v\in V$. Then $G^{+}$ is locally-primitive on ${\it\Gamma}$. Suppose that ${\it\Gamma}\not\cong{\sf K}_{d,d}$. Then, by ****[23]****, $G^{+}$ is faithful on both $V_{1}$ and $V_{2}$, and either (2) of this lemma holds, or the unique minimal normal subgroup of $G$ is a direct product $M_{1}\times M_{2}$, where $M_{1}$ and $M_{2}$ are normal in $G^{+}$ and conjugate in $G$, and $M_{i}$ is intransitive on $V_{i}$ for $i=1,2$. For the latter case, if $M_{1}$ is intransitive on $V_{2}$ then $M_{1}$ is semiregular on $V$ by ****[8**, Lemma 5.1]****; if $M_{1}$ is transitive on $V_{2}$ then $M_{2}$ is semiregular on $V_{2}$. These two cases all contradict Lemma 3.4. Thus $G^{+}$ is quasiprimitive on both $V_{1}$ and $V_{2}$. **$\square$

As permutation groups on $V$ and on $E$, the types of $G$ (and $G^{+}$) have been determined in ****[9]****. Then by Lemma 3.5 and combining with the reduction theorems for 2-arc-transitive graphs given by Preager ****[22, 23]****, we get the following result.

Lemma 3.6**.**

Assume that the graph ${\it\Gamma}=(V,E)$ is $(G,2)$-arc-transitive. Suppose that ${\it\Gamma}\not\cong{\sf K}_{d,d}$. If $G$ is not almost simple, then $G$ has type PA on $E$ and either

1. (1)

$G$* is quasiprimitive and of type PA on $V$; or* 2. (2)

${\it\Gamma}$* is bipartite, $G^{+}$ is faithful and quasiprimitive on each bipartite half of ${\it\Gamma}$ with type PA.*

Now we are ready to give a proof of Theorem 1.1.

Theorem 3.7**.**

Let ${\it\Gamma}=(V,E)$ be a connected $d$-regular graph for some $d\geq 3$, and let $G\leq{\sf Aut}{\it\Gamma}$. Assume that ${\it\Gamma}$ is both $G$-edge-primitive and $(G,2)$-arc-transitive. Then either ${\it\Gamma}\cong{\sf K}_{d,d}$, or $G$ is almost simple.

*Proof.  *Assume that ${\it\Gamma}\not\cong{\sf K}_{d,d}$, and let $\{u,v\}\in E$. By the 2-arc-transitivity of $G$ on ${\it\Gamma}$, we know that $G_{v}^{{\it\Gamma}(v)}$ is a $2$-transitive permutation group of degree $d$.

Let $M={\rm soc}(G)=T_{1}\times\cdots\times T_{k}$, where $T_{i}$ are isomorphic nonabelian simple groups. Then $M_{v}\unlhd G_{v}$, and $M_{v}\neq 1$ by Lemma 3.1 or 3.4. Thus $M_{v}^{{\it\Gamma}(v)}$ is a transitive normal subgroup of $G_{v}^{{\it\Gamma}(v)}$.

Assume that $M_{v}^{{\it\Gamma}(v)}$ is primitive on ${\it\Gamma}(v)$. Noting that $G$ is transitive on $V$, we conclude that $M_{w}^{{\it\Gamma}(w)}$ is primitive for every $w\in V$. Thus ${\it\Gamma}$ is $M$-locally primitive. Then, by Lemma 3.4 and ****[8, Lemma 5.1]****, we conclude that $k=1$, and so $G$ is almost simple.

Next assume that $M_{v}^{{\it\Gamma}(v)}$ is imprimitive on ${\it\Gamma}(v)$.

Note that every non-trivial normal subgroup of an almost simple $2$-transitive group is primitive. Then $G_{v}^{{\it\Gamma}(v)}$ is an affine $2$-transitive group, and by Lemma 2.2, $M_{v}^{{\it\Gamma}(v)}$ is a soluble Frobenius group and $(M_{v}^{{\it\Gamma}(v)})_{u}$ is cyclic. Set $(M_{v}^{{\it\Gamma}(v)})_{u}\cong\mathbb{Z}_{e}$ and ${\rm soc}(G_{v}^{{\it\Gamma}(v)})\cong\mathbb{Z}_{r}^{l}$ for a prime $r$ and integer $l\geq 1$ with $d=r^{l}$. Then $e$ is a divisor of $r^{l}-1$, and $e<r^{l}-1$.

Assume that $e=1$. Then $M_{v}^{{\it\Gamma}(v)}={\rm soc}(G_{v}^{{\it\Gamma}(v)})\cong\mathbb{Z}_{r}^{l}$, and so $M_{v}^{{\it\Gamma}(v)}$ is regular on ${\it\Gamma}(v)$. By ****[17, Lemma 2.3]****, $M_{v}$ is faithful and hence regular on ${\it\Gamma}(v)$, and thus $M_{uv}=1$, which contradicts Corollary 3.2. Thus $e\neq 1$.

Note that $e$ is a proper divisor of $d-1=r^{l}-1$. Neither $d$ nor $d-1$ is a prime, in particular, $l>1$ and $d=r^{l}\geq 9$. Thus $G_{v}^{{\it\Gamma}(v)}$ has no normal subgroup isomorphic to a projective special linear group of dimension $\geq 2$. By Theorem 2.5, $G_{uv}^{[1]}=1$, and so $M_{uv}^{[1]}=1$.

Let $x\in G_{\{u,v\}}\setminus G_{uv}$. Then $(u,v)^{x}=(v,u)$, this implies that $M_{v}^{{\it\Gamma}(v)}$ and $M_{u}^{{\it\Gamma}(u)}$ are permutation isomorphic. In particular, $(M_{u}^{{\it\Gamma}(u)})_{v}\cong(M_{v}^{{\it\Gamma}(v)})_{u}=\mathbb{Z}_{e}$. Since $M_{v}^{[1]}\cap M_{u}^{[1]}=M_{uv}^{[1]}=1$, we know that $M_{uv}$ is isomorphic to a subgroup of $(M_{uv}/M_{u}^{[1]})\times(M_{uv}/M_{v}^{[1]})$. Note that $M_{uv}/M_{v}^{[1]}\cong(M_{v}^{{\it\Gamma}(v)})_{u}$ and $M_{uv}/M_{u}^{[1]}\cong(M_{u}^{{\it\Gamma}(u)})_{v}$. Then $M_{uv}$ is isomorphic to a subgroup of $\mathbb{Z}_{e}\times\mathbb{Z}_{e}$. In particular, $M_{uv}$ is abelian. Then, by Theorem 3.3, $M$ is transitive on the arcs of ${\it\Gamma}$, and so $M_{\{u,v\}}=M_{uv}.2$.

If $e$ is a power of $2$ then, by Theorem 3.3, $M\cong{\rm PSL}(2,r^{l})$, ${\it\Gamma}\cong{\sf K}_{r^{l}+1}$; however, in this case, $M$ is locally primitive on ${\it\Gamma}$, a contradiction. Thus $e$ has odd prime divisors. Let $s$ be an odd prime divisor of $e$, and $S$ be a Sylow $s$-subgroup of $M_{uv}$. Then, noting that $M_{\{u,v\}}=M_{uv}.2$, we know that $S$ is also a Sylow $s$-subgroup of $M$ by Theorem 3.3. Thus $S=S_{1}\times\cdots\times S_{k}$, where $S_{i}$ is a Sylow $s$-subgroup of $T_{i}$ for $1\leq i\leq k$. Since $M_{uv}$ is isomorphic to a subgroup of $\mathbb{Z}_{e}\times\mathbb{Z}_{e}$, we know that $M_{uv}$ has no subgroup isomorphic to $\mathbb{Z}_{s}^{3}$. It follows that $k\leq 2$.

Now we deduce a contradiction by supposing that $k=2$.

Let $k=2$. Since $G\leq({\sf Aut}(T_{1})\times{\sf Aut}(T_{1})){:}2$, we have

[TABLE]

It follows that $G_{\{u,v\}}/M_{\{u,v\}}$ is soluble, and so $G_{\{u,v\}}$ is soluble as $M_{\{u,v\}}$ is soluble. Thus $(G_{v}^{{\it\Gamma}(v)})_{u}$ is soluble, and $G_{v}^{{\it\Gamma}(v)}={\rm soc}(G_{v}^{{\it\Gamma}(v)}){:}(G_{v}^{{\it\Gamma}(v)})_{u}$ is also soluble. Checking the soluble affine $2$-transitive groups, by Lemma 2.2, $(G_{v}^{{\it\Gamma}(v)})_{u}\leq{\rm\Gamma L}(1,r^{l})$ or $\mathbb{Z}_{e}\cong(M_{v}^{{\it\Gamma}(v)})_{u}\leq{\bf Z}((G_{v}^{{\it\Gamma}(v)})_{u})\cong\mathbb{Z}_{2}$. Note that $(M_{v}^{{\it\Gamma}(v)})_{u}$ is a reducible subgroup of $(G_{v}^{{\it\Gamma}(v)})_{u}$. Recalling that $e$ is not a power of $2$, the latter case does not occur.

Since $|M_{\{u,v\}}:M_{uv}|=2$, we have $M_{\{u,v\}}\not\leq G_{uv}$, and so $G_{uv}\neq M_{\{u,v\}}G_{uv}\leq G_{\{u,v\}}$. Then $M_{\{u,v\}}G_{uv}=G_{\{u,v\}}$, and $G=MG_{\{u,v\}}=MG_{uv}$. Recalling that $M=T_{1}\times T_{2}$, it follows that $G_{uv}$ acts transitively on $\{T_{1},T_{2}\}$ by conjugation. Let $H$ be the kernel of this action. Then $|G_{uv}:H|=2$, and each $T_{i}$ is normalized by $H$. For $h\in H$,

[TABLE]

This implies that $H$ normalizes each $(T_{i})_{v}$. Then $(T_{i})_{v}^{{\it\Gamma}(v)}$ is normalized by $H^{{\it\Gamma}(v)}$. Note that $(T_{i})_{v}^{{\it\Gamma}(v)}$ is a normal subgroup of $M_{v}^{{\it\Gamma}(v)}={\rm soc}(G_{v}^{{\it\Gamma}(v)}){:}(M_{v}^{{\it\Gamma}(v)})_{u}$, and $e=|(M_{v}^{{\it\Gamma}(v)})_{u}|$ is a proper divisor of $r^{l}-1$. Let $K_{i}$ be the Sylow $r$-subgroup of $(T_{i})_{v}^{{\it\Gamma}(v)}$. Then $K_{i}$ is normalized by $H^{{\it\Gamma}(v)}$, of course, $K_{i}\leq{\rm soc}(G_{v}^{{\it\Gamma}(v)})$ and $K_{1}\cap K_{2}=1$.

Recalling that $|G_{uv}:H|=2$, we have $|(G_{v}^{{\it\Gamma}(v)})_{u}:H^{{\it\Gamma}(v)}|\leq 2$. Since $G_{v}^{{\it\Gamma}(v)}$ is $2$-transitive, $|(G_{v}^{{\it\Gamma}(v)})_{u}|$ is divisible by $r^{l}-1$, and so $|H^{{\it\Gamma}(v)}|$ is divisible by ${r^{l}-1}\over 2$. Note that ${{r^{l}-1}\over 2}>{r^{l}\over 2}-1\geq r^{l-1}-1$. Then $|H^{{\it\Gamma}(v)}|$ is not a divisor of $r^{b}-1$ for all $1\leq b<l$. Then, by ****[13, Lemma 5.1]****, $H^{{\it\Gamma}(v)}$ is irreducible on ${\rm soc}(G_{v}^{{\it\Gamma}(v)})$. It implies that $K_{1}=K_{2}=1$, and thus $(T_{i})_{v}^{{\it\Gamma}(v)}\leq(M_{v}^{{\it\Gamma}(v)})_{u}$ for $i=1,2$. Let $u$ run over ${\it\Gamma}(v)$. It follows that $(T_{i})_{v}^{{\it\Gamma}(v)}=1$, and hence $(T_{i})_{v}\leq M_{v}^{[1]}$, $i=1,2$. Since $M$ is transitive on $V$, by ****[17**, Lemma 2.3]****, we have $(T_{1})_{v}=(T_{2})_{v}=1$, which contradicts Lemma 3.4. This completes the proof. **$\square$

As consequence of Theorems 3.3 and 3.7, an edge-primitive graph of prime valency is $2$-arc-transitive, and then it has almost simple automorphism group if it is not a complete bipartite graph. See also ****[21]****.

Corollary 3.8**.**

Assume that $d$ is a prime and ${\it\Gamma}\not\cong{\sf K}_{d,d}$. Then $G$ is almost simple, and either $G={\rm PSL}(2,d)$ with $d>11$ and ${\it\Gamma}\cong{\sf K}_{d+1}$ or $G$ is transitive on the $2$-arcs of ${\it\Gamma}$.

*Proof.  *Note that $G$ is transitive on the arc set of ${\it\Gamma}$. Let $\{u,v\}\in E$. By Theorem 3.7, it suffices to deal with the case where $G_{v}^{{\it\Gamma}(v)}$ is not $2$-transitive.

Suppose that $G_{v}^{{\it\Gamma}(v)}$ is not $2$-transitive. Then $G_{v}^{{\it\Gamma}(v)}\cong\mathbb{Z}_{d}{:}\mathbb{Z}_{l}$ with $l<d-1$ and $l$ a divisor of $d-1$. If $l=1$ then $G_{v}\cong\mathbb{Z}_{d}$ by ****[17**, Lemma 2.3]****, and so $G_{uv}=1$, which contradicts Corollary 3.2. Then $l>1$, and so $d\geq 5$. By Theorem 2.5, $G_{uv}^{[1]}=1$. Then $G_{uv}$ is isomorphic to a subgroup of $(G_{u}^{{\it\Gamma}(u)})_{v}\times(G_{v}^{{\it\Gamma}(v)})_{u}\cong\mathbb{Z}_{l}\times\mathbb{Z}_{l}$. Thus $G_{uv}$ is abelian. By Theorem 3.3, ${\it\Gamma}\cong{\sf K}_{d+1}$, ${\rm soc}(G)\cong{\rm PSL}(2,d)$, ${\rm soc}(G)_{v}\cong\mathbb{Z}_{d}{:}\mathbb{Z}_{d-1\over 2}$ and ${\rm soc}(G)_{\{u,v\}}\cong{\rm D}_{d-1}$. If $G\cong{\rm PGL}(2,d)$ then $G$ is transitive on the $2$-arcs of ${\it\Gamma}$, which is not the case. Thus $G\cong{\rm PSL}(2,d)$, and so $d>11$ by the maximality of $G_{\{u,v\}}$. **$\square$

4. Examples

Let ${\it\Gamma}=(V,E)$ be a connected $d$-regular graph, where $d\geq 3$. Let $v\in V$ and $G\leq{\sf Aut}{\it\Gamma}$. Assume that ${\it\Gamma}$ is $(G,2)$-arc-transitive. Choose an integer $s\geq 2$ such that ${\it\Gamma}$ is $(G,s)$-arc-transitive but not $(G,s+1)$-arc-transitive; in this case, we call ${\it\Gamma}$ a $(G,s)$-transitive graph. Then $s\leq 7$ by ****[28]****. If $G_{v}$ is faithful on ${\it\Gamma}(v)$ then $s\leq 3$ by Theorem 2.5, and $s=3$ yields that $d=7$ and $G_{v}\cong{\rm A}_{7}$ or ${\rm S}_{7}$, see ****[16, Proposition 2.6]****. This leads to the following interesting problem: Do there exist $3$-arc-transitive graphs with faithful stabilizers? We next answer this problem by giving several examples of edge-primitive graphs which are $3$-transitive and have faithful stabilizers.

The first example is the Hoffman-Singleton graph, which has valency $7$, order $50$ and automorphism group $G=T.2$, where $T={\rm PSU}(3,5)$. Let $X=T$ or $G$. For an edge ${\{u,v\}}$ of this graph, $X_{v}\cong{\rm A}_{7}$ or ${\rm S}_{7}$ and $X_{\{u,v\}}\cong{\rm M}_{10}$ or ${\rm P\Gamma L}(2,9)$, which are maximal subgroups of $X$. Thus the Hoffman-Singleton graph is both $X$-edge-primitive and $(X,2)$-arc-transitive. To see the $3$-arc-transitivity, we fix an edge $\{u,v\}$ and consider the action of the arc-stabilizer $X_{uv}$ ($\cong{\rm A}_{6}$** or ${\rm S}_{6}$) on ${\it\Gamma}(u)\cup{\it\Gamma}(v)$. By the $2$-arc-transitivity of $X$, we have two faithful transitive actions of $X_{uv}$ on ${\it\Gamma}(u)$ and ${\it\Gamma}(v)$, respectively. Let $v_{1}\in{\it\Gamma}(v)\setminus\{u\}$ and $x\in X_{\{u,v\}}\setminus X_{uv}$. Then $u_{1}:=v_{1}^{x}\in{\it\Gamma}(u)\setminus\{v\}$, and**

[TABLE]

By the choice of $x$, we know that $(X_{uv})_{v_{1}}$ and $((X_{uv})_{v_{1}})^{x}$ are not conjugate in $X_{uv}$, and so do for $(X_{uv})_{v_{1}}$ and $(X_{uv})_{u_{1}}$. This implies that the actions of $X_{uv}$ on ${\it\Gamma}(u)$ and ${\it\Gamma}(v)$ are not equivalent. Thus $(X_{uv})_{v_{1}}$ acts on ${\it\Gamma}(u)\setminus\{v\}$ without fixed-points, this yields that $(X_{uv})_{v_{1}}$ is transitive on ${\it\Gamma}(u)\setminus\{v\}$. It follows that the Hoffman-Singleton graph is $(X,3)$-arc-transitive.

In general, combining with ****[16, Proposition 2.6]****, a similar argument as above yields the following result.

Lemma 4.1**.**

Let ${\it\Gamma}=(V,E)$ be a connected $d$-regular graph for $d\geq 3$, $\{u,v\}\in E$ and $G\leq{\sf Aut}{\it\Gamma}$. If ${\it\Gamma}$ is $(G,2)$-arc-transitive and $G_{v}$ is faithful on ${\it\Gamma}(v)$, then ${\it\Gamma}$ is $(G,3)$-arc-transitive if and only if $d=7$, ${\rm soc}(G_{v})\cong{\rm A}_{7}$ and $G_{\{u,v\}}\not\cong{\rm S}_{6}$, i.e. $G_{\{u,v\}}\cong{\rm PGL}(2,9)$, ${\rm M}_{10}$ or ${\sf Aut}({\rm A}_{6})$.

We next give another example.

Example 4.2**.**

By the information given in the Atlas [3] for the O’Nan simple group ${\rm O^{\prime}N}$, there exactly two conjugacy classes $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ of (maximal) subgroups isomorphic to ${\rm A}_{7}$, which are merged into one class in ${\rm O^{\prime}N}.2$. Further, there are $H\in\mathcal{C}_{1}$ and involutions $x_{1},x_{2}\in{\rm O^{\prime}N}.2\setminus{\rm O^{\prime}N}$ such that $(H\cap H^{x_{i}}){:}\langle x_{i}\rangle$ all are maximal subgroups of ${\rm O^{\prime}N}.2$ with $(H\cap H^{x_{1}}){:}\langle x_{1}\rangle\cong{\rm PGL}(2,9)$ and $(H\cap H^{x_{2}}){:}\langle x_{2}\rangle\cong{\rm PSL}(2,7){:}2$. Define two bipartite graphs ${\it\Gamma}_{1}=(V,E_{1})$ and ${\it\Gamma}_{2}=(V,E_{2})$ with vertex set $V=\mathcal{C}_{1}\cup\mathcal{C}_{2}$ and edge sets

1. $E_{1}=\{\{H_{1},H_{2}\}\mid H_{1}\in\mathcal{C}_{1},H_{2}\in\mathcal{C}_{2},H_{1}\cap H_{2}\cong{\rm A}_{6}\}$;
2. $E_{2}=\{\{H_{1},H_{2}\}\mid H_{1}\in\mathcal{C}_{1},H_{2}\in\mathcal{C}_{2},H_{1}\cap H_{2}\cong{\rm PSL}(2,7)\}$.

Then ${\it\Gamma}_{1}$ and ${\it\Gamma}_{2}$ are both ${\rm O^{\prime}N}.2$-edge-primitive and $({\rm O^{\prime}N}.2,2)$-arc-transitive, which have valency $7$ and $15$ respectively. By Lemma 4.1, only ${\it\Gamma}_{1}$ is $({\rm O^{\prime}N}.2,3)$-arc-transitive.* $\square$*

Lemma 4.3**.**

Let ${\it\Gamma}_{1}$ be as in Example 4.2. Then ${\sf Aut}{\it\Gamma}_{1}={\rm O^{\prime}N}.2$.

Proof.  Let $G={\sf Aut}{\it\Gamma}_{1}$. Then $G\geq{\rm O^{\prime}N}.2$. By Theorem 1.1, $G$ is almost simple, and so ${\rm O^{\prime}N}\leq{\rm soc}(G)$. Let $\{u,v\}$ be an edge of ${\it\Gamma}_{1}$. Then $G_{v}^{{\it\Gamma}(v)}\cong{\rm A}_{7}$ or ${\rm S}_{7}$, and $G_{uv}^{[1]}=1$ by Theorem 2.5. Thus, by the group extensions ($\circledast$) in Section 2, we conclude that $|G_{v}|$ has no prime divisor other than $2$, $3$, $5$ and $7$. Since ${\rm O^{\prime}N}.2$ is transitive on the vertices of ${\it\Gamma}_{1}$, we have $G=({\rm O^{\prime}N}.2)G_{v}$. It follows that $|{\rm O^{\prime}N}|$ and $|{\rm soc}(G)|$ have the same prime divisors. Using ****[19**, Corollary 5]**, we get ${\rm soc}(G)={\rm O^{\prime}N}$, and so $G={\rm O^{\prime}N}.2$. **$\square$