On the structure of vertex stabilizers of arc-transitive locally quasiprimitive graphs

Abstract

Let $\Delta$ be a connected arc-transitive G-graph which is locally finite and locally quasiprimitive. Let $\{x,y\}$ be an edge of $\Delta$. A relation between $G_x^{[1]}/O_p(G_x^{[1]})$ and the existence of certain normal subgroups of $G_x^{\Delta (x)}$ and $G_{x,y}^{\Delta (x)}$ is established. This is then used to determine the vertex stabilizers of a class of 2-arc-transitive graphs with trivial edge kernel.

1 Introduction

In this article we study the action of groups on graphs. We will only consider graphs that are connected, undirected and without loops or multiple edges. A G-graph is a graph $\Delta$ together with a subgroup $G \le \textrm{Aut}(\Delta )$. Our focus is on the relation between local properties of the graph $\Delta$ and subgroups of G. We start with introducing some notation and terminology. Let $\Delta$ be a G-graph. The vertex set and edge set of $\Delta$ will be denoted with $V\Delta$ and $E\Delta$, respectively. For any vertex $z\in V\Delta$, we define the vertex stabilizer in G of z by $G_z:=\{g \in G \mid z^g=z\}$ and the set of neighbours of z in $\Delta$ by $\Delta (z):=\{v \mid \{z, v\} \in E\Delta \}$. The group $G_z$ acts on the set $\Delta (z)$. The group induced by $G_z$ on $\Delta (z)$ will be denoted by $G_z^{\Delta (z)}$, and its kernel of action by $G_z^{[1]}$. Thus, $G_z^{\Delta (z)}\cong G_z/G_z^{[1]}$. An s-arc emanating from $x_0 \in V \Delta$ is a path $(x_0, x_1, \ldots , x_s)$ with $x_{i-1} \ne x_{i+1}$ for $1 \le i\le s-1$. A permutation group, acting on a set $\Omega$, is called quasiprimitive if each of its non-trivial normal subgroups is transitive on $\Omega$, and is called 2-transitive if it is transitive on the set of ordered pairs of distinct elements of $\Omega$. The G-graph $\Delta$ is called

• arc-transitive if G acts transitively on the set of 1-arcs of $\Delta$;

• locally finite if for each $z \in V\Delta$ the stabilizer $G_z$ is a finite group;

• locally quasiprimitive if for each $z \in V\Delta$ the group $G_z^{\Delta (z)}$ is a quasiprimitive permutation group on $\Delta (z)$;

• locally 2-transitive if for each $z \in V\Delta$ the group $G_z^{\Delta (z)}$ is a 2-transitive permutation group on $\Delta (z)$;

• locally s-arc transitive if for each $z \in V\Delta$ the group $G_z$ is transitive on the set of s-arcs emanating from z.

Note that, if $\Delta$ is a G-graph in which there are no vertices of valency 1, then $\Delta$ is locally 2-transitive if and only if it is locally s-arc transitive for some $s \ge 2$.

Let $\Delta$ be a locally finite and locally quasiprimitive G-graph, and let $\{x,y\}$ be an edge. Then G is transitive on the set of edges $E \Delta$ and, consequently, has at most two orbits on the vertex set $V \Delta$. The triple $(G_{x},G_{y};G_{x,y})$ is called the vertex stabilizer amalgam with respect to the 1-arc (x, y). The vertex stabilizer amalgam describes the group and graph locally. The central problem in the theory of locally finite and locally quasiprimitive G-graphs is to determine the structure of the vertex stabilizer amalgam. Indeed, since it is impossible to determine $\Delta$ only from its local structure, the best one can hope for is a classification of the vertex stabilizer amalgams up to isomorphism.

The study of G-graphs by means of their vertex stabilizers goes back to the seminal work of Tutte [13, 14] on finite trivalent graphs, where the order of $G_x$ was bounded when G is transitive on $V\Delta$. This result was later generalized by Goldschmidt [3] who bounded the order of $G_x$ for locally finite trivalent G-graphs, and determined their vertex stabilizer amalgams when they are locally primitive (there are 15 isomorphism types). For locally finite and locally s-arc transitive graphs with $s \ge 6$, in which each vertex has valency at least three, the vertex stabilizer amalgams are weak BN-pairs, see [16]. Therefore, their vertex stabilizer amalgams are known by Delgado and Stellmacher [2]. For $s\ge 4$ a partial classification of the vertex stabilizer amalgams was recently obtained in [17].

One of the few general results on the structure of vertex stabilizers of G-graphs finds their origin in work by Thompson and Wielandt [11, 21] and is known under the name of the Thompson–Wielandt Theorem. For a locally finite and locally quasiprimitive G-graph with G transitive on $V\Delta$, the Thompson–Wielandt Theorem implies that the group $G_{x,y}^{[1]}:=G_{x}^{[1]}\cap G_{y}^{[1]}$ is a (possibly trivial) p-group, see [15]. We will discuss the case where G is transitive on the vertex set of $\Delta$ in some more detail.

Let $\Delta$ be an arc-transitive locally finite and locally quasiprimitive G-graph. Since $\Delta$ is also vertex-transitive, we have $G_x \cong G_y$ and, by the Thompson–Wielandt Theorem, $G_{x,y}^{[1]}$ is a (possibly trivial) p-group. When $G_{x,y}^{[1]}=1$, the group $G_x^{[1]}$ is isomorphic to a subgroup of $G_x^{\Delta (x)}$ and so its order is bounded by $|\Delta (x)|!$. In [19] Weiss conjectured that the order of $O_p(G_x^{[1]})$ is also bounded by a function of $|\Delta (x)|$ when $G_{x,y}^{[1]}\ne 1$ and $G_x^{\Delta (x)}$ is a primitive permutation group on $\Delta$. In the study of this type of G-graphs, one often assumes knowledge about $G_x^{\Delta (x)}$ and its action on $\Delta (x)$. Using this information, one tries to obtain the structure of $O_p(G_x^{[1]})$ or bound its order. The vertex stabilizer amalgams of (finite) arc-transitive locally 2-transitive graphs with $G_{x,y}^{[1]}\ne 1$ have been studied in a series of papers by Trofimov and by Weiss. We refer the reader to [12, 19, 20] for more information about what is known in that case.

Recently, there is also interest in determining the vertex stabilizer amalgams of arc-transitive G-graphs of a given valency [4,5,6,7,8]. In these studies the case of $G_{x,y}^{[1]}=1$ is quite often analysed using ad hoc arguments.

This raises the question about what can be said about $G_x^{[1]}$ when $G_{x,y}^{[1]}=1$, and about $G_x^{[1]}/O_p(G_x^{[1]})$ in general. In this article we will have a closer look at the structure of $G_x^{[1]}/O_p(G_x^{[1]})$ and relate it to the existence of certain normal subgroups of $G_x^{\Delta (x)}$ and $G_{x,y}^{\Delta (x)}$. Before stating the main theorem, we need the following definition. Let K be a finite group, and N and M be normal subgroups of K with $N \trianglelefteq M$. We define

$$\begin{aligned} C_K(M/N):=\langle k \in K \mid [\langle k \rangle ,M] \le N\rangle . \end{aligned}$$

Observe that this is a normal subgroup of K containing N. Moreover, for $a,b \in K$ and $c\in M$ we have $[ab,c]=[a,c]^b[b,c]$. Since $N \trianglelefteq K$, we get

$$\begin{aligned} C_K(M/N)= \{k \in K \mid [k,c]\in N \hbox { for all } c\in M\}. \end{aligned}$$

Theorem 1.1

Let $\Delta$ be a connected arc-transitive locally finite and locally quasiprimitive G-graph of valency at least 2. Let $\{x,y\}$ be an edge of $\Delta$. Let p be a prime such that $O_p(G_{x,y}^{[1]})=G_{x,y}^{[1]}$ and let $A:=C_{G_x} (G_x^{[1]}/O_p(G_x^{[1]}))$. Let $H:=G_x^{\Delta (x)}$, $E:=A^{\Delta (x)}$, $Y:=(G_y^{[1]})^{\Delta (x)}$, $F:=E C_{H_y}(Y/O_p(Y))$ and $N:=YC_{E_y}(Y/O_p(Y))$. Assume that H is not regular on $\Delta (x)$. Then $1 \ne E \trianglelefteq F \trianglelefteq H$ with $E\trianglelefteq H$, $Y \trianglelefteq N \trianglelefteq E_y$ with $Y \trianglelefteq E_y$, and the following hold:

• 1. $F/E \cong E_y/N$. In particular, $|F:E|=|E_y:N|$;

• 2. $E_y/N$ is isomorphic to a subgroup of $\textrm{Out}(Y/O_p(Y))$;

• 3. F/E and H/E are isomorphic to subgroups of $\textrm{Out}(E)$ and of $\textrm{Out}(G_x^{[1]}/O_p(G_x^{[1]}))$;

• 4. $C_{H_y}(Y/O_p(Y)) \cong C_{ F_y}(Y/O_p(Y))$.

In particular, for $R:=N/O_p(Y)$, $R_1:=Y/O_p(Y)$ and $R_2:=C_{E_y}(Y/O_p(Y))/O_p(Y)$, we have $R=R_1R_2$ with $[R_1,R_2]=1$ and $R_1 \cap R_2=Z(R_1) \le Z(R_2)=Z(R)$.

Remarks.

1. H is regular on $\Delta (x)$ if and only if $G_x$ is regular on $\Delta (x)$. In this case $G_x \cong S$, with S a finite simple group.

2. If $G_{x,y}^{[1]}=1$, then we can choose p to be a prime not dividing $|G_x|$ and thus $O_p(G_x^{[1]})=O_p(Y)=1$. In particular, $G_y^{[1]} \cong Y$ and $N=YC_{E_y}(Y)$.

3. If $C_{H_y}(Y/O_p(Y))\le Y$, then $Y=N$ and $E=F$, and therefore $Y=E_y$.

The following theorem describes the structure of $G_x/O_p(G_x^{[1]})$ in terms of $G_x^{[1]}$ and A.

Theorem 1.2

Let $\Delta$ be a connected arc-transitive locally finite and locally quasiprimitive G-graph of valency at least 2. Let $\{x,y\}$ be an edge of $\Delta$. Let p be a prime with $O_p(G_{x,y}^{[1]})= G_{x,y}^{[1]}$. Let $A:=C_{G_x} (G_x^{[1]}/O_p(G_x^{[1]}))$, $E:=A^{\Delta (x)}$ and $H:=G_x^{\Delta (x)}$. For any subgroup $K \le G_x$ with $O_p(G_x ^{[1]})\le K$, let $\overline{K}:= K/O_p(G_x^{[1]})$. Then one of the following holds:

• (i) $G_{x}^{[1]}=1$ and $G_x \cong H$;

• (ii) $G_x^{[1]}\ne 1$, $Z(\overline{G_x^{[1]}}) = Z(\overline{A})$, $\overline{A}/Z(\overline{A}) \cong E$ and $\overline{G_x} \cong (\overline{G_x^{[1]}} \circ \overline{A}).D$ with $C_{\overline{G_x}} (\overline{G_x^{[1]}}) \cong \overline{A}$, $C_{\overline{G_x}} (\overline{A}) = \overline{G_x^{[1]}}$ and $D\cong H/E \le \textrm{Out}(E)$.

As an application of 1.1 and 1.2, we prove the following proposition to obtain the structure of vertex stabilizers of a class of arc-transitive locally finite and locally 2-transitive G-graphs with trivial edge kernel.

Proposition 1.3

Let $\Delta$ be a connected arc-transitive locally finite and locally 2-transitive G-graph of valency at least 2. Let $\{x,y\}$ be an edge, and let $H:=G_x^{\Delta (x)}$. Suppose that $G_{x,y}^{[1]}=1$ and that $F^*(H)$ is a non-abelian simple group. Then either $G_x\cong H$ or there exists a group E with $F^*(H) \trianglelefteq E \trianglelefteq H$ such that $G_x^{[1]}\cong E_y$ and $G_x \cong (E_y \times E).D$, with $C_{G_x}(E_y)\cong E$, $C_{G_x}(E)\cong E_y$ and $D \cong H/E$.

Remark

The amalgams with $G_x\not \cong H$ appearing in the conclusion can be found inside the action of $H\wr C_2$ acting on the complete bipartite graph with vertex set $V=V_1 \dot{\cup }V_2$ where $V_1=V_2=\Delta (x)$.

The organization of this paper is as follows. In Section 2 we will define some special subgroups of a vertex stabilizer and derive some properties of them. In Section 3 we will then use these subgroups to prove Theorems 1.1 and 1.2. In Section 4 we will then apply these results to prove Proposition 1.3. In Section 5 we will illustrate how these results can be used to obtain restrictions on the structure of vertex stabilizers of arc-transitive locally quasiprimitive graphs. The notation used in this paper is standard in the theory of (locally) quasiprimitive G-graphs. We will use the following isomorphism theorems.

Isomorphism Theorems. Let G be a group, and let K, N and M be subgroups of G with $M \trianglelefteq G$. Then

• 1. $KM/M \cong K/K\cap M$;

• 2. If $N \trianglelefteq G$ and $N \le M$, then $G/M \cong (G/N)/(M/N)$;

• 3. If $N \trianglelefteq K$ with $K \cap M=N\cap M$, then $K/N \cong KM/NM$.

2 Locally quasiprimitive graphs

In this section, we will define some subgroups of a vertex stabilizer in a locally finite G-graph and derive some properties of them. Throughout this section, we assume:

Hypothesis 1. Assume $\Delta$ to be a connected locally finite and locally quasiprimitive G-graph. Furthermore, assume that for any edge $\{x,y\}$ the group $G_{x,y}^{[1]}$ is a (possibly trivial) p-group and that the valencies of x and y are at least 2.

The hypothesis implies that G has at most two orbits on the vertex set of $\Delta$ and one orbit on the set of edges. In particular, we do not assume that G is vertex-transitive in this section.

For any vertex $z\in V\Delta$, we define $Q_z:=O_p(G_z^{[1]})$. For a subgroup $N \le G_x$, we denote with $\overline{N}:= NQ_x/Q_x$ and with $\widehat{N}:=N^{\Delta (x)}=NG_x^{[1]}/G_x^{[1]}$. Similarly, for an element $g\in G_x$ we denote with $\overline{g}$ and $\widehat{g}$ its image in $\overline{G_x}$ and $\widehat{G_x}$, respectively.

Let

$$\begin{aligned}{} & {} A:=\langle g \in G_x \mid [\langle g \rangle ,G_x^{[1]}] \le Q_x\rangle , \,\, B:=\langle g \in G_{x,y} \mid [\langle g \rangle ,G_x^{[1]}] \le Q_x\rangle , \,\\{} & {} \quad Z:=\langle g \in G_x^{[1]} \mid [\langle g \rangle ,G_x^{[1]}] \le Q_x\rangle ,\\{} & {} \quad D:=\langle g \in G_{x,y} \mid [\langle g \rangle ,G_y^{[1]}] \le Q_y \rangle , \, C\!:=\!\langle g \in G_{x,y} \mid [\langle g \rangle ,G_x^{[1]}G_y^{[1]}] \le Q_xQ_y \rangle ,\\{} & {} \quad H:=\widehat{G_x}, \,\, Y:= \widehat{G_y^{[1]}} \hbox { and } Q:=\widehat{Q_y}. \end{aligned}$$

Note that $Q_x$ is a subgroup of A, B, C, D and Z, and that $Z \le B \le A$. Moreover, H acts on $\Delta (x)$, and Q and Y are normal subgroups of $H_y$, with $Q \trianglelefteq Y$.

Lemma 2.1

$G_{x,y}^{[1]}= Q_x\cap Q_y$. In particular, $G_x^{[1]} \cap Q_y =Q_x\cap Q_y =G_y^{[1]} \cap Q_x$ and $[ G_x^{[1]}, G_y^{[1]}] \le Q_x\cap Q_y$.

Proof

Observe that $G_{x,y}^{[1]}$ is a normal subgroup of $G_x^{[1]}$ and $G_y^{[1]}$. Since $G_{x,y}^{[1]}$ is a p-group, we have $G_{x,y}^{[1]}\le Q_x\cap Q_y$. Thus, $Q_x \cap Q_y \le G_x^{[1]} \cap Q_y \le G_{x,y}^{[1]} \le Q_x \cap Q_y$. Hence, $G_{x,y}^{[1]}= Q_x\cap Q_y$ and $G_x^{[1]} \cap Q_y =Q_x\cap Q_y$. Similarly, $G_y^{[1]} \cap Q_x =Q_x\cap Q_y$. Finally, $[G_x^{[1]}, G_y^{[1]}]\le G_{x,y}^{[1]}\le Q_x\cap Q_y$. $\square$

Lemma 2.2

$O_p(Y)= Q$.

Proof

We have $Q \le O_p(Y)$ for $Q_y=O_p(G_y^{[1]})$. Let $Q_1$ be the pre-image of $O_p(Y)$ in $G_x$. Then $Q_yG_x^{[1]} \le Q_1 \le G_x^{[1]}G_y^{[1]}$ and $Q_1$ is normalized by $G_y^{[1]}$. Let $N:=Q_1\cap G_y^{[1]}$. Then $N \trianglelefteq G_y^{[1]}$ and N is normalized by $G_x^{[1]}$. Since $G_x^{[1]} \le Q_1 \le G_x^{[1]}G_y^{[1]}$ we have $Q_1=Q_1 \cap G_x^{[1]}G_y^{[1]} = G_x^{[1]}( Q_1 \cap G_y^{[1]})= G_x^{[1]}N$. Hence, $O_p(Y)=NG_x^{[1]}/G_x^{[1]} \cong N/N\cap G_x^{[1]}$. Since $N\cap G_x^{[1]}\le G_{x,y}^{[1]}= Q_x\cap Q_y$, both $N\cap G_x^{[1]}$ and $O_p(Y)$ are p-groups. It follows that N is a p-group too, and thus, $N\le Q_y$. Hence, $N=Q_y$ and $Q_1=Q_yG_x^{[1]}$. $\square$

We derive some properties of the groups A, B, C and D.

Lemma 2.3

The following hold:

• (i) $\overline{A}=C_{\overline{G_x}} (\overline{G_x^{[1]}})$;

• (ii) $G_y^{[1]}\le A$ and $A \cap G_x^{[1]}= Z$;

• (iii) Either $\widehat{A}=1$ or A is transitive on $\Delta (x)$ and $A_y=B$;

• (iv) If $\widehat{A} =1$, then $G_y^{[1]}=G_{x,y}^{[1]}$ and $\widehat{G_x}$ is isomorphic to a subgroup of $\textrm{Out}(\overline{G_x^{[1]}})$;

• (v) If $\widehat{A}$ is regular on $\Delta (x)$, then $G_y^{[1]}=G_{x,y}^{[1]}$;

• (vi) If $\widehat{A} \ne 1$ and $\widehat{A}$ is not regular on $\Delta (x)$, then $C_{H} (\widehat{A})=1$ and $C_{\overline{G_x}} (\overline{A})= \overline{G_x^{[1]}}$.

Proof

(i):

This follows directly from the definition.

(ii):

By 2.1$[ G_x^{[1]}, G_y^{[1]}] \le Q_x$, which yields $G_y^{[1]} \le A$. The fact that $A \cap G_x^{[1]}= Z$ follows from the definition of the groups A and Z.

(iii):

Since $A\trianglelefteq G_x$, we have that either A is transitive on $\Delta (x)$ or $A \le G_x^{[1]}$. In the first case $G_x=AG_{x,y}$ and $A_y=A \cap G_{x,y}=B$. In the latter $\widehat{A} =1$.

(iv):

We have $G_y^{[1]} \le A \le G_x^{[1]}$, since $\widehat{A}=1$ and $G_y^{[1]} \le A$ by (ii). Thus, $G_y^{[1]}=G_{x,y}^{[1]}$. By (i) $\overline{A}=C_{\overline{G}_x} (\overline{G_x^{[1]}})$ and so $C_{\overline{G_x}} (\overline{G_x^{[1]}}) \le \overline{G_x^{[1]}}$. As $\widehat{G_x} \cong G_x/G_x^{[1]}\cong \overline{G_x}/ \overline{G_x^{[1]}}$, we conclude that $\widehat{G_x}$ is isomorphic to a subgroup of $\textrm{Out}(\overline{G_x^{[1]}})$.

(v):

By (ii) $G_y^{[1]} \le A \cap G_{x,y}$, so $Y \le \widehat{A} \cap H_y$. Hence, $Y=1$, since $\widehat{A}$ is regular on $\Delta (x)$. But then $G_y^{[1]} \le G_{x}^{[1]}$, and thus, $G_y^{[1]}=G_{x,y}^{[1]}$.

(vi):

Let $N:=\langle g \in G_{x} \mid [\langle g \rangle , A] \le G_x^{[1]} \rangle$ and $M:=\langle g \in G_{x} \mid [\langle g \rangle , A] \le Q_x \rangle$. Then $\widehat{N}=C_{H} (\widehat{A})$, $\overline{M}= C_{\overline{G_x}} (\overline{A})$ and both N and M are normal subgroups of $G_x$. Note that $G_x^{[1]}\le M \le N$. Since $\Delta$ is locally quasiprimitive, either $N \le G_x^{[1]}$ or $\widehat{N}$ is transitive on $\Delta (x)$. Suppose $\widehat{N}$ is transitive on $\Delta (x)$. Since $[\widehat{A}, \widehat{N}]=1$ and $\widehat{A}$ is transitive on $\Delta (x)$ too, we must have that $\widehat{A}$ is regular on $\Delta (x)$, which is a contradiction. Hence, $N \le G_x^{[1]}$ and thus $G_x^{[1]} \le M \le N \le G_x^{[1]}$. We conclude that $M =N =G_x^{[1]}$ from which (vi) follows.

$\square$

Lemma 2.4

The following hold:

• (i) $\overline{B}=C_{\overline{G_{x,y}}} (\overline{G_x^{[1]}})$ and $\overline{Z}= Z (\overline{G_x^{[1]}})$;

• (ii) $G_x^{[1]}\le D$ and $G_y^{[1]}\le B$;

• (iii) $B \cap G_x^{[1]}=Z$.

Proof

These follow directly from the definitions and 2.1. $\square$

Lemma 2.5

The following hold:

• (i) $C=\langle g \in G_{x,y} \mid [\langle g \rangle ,G_x^{[1]}]\le Q_x \hbox { and } [\langle g \rangle ,G_y^{[1]}] \le Q_y \rangle$;

• (ii) $B \cap D=C$;

• (iii) $C \cap G_x^{[1]} = Z$;

• (iv) $A\cap D= C$;

• (v) $G_y^{[1]}C \cap G_x^{[1]} = Z$.

Proof

(i):

Let $N:= \langle g \in G_{x,y} \mid [\langle g \rangle ,G_x^{[1]}] \le Q_x \hbox { and } [\langle g\rangle ,G_y^{[1]}] \le Q_y\rangle$. Let $g \in G_{x,y}$ and let $a \in G_x^{[1]}$ and $b \in G_y^{[1]}$. By the commutator formulas, $[g,ab]=[g,b][g,a][g,a,b]$. By 2.1$[g,a,b] \in Q_x\cap Q_y$, which implies that $[g,ab] \in [g,a][g,b]Q_x$. Thus, $[g,ab] \in Q_xQ_y$ if and only if $[g,a][g,b]\in Q_xQ_y$. This, in particular, implies that $N \le C$. On the other hand, let $g\in C$, $g_1 \in G_x^{[1]}$ and $g_2 \in G_y^{[1]}$. Let $c=[g,g_1]$ and $d=[g,g_2]$. Then $c \in G_x^{[1]}$, $d\in G_y^{[1]}$ and, by the above, $cd\in Q_xQ_y$. In particular, $c \in Q_xG_y^{[1]}$ and thus $c \in Q_x$ since $G_x^{[1]} \cap Q_xG_y^{[1]}=Q_x(G_x^{[1]} \cap G_y^{[1]})=Q_x$. Similarly, $d \in Q_yG_x^{[1]}$ and thus $d \in Q_y$, since $G_y^{[1]} \cap Q_yG_x^{[1]}=Q_y(G_x^{[1]} \cap G_y^{[1]})=Q_y$. It follows that $g\in N$ and consequently $C \le N$.

(ii):

This follows from (i).

(iii):

By (i), 2.1 and 2.3(ii), we have $Z \le C \cap G_x^{[1]} \le A \cap G_x^{[1]}=Z$.

(iv):

We have $A \cap D=\langle g \in G_{x,y} \mid [\langle g \rangle ,G_x^{[1]}] \le Q_x \hbox { and } [\langle g\rangle ,G_y^{[1]}] \le Q_y \rangle$. So by (i), $A \cap D=C$.

(v):

Note that $Z \le G_y^{[1]}C \cap G_x^{[1]}$ since $Z \le C$. On the other hand $[G_y^{[1]}C, G_x^{[1]}]\le Q_x$; hence, $G_y^{[1]}C \cap G_x^{[1]} \le Z$.

$\square$

Lemma 2.6

The following hold:

• (i) $\widehat{C} = C_{\widehat{B}}(Y/ Q)$;

• (ii) $\widehat{D} = C_{H_y}(Y/Q)$;

• (iii) $Y\cong \overline{ G_y ^{[1]}}$ and $G_y^{[1]}/Q_y \cong Y/Q$.

Proof

(i):

Since $C \le B$, we have $\widehat{C} \le \widehat{B}$. Let $g\in G_{x,y}$ with $\widehat{g} \in \widehat{C}$. We can write $g=g_1g_2$ for some $g_1 \in C$ and $g_2\in G_x^{[1]}$. Since $[\langle g_1 \rangle , G_y^{[1]}] \le Q_y$ and $[\langle g_2\rangle , G_y^{[1]}] \le Q_x$, we have $[\langle g \rangle , G_y^{[1]}] \le Q_xQ_y$. Hence, $[\langle \widehat{g} \rangle , Y] \le \widehat{Q_xQ_y}=Q$. It follows that $\widehat{g} \in C_{\widehat{B}}(Y/Q)$. On the other hand, let $g\in G_{x,y}$ with $\widehat{g} \in C_{\widehat{B}}(Y/Q)$. Then $g \in BG_x^{[1]}$ and $[\langle \widehat{g} \rangle , Y] \le Q$ so $[\langle g \rangle , G_y^{[1]}] \le Q_yG_x^{[1]}$. Since $[\langle g\rangle , G_y^{[1]}]\le G_y^{[1]}$ we obtain by 2.1$[\langle g \rangle , G_y^{[1]}] \le Q_yG_x^{[1]}\cap G_y^{[1]} =Q_y(G_x^{[1]}\cap G_y^{[1]}) \le Q_y$. Since $g=g_1g_2$, for some $g_1 \in B$ and $g_2\in G_x^{[1]},$ and $[\langle g_2\rangle ,G_y^{[1]}]\le Q_x\cap Q_y \le Q_y$, we must have $[\langle g_1 \rangle , G_y^{[1]}] \le Q_y$. Hence, $g_1 \in D$ and thus by 2.5(ii) $g_1\in B\cap D=C$. Hence, $\widehat{g} \in \widehat{C}$.

(ii):

Let $g \in G_{x,y}$ with $\widehat{g} \in \widehat{D}$. Then $g \in D$, for $G_x^{[1]}\le D$. Hence, $[g, G_y^{[1]}] \le Q_y$, and thus $\widehat{g} \in C_{H_y}(Y/Q)$. Assume now that $g \in G_{x,y}$ with $\widehat{g}\in C_{H_y}(Y/Q)$. Then $[\langle g \rangle , G_y^{[1]}] \le Q_yG_x^{[1]}$. Since g normalizes $G_y^{[1]}$ and, by 2.1, $Q_yG_x^{[1]} \cap G_y^{[1]}=Q_y(G_x^{[1]} \cap G_y^{[1]})=Q_y$, we have that $g \in D$. Whence $\widehat{g} \in \widehat{D}$.

(iii):

By 2.1, $G_{x,y}^{[1]}=Q_y\cap Q_x= G_y^{[1]}\cap Q_x$ and $G_y^{[1]}\cap G_x^{[1]}=Q_y \cap G_x^{[1]}$. The isomorphism theorems now give $Y = G_y^{[1]}G_x^{[1]}/G_x^{[1]} \cong G_y^{[1]}/Q_x\cap Q_y \cong G_y^{[1]}Q_x/Q_x \cong \overline{G_y^{[1]}}$ and $G_y^{[1]}/Q_y\cong G_y^{[1]}G_x^{[1]}/Q_yG_x^{[1]} \cong Y/Q$.

$\square$

3 Arc-transitive locally quasiprimitive graphs

In this section, we assume

Hypothesis 2. Assume $\Delta$ to be a connected arc-transitive locally finite and locally quasiprimitive G-graph of valency at least 2.

By the Thompson–Wielandt theorem [15] Hypothesis 1 holds; thus, we can apply the results of Section 2. We will retain the notation of Section 2. Moreover, let

$$\begin{aligned} k:=|\Delta (x)|, \, K:=AD, \,\,L:=AG_x^{[1]} \hbox { and } M:=G_x^{[1]}G_y^{[1]}C. \end{aligned}$$

Lemma 3.1

$\widehat{L} \ne 1$ and $H=\widehat{L} H_y$. Moreover, $\widehat{L}$ is regular on $\Delta (x)$ if and only if $G_x$ is regular on $\Delta (x)$.

Proof

Suppose $\widehat{L} =1$. Then $L=G_x^{[1]}$ and $G_y^{[1]} \le A \le G_x^{[1]}$. Hence, by arc-transitivity, $G_y^{[1]}=G_x^{[1]}=1$. But then $A=G_x$ and so $G_x=G_x^{[1]}=1$, which is a contradiction since $k \ge 2$. Hence, $\widehat{L} \ne 1$. Now $G_x=LG_{x,y}=AG_{x,y}$, since $\widehat{L} \trianglelefteq H$. It follows that $H=\widehat{L} H_y$. Suppose $\widehat{L}$ is regular on $\Delta (x)$. Then $\widehat{A}$ is regular on $\Delta (x)$ too and now 2.3(v) gives that $G_y^{[1]}= G_{x,y}^{[1]}$. By arc-transitivity we obtain $G_y^{[1]}=G_x^{[1]}=1$. But then $L=A=G_x$ and so $G_x$ is regular in $\Delta (x)$. The other implication is obvious. $\square$

Since L acts non-trivially on $\Delta (x)$ and $L \le K$, we have $|L:L_y|=|K:K_y|=k$. Note that $K\trianglelefteq G_x$ since $G_x=LG_{x,y}=AG_{x,y}$ and $D \trianglelefteq G_{x,y}$. Recall that $A_y=B$, so $L_y=BG_x^{[1]}$. It follows from the definitions that $L \trianglelefteq K \trianglelefteq G_x$, $L \trianglelefteq G_x$, $M \trianglelefteq L_y \trianglelefteq G_{x,y}$ and $M \trianglelefteq G_{x,y}$. We have the following chains of subgroups:

$$\begin{aligned} G_x^{[1]} \trianglelefteq L \trianglelefteq K \trianglelefteq G_x \hbox { and } G_x^{[1]} \trianglelefteq M \trianglelefteq L_y \trianglelefteq G_{x,y}. \end{aligned}$$

Lemma 3.2

The following hold:

• (i) $B/CG_y^{[1]} \cong L_y/M$;

• (ii) $D/CG_x^{[1]} \cong K/L$;

• (iii) $K/L \cong L_y/M$; in particular, $|K:L|=|L_y:M|$;

• (iv) $C_{G_{x,y}}(G_y^{[1]}/ Q_y) = C_{K_y}(G_y^{[1]}/ Q_y)$.

Proof

(i):

By 2.4(iii) and 2.5(v), $G_x^{[1]}\cap B=Z=G_x^{[1]}\cap G_y^{[1]}C$. Since $B=A_y$, we have $L_y=BG_x^{[1]}$. The isomorphism theorems now give $B/G_y^{[1]}C \cong BG_x^{[1]}/G_y^{[1]}CG_x^{[1]} = L_y/M$.

(ii):

By 2.5(iv), $D \cap AG_x^{[1]} = (D \cap A)G_x^{[1]} = CG_x^{[1]}$. Hence, $D/CG_x^{[1]} = D/(D\cap AG_x^{[1]}) \cong DAG_x^{[1]}/AG_x^{[1]} = K/L$, for $G_x^{[1]} \le D$.

(iii):

Since $\Delta$ is arc-transitive, there exists an element g in G interchanging x and y. Thus, g induces an automorphism $\phi$ of $G_{x,y}$ inducing the isomorphisms $B \cong D$ and $G_x^{[1]} \cong G_y^{[1]}$. Moreover, by 2.5(i), $\phi$ induces an automorphism of C. Hence, we also have $B/CG_y^{[1]} \cong D/CG_x^{[1]}$. It now follows from (i) and (ii) that $K/L \cong L_y/M$.

(iv):

Since $C_{G_{x,y}}(G_y^{[1]}/ Q_y)= D \le K_y$ we have that $C_{G_{x,y}}(G_y^{[1]}/ Q_y) = C_{K_y}(G_y^{[1]}/ Q_y).$

$\square$

Lemma 3.3

The following hold:

• (i) $\widehat{K}/ \widehat{L} \cong \widehat{L}_y/ \widehat{M}$; in particular $|\widehat{K}: \widehat{L}|=| \widehat{L}_y: \widehat{M}|$;

• (ii) $\widehat{K}_y = \widehat{B} \widehat{D}$ and $\widehat{L}_y = \widehat{A}_y = \widehat{B}$;

• (iii) $\widehat{M} = YC_{\widehat{L}_y}(Y/Q)$;

• (iv) $\widehat{K} = \widehat{L} C_{H_y}(Y/Q)$.

Proof

(i).:

Note that $G_x^{[1]}$ is a subgroup of K, L, $L_y$ and M. The isomorphism theorems give $\widehat{K} / \widehat{L} \cong K/L$ and $\widehat{L}_y / \widehat{M} \cong L_y/M$. The claim now follows from 3.2(iii).

(ii).:

This follows from the fact that $K=AD$, $L=AG_x^{[1]}$ and, by 3.1 and 2.3(iii), $L_yG_x^{[1]}=A_yG_x^{[1]}=BG_x^{[1]}$.

(iii).:

By (ii) and 2.6(i), $\widehat{L}_y=\widehat{B}$ and $\widehat{C} = C_{\widehat{B}}(Y/Q)$. This yields $\widehat{M}=Y\widehat{C}= Y C_{\widehat{L}_y}(Y/Q)$.

(iv).:

We have $\widehat{A} = \widehat{L}$ and, by 2.6(ii), $\widehat{D} = C_{H_y}(Y/ Q)$. Hence, $\widehat{K} = \widehat{A} \widehat{D} = \widehat{L} C_{H_y}(Y/Q)$.

Lemma 3.4

Suppose $\widehat{L}$ is not regular on $\Delta (x)$, then the following hold:

• (i) $H / \widehat{L}$ is isomorphic to a subgroup of $\textrm{Out}(\widehat{L})$;

• (ii) $L_y/M$ is isomorphic to a subgroup of $\textrm{Out}(G_y^{[1]}/Q_y)$;

• (iii) K/L is isomorphic to a subgroup of $\textrm{Out}(G_x^{[1]}/Q_x)$.

Proof

(i).:

Since $\widehat{L} = \widehat{A}$ is not regular on $\Delta (x)$, it follows from 2.3(vi) that $C_{H} (\widehat{A}){=}1$.Hence, H is isomorphic to a subgroup of $\textrm{Aut}(\widehat{A})$ and $G_x/AG_x^{[1]} \cong H/ \widehat{A}$ is isomorphic to a subgroup of $\textrm{Out}(\widehat{A})$. Therefore, $H / \widehat{L}$ is isomorphic to a subgroup of $\textrm{Out}(\widehat{L})$.

(ii).:

By 2.5(ii) $B \cap D =C$. Hence, B/C is isomorphic to a subgroup of $\textrm{Aut}(G_y^{[1]}/Q_y)$. So, $B/CG_y^{[1]}$ is isomorphic to a subgroup of $\textrm{Out}(G_y^{[1]}/Q_y)$. Thus, by 3.2(i), $L_y/M$ is isomorphic to a subgroup of $\textrm{Out}(G_y^{[1]}/Q_y)$.

(iii).:

By 2.5(ii) $B \cap D=C$. Hence, D/C is isomorphic to a subgroup of $\textrm{Aut}(G_x^{[1]}/Q_x)$. So, $D/CG_x^{[1]}$ is isomorphic to a subgroup of $\textrm{Out}(G_x^{[1]}/Q_x)$. Thus, by 3.2(ii), K/L is isomorphic to a subgroup of $\textrm{Out}(G_x^{[1]}/Q_x)$. $\square$

Lemma 3.5

$C_{H_y}(Y/Q) = C_{\widehat{K}_y}(Y/Q)$.

Proof

Let $g \in G_{x,y}$ with $\widehat{g} \in C_{H_y}(Y/Q)$. We have $[\langle g \rangle ,G_y^{[1]}]\le Q_yG_x^{[1]}$. By 2.1$G_y^{[1]} \cap Q_yG_x^{[1]}=Q_y(G_y^{[1]} \cap G_x^{[1]})=Q_y$. Since $G_y^{[1]}\trianglelefteq G_{x,y}$, we have $[\langle g \rangle ,G_y^{[1]}]\le Q_yG_x^{[1]} \cap G_y^{[1]}=Q_y$. Hence, $g \in C_{G_{x,y}}({G_y^{[1]}}/Q_y)=D$. It follows that $g \in C_{K_y}( {G_y^{[1]}}/ Q_y)$ and thus $\widehat{g} \in C_{\widehat{K}_y}(Y/ Q)$. The other inclusion is obvious. $\square$

We are now ready to prove the two main theorems of this paper.

Proof of Theorem 1.1

Recall that $H=\widehat{G_x}$ and $Y=\widehat{G_y^{[1]}}$. Let $E:=\widehat{A}$, $F:=\widehat{K}$, and $N:=\widehat{M}$. Then $E = \widehat{L}$ and, by 2.2, 3.3(iii) and 3.3(iv), $O_p(Y)=Q$, $N= YC_{\widehat{L}_y}(Y/Q) = YC_{E_y}(Y/O_p(Y))$ and $F = \widehat{L} C_{H_y}(Y/Q) = E C_{H_y}(Y/O_p(Y))$.

(1) This follows from 3.3(i).

(2) This follows from 3.4(ii) since $E_y/N = \widehat{L}_y /\widehat{M} \cong L_y/M$.

(3) Since $F/E = \widehat{K} /\widehat{L} \cong K/L$, it follows from 3.4(iii) that F/E is isomorphic to a subgroup of $\textrm{Out}(G_x^{[1]}/Q_x)$. Since $H/E \cong G_x/AG_x^{[1]}$, it is isomorphic to subgroup of $\textrm{Out}(G_x^{[1]}/Q_x)$. Moreover, by 2.3(vi), $C_H (E) =1$. Hence, F/E and H/E are also isomorphic to subgroups of $\textrm{Out}(E)$.

(4) This follows from 3.5.

Proof of Theorem 1.2

Suppose $G_x^{[1]}=1$. Then $G_x \cong H$ and (i) holds. Assume now that $G_x^{[1]} \ne 1$. Note that $Z(\overline{G_x^{[1]}}) = \overline{Z}$ and $\widehat{A} \ne 1$. Let $Q_x \le U \le A$ with $\overline{U}=Z(\overline{A})$. Then $U \trianglelefteq G_x$ and, since H is quasiprimitive on $\Delta (x)$, we either have that U is transitive on $\Delta (x)$ or $\widehat{U} =1$. In the first case $[\widehat{U}, \widehat{A}]=1$, and thus $\widehat{A}$ is regular on $\Delta (x)$. But then H is regular on $\Delta (x)$ too by 3.1. In particular, $G_y^{[1]}=G_{x,y}^{[1]}$, which contradicts $G_x^{[1]} \ne 1$. Thus, $\widehat{U} =1$. Now $U \le G_x^{[1]}$ and, since $U \le A$, we have $\overline{U} = \overline{Z}$. We have shown that $Z( \overline{A}) = \overline{Z}$. By 2.3(ii), $A \cap G_x^{[1]}=Z$. Thus, we have $\overline{A} \cap \overline{G_x^{[1]}}= \overline{Z}$ and it follows that $\overline{A} \overline{G_x^{[1]}}= \overline{G_x^{[1]}} \circ \overline{A}$. By 2.3(i) and 2.3(vi), $\overline{A} =C_{\overline{G_x}} (\overline{G_x^{[1]}})$ and $\overline{G_x^{[1]}} = C_{\overline{G_x}} (\overline{A})$. Thus, $\overline{G_x} \cong (\overline{G_x^{[1]}} \circ \overline{A}).D$ with $C_{\overline{G_x}} (\overline{G_x^{[1]}}) = \overline{A}$, $C_{\overline{G_x}} (\overline{A})=\overline{G_x^{[1]}}$ and $D \cong G_x/AG_x^{[1]} \cong H/E$ which, by 2.3(vi) and 3.4, is isomorphic to a subgroup of $\textrm{Out}(\overline{G_x^{[1]}})$, $\textrm{Out}(\overline{A})$ and $\textrm{Out}(E)$.

4 An application

In this section we prove Proposition 1.3. Before doing so, we need first to prove the following corollary of 1.1 and 1.2 and collect some properties of groups of Lie type of rank 1.

Corollary 4.1

Let $\Delta$ be a connected arc-transitive locally finite and locally quasiprimitive G-graph such that there exists an edge $\{x,y\}$ of $\Delta$ with $G_{x,y}^{[1]}=1$ and $G_x^{[1]}\ne 1$. Let $A:=C_{G_x} (G_x^{[1]})$, $H:=G_x^{\Delta (x)}$, $E:=A^{\Delta (x)}$ and $Y:=(G_y^{[1]})^{\Delta (x)}$. Suppose $F^*(H_y) \le Y$ then the following hold:

• (i) $G_x^{[1]} \cong Y = E_y$, $Z(A)\cong Z(E_y)$ and $A/Z(A) \cong E$;

• (ii) $G_x \cong (E_y \circ A).D$, with $C_{G_x}(E_y)\cong A$, $C_{G_x}(A)\cong E_y$ and $D \cong H/E \le \textrm{Out}(E)$.

In particular, if $Z(E_y)=1$ then $G_x \cong (E_y \times E).D$, with $C_{G_x}(E_y)\cong E$, $C_{G_x}(E)\cong E_y$ and $D \cong H/E \le \textrm{Out}(E)$.

Proof

Let p be a prime number not dividing |H|. Then $O_p(G_{x,y}^{[1]})=G_{x,y}^{[1]}=1$ and $O_p(Y)=1$. In the notation of 1.1 we get $A:=C_{G_x}(G_x^{[1]})$, $E:=A^{\Delta (x)}$, $F:=EC_{H_y}(Y)$ and $N:=YC_{E_y}(Y)$. Since $\Delta$ is an arc-transitive G-graph with $G_{x,y}^{[1]}=1$, we have $G_x^{[1]} \cong G_{y}^{[1]} \cong Y$. Hence, $1 \ne G_{y}^{[1]} \le A$, and thus $1\ne Y \le E_y$. In particular, H is not regular on $\Delta (x)$ and thus we can apply 1.1.

Suppose $F^*(H_y) \le Y$. Then $C_{H_y}(Y) \le C_{H_y} (F^*(H_y)) \le F^*(H_y) \le Y$. Since $C_{H_y}(Y) \le Y$, we have $C_{E_y}(Y) \le Y$ and $C_{H_y}(Y) \le E$. It follows that $N=Y$ and $E=F$. Since $E=F$, 1.1 yields $E_y=N$. Hence, $G_x^{[1]} \cong Y = E_y$ and thus $Z(G_x^{[1]})\cong Z(Y) = Z(E_y)$. Now 1.2 gives $Z(A)\cong Z(E_y)$, $A/Z(A) \cong E$ and $G_x \cong (E_y \circ A).D$, with $C_{G_x}(E_y)\cong A$, $C_{G_x}(A)\cong E_y$ and $D \le \textrm{Out}(E)$. This completes the proof of the corollary. $\square$

Lemma 4.2

Let $K \trianglelefteq G \le \textrm{Aut}(K)$ with K be a simple rank 1 group of Lie type in its natural permutation representation in characteristic p. Let P be a point stabilizer of G and $T_0 \le P$ be a torus of G. Assume $K \not \cong \textrm{Ree}(3)^\prime$. Then

• (a) $O_p(P) \le K$ and $O_p(P)$ is regular on the points not fixed by P;

• (b) $[O_p(P), \langle t \rangle ] =O_p(P)$ for all $1 \ne t \in T_0$;

• (c) $Z(O_p(P))=Z(O_p(P \cap K))=\Omega _1(Z(O_p(P\cap K)))$ and $|Z(O_p(P\cap K))|=q$;

• (d) $[Z(O_p(P)), \langle t \rangle ] =Z(O_p(P))$ for all $1 \ne t \in T_0$, or $K \cong \textrm{PSU}_3(q)$ and $[Z(O_p(P)), \langle t \rangle ]=1$ for all $t \in T_0$ with $t^{q+1}=1$;

• (e) $T_0\cap K$ is irreducible on $\Omega _1 (Z(O_p(P)))$;

• (f) $C_G(Z(O_p(P)))= O_p(P)$ unless $K \cong \textrm{PSU}_3(q)$;

• (g) If $K \cong \textrm{PSU}_3(q)$, then $T_0$ acts irreducibly on $O_p(P)/Z(O_p(P))$, and for $L=(G\cap K)T_0$

  • (1) $[\langle g \rangle , O_p(P)]\not \le Z(O_p(P))$ for all $g \in G \backslash L$;

  • (2) $|C_G(Z(O_p(P)))/C_L(Z(O_p(P)))|\le 2$.

Proof

(a)–(e). This can be deduced from the description of these groups as given in [9, 10, 18, 22]. See also [17, 4.3].

(f).:

For $K \cong \textrm{PSL}_2(q)$, $q \not \in \{2,3\}$, this is well known. Therefore, we may assume that $K \cong \textrm{Sz}(q)$, with $q >2$, or $K \cong \textrm{Ree}(q)$, with $q >3$. In both cases $Z(O_p(P))= \Omega _1(O_p(P))$ is a group of order q and $T_0\cong GL_1(q)$. From (c) and (d) we get $C_{Z(O_p(P))}(t)=1$ for all $t \in T_0$ with $t\ne 1$. Hence, $T_0$ acts regularly on the non-trivial elements of $Z(O_p(P))$. In both cases $P=O_p(P)N_G(T_0)$, with $N_G(T_0)$ isomorphic to a subgroup of $\mathrm{\Gamma L}_1(q)$ containing $GL_1 (q)$. Let $g \in C_P(Z(O_p(P)))$, then $g=g_1g_2$ for some $g_1 \in O_p(P)$ and $g_2 \in N_G(T_0)$. Now $g_2 \in C_P(Z(O_p(P)))$ and thus $[\langle g_2 \rangle , T_0] \le C_{T_0}(Z(O_p(P)))=1$. Whence $g_2=1$ and $g \in O_p(P)$.

(g).:

Let $V=O_p(P)/Z(O_p(P))$. Then $|V|=q^2$ and $T_0$ acts semiregular on V. Since $|T_0|=a^{-1}(q^2-1)$ with $a=(3,q+1)$, the group $T_0$ has a orbits of length $a^{-1}(q^2-1)$ on the non-trivial elements of V. Let $1 \ne W \le V$, be a $T_0$-invariant subspace. Then either $W=V$, or $W < V$ and $a=3$. Assume that we are in the latter case. Then either $|W|-1= \frac{1}{3}(q^2-1)$ or $|V/W|-1= \frac{1}{3}(q^2-1)$. Since $a=3$, we have $p\ne 3$ and thus must have $3 \equiv 1 \, (\hbox {mod }p)$. We conclude that $p=2$. Suppose $\frac{1}{3}(q^2-1)+1 > 2$, then $\frac{1}{3}(q^2-1)+1 \equiv 0 \, (\hbox {mod }4)$, which is impossible since $q^2 \equiv 0 \, (\hbox {mod }4 )$. Thus, $\frac{1}{3}(q^2-1) \le 2$, which implies $q=2$. This is a contradiction with K being simple. Hence, $W=V$ and $T_0$ is irreducible on V. Let $L=(G\cap K)T_0$. (1) follows from the description of this group given in [10]. To prove (2) let $R:=\mathrm{P\Gamma U}_3(q)$, S be the stabilizer in R of a point and $S_0$ be a torus in S. Then $S/O_p(S) \cong \mathrm{\Gamma L}_1(q^2)$. Let $g \in C_{S}(Z(O_p(S)))$. Then $[\langle g \rangle , O_p(S)S_0] \le O_p(S)C_{S_0}(Z(O_p(S)))$ and, by (d), $C_{S_0}(Z(O_p(S)))= \langle t\in S_0 \mid t^{q+1}=1 \rangle$. This implies that $g \in S_0$ or g induces the map $x \mapsto x^q$ on $S_0$. Hence, $g^2 \in O_p(S)S_0 \le KS_0$. In particular, $|C_G(Z(O_p(P)))/C_L(Z(O_p(P)))|\le 2$. $\square$

Proof of Proposition 1.3

Let $H:=G_x^{\Delta (x)}$ and $Y:=(G_y^{[1]})^{\Delta (x)}$. Suppose $G_x^{[1]}=1$. Then $G_x \cong H$ and the proposition holds. From now on we will assume that $G_x^{[1]}\ne 1$. Now H is not regular on $\Delta (x)$ and $Y \ne 1$, since $Y \cong G_y^{[1]} \cong G_x^{[1]} \ne 1$. Let d be a prime not dividing |H|. Then $O_d(G_{x,y}^{[1]})=G_{x,y}^{[1]}=1$ and $O_d(Y)=1$. Let $A:=C_{G_x}(G_x^{[1]})$, $E:=A^{\Delta (x)}$, $F:=EC_{H_y}(Y)$ and $N:=YC_{E_y}(Y)$. Then 1.1 yields $E/F \cong E_y/N$.

Let $S:=F^*(H)$. Then S is a non-abelian simple group and $S \trianglelefteq E \trianglelefteq H \le \textrm{Aut}(S)$, for $1 \ne Y \trianglelefteq E_y$. By the classification of 2-transitive groups, see [1] and [16, 2.5] for a list, $F^*(H_y)$ is either a non-abelian simple group or a p-group. The classification also yields that either $F^*(E_y)=F^*(S_y)$, or $E \ne S$ and $S\cong \textrm{Ree}(3)^\prime$. Suppose first that $S\not \cong \textrm{Ree}(3)^\prime$. Then $Z(E_y) \le F^*(E_y)=F^*(S_y)$ and so $Z(E_y) \le Z(S_y)$. It follows that either $Z(E_y)=1$ or $F^*(S_y)$ is a p-group. In the latter case either S is a rank 1 group of Lie type or $S\cong \textrm{PSL}_n(q)$ with $n \ge 3$. We claim that for these groups $Z(S_y)=1$. In case S is a rank 1 group of Lie type it follows from 4.2(e), whereas when $S\cong \textrm{PSL}_n(q)$ with $n \ge 3$ this follows from the irreducible action of $S_y$ on $O_p(S_y)$. Since $Z(E_y) \le Z(S_y)=1$ we conclude that $Z(E_y)=1$ too in these cases. Now assume $S \cong \textrm{Ree}(3)^\prime$. Then either $E_y \cong C_9 \rtimes C_6$ or $E_y \cong \textrm{Dih}_{18}$. In both cases $Z(E_y)=1$. We have shown that $Z(E_y)=1$. If $F^*(H_y) \le Y$, then the proposition follows from 4.1. Thus, we may assume that

1. $Z(E_y)=1$ and $F^*(H_y) \not \le Y$.

Suppose $F^*(H_y)$ is a non-abelian simple group. Then $[F^*(H_y), Y]\le F^*(H_y)\cap Y \trianglelefteq F^*(H_y)$. Thus, either $[F^*(H_y), Y]= F^*(H_y) \le Y$ or $[F^*(H_y), Y]=1$. Since $F^*(H_y) \not \le Y$, we must have $[F^*(H_y), Y]=1$ and thus $Y \le F^*(H_y)$. But then $Y= F^*(H_y)$ and so $F^*(H_y)$ is abelian, which is a contradiction. Hence,

2. $F^*(H_y)$ is a p-group.

Suppose $S \cong \textrm{Ree}(3)^\prime \cong L_2(8)$. Then $H \cong \mathrm{P\Gamma L}_2(8)$, for H is 2-transitive on $\Delta (x)$. Since $E \le F \le H$ we have either $H=F=E$ or $|H:E|= 3$. If $E=H$, then $E_y \cong C_9 \rtimes C_6$. By 1.1$N=E_y$, and thus $Z(N)=1$. If $E \ne H$, then $E_y \cong \textrm{Dih}_{18}$. By 1.1 we must have either $N=E_y$ or $|E_y:N|=3$. Hence, $N\cong \textrm{Dih}_{18}$ or $N \cong \textrm{Dih}_6$, respectively. In both cases $Z(N)=1$ too. Thus, $Z(N)=1$, which implies that $N= Y \times C_{E_y}(Y)$ and $Z(Y)=Z(C_{E_y}(Y))=1$. In particular, Y is not a 3-group; therefore, Y contains a Sylow 2-subgroup of $E_y$. We conclude that $O^3(E_y)=O^{2^\prime }(Y) \le Y\le E_y$. Moreover, since $N= Y \times C_{E_y}(Y)$ and Y contains a Sylow 2-subgroup of $E_y$, we have that $C_{E_y}(Y)$ is a 3-group. Since $Z(C_{E_y}(Y))=1$ we get $N=Y$. Suppose $F=E$, then, by 1.1, $E_y=N=Y$ and the proposition follows from 1.2. Thus, we may assume $F \ne E$. But then $F=H$, $E\cong PSL_2(8)$ and $E_y \cong \textrm{Dih}_{18}$. Hence, $O^3(E_y)=E_y$ and thus $N=Y=E_y$. However, by 1.1 we have $|F:E|=|E_y:N|=3$, which is a contradiction with $E_y=N$. We have shown that $S \not \cong \textrm{Ree}(3)^\prime$. Suppose $S \cong \textrm{PSL}_n(q)$, with $(n,q)\not \in \{(2,2), (2,3)\}$. Then $F^*(H_y)=O_p(H_y)=O_p(S_y)$ and $O_p(Y) \le O_p(H_y)$. Since $F^*(H_y) \not \le Y$ and $S_y$ is irreducible on $O_p(S_y)$, we must have that $O_p(Y)=1$. But then $[Y,O_p(H_y)]=1$ and thus $Y \le O_p(H_y)$. Hence, $Y=O_p(Y)$, which is a contradiction with $O_p(Y)=1$ and $Y \ne 1$. We conclude that $S \not \cong \textrm{PSL}_n(q)$, with $(n,q)\not \in \{(2,2), (2,3)\}$. By the classification of finite 2-transitive groups, we may assume that

3. S is a rank 1 group of Lie type in its natural representation in characteristic p, with S isomorphic to one of $\textrm{PSU}_3(q)$ with $q >2$, $\textrm{Ree}(q)$ with $q >3$, or $\textrm{Sz}(q)$ with $q >2$.

By 4.2(a), $F^*(H_y)=F^*(E_y)=F^*(S_y)=O_p(H_y)=O_p(E_y)=O_p(S_y)$. If $[Y, F^*(E_y)]=1$, then $Y \le F^*(E_y)$ and Y is p-group. If $[Y, F^*(E_y)] \ne 1$, then $[Y, F^*(E_y)] \le O_p(Y)$ and thus $O_p(Y) \ne 1$. In any case $O_p(Y) \ne 1$. Hence, $1 \ne O_p(Y) \le O_p(E_y)$, since $Y \trianglelefteq E_y$. If $O_p(Y)=O_p(E_y)$, then $F^*(H_y) \le Y$, which contradicts $F^*(H_y) \not \le Y$. Therefore,

4. $1 \ne O_p(Y) < O_p(E_y)$ and $F^*(H_y)=O_p(H_y)= O_p(E_y)=O_p(S_y)$.

Since $O_p(E_y)=O_p(S_y)$ and $Y \trianglelefteq E_y$, we have that $O_p(Y) \le O_p(S_y)$ and that $O_p(Y)$ is normalized by $S_y$. In particular, $O_p(Y) \cap Z(O_p(S_y)) \ne 1$. Let $z \in \Delta (x)\backslash \{y\}$. Then $S=O_p(S_y)S_{y,z}$ and $S_{y,z}$ is a torus of S. Since $O_p(S_y)=O_p(E_y)$ acts transitively on the set $\Delta (x) \backslash \{y\}$, we have that $E_y=O_p(E_y)E_{y,z}$ too. By 4.2(c) $Z(O_p(E_y))=Z(O_p(S_y))=\Omega _1(Z(O_p(S_y)))=\Omega _1(Z(O_p(E_y)))$ is of order q, and by 4.2(e) $S_{y,z}$ is irreducible on $\Omega _1(Z(O_p(E_y)))$. It follows that $Z(O_p(E_y))=\Omega _1(Z(O_p(S_y))) \le O_p(Y)$ since $S_{y,z}$ normalizes $O_p(Y)$. We have shown that

5. $1 \ne Z(O_p(E_y)) \le O_p(Y)$.

Assume S is isomorphic to one of $\textrm{Ree}(q)$ with $q >3$, or $\textrm{Sz}(q)$ with $q >2$. By 4.2(f) we have $C_{H_y}(Z(O_p(H_y)))\le O_p(H_y)$. Let $M=C_{E_y}(Y)$. We obtain $M \le C_{H_y}(Y) \le C_{H_y}(Z(O_p(E_y)))= C_{H_y}(Z(O_p(H_y)))\le O_p(H_y) \le E_y$. Whence $E=F$ and $M \le O_p(E_y)$. From 1.1 we conclude that $E_y=N$. Note that $Y \cap S_{y,z}\ne 1$, for $M \le O_p(E_y)=O_p(S_y)$ and $YM=N=E_y$. Let $t \in Y \cap S_{y,z}$ with $t\ne 1$. By 4.2(b) $[O_p(S_y), \langle t \rangle ]=O_p(S_y)$ so we obtain $O_p(S_y) \le Y$, for $Y \trianglelefteq E_y$. This is a contradiction with $O_p(S_y)=F^*(H_y) \not \le Y$. It follows that

6. $S \cong \textrm{PSU}_3(q)$ with $q > 2$.

Let $z \in \Delta (x)\backslash \{y\}$ and T be a torus of $H_y$. Since, by 4.2(g), T acts irreducibly on $O_p(H_y)/Z(O_p(H_y))$ and $O_p(E_y)=O_p(H_y) \not \le Y$, we must have that $1 \ne Z(O_p(E_y))= O_p(Y)$. Let $L=(H\cap S)T$. Suppose there exists a $g \in Y \backslash L$. Then $[\langle g \rangle , O_p(E_y)]\le Y \cap O_p(E_y)=O_p(Y)=Z(O_p(E_y))$. This contradicts 4.2(g)(1). Hence, $Y \le L$. Suppose $T \cap Y \ne 1$. Let $t \in T \cap Y$. By 4.2(b) $[O_p(E_y), \langle t \rangle ]=O_p(E_y)$ and thus $O_p(E_y) \le Y$, for $Y \trianglelefteq E_y$. This is a contradiction with $O_p(E_y)=F^*(H_y) \not \le Y$. Hence, $T \cap Y =1$. Since $Y \trianglelefteq L_y=O_p(S_y)T$, we conclude that $Y \le O_p(S_y)$, which gives $Y=O_p(Y)$. Hence,

7. $Y=O_p(Y)=Z(O_p(E_y))$.

Now $N=YC_{E_y}(Y)=C_{E_y}(Z(O_p(E_y)))$ and $F=EC_{H_y}(Z(O_p(H_y)))$. By 4.2(g)(2) $|F:ST| \le 2$. Hence,

$$\begin{aligned} |F:ST||ST:S|=|F:S|=|F:E||E:S| \hbox { divides }2a \hbox { where } a=(q+1,3), \end{aligned}$$

and thus |F : E| divides 2a too. By 1.1(1) and 4.2(d) we have $F/E \cong E_y/N$ and $|S_y:C_{S_y}(Z(O_p(S_y)))|=(q-1)$. Therefore,

$$\begin{aligned} 6 \ge 2a \ge |F:E|=|E_y:N|\ge |S_y:N\cap S_y|=|S_y:C_{S_y}(Z(O_p(S_y)))|=(q-1). \end{aligned}$$

Since $q > 2$, this yields $q=3$ or $q=5$. If $q=5$, then we must have $|F:E|=|E_y:N|=6$. Since $Y=Z(O_p(E_y))\cong C_5$, we get $\textrm{Out}(Y) \cong C_4$. Thus, $E_y/N$ is not isomorphic to a subgroup of $\textrm{Out}(Y)$, which contradicts 1.1(2). Hence, $q=3$ and thus $|F:E|=|E_y:N|=2$. Consequently, $H=F\cong \mathrm{P\Gamma U}_3(3)$, $E\cong \textrm{PSU}_3(3)$ and $G_x^{[1]}\cong Y\cong C_3$. Now $C_{G_x}(G_x^{[1]}) \cong C_3 \times PSU_3(3)$ and thus Sylow 2-subgroups of $C_{G_{x,y}}(G_x^{[1]})$ are isomorphic to $C_8$. But Sylow 2-subgroups of $C_{H_y}(Z(O_p(H_y)))$ are isomorphic to $Q_8$. Hence, also Sylow 2-subgroups of $C_{G_{x,y}}(G_y^{[1]})$ are isomorphic to $Q_8$. Therefore, there can’t be an element in G interchanging x and y, which is a contradiction with $\Delta$ being arc-transitive. This final contradiction completes the proof of the proposition. $\square$

Remark

The group $G_x\cong A_6 \times M_{11}$ appears in the conclusion of [8, Corollary 1.3] but does not occur in the conclusion of 1.3 with $H=M_{11}$. We give an argument which is independent of the proof of 1.3, but which uses the idea of the proof of 1.1. For a locally quasiprimitive graph with $H\cong M_{11}$ and $G_x\cong A_6 \times M_{11}$, we have $G_x^{[1]}\cong A_6$ and $G_x = G_x^{[1]} \times M$, with $M=C_{G_x}(G_x^{[1]})$. Hence, $G_y^{[1]} \le M$ and so $C_{G_{x,y}} (G_y^{[1]})= G_x^{[1]} \times C_{M\cap G_{x,y}}(G_y^{[1]}) \cong G_x^{[1]}\cong A_6$. But $C_{G_{x,y}} (G_x^{[1]})= M \cap G_{x,y} \cong M_{10}$. Whence x and y are in different G-orbits. Thus, such a graph cannot be arc-transitive.

5 An example

In [8, Theorem 1.2] the possible vertex stabilizers for finite arc-transitive G-graphs with non-solvable vertex stabilizers were determined when the valency of the graph is at least 11 and a prime number. In this section we will reprove part of this theorem using the results of this paper. To be more precise, we will limit the possible structure of a vertex stabilizer $G_x$ of a locally finite arc-transitive G-graph $\Delta$ in the case that $F^*(G_x^{\Delta (x)}) \cong \textrm{PSL}_n(q)$ with $\textrm{PSL}_n(q)$ acting in its natural 2-transitive representation on $\Delta (x)$. For the remaining cases of [8, Theorem 1.2], the possible vertex stabilizers follow almost immediately from 1.3.

Proposition 5.1

Let p be a prime number, $r,n\in \textrm{Z}\hspace{-3.8pt}\textrm{Z} \hspace{2.0pt}$ with $r\ge 1$ and $n \ge 2$, and let $q=p^r$. Let $\Delta$ be a connected, arc-transitive and locally finite G-graph. Let $\{x,y\}$ be an edge of $\Delta$, $H:=G_x^{\Delta (x)}$, $M:=F^*(H)$ and $k:=|\Delta (x)|$. Suppose $M \cong \textrm{PSL}_n(q)$ and $k=\frac{q^n-1}{q-1}$ is a prime number with $k \ge 11$. Then n is a prime number and either $G_x \cong H$ or there exists a group E, with $M\trianglelefteq E \trianglelefteq H$, such that, with $d=|H:E|$, one of the following holds:

• (i) $G_{x,y}^{[1]} = 1$ and $G_x \cong (E_y \times E).C_d$ with $C_{G_x}(E_y)=E$ and $C_{G_x}(E)=E_y$;

• (ii) $G_{x,y}^{[1]} \ne 1$ and there exist groups K and L, with $L \trianglelefteq E_y/O_p(E_y)$, $K/Z(K) \cong E$ and $Z(K)\cong Z(L)$, such that $G_x/O_p(G_x) \cong (L \circ K).C_d$ with $C_{G_x/O_p(G_x)}(L)=K$ and $C_{G_x/O_p(G_x)}(K)=L$. Moreover, $O_p(E_y)=O_p(H_y)$ and one of the following holds:

  • (a) $LZ(M_y/O_p(E_y))=E_y/O_p(E_y)$;

  • (b) $C_{H_y/O_p(H_y)} (L) =E_y/O_p(E_y)$.

Proof

Suppose n is not a prime number, then $n=ab$ for some $a,b \in \textrm{Z}\hspace{-3.8pt}\textrm{Z} \hspace{2.0pt}$ with $1< a \le b <n$. Let $l=\frac{q^a-1}{q-1}$, then $l \in \textrm{Z}\hspace{-3.8pt}\textrm{Z} \hspace{2.0pt}$, $1< l < k$ and $\frac{k}{l}=\frac{q^{ab}-1}{q^a-1} \in \textrm{Z}\hspace{-3.8pt}\textrm{Z} \hspace{2.0pt}$, which is a contradiction with k being a prime number. Hence, n is a prime number. If n divides $q-1$, then $q \equiv 1 \hbox { ({mod }}$n) and thus $k=q^{n-1}+\cdots +1 \equiv 0 \hbox { ({mod }}$n). Hence, $k=n$, but then $q< k=n <q$, which is a contradiction. It follows that $(n,q-1)=1$. We conclude that $M \cong \textrm{SL}_n(q) \cong \textrm{PSL}_n(q) \cong \textrm{PGL}_n(q)$ and that $H/M \cong C_a$, where a divides r.

If $G_{x}^{[1]}=1$, then $G_x \cong H$ and the proposition holds. If $G_{x}^{[1]}\ne 1$ and $G_{x,y}^{[1]}=1$, then we can apply 1.3 and we obtain that there exists a group E with $M \trianglelefteq E \trianglelefteq H$ such that, $G_x \cong (E_y \times E).C$ with $C_{G_x}(E_y)=E$, $C_{G_x}(E)=E_y$ and $C\cong C_d$, where $d=|H:E|$. Thus, (i) holds.

Therefore, we may assume that $G_{x,y}^{[1]} \ne 1$. By the Thompson–Wielandt Theorem [15] $G_{x,y}^{[1]}$ is a non-trivial p-group. We use the notation used in 1.1. Let $A:=C_{G_x} (G_x^{[1]}/O_p(G_x^{[1]}))$ and $E:=A^{\Delta (x)}$. Then $E\cong \textrm{PSL}_n (q)\rtimes C_b$, for some integer b|a, $E_y \cong q^n\rtimes \textrm{GL}_{n-1}(q)\rtimes C_b$, and for $Y:=(G_y^{[1]})^{\Delta (x)}$ we have $Y \trianglelefteq E_y$. Let $Q:=O_p(Y)$. Since $1 \ne Y \trianglelefteq E_y$ we have $Q\le O_p(E_y)$. We have $Q=O_p(E_y)=O_p(H_y)$, since $F^*(E_y)=O_p(E_y)=O_p(H_y)$ and $E_y/O_p(E_y)$ acts irreducible on $O_p(E_y)$. Let $L:=Y/Q$ and $K:=A/O_p(G_x^{[1]})$. Then 1.2 gives $Z(K)\cong Z(L)$, and $G_x/O_p(G_x) \cong (L \circ K).C_d$ with $C_{G_x/O_p(G_x)}(L)=K$ and $C_{G_x/O_p(G_x)}(K)=L$. To finish the proof, it remains to determine the structure of Y/Q.

Assume first that $n=2$. Then $k=q+1$ and so $p=2$, for k is an odd prime. Since $k=2^r+1$ is a prime, we have $r=2^l$ for some $l \in \textrm{Z}\hspace{-3.8pt}\textrm{Z} \hspace{2.0pt}$, where $l \ge 2$ since $k \ge 11$. Since $p=2$, $M_y/O_p(M_y)$ is abelian of odd order. Hence, r is coprime to $M_y/O_p(M_y)$. Suppose first that $Y \le M_y$. Then $Y \le M_y \le C_{E_y}(Y/Q)$ and thus $N=C_{E_y}(Y/Q)$. Since H/M is cyclic, there exists a $h \in C_{H_y}(Y/Q)$ such that hM is a generator of F/M. Then for $d= \frac{a}{b}$ we have that $h^dM$ is a generator of E/M. Now $h^d \in C_{E_y}(Y/Q)$ and $E_y=\langle M_y, h^d \rangle \le N$. Whence $N=E_y$. By 1.1$F=E$ and $C_{H_y}(Y/Q) = C_{F_y}(Y/Q)$. Thus, $C_{H_y}(Y/Q) = C_{E_y}(Y/Q)=E_y$. We conclude that (ii)(b) holds. Hence, we may assume that $Y \not \le M_y$. Let $T\in Syl_2(Y)$. Then $T\not \le M_y, M_y \cap T=Q$ and $Y=\langle M_y \cap Y, T\rangle$. For $U \le H_y$ we denote with $\overline{U}$ the image of U in $H_y/Q$. We have $\overline{Y}=\langle \overline{M_y \cap Y}, \overline{T} \rangle$, so $C_{\overline{M_y}} (\overline{Y})=C_{\overline{M_y}}(\overline{T})$, since ${\overline{M_y}}$ is abelian. By coprime action $\overline{M_y}=[\overline{T},\overline{M_y}] \times C_{\overline{M_y}}(\overline{T})$. Hence, $\overline{M_y}\le \overline{Y} C_{\overline{M_y}} (\overline{Y})\le \overline{N}$. It follows that $M_y \le N$ and thus $M_yY\le N$. Moreover, $TM \cong \textrm{PSL}_2(q) \rtimes C_e$ for some e dividing b. Hence, $T\cong Q \rtimes T_1$ with $T_1 \cong C_e$ and $C_{T_1} (\overline{M_y}) =1$. Let $t \in T_1$ be an involution and $L:=[\langle t \rangle ,\overline{M_y}]=[\langle t \rangle \overline{M_y},\overline{M_y}]$. Then $1 \ne L \le Y$ and $[\langle t \rangle ,L]=L$. Whence $t \not \in C_{Y}(\overline{Y})$, and since $T_1 \cong C_e$, we conclude that $T_1 \cap C_{Y}(\overline{Y})=1$. Hence, $C_{Y}(\overline{Y}) \le M_y$. Let $R \in Syl_2(C_{H_y}(\overline{Y}))$. We have $Q \le R$, $[R,Y]\le Q$ and $R\cap Y=Q$. Therefore, $[\overline{R}, \overline{Y}]=1$ and $\overline{R} \cap \overline{Y}=1$. Recall that $\overline{Y}$ has even order. If $\overline{R}$ has even order too, then a Sylow 2-subgroup of $\overline{H_y}$ contains an elementary abelian subgroup of order 4. But $H_y/M_y \cong C_a$ and $\overline{M_y}$ has odd order, whence the Sylow 2-groups of $\overline{H_y}$ are cyclic. It follows that $\overline{R} =1$, and so $R= Q$ and $C_{H_y}(\overline{Y}) \le M_y$. We conclude that $C_{H_y}(\overline{Y})=C_{E_y}(\overline{Y})=C_{M_y}(\overline{Y})$. It follows that $E=F$ and $N\le M_yY$. The latter implies $N=M_yY$, whereas, by 1.1, the former gives $E_y=N$. Hence, $E_y=M_yY$ and we have that $\overline{E_y}=\overline{M_y}\overline{Y}=Z(\overline{M_y})\overline{Y}$. Thus, (ii)(a) holds. This finishes the proof for $n=2$.

For the remainder of the proof, we assume that $n\ge 3$. Since $k \ge 11$ we have $q \ge 3$. Let B be the pre-image of $Z(E_y/Q)$ in $E_y$.

Suppose first that $n=q=3$. Then $H=E=M=GL_3(3)$ and $E_y/Q \cong Q_8 \rtimes S_3$. Since $H=E=M$ we have, by 1.1, $N=E_y$. Moreover, $N/B =E_y/B \cong S_4$. By 1.1 we have for $R:=N/O_p(Y)$, $R_1:=Y/Q$ and $R_2:=C_{E_y}(Y/Q)/Q$ that $R=R_1R_2$, with $[R_1,R_2]=1$ and $R_1 \cap R_2=Z(R_1) \le Z(R_2)=Z(R)$. Thus, either $YB/B=1$ or $YB/B \cong S_4$. Whence $Y \le B$ or $Y=E_y$. We conclude that either $Y/Q \le Z(E_y/Q)$ or $Y/Q=E_y/Q$, and thus, either (ii)(b) or (ii)(a) hold.

From now on we may assume that $q \ge 4$ when $n=3$. In particular, $M_y\cong O_p(M_y) \rtimes GL_{n-1}(q)$, $Q=O_p(M_y)$ and $O^{p^\prime }(M_y/Q)$ is the unique component of $M_y/Q$ and of $E_y/Q$. Let S be the pre-image in $E_y$ of the component of $E_y/Q$, then $S/Q \cong \textrm{SL}_{n-1}(q)$. We have $[S/Q, Y/Q] \le S/Q \cap Y/Q$ and thus either $[S/Q, Y/Q]=1$ or $S \le Y$. Suppose $S \le Y$. Then $C_{H_y}(Y/Q) \le C_{H_y}(S/Q) \le M_y$. So $C_{H_y}(Y/Q) \le C_{E_y}(Y/Q)$. By 1.1$F=E$, and thus $N=E_y$. Let $g \in E_y\backslash Y$. Since $g \in N$ we must have $gb \in Y$, for some $b\in C_{E_y}(Y/Q) \le C_{E_y}(S/Q) \le B$. Hence, $g=b^{-1}(bg)\in BY$. It follows that $BY=E_y$, and thus $BY/Q=E_y/Q$. We conclude that (ii)(a) holds.

Finally assume $S \not \le Y$. Now $S \le C_{E_y}(Y/Q)\le N$ and so $Y \le C_{E_y}(S/Q)=C_{M_y}(S/Q) \le B$. Hence, $Y \le B$, and thus $M_y \le C_{E_y}(Y/Q)= N$. Let $X:=C_{H_y}(Y/Q)$. Then $M_y \le X\le H_y$ and $F=XE$. Since E is transitive on $\Delta (x)$, we have $F_y=E_yX$. Let t be a prime dividing $|F:E|=|F_y:E_y|$. Let $T_1 \in Syl_t(X)$ and $T_2\in Syl_t(E_y)$. Since $H_y/M_y$ is cyclic, we either have $T_1M_y \le T_2M_y$ or $T_2M_y \le T_1M_y$. Since $F_y=XE_y$, we cannot have $T_1M_y \le T_2M_y \le E_y$. Whence $T_2M_y \le T_1M_y \le X$ and thus $T_2 \le X \cap E_y = C_{E_y}(Y/Q)\le N$. It follows that for any prime t dividing |F : E| the Sylow t-subgroups of $E_y$ are in N. Whence |F : E| and $|E_y:N|$ are coprime. From 1.1 we now conclude that $F=E$, $E_y=N$ and $C_{H_y}(Y/Q) = C_{E_y}(Y/Q)$. It follows that $E_y=C_{H_y}(Y/Q)$ and thus (ii)(b) holds. $\square$

Remark

The proof of [8, Theorem 1.2] relies in part on work by Trovimof and Weiss, whereas that of 5.1 does not. This accounts for some of the extra groups appearing in the conclusion of 5.1. However, in the other cases 5.1 is more restrictive. In particular, the groups listed under (a.2) of [8, Theorem 1.2] do not appear in the conclusion of 5.1. We give a reason for this without using 1.1. Let $\Delta$ be a G-graph and $\{x,y\}\in E\Delta$ be an edge. Assume that $G_{x,y}^{[1]}=1$ and let $M:=F^*(G_x^{\Delta (x)}) \cong PSL_n(q)$, with $n \ge 3$. Let B be the pre-image of $Z(M_y/O_p(M_y))$ in $M_y$. The vertex stabilizers listed in (a.2) can be described as $G_x \cong (N \times M).D$, where N is a subgroup of $M_y$ with $O_p(M_y)\trianglelefteq N \le B$ and $MD \le \mathrm{P\Gamma L}_n(q)$. For such a group we have $O_p(G_x^{[1]})=O_p(N)$, $M \le C_{G_x}(G_x^{[1]})$ and $G_y^{[1]} \le C_{G_x}(G_x^{[1]})$. Since $O_p(G_y^{[1]})$ is a normal subgroup of $O_p(N) \times M_y$ and of order $|O_p(N)|$, we must have $O_p(G_y^{[1]})=O_p(N)$ or $O_p(M_y)$. As $O_p(G_x^{[1]})\ne 1$, we must have $O_p(G_y^{[1]})=O_p(M_y)$. It follows that $C_{G_{x,y}}(G_y^{[1]})=N \times C_{M_{y}}(G_y^{[1]})$ with $C_{M_{y}}(G_y^{[1]}) \le C_{M_{y}}(O_p(G_y^{[1]})) \le O_p(M_y)$. Hence, $C_{G_{x,y}}(G_y^{[1]})$ does not contain a subgroup isomorphic to $M_y$. However, $C_{G_{x,y}}(G_x^{[1]}) \ge M_y$. We conclude that x and y are in different G-orbits.