Arc-transitive graphs of valency twice a prime admit a semiregular automorphism

TL;DR

This paper proves that all finite arc-transitive graphs with valency twice a prime have a nontrivial automorphism acting regularly on the vertices, supporting the broader Polycirculant Conjecture.

Contribution

It establishes the existence of semiregular automorphisms in a specific class of highly symmetric graphs, advancing the understanding of automorphism structures.

Findings

Every finite arc-transitive graph of valency twice a prime admits a semiregular automorphism.

Supports the Polycirculant Conjecture for a new class of graphs.

Provides structural insights into automorphisms of symmetric graphs.

Abstract

We prove that every finite arc-transitive graph of valency twice a prime admits a nontrivial semiregular automorphism, that is, a non-identity automorphism whose cycles all have the same length. This is a special case of the Polycirculant Conjecture, which states that all finite vertex-transitive digraphs admit such automorphisms.

Equations

$$|M|⩽|M_v||\mathrm{V}(\Gamma /P)|.$$

$$|M_{v_0}:M_{\alpha }|⩽2^s.$$

Full text

1. Introduction

All graphs in this paper are finite. In 1981, Marušič asked if every vertex-transitive digraph admits a nontrivial semiregular automorphism [11], that is, an automorphism whose cycles all have the same length. This question has attracted considerable interest and a generalisation of the affirmative answer is now referred to as the Polycirculant Conjecture [3]. One line of investigation of this question has been according to the valency of the graph or digraph. Every vertex-transitive graph of valency at most four admits such an automorphism [5, 12], and so does every vertex-transitive digraph of out-valency at most three [7]. Every arc-transitive graph of prime valency has a nontrivial semiregular automorphism [8] and so does every arc-transitive graph of valency 8 [15]. Partial results were also obtained for arc-transitive graphs of valency a product of two primes [16]. We continue this theme by proving the following theorem.

Theorem 1.1**.**

Arc-transitive graphs of valency twice a prime admit a nontrivial semiregular automorphism.

2. Preliminaries

If $G$ is a group of automorphisms of a graph $\Gamma$ and $v$ is a vertex of $\Gamma$, we denote by $G_{v}$ the stabiliser in $G$ of $v$, by $\Gamma(v)$ the neighbourhood of $v$, and by $G_{v}^{\Gamma(v)}$ the permutation group induced by $G_{v}$ on $\Gamma(v)$. We will need the following well-known results.

Lemma 2.1**.**

Let $\Gamma$ be a connected graph and $G\leqslant{\rm Aut}(\Gamma)$ such that all orbits of $G$ on $\mathrm{V}(\Gamma)$ have the same size. If a prime $p$ divides $|G_{v}|$ for some $v\in\mathrm{V}(\Gamma)$, then $p$ divides $|G_{v}^{\Gamma(v)}|$.

Lemma 2.2**.**

Let $G$ be a permutation group and let $K$ be a normal subgroup of $G$ such that $G/K$ acts faithfully on the set of $K$-orbits. If $G/K$ has a semiregular element $Kg$ of order $r$ coprime to $|K|$, then $G$ has a semiregular element of order $r$.

Proof.

See for example [15, Lemma 2.3]. ∎

Lemma 2.3**.**

A transitive group of degree a power of a prime $p$ contains a semiregular element of order $p$.

Proof.

In a transitive group of degree a power of a prime $p$, every Sylow $p$-subgroup is transitive. A non-trivial element of the center of this subgroup must be semiregular. ∎

Recall that a permutation group is quasiprimitive if every non-trivial normal subgroup is transitive, and biquasiprimitive if it is not quasiprimitive and every non-trivial normal subgroup has at most two orbits.

3. Arc-transitive graphs of prime valency

In the most difficult part of our proof, the arc-transitive graph of valency twice a prime will have a normal quotient with prime valency. We will thus need a lot of information about arc-transitive graphs of prime valency, which we collect in this section. We start with the following result, which is [2, Theorem 5]:

Theorem 3.1**.**

Let $\Gamma$ be a connected $G$-arc-transitive graph of prime valency $p$ such that the action of $G$ on $\mathrm{V}(\Gamma)$ is either quasiprimitive or biquasiprimitive. Then one of the following holds:

1. (1)

$G$* contains a semiregular element of odd prime order;* 2. (2)

$|\mathrm{V}(\Gamma)|$* is a power of $2$;* 3. (3)

$\Gamma=\mathrm{K}_{12}$, $G=\mathrm{M}_{11}$ and $p=11$; 4. (4)

$|\mathrm{V}(\Gamma)|=(p^{2}-1)/2s$* and $G={\rm PSL}_{2}(p)$ or ${\rm PGL}_{2}(p)$, where $p$ is a Mersenne prime and $\mathrm{C}_{r}\leqslant\mathrm{C}_{s}<\mathrm{C}_{(p-1)/2}$ , where $r$ is the product of the distinct prime divisors of $(p-1)/2$;* 5. (5)

$|\mathrm{V}(\Gamma)|=(p^{2}-1)/s$* and $G={\rm PGL}_{2}(p)$, where $p$ and $s$ are as in part (4), and $G$ is the standard double cover of the graph given in (4).*

We note that in Cases (4) and (5), we must have $p\geqslant 127$, since this is the smallest Mersenne prime $p$ such that $(p-1)/2$ is not squarefree. This fact will be used at the end of Section 4.

Corollary 3.2**.**

Let $\Gamma$ be a connected $G$-arc-transitive graph of prime valency. Then one of the following holds:

1. (1)

$G$* contains a semiregular element of odd prime order;* 2. (2)

$|\mathrm{V}(\Gamma)|$* is a power of $2$;* 3. (3)

$G$* contains a normal $2$-subgroup $P$ such that $(\Gamma/P,G/P)$ is one of the graph-group pairs in (3–5) of Theorem 3.1.*

Proof.

Suppose that $|\mathrm{V}(\Gamma)|$ is not a power of $2$ and let $P$ be a normal subgroup of $G$ that is maximal subject to having at least three orbits on $\mathrm{V}(\Gamma)$. In particular, $P$ is the kernel of the action of $G$ on the set of $P$-orbits. Hence $G/P$ acts faithfully, and quasiprimitively or biquasiprimitively on $\mathrm{V}(\Gamma/P)$. Since $\Gamma$ has prime valency and $G$-arc-transitive, [10, Theorem 9] implies that $P$ is semiregular. We may thus assume that $P$ is a $2$-group. (Otherwise $P$ contains a semiregular element of odd prime order.) If $G/P$ contains a semiregular element of odd prime order, then Lemma 2.2 implies that so does $G$. We may assume that this is not the case. Similarly, we may assume that $|\mathrm{V}(\Gamma/P)|$ is not a power of $2$. (Otherwise, $|\mathrm{V}(\Gamma)|$ is a power of $2$.) It follows that $\Gamma/P$ and $G/P$ are as is (3–5) of Theorem 3.1. ∎

We will now prove some more results about the graphs that appear in (3–5) of Theorem 3.1. Let us first recall the notion of coset graphs. Let $G$ be a group with a subgroup $H$ and let $g\in G$ such that $g^{2}\in H$ but $g\notin\mathrm{N}_{G}(H)$. The graph ${\rm Cos}(G,H,HgH)$ has vertices the right cosets of $H$ in $G$, with two cosets $Hx$ and $Hy$ adjacent if and only if $xy^{-1}\in HgH$. Observe that the action of $G$ on the set of vertices by right multiplication induces an arc-transitive group of automorphisms such that $H$ is the stabiliser of a vertex. Moreover, every arc-transitive graph can be constructed in this way [14].

Lemma 3.3**.**

The graphs in (3) and (4) of Theorem 3.1 have a triangle.

Proof.

Clearly $K_{12}$ has a triangle so suppose that $\Gamma$ is one of the graphs given in (4). Since $G$ is quasiprimitive on $\mathrm{V}(\Gamma)$ and $\Gamma$ has prime valency, it follows that ${\rm PSL}_{2}(p)$ is arc-transitive on $\Gamma$ and so $\Gamma={\rm Cos}(G,H,HgH)$ where $G={\rm PSL}_{2}(p)$, $H\cong\mathrm{C}_{p}\rtimes\mathrm{C}_{s}$ and $g\in G\backslash\mathrm{N}_{G}(H)$ such that $HgH$ is a union of $p$ distinct right cosets of $H$.

Let $x\in\mathrm{N}_{G}(H)\cong\mathrm{C}_{p}\rtimes\mathrm{C}_{(p-1)/2}$. Note that $G$ acts 2-transitively on the set of right cosets of $\mathrm{N}_{G}(H)$ with the stabiliser of any two points being isomorphic to $\mathrm{C}_{(p-1)/2}$. Now the coset $Hx\in\mathrm{V}(\Gamma)$ is fixed by $H$ and in particular $H$ has $|\mathrm{N}_{G}(H):H|=(p(p-1)/2)/ps=(p-1)/2s$ fixed points on $\mathrm{V}(\Gamma)$. If $y\notin\mathrm{N}_{G}(H)$, then the orbit of $Hy$ under $H$ has length $|H:H\cap H^{y}|$ and since $|\mathrm{N}_{G}(H):(\mathrm{N}_{G}(H)\cap\mathrm{N}_{G}(H)^{y})|=p$, it follows that $|H:H\cap H^{y}|=p$. Thus the points of $\mathrm{V}(\Gamma)$ that are not fixed by $H$ are permuted by $H$ in orbits of size $p$ and so for any $g\notin\mathrm{N}_{G}(H)$ we have that $HgH$ is a union of $p$ distinct right cosets of $H$.

For each $x\in\mathrm{N}_{G}(H)$ define the bijection $\lambda_{x}$ of $\mathrm{V}(\Gamma)$ by $Hy\mapsto x^{-1}Hy=Hx^{-1}y$. Since $G$ acts on $\mathrm{V}(\Gamma)$ by right multiplication, we see that $\lambda_{x}$ commutes with each element of $G$. Moreover, $\lambda_{x}$ is nontrivial if and only if $x\notin H$. Let $C=\{\lambda_{x}\mid x\in\mathrm{N}_{G}(H)\}\leqslant{\rm Sym}(\mathrm{V}(\Gamma))$. Since $G$ acts transitively on $\mathrm{V}(\Gamma)$ and $C$ centralises $G$, it follows from [4, Theorem 4.2A] that $C$ acts semiregularly on $\mathrm{V}(\Gamma)$. Now $|C|=|\mathrm{N}_{G}(H):H|=(p-1)/2s$ and $G\times C\leqslant{\rm Sym}(\mathrm{V}(\Gamma))$. Since $C\trianglelefteq G\times C$, the set of orbits of $C$ forms a system of imprimitivity for $G\times C$ and hence for $G$. Moreover, since $C$ is semiregular, comparing orders yields that $C$ has $|\mathrm{V}(\Gamma)|/|C|=p+1$ orbits. One of these orbits is the set of fixed points of $H$ and $H$ transitively permutes the remaining $p$ orbits of $C$. In particular, it follows that $C$ transitively permutes the nontrivial orbits of $H$ and so the isomorphism type of $\Gamma$ does not depend on the choice of the double coset $HgH$.

Let $Z$ be the subgroup of scalar matrices in ${\rm SL}_{2}(p)$ and let $\hat{H}$ be the subgroup of the stabiliser in ${\rm SL}_{2}(p)$ of the 1-dimensional subspace $\langle(1,0)\rangle$ such that $H=\hat{H}/Z$. Note that $\hat{h}=\begin{pmatrix}1&0\\ 1&1\end{pmatrix}\in\hat{H}$ and let $\hat{g}=\begin{pmatrix}0&1\\ -1&0\end{pmatrix}$. In particular, letting $h=\hat{h}Z$ and $g=\hat{g}Z$ we have that $g\notin\mathrm{N}_{G}(H)$ and so we may assume that $\Gamma={\rm Cos}(G,H,HgH)$. Now $Hg$ and $Hgh$ are both adjacent to $H$ and one can check that $g(gh)^{-1}=hgh\in HgH$ and so $\{H,Hg,Hgh\}$ is a triangle in $\Gamma$. ∎

Definition 3.4**.**

Let $\Gamma$ be a graph and let $S_{0}$ be a subset of $\mathrm{V}(\Gamma)$. Let $S=S_{0}$. If a vertex $u$ outside $S$ has at least two neighbours in $S$, add $u$ to $S$. Repeat this procedure until no more vertex outside $S$ have this property. If at the end of the procedure, we have $S=\mathrm{V}(\Gamma)$, then we say that $\Gamma$ is dense with respect to $S_{0}$.

Corollary 3.5**.**

Let $\Gamma$ be a graph in (3) or (4) of Theorem 3.1 and let $S_{0}=\{u,v\}$ be an edge of $\Gamma$. Then $\Gamma$ is dense with respect to $S_{0}$.

Proof.

Since $\Gamma$ is arc-transitive of prime valency $p$, the local graph at $v$ (that is, the subgraph induced on $\Gamma(v)$) is a vertex-transitive graph of order $p$ and thus vertex-primitive. By Lemma 3.3, $\Gamma$ has a triangle so the local graph has at least one edge and thus must be connected. It follows that, running the process described in Definition 3.4 starting at $S=S_{0}$, eventually $S$ will contain all neighbours of $v$. Repeating this argument and using connectedness of $\Gamma$ yields the desired conclusion. ∎

The following is immediate from Definition 3.4.

Lemma 3.6**.**

Let $\Gamma$ be a graph and $S_{0}$ be a set of vertices such that $\Gamma$ is dense with respect to $S_{0}$. Then the standard double cover of $\Gamma$, with vertex-set $\mathrm{V}(\Gamma)\times\{0,1\}$, is dense with respect to $S_{0}\times\{0,1\}$.

4. Proof of Theorem 1.1

Let $p$ be a prime, let $\Gamma$ be an arc-transitive graph of valency $2p$ and let $G={\rm Aut}(\Gamma)$. We may assume that $\Gamma$ is connected. If $G$ is quasiprimitive or bi-quasiprimitive, then $G$ contains a nontrivial semiregular element, by [6, Theorem 1.1] and [8, Theorem 1.4]. We may thus assume that $G$ has a minimal normal subgroup $N$ such that $N$ has at least three orbits. In particular, $\Gamma/N$ has valency at least $2$.

If $N$ is nonabelian, then $G$ has a nontrivial semiregular element by [16, Theorem 1.1]. We may therefore assume that $N$ is abelian and, in particular, $N$ is an elementary abelian $q$-group for some prime $q$.

We may also assume that $N$ is not semiregular that is, $N_{v}\neq 1$ for some vertex $v$. It follows from Lemma 2.1 that $1\neq N_{v}^{\Gamma(v)}\trianglelefteq G_{v}^{\Gamma(v)}$. As $\Gamma$ is $G$-arc-transitive, we have that $G_{v}^{\Gamma(v)}$ is transitive and so the orbits of $N_{v}^{\Gamma(v)}$ all have the same size, either $2$ or $p$. Since $N$ is a $q$-group, this size is equal to $q$. Writing $d$ for the valency of $\Gamma/N$, we have that either $(d,q)=(2,p)$ or $(d,q)=(p,2)$.

If $d=2$ and $q=p$, then it follows from [13, Theorem $1$] that $\Gamma$ is isomorphic to a graph denoted by $\mathrm{C}(p,r,s)$ in [13]. By [13, Theorem $2.13$], ${\rm Aut}(\mathrm{C}(p,r,s))$ contains the nontrivial semiregular automorphism $\varrho$ defined in [13, Lemma $2.5$].

We may thus assume that $d=p$ and $q=2$. In this case, if $u$ is adjacent to $v$, then $u$ has exactly $2=2p/d$ neighbours in $v^{N}$. Let $K$ be the kernel of the action of $G$ on $N$-orbits. By the previous observation, the orbits of $K_{v}^{\Gamma(v)}$ have size $2$ and thus it is a $2$-group. It follows from Lemma 2.1 that $K_{v}$ is a $2$-group and thus so is $K=NK_{v}$.

Now, $G/K$ is an arc-transitive group of automorphisms of $\Gamma/N$, so we may apply Corollary 3.2. If $G/K$ has a semiregular element of odd prime order, then so does $G$, by Lemma 2.2. If $|\mathrm{V}(\Gamma/N)|$ is a power of $2$, then so is $|\mathrm{V}(\Gamma)|$ and, in this case, $G$ contains a semiregular involution by Lemma 2.3. We may thus assume that we are in case (3) of Corollary 3.2, that is, $G/K$ contains a normal $2$-subgroup $P/K$ such that $(\Gamma/P,G/P)$ is one of the graph-group pairs in (3–5) of Theorem 3.1. Note that $P$ is a $2$-group. Let $M$ be a minimal normal subgroup of $G$ contained in the centre of $P$. We may assume that $M$ is not semiregular hence $M_{v}\neq 1$ and so by Lemma 2.1, $M_{v}^{\Gamma(v)}\neq 1$. Moreover, $|M|\neq 2$ as otherwise $M_{v}=M$ and we would deduce that $M$ fixes each element of $\mathrm{V}(\Gamma)$, a contradicition. Since $M$ is central in $P$, $M_{v}$ fixes every vertex in $v^{P}$.

Note that the $G$-conjugates of $M_{v}$ must cover $M$, otherwise $M$ contains a nontrivial semiregular element. By the previous paragraph, the number of conjugates of $M_{v}$ is bounded above by the number of $P$-orbits, that is $|\mathrm{V}(\Gamma/P)|$, so we have

[TABLE]

Since $\Gamma$ is connected and $G$-arc-transitive, there are no edges within $P$-orbits. As $M_{v}^{\Gamma(v)}\neq 1$, there exists $g\in M_{v}$ such that $w$ and $w^{g}$ are distinct neighbours of $v$. Let $u$ be the other neighbour of $w$ in $v^{P}$. Since $M_{v}$ fixes every element of $v^{P}$ it follows that $u$ is also a neighbour of $w$ and $w^{g}$ and so $\{v,w,u,w^{g}\}$ is a 4-cycle in $\Gamma$. Thus the graph induced between adjacent $P$-orbits is a union of $\mathrm{C}_{4}$’s.

If $x$ is a vertex and $y^{P}$ is a $P$-orbit adjacent to $x$, then there is a unique $\mathrm{C}_{4}$ containing $x$ between $x^{P}$ and $y^{P}$, and thus a unique vertex $z$ antipodal to $x$ in this $\mathrm{C}_{4}$. We say that $z$ is the buddy of $x$ with respect to $y^{P}$. The set of buddies of $v$ is equal to $\Gamma_{2}(v)\cap v^{P}$, which is clearly fixed setwise by $G_{v}$. Moreover, each vertex has the same number of buddies. Furthermore, since $G_{v}$ transitively permutes the set of $p$ $P$-orbits adjacent to $v^{P}$, either $v$ has a unique buddy or has exactly $p$ buddies.

If $v$ has a unique buddy $z$, then $\Gamma(v)=\Gamma(z)$, and so swapping every vertex with its unique buddy is a nontrivial semiregular automorphism. Thus it remains to consider the case where $v$ has $p$ buddies. We first prove the following.

Claim: If $X$ is a subgroup of $M$ that fixes pointwise both $a^{P}$ and $b^{P}$, and $c^{P}$ is a $P$-orbit adjacent to both $a^{P}$ and $b^{P}$, then $X$ fixes $c^{P}$ pointwise.

Proof: Suppose that some $x\in X$ does not fix $c$. Now $x$ fixes $a^{P}$ pointwise, so $c^{x}$ must be the buddy of $c$ with respect to $a^{P}$. Similarly, $c^{x}$ must be the buddy of $c$ with respect to $b^{P}$. These are distinct, which is a contradiction. It follows that $X$ fixes $c$ and, since $X\leqslant M$, also $c^{P}$.

Let $s\geqslant 1$, let $\alpha=(v_{0},\ldots,v_{s})$ be an $s$-arc of $\Gamma$ and let $\alpha^{\prime}=(v_{0},\ldots,v_{s-1})$. Now $|{v_{s}}^{M_{v_{s-1}}}|=2$, so $|M_{v_{s-1}}:M_{v_{s-1}v_{s}}|=2$ and $|M_{\alpha^{\prime}}:M_{\alpha}|\leqslant 2$. Applying induction yields that

[TABLE]

We first assume that $\Gamma/P$ and $G/P$ are as in (3) or (4) of Theorem 3.1. Let $\{u,v\}$ be an edge of $\Gamma$. By the previous paragraph, we have $|M_{v}:M_{uv}|\leqslant 2$. Recall that $M_{v}$ fixes all vertices in $v^{P}$, so $M_{uv}$ fixes all vertices in $v^{P}\cup u^{P}$. Combining the claim with Corollary 3.5 yields that $M_{uv}=1$ and thus $|M_{v}|=2$. It follows that $|M|\leqslant|M_{v}||\mathrm{V}(\Gamma/P)|$ so $|M|\leqslant 2|\mathrm{V}(\Gamma/P)|$. On the other hand, $M$ is an irreducible $(G/P)$-module over ${\mathrm{GF}}(2)$ of dimension at least two. Since $G/P$ is nonabelian simple or has a nonabelian simple group as an index two subgroup, this implies that $M$ is a faithful irreducible $(G/P)$-module over ${\mathrm{GF}}(2)$. If $G/P=\mathrm{M}_{11}$, then $|M|\geqslant 2^{10}$ [9], contradicting $|M|\leqslant 2\cdot 12=24$. If $G/P={\rm PSL}_{2}(p)$ or ${\rm PGL}_{2}(p)$ then by [1], $|M|\geqslant 2^{(p-1)/2}$. Recall that $p\geqslant 127$ and so this contradicts $|M|\leqslant 2(p^{2}-1)/2s<p^{2}-1$.

Finally, we assume that $\Gamma/P$ is in (5) of Theorem 3.1, that is, $\Gamma/P$ is the standard double cover of a graph $\Gamma^{\prime}$ which appears in (4) of Theorem 3.1. In particular, $\mathrm{V}(\Gamma/P)=\mathrm{V}(\Gamma^{\prime})\times\{0,1\}$. By Lemma 3.3, $\Gamma^{\prime}$ has a triangle, say $\{u,v,w\}$. By Corollary 3.5 and Lemma 3.6, $\Gamma/P$ is dense with respect to $\{u,v\}\times\{0,1\}$. Now, let $\overline{\alpha}=((u,0),(v,1),(w,0),(u,1),(v,0))$. Since $\overline{\alpha}$ contains $\{u,v\}\times\{0,1\}$, $\Gamma/P$ is dense with respect to $\overline{\alpha}$. Note that $\overline{\alpha}$ is a $4$-arc of $\Gamma/P$. Let $\alpha$ be a $4$-arc of $\Gamma$ that projects to $\overline{\alpha}$. Since $\Gamma/P$ is dense with respect to $\overline{\alpha}$, arguing as in the last paragraph yields $M_{\alpha}=1$. On the other hand, if $v\in\mathrm{V}(\Gamma)$ is the the initial vertex of $\alpha$, then by (1), we have $|M_{v}:M_{\alpha}|\leqslant 2^{4}$ and thus $|M_{v}|\leqslant 2^{4}$. Since $|M|\leqslant|M_{v}||\mathrm{V}(\Gamma/P)|$ it follows that $|M|\leqslant 2^{4}(p^{2}-1)/s$. As above, $M$ is a faithful irreducible $(G/P)$-module over ${\mathrm{GF}}(2)$ of dimension at least two. Since $G/P={\rm PGL}_{2}(p)$ we have from [1] that $|M|\geqslant 2^{(p-1)/2}$, which again contradicts $|M|\leqslant 2^{4}(p^{2}-1)/s<2^{4}(p^{2}-1)$.