On cubic arc-transitive k-multicirculants with soluble groups

Abstract

A finite simple graph is called a k-multicirculant if its automorphism group contains a cyclic semiregular subgroup having k orbits on the vertex set. It was shown by Giudici et al. that, if k is squarefree and coprime to 6, then a cubic arc-transitive k-multicirculant has at most $6k^2$ vertices (J. Combin. Theory Ser. B, 2017). In this paper, we classify the latter graphs under the assumption that their semiregular cyclic subgroups are contained in a soluble group of automorphisms acting transitively on the arc set of the graphs. As an application, cubic arc-transitive p-multicirculants are completely described for each odd prime p.

1 Introduction

All graphs in this paper are finite, simple and connected. For an integer $k \geqslant 1$, a graph is called a k-multicirculant if its automorphism group contains a cyclic semiregular subgroup with k orbits on the vertex set (the terms circulant, bicirculant, tricirculant, etc. are also in use instead of 1-, 2-, 3-multicirculant, etc.). It is a famous open problem whether each vertex-transitive graph is a k-multicirculant for some k, which is smaller than the order of the graph (see [3, 20]). For more information on this problem, we refer to the recent survey [1]. In particular, this has been answered in the positive for regular graphs of valency 3 by Marušič and Scapellato [21]. In what follows, we use the term cubic graph for a regular graph of valency 3.

In this paper, we focus on cubic k-multicirculants which are not only vertex- but also arc-transitive. A graph is called arc-transitive when its automorphism group is transitive on the set of its arcs (i.e., ordered pairs of adjacent vertices). These graphs attracted considerable attention (see [13,14,15, 17, 22, 23]). It follows from [18, Theorem 1.2 and Corollary 5.8] that, if k is even, then there exist infinitely many cubic arc-transitive k-multicirculants. On the other hand, Giudici et al. [15] proved the following theorem:

Theorem 1.1

( [15, Theorem 1.2]) If k is a squarefree positive integer and coprime to 6, then a cubic arc-transitive k-multicirculant has at most $6k^2$ vertices.

It is also conjectured that the theorem above holds for each odd number k (see [15, Conjecture 1.3]).

Besides the examples and the bound, cubic arc-transitive k-multicirculants have been also determined for $k \leqslant 5$. It is a rather easy exercise to show that $K_4$ and $K_{3,3}$ are the only graphs for $k=1$. The classification for $k=2$ follows from the work in [14, 22, 23], and it was obtained for $k=3$ in [17] and for $k=4$ and 5 in [13].

In this paper, we aim to extend these classification results for larger values of k. Motivated by Theorem 1.1, we consider the case when k is squarefree and coprime to 6. Our main result is the classification of such graphs under the additional assumption that their semiregular cyclic subgroups are contained in a soluble group of automorphisms acting transitively on the arc set of the graphs.

All the graphs in our main theorem below are s-arc-regular for some $s \leqslant 5$ and also Cayley graphs. A graph is called s-arc-regular when its automorphism group is regular on the set of its s-arcs (for the definition of an s-arc, see Sect. 2). Let H be a group with identity element 1 and $X \subset H$ be a subset such that $1 \notin X$ and $x^{-1} \in X$ whenever $x \in X$. The Cayley graph $\Gamma (H,X)$ is defined to have vertex set H and edges $\{h,xh\}$, $h \in H$ and $x \in X$. Our Cayley graphs will be constructed over the groups $H_1(n)$, $H_2(k)$, $H_3(k)$ and $H_4(k, d)$ defined as follows:

$$\begin{aligned} H_1(n)= & {} \langle u, y \mid u^n=y^2=1, u^y=u^{-1} \rangle \cong D_n,~ n\geqslant 2,\\ H_2(k)= & {} \langle u, v \rangle \rtimes \langle y \rangle \cong (\mathbb {Z}_k \times \mathbb {Z}_k) \rtimes \mathbb {Z}_2,~k \geqslant 2, \end{aligned}$$

where $u^k=v^k=1$, $y^2=1$, $u^y=u^{-1}$ and $v^y=v^{-1}$, and

$$\begin{aligned} H_3(k)=\langle u, v \rangle \rtimes \langle y \rangle \cong (\mathbb {Z}_{3k} \times \mathbb {Z}_k) \rtimes \mathbb {Z}_2,~k \geqslant 2, \end{aligned}$$

where $u^{3k}=v^{3k}=1$, $u^k=v^k$, $y^2=1$, $u^y=u^{-1}$ and $v^y=v^{-1}$. Note that, $H_1(n)$ is the dihedral group of order 2n, and $H_2(k)$ and $H_3(k)$ are examples of so called generalized dihedral groups (for the definition, see Section 2). Finally, for a divisor d of k such that $1< d < k$,

$$\begin{aligned} H_4(k,d)=\big (\langle u, v \rangle \times \langle w \rangle \big ) \rtimes \big \langle x, y \big \rangle \cong ((\mathbb {Z}_{k/d})^2 \times \mathbb {Z}_d) \rtimes {{\,\mathrm{Sym}\,}}(3), \end{aligned}$$

where $u^{k/d}=v^{k/d}=w^d=1$, $x^3=y^2=1$, $x^y=x^{-1}$, $u^x=v$, $v^x=u^{-1}v^{-1}$, $w^x=w$, $u^y=u^{-1}$, $v^y=uv$ and $w^y=w^{-1}$.

We are ready to state the main result of this paper.

Theorem 1.2

Let $k \geqslant 5$ be a squarefree integer coprime to 6 and $\Gamma$ be a cubic graph such that ${{\,\mathrm{\mathsf{Aut}}\,}}\Gamma$ contains a semiregular cyclic group C with k orbits on the vertex set. Then there exists a soluble group $G \leqslant {{\,\mathrm{\mathsf{Aut}}\,}}\Gamma$ such that $C \leqslant G$ and G acts transitively on the arcs of $\Gamma$ if and only if $\Gamma$ is s-arc-regular and $\Gamma \cong {{\,\mathrm{\Gamma }\,}}(H,X)$, where s, H, and X are listed in Table 1.

Table 1 Cubic s-arc-regular k-multicirculants $\Gamma ={{\,\mathrm{\Gamma }\,}}(H,X)$, $k \ge 5$ is squarefree and coprime to 6

Remark 1.3

Let n be a number given in (1) of Table 1. Notice that, there exists an integer $\lambda$ with $1+\lambda +\lambda ^2\equiv 0\pmod n$ if and only if $p \equiv 1\pmod 3$ for every prime divisor p of n such that $p > 3$. As $n=k$ or 3k and k is squarefree and coprime to 6, we also have $k \equiv 1\pmod 3$. This property of k will be needed later. Clearly, the above argument works also for the numbers d and $\lambda$ given in (4) of Table 1.

Furthermore, distinct choices of $\lambda$ might give rise to nonisomorphic graphs. The smallest such order is $2n=2 \cdot 7 \cdot 13=182$. There are four cubic arc-transitive graphs of order 182, namely,

$$\begin{aligned} \mathrm {C}182.1,~\mathrm {C}182.2,~\mathrm {C}182.3~\text {and}~\mathrm {C}182.4 \end{aligned}$$

in Marston Conder’s database [5]. This database contains all cubic arc-transitive graphs of order up to 10000, in which $\mathrm {C}n.i$ denotes the i-th graph of order n. A computation with the the computer algebra system Magma [2] shows that $\mathrm {C}182.1$ and $\mathrm {C}182.2$ belong to (1) of Table 1, whereas $\mathrm {C}182.3$ has $\mathrm {C}182.4$ admit no soluble group of automorphisms acting transitively on the arcs of the graph (the full groups are $\mathrm {PSL}(2,13)$ and $\mathrm {PGL}(2,13)$, respectively).

Remark 1.4

All graphs in (1)-(3) of Table 1 are edge-transitive bi-Cayley graphs of abelian groups. For recent development on this class of graphs, we refer to [7].

As an application of Theorem 1.2, we also classify the cubic arc-transitive p-multicirculants for each odd prime p, and by this extend the previous result for $p=3$ [17] and for $p=5$ [13].

Theorem 1.5

Let $p \geqslant 3$ be a prime. The cubic arc-transitive p-multicirculants are exactly the graphs listed in Table 2.

Table 2 Cubic s-arc-regular p-multicirculants $\Gamma$ for an odd prime p. In (7) $3_{+}^{1+2}$ denotes the unique nonabelian group of order 27 and of exponent 3

The paper is organized as follows: Section 2 contains the preliminary results needed in this paper. Following several preparatory lemmas derived in Sect. 3, the proof of Theorem 1.2 is presented in Sect. 4. Finally, we prove Theorem 1.5 in Sect. 5.

2 Preliminaries

If $\Gamma$ is a graph, then ${{\,\mathrm{\mathsf{V}}\,}}\Gamma$, ${{\,\mathrm{\mathsf{E}}\,}}\Gamma$ and ${{\,\mathrm{\mathsf{Aut}}\,}}\Gamma$ denote its vertex set, edge set and the automorphism group, respectively. The number $|{{\,\mathrm{\mathsf{V}}\,}}\Gamma |$ is called the order of $\Gamma$.

If $G \leqslant {{\,\mathrm{\mathsf{Aut}}\,}}\Gamma$ and $v \in {{\,\mathrm{\mathsf{V}}\,}}\Gamma$, then the stabilizer of v in G is denoted by $G_v$ and the orbit of v under G by $v^G$.

The set of neighbors of a vertex v is denoted by $\Gamma (v)$.

Let $s \geqslant 1$ be an integer. An s-arc of $\Gamma$ is an $(s+1)$-tuple $(v_1,\ldots ,v_{s+1})$ of its vertices such that $v_i$ is adjacent with $v_{i+1}$ for each $1 \leqslant i \leqslant s$, and $v_i \ne v_{i+2}$ for each $1 \leqslant i \leqslant s-1$.

$\Gamma$ is said to be (G, s)-transitive/regular if $G \leqslant {{\,\mathrm{\mathsf{Aut}}\,}}\Gamma$ and G is transitive/regular on the set of all s-arcs of $\Gamma$. An $({{\,\mathrm{\mathsf{Aut}}\,}}\Gamma ,s)$-regular graph is also called s-arc-regular, and an $({{\,\mathrm{\mathsf{Aut}}\,}}\Gamma ,1)$-transitive graph is simply called arc-transitive.

Let $\Gamma$ be a graph and $N \leqslant {{\,\mathrm{\mathsf{Aut}}\,}}\Gamma$ be a subgroup. The quotient graph $\Gamma _N$ is defined to have vertex set the orbits of N, and edges $\{ u^N, v^N \}$, where $\{u,v\} \in {{\,\mathrm{\mathsf{E}}\,}}\Gamma$ and $u^N \ne v^N$. If the natural projection $\pi : {{\,\mathrm{\mathsf{V}}\,}}\Gamma \rightarrow {{\,\mathrm{\mathsf{V}}\,}}\Gamma _N$ is a local bijection, i.e., the restriction $\pi \mid _{\Gamma (v)}$ induces a bijection

$$\begin{aligned} \pi \mid _{\Gamma (v)} : \Gamma (v) \rightarrow \Gamma _N(v^N) \end{aligned}$$

for each vertex $v \in {{\,\mathrm{\mathsf{V}}\,}}\Gamma$, then $\Gamma$ is called a regular N-cover of $\Gamma _N$.

If G is a group and $g \in G$, then the right multiplication $\rho _g$ is the permutation of G defined by $x^{\rho _g}=xg$, $x \in G$. For a subgroup $H \leqslant G$, let $H_{\mathrm {right}}=\{\rho _h : h \in H\}$.

By a generalized dihedral group, we mean a semidirect product $A \rtimes \langle y \rangle$ such that A is abelian, $\langle y \rangle \cong \mathbb {Z}_2$ and $a^y=a^{-1}$ for each $a \in A$.

2.1 Cubic arc-transitive graphs

In this subsection, we collect several results on cubic arc-transitive graphs. According to Tutte [25, 26], each cubic arc-transitive graph is s-arc-regular for some $s \leqslant 5$. The structure of a vertex-stabilizer is also known (see, e.g., [9]), and these are:

$$\begin{aligned} \mathbb {Z}_3,~{{\,\mathrm{Sym}\,}}(3),~{{\,\mathrm{Sym}\,}}(3) \times \mathbb {Z}_2,~{{\,\mathrm{Sym}\,}}(4)~\text {and}~ {{\,\mathrm{Sym}\,}}(4) \times \mathbb {Z}_2. \end{aligned}$$

A couple of results on graphs of particular order are invoked next, all of them were obtained by Feng and Kwak [11, 12].

Proposition 2.1

( [11, Proposition 2.8]) Let $\Gamma$ be a cubic arc-transitive graph of order 2p for a prime p. Then $\Gamma$ is s-arc-regular with $s \leqslant 4$ and one of the following holds:

• (i) $s=1$ and $\Gamma \cong {{\,\mathrm{\Gamma }\,}}(H,X)$, where H and X are given in (1) of Table 1 with $n=p$ and $p \geqslant 13$.

• (ii) $s=2$ and $\Gamma \cong \mathrm {C}4.1$.

• (iii) $s=3$ and $\Gamma \cong \mathrm {C}6.1$ or $\mathrm {C}10.1$.

• (iv) $s=4$ and $\Gamma \cong \mathrm {C}14.1$.

Proposition 2.2

( [11, Theorem 3.5]) Let $\Gamma$ be a cubic arc-transitive graph of order $2p^2$ for a prime p. Then $\Gamma$ is s-arc-regular with $s \leqslant 3$ and one of the following holds:

• (i) $s=1$ and $\Gamma \cong {{\,\mathrm{\Gamma }\,}}(H,X)$, where H and X are given in (1) of Table  1 with $n=p^2$.

• (ii) $s=2$ and $\Gamma \cong \mathrm {C}8.1$ or $\Gamma \cong {{\,\mathrm{\Gamma }\,}}(H,X)$, where H and X are given in (2) of Table 1 with $k=p$ and $p \geqslant 5$.

• (iii) $s=3$ and $\Gamma \cong \mathrm {C}18.1$.

Proposition 2.3

( [12, Theorem 6.2]) The cubic arc-transitive graphs of order 4p or $4p^2$ for a prime p are the following graphs: $\mathrm {C}8.1$, $\mathrm {C}16.1$, $\mathrm {C}20.1$, $\mathrm {C}20.2$ and $\mathrm {C}28.1$.

Now, we turn to the graphs of order 6p and $6p^2$, where p is a prime. These were classified in [12]; in particular, the 1- and 2-arc-regular graphs were described as voltage graphs in [12, Examples 3.3 and 3.4]. However, each of the latter graphs can also be represented as a Cayley graph. We prefer to use the Cayley graph representations.

Proposition 2.4

( [12, Theorem 5.2]) Let $\Gamma$ be a cubic arc-transitive graph of order 6p for a prime p. Then $\Gamma$ is s-arc-regular with $s \in \{1,3,4,5\}$ and one of the following holds:

• (i) $s=1$ and $\Gamma \cong {{\,\mathrm{\Gamma }\,}}(H,X)$, where H and X are given in (1) of Table 1 with $n=3p$.

• (ii) $s=3$ and $\Gamma \cong \mathrm {C}18.1$.

• (iii) $s=4$ and $\Gamma \cong \mathrm {C}102.1$.

• (iv) $s=5$ and $\Gamma \cong \mathrm {C}30.1$.

Proposition 2.5

( [12, Theorem 5.3]) Let $\Gamma$ be a cubic arc-transitive graph of order $6p^2$ for a prime p. Then $\Gamma$ is s-arc-regular with $s \leqslant 2$, and

• (i) $s=1$ and $\Gamma \cong {{\,\mathrm{\Gamma }\,}}(H,X)$, where H and X are given in (1) of Table 1 with $n=3p^2$.

• (ii) $s=2$ and $\Gamma \cong {{\,\mathrm{\Gamma }\,}}(H,X)$, where H and X are given in (3) of Table 1 with $k=p$.

We will also make use of the classification of cubic arc-transitive graphs of girth 6. The girth of a graph is the length of its smallest cycles (if any). Conder and Nedela [6] proved that, if a cubic arc-transitive graph has girth 6 and is not isomorphic to $\mathrm {C}14.1$, $\mathrm {C}18.1$ nor $\mathrm {C}20.2$, then it is s-arc-regular with $s \leqslant 2$. The 1- and 2-arc-regular graphs were classified later by Kutnar and Marušič [18]. The statement below is a direct consequence of their classification stated in [18, Theorems 1.1 and 1.2].

Proposition 2.6

Let $\Gamma$ be a cubic arc-transitive graph of girth 6. Then ${{\,\mathrm{\mathsf{Aut}}\,}}\Gamma$ contains a regular normal subgroup D isomorphic to a generalized dihedral group, unless $\Gamma \cong \mathrm {C}14.1$, $\mathrm {C}16.1$, $\mathrm {C}18.1$ or $\mathrm {C}20.2$.

2.2 Cubic arc-transitive $\mathbf {k}$-multicirculants

In this subsection, we review a few results on the graphs in the title, all of which were obtained in [15]. Throughout the paper,, we work with the notation

$$\begin{aligned} (\Gamma ,G,C,k), \end{aligned}$$

where $\Gamma$ is a cubic (G, 1)-transitive graph, $C < G$ is a cyclic and semiregular subgroup having k orbits. The set of all such quadruples is denoted by $\mathcal {Q}$.

Proposition 2.7

( [15, Lemmas 2.3 and 2.4]) Let $(\Gamma ,G,C,k) \in \mathcal {Q}$ and suppose that k is odd and N is an intransitive normal subgroup of G. Then $|N \cap C|$ is odd.

Let $(\Gamma ,G,C,k) \in \mathcal {Q}$ and suppose that $N \lhd G$ such that N has more than 2 orbits. It is known that in this case $\Gamma$ is a regular N-cover of $\Gamma _N$, and $\Gamma _N$ is a cubic (G/N, 1)-transitive graph (see, e.g., [19, Theorem 9]). The following result gives us information about the group NC/N.

Proposition 2.8

( [15, Lemma 2.5]) Let $(\Gamma ,G,C,k) \in \mathcal {Q}$ and suppose that $N \lhd G$ such that |N| is coprime to 6 and N has more than 2 orbits. Then

$$\begin{aligned} (\Gamma _N,G/N,NC/N,k') \in \mathcal {Q}, \end{aligned}$$

where $k=k' \cdot |N:C \cap N|$.

Proposition 2.9

( [15, Lemma 3.4]) Let $(\Gamma ,G,C,k) \in \mathcal {Q}$ and suppose that k is squarefree and coprime to 6, and $P \lhd G$ is a normal Sylow p-subgroup of G for a prime $p \geqslant 5$. Then

• (i) $|C_C(P)|$ is odd.

• (ii) $|P: C \cap P|=p$ and $|C \cap P| \leqslant p$.

Corollary 2.10

Assuming the hypothesis in Proposition 2.9, $P\cong \mathbb {Z}_p$ or $\mathbb {Z}_p^2$.

Proof

It follows from Proposition 2.9(ii) that $|P| \leqslant p^2$. Assume on the contrary that $P \cong \mathbb {Z}_{p^2}$. Note that, then $P\cap C \ne 1$. Clearly, |C| is even. Let $t_c$ be the unique involution in C and let us consider its action on P by conjugation. Then ${{\,\mathrm{\mathsf{Aut}}\,}}(P) \cong \mathbb {Z}_{p(p-1)}$, hence the only automorphism of P of order two is the one inverting its elements. Using also that $P \cap C \ne 1$, this shows that $t_c \in C_C(P)$, a contradiction to Proposition 2.9(i). $\square$

In the case when G is also soluble we have the following additional property.

Proposition 2.11

( [15, Lemma 3.3]) Let $(\Gamma ,G,C,k) \in \mathcal {Q}$ and k be squarefree and coprime to 6. If G is a soluble group, then $\Gamma$ is (G, s)-regular with $s \leqslant 2$.

The soluble radical of a group is its largest soluble normal subgroup. Part (i) in the following proposition is [15, Lemma 2.10], part (ii) is deduced from [15, Lemma 3.6] and part (iii) is contained [15, Lemma 2.12(1)].

Proposition 2.12

( [15]) Let $(\Gamma ,G,C,k) \in \mathcal {Q}$ and k be squarefree and coprime to 6. Suppose that $\Gamma$ is (G, s)-regular, G is an insoluble group and S is the soluble radical of G. Then

• (i) S is semiregular and has more than 2 orbits.

• (ii) The possibilities for G/S, |CS/S|, s and $\log _2(|S|_2)$ are as given in Table 3.

• (iii) $|C \cap S|$ is odd.

Table 3 The group G/S, the order |SC/S|, and upper bounds on s and $\log _2(|S|_2)$, where r is a prime

3 Preparatory lemmas

For this section we set the following hypothesis.

Hypothesis 3.1

Let $(\Gamma ,G,C,k) \in \mathcal {Q}$, where k is squarefree and coprime to 6 and G is a soluble group. Let N be a Hall $\{2,3\}^\prime$-subgroup of G and for $1 \leqslant i \leqslant m$, let $P_i$ be a Sylow $p_i$-subgroup of N, where the primes $p_i$ comprise the distinct prime divisors of |N|.

Notice that, each Sylow p-subgroup of G is metacyclic. This is clear when $p > 2$, and it follows for $p=2$ from Proposition 2.11. Since G is soluble, by [4, Theorem 1], the group G is equal to the semidirect product $N \rtimes A$, where A is a Hall $\{2,3\}$-subgroup of G and N has a normal series

$$\begin{aligned} 1=N_0 \lhd N_1 \lhd N_2 \lhd \cdots \lhd N_m=N, \end{aligned}$$

(1)

where for each $1 \leqslant i \leqslant m$, $N_i/N_{i-1}$ is isomorphic to $P_i$.

Since |N| is coprime to 6, it follows that N is semiregular. Observe that, if N has more than 2 orbits, then it follows from Proposition 2.8 that CN/N is regular. In other words, $\Gamma _N$ is a cubic arc-transitive circulant, and so $\Gamma _N \cong K_4$ or $K_{3,3}$. Our first lemma excludes the former possibility.

Lemma 3.2

Assuming Hypothesis 3.1, suppose that N has more than 2 orbits. Then $\Gamma _N \cong K_{3,3}$.

Proof

By Proposition 2.8,

$$\begin{aligned} (\Gamma _{N_{m-1}},G/N_{m-1},CN_{m-1}/N_{m-1},k_{m-1}) \in \mathcal {Q}, \end{aligned}$$

where $k_{m-1}$ is a divisor of k. The group $N/N_{m-1}$ is a normal Sylow $p_m$-subgroup of $G/N_{m-1}$, and it follows from Proposition 2.9(ii) that $|N/N_{m-1}| \leqslant p_m^2$.

Assume on the contrary that $\Gamma _N \cong K_4$. Then $|{{\,\mathrm{\mathsf{V}}\,}}\Gamma _{N_{m-1}}|=4p_m$ or $4p_m^2$, and by Proposition 2.3,

$$\begin{aligned} \Gamma _{N_{m-1}} \cong \mathrm {C}8.1,~\mathrm {C}16.1,~\mathrm {C}20.1,~\mathrm {C}20.2~\text {or}~ \mathrm {C}28.1. \end{aligned}$$

The group $CN_{m-1}/N_{m-1}$ is a semiregular cyclic subgroup of $GN_{m-1}/N_{m-1}$ having $k_{m-1}$ orbits. Since $k_{m-1}$ is odd, it follows immediately that $\Gamma _{N_{m-1}} \ncong \mathrm {C}8.1$ nor $\mathrm {C}16.1$. To finish the proof, it is sufficient to show that none of the graphs $\mathrm {C}20.1$, $\mathrm {C}20.2$ and $\mathrm {C}28.1$ has a cyclic group of automorphisms of order 4, which is semiregular on the vertices. We checked this with the help of Magma [2]. $\square$

Lemma 3.3

Assuming Hypothesis 3.1, $C_G(N_1)$ is semiregular, where $N_1$ is defined in (1). Furthermore, if $\Gamma$ is (G, 1)-regular, then $|C_G(N_1)|$ is odd.

Proof

Let $M=C_G(N_1)$. Then . Assume on the contrary that $M_v \ne 1$ for a vertex v. There exists $L \leqslant M$ such that $M=N_1 \times L$. Since $N_1$ and L have coprime order, $M_v=(N_1)_v \times L_v$, implying that $(N_1)_v=1$ and so $L_v \ne 1$. Since L is characteristic in M and , we have $L \lhd G$. Also, as $p_1$ does not divide |L|, L has more than 2 orbits, a contradiction to $L_v \ne 1$.

Suppose that $\Gamma$ is (G, 1)-regular. Then $|G|=3k \cdot |C|$, hence G has a cyclic Sylow 2-subgroup, and therefore, all involutions are conjugate to each other. Assume on the contrary that |M| is even, and let $t \in M$ be an involution. By the previous remark $t^g \in C$ for some $g \in G$, and as , we have $t^g \in M$, and by this $|C_C(N_1)|$ is even. This contradicts Proposition 2.9(i). $\square$

In the next lemma, we consider the case when |N| is squarefree.

Lemma 3.4

Assuming Hypothesis 3.1, suppose that |N| is squarefree. Then

• (i) N is a cyclic group.

• (ii) $\Gamma$ is (G, 1)-regular.

• (iii) G contains a regular normal subgroup H isomorphic to $D_n$ with $n=k$ or 3k.

Proof

Clearly, |C| is even. Let $t_c$ be the unique involution contained in C.

(i): Assume for the moment that the commutator $[t_c,x]=1$ for some $x \in N$ and $x \ne 1$. Let i be the unique index such that $x \in N_i$ and $x \notin N_{i-1}$. Write $\bar{g}$ for the image of $g \in G$ under the canonical epimorphism $G \rightarrow G/N_{i-1}$, and $\bar{K}$ for the image of a subgroup $K \leqslant G$. Since $\bar{N_i} \cong \mathbb {Z}_{p_i}$, we have $\bar{t}_c \in C_{\bar{C}}(\bar{N_i})$, and thus $|C_{\bar{C}}(\bar{N_i})|$ is even. On the other hand, $\bar{N_i}$ is a normal Sylow $p_i$-subgroup of $\bar{G}$, and then one can apply Proposition 2.9(i) to $(\Gamma _{N_{i-1}},\bar{G},\bar{C},k_{i-1})$, and obtain that $|C_{\bar{C}}(\bar{N_i})|$ is odd, a contradiction.

It follows that $C_N(t_c)=1$, and thus $t_c$ acts on N by conjugation as a fixed-point-free automorphism. By [16, Theorem 10.1.4], N is abelian and

$$\begin{aligned} x^{t_c}=x^{-1},~x \in N. \end{aligned}$$

(2)

As |N| is squarefree, N is a cyclic group.

(ii): For $1 \leqslant i \leqslant m$, $P_i$ is characteristic in N, hence $P_i \lhd G$. By Proposition 2.9(i), $|C_C(P_i)|$ is odd, implying that $p_i$ does not divide |C|, and so C is a $\{2,3\}$-group. Using also that $|G|=|C| \cdot k \cdot |G_v|$, we find

$$\begin{aligned} k=|G_{\{2,3\}^\prime }|=|N|. \end{aligned}$$

(3)

Let $M=C_G(N_1)$. Then . The group G/M is isomorphic to a subgroup of ${{\,\mathrm{\mathsf{Aut}}\,}}(N_1)$, and hence it is a cyclic group. By Lemma 3.3, M is semiregular, hence $G/M > 1$. Furthermore, $G_vM/M \cong G_v/(G_v \cap M) \cong G_v$. Thus, $G_v$ is also a cyclic group, by which it follows that $G_v \cong \mathbb {Z}_3$, and (ii) is proved.

(iii): Note that, as $\Gamma$ is (G, 1)-regular, |M| is odd by Lemma 3.3. Since $|G_v|=3$ and $\Gamma _N \cong K_2$ or $K_{3,3}$ by Lemma 3.2, it follows that $|A|=6$ or 18.

Case 1. $|A|=6$.

In this case $M=N$ and $|C|=2$. We claim that the group $H:=\langle N, t_c \rangle$ satisfies all conditions in part (iii) of the lemma. By Eqs. (2) and (3), $H \cong D_k$ and $|H|=|{{\,\mathrm{\mathsf{V}}\,}}\Gamma |$. ${{\,\mathrm{\mathsf{V}}\,}}\Gamma$ splits into two N-orbits. If $t_c$ fixes setwise these orbits, then it is contained in an intransitive normal subgroup of G. This is impossible by Proposition 2.7, hence H is regular. Finally, since $M < H$ and G/M is a cyclic group, $H \lhd G$ also holds.

Case 2. $|A|=18$.

In this case $|C|=6$. Thus G contains a semiregular subgroup of order 3. If a Sylow 3-subgroup of G was cyclic, then any two subgroups of order 3 in G would be conjugate, which is impossible. Thus a Sylow 3-subgroup of G is isomorphic to $\mathbb {Z}_3^2$. Since G/M is cyclic, it follows that 3 divides |M|. Using also that M is semiregular and |M| is odd, see Lemma 3.3, we conclude that $|M : N|=3$. As C is contained in some Hall $\{2,3\}$-subgroup of G, we may assume that $C < A$. Since G/M is a cyclic group, $|M \cap A|=3$, and $A/(M \cap A) \cong \mathbb {Z}_6$. The group A cannot be abelian because A has a faithful arc-transitive action on $\Gamma _N \cong K_{3,3}$.

It follows that $M \cap A=A^\prime$, the derived subgroup of A, so $C_A(N_1)=A^\prime$. Notice that, one can prove in the same way that $C_A(N_i)=A^\prime$ for each $1 \leqslant i \leqslant m$. This shows that $M=N \times (M \cap A)$, in particular, M is a cyclic group of order 3k.

Also, as $M \cap A=A^\prime$, $\langle M \cap A, t_c \rangle \lhd A$ and $|\langle M \cap A, t_c \rangle |=6$. If $\langle M \cap A, t_c \rangle$ is abelian, then $A^\prime =M \cap A \leqslant Z(A)$, implying that A is abelian, which is not the case. We obtain that $m^{t_c}=m^{-1}$ for each $m \in M \cap A$. Since $M=N \times (M \cap A)$, it follows from Eq. (2) that the group $H:=\langle M, t_c \rangle$ is isomorphic to $D_{3k}$. One can copy the argument used by Case 1 to derive that H is also regular and $H \lhd G$. $\square$

We would like to remark that to derive Lemma 3.4(iii), one could have used [7, Proposition 4.1] in Case 1 and [10] in Case 2.

One of the key steps toward Theorem 1.2 is to show that N is always abelian.

Lemma 3.5

Assuming Hypothesis 3.1, N is an abelian group.

Proof

We proceed by induction on the order of N. The lemma holds trivially if $N=N_1$. Let $N > N_1$ and assume that the lemma holds for $(\Gamma ^*,G^*,C^*,k^*)\in \mathcal {Q}$ whenever $k^*$ is squarefree and coprime to 6 and a Hall $\{2,3\}^\prime$-subgroup of $G^*$ has order less than |N|. By Proposition 2.8, $(\Gamma _{N_1},G/N_1,N_1C/N_1,k_1) \in \mathcal {Q}$, and in view of what has been just said $N/N_1$ is abelian, and thus $N^\prime \leqslant N_1$.

Let $M=C_G(N_1)$. Then . If $M \cap N > N_1$, then $M \cap N=N_1 \times L$, where |L| is coprime to $|N_1|$. Then $L \lhd G$. As |L| is coprime to 6 and L is semiregular having more than 2 orbits. By Proposition 2.8, $(\Gamma _L,G/L,LC/L,k^\prime ) \in \mathcal {Q}$ for some divisor $k^\prime$ of k. We obtain that N/L is abelian, and hence $N^\prime \leqslant L$. By the previous paragraph, $N^\prime \leqslant N_1 \cap L=1$, so N is abelian.

For the rest of the proof let $M \cap N=N_1$, in other words, $C_N(N_1)=N_1$. We finish the proof by showing that this leads to a contradiction. Note that, G/M is isomorphic to a subgroup of ${{\,\mathrm{\mathsf{Aut}}\,}}(N_1)$. Since $P_1=N_1 \lhd G$, Corollary 2.10 can be applied to $N_1$ and thus $N_1 \cong \mathbb {Z}_{p_1}$ or $\mathbb {Z}_{p_1}^2$.

Suppose that $N_1 \cong \mathbb {Z}_{p_1}$. Then ${{\,\mathrm{\mathsf{Aut}}\,}}(N_1) \cong \mathbb {Z}_{p_1-1}$, and G/M is a cyclic group. As $NM/N \cong N/(M\cap N)=N/N_1$, $N/N_1$ is also a cyclic group. Then $|N/N_1|$ is squarefree by Corollary 2.10, and hence so is |N|. By Lemma 3.4, N is then a cyclic group, which contradicts that $C_N(N_1)=N_1$.

Suppose now that $N_1 \cong \mathbb {Z}_{p_1}^2$. Denote by $N_{(p_2)^\prime }$ the Hall $(p_2)^\prime$-subgroup of N. Since $N/N_1$ is abelian, $N_{(p_2)^\prime }/N_1$ is characteristic in $N/N_1$, and as $N/N_1 \lhd G/N_1$, it follows that , hence . By Proposition 2.8, $(\Gamma _{N_{(p_2)^\prime }},G/N_{(p_2)^\prime },CN_{(p_2)^\prime }/N_{(p_2)^\prime },k^\prime ) \in \mathcal {Q}$ for some divisor $k^\prime$ of k. Furthermore, $N/N_{(p_2)^\prime } \cong \mathbb {Z}_{p_2}$ or $\mathbb {Z}_{p_2}^2$, see Corollary 2.10.

Let K be a Hall $\{2,3,p_2\}$-subgroup of G. We claim that, there exists a subgroup $L < K$ such that

$$\begin{aligned} |L|=3 \cdot |N/N_{(p_2)^\prime }|~\text {and}~Z(L)=1. \end{aligned}$$

We are going to use the fact that K has a faithful arc-transitive action on $\Gamma _{N_{(p_2)^\prime }}$. Since $|{{\,\mathrm{\mathsf{V}}\,}}\Gamma _{N_{(p_2)^\prime }}|= 2|N/N_{(p_2)^\prime }|$ or $6|N/N_{(p_2)^\prime }|$ and $|N/N_{(p_2)^\prime }|=p_2$ or $p_2^2$, the graph $\Gamma _{N_{(p_2)^\prime }}$ is described in Propositions 2.1–2.5. We deal with each case separately.

- $|{{\,\mathrm{\mathsf{V}}\,}}\Gamma _{N_{(p_2)^\prime }}|=2p_2$. According to Proposition 2.1, $\Gamma _{N_{(p_2)^\prime }}={{\,\mathrm{\Gamma }\,}}(H,X)$, where H and X are given in (1) of Table 1 with $n=p_2 \geqslant 13$, or

$$\begin{aligned} \Gamma _{N_{(p_2)^\prime }} \cong \mathrm {C}4.1,~\mathrm {C}6.1,~\mathrm {C}10.1~\text {or}~\mathrm {C}14.1. \end{aligned}$$

Since $p_2 \geqslant 5$, $\mathrm {C}4.1$ and $\mathrm {C}6.1$ are impossible. A computation with Magma [2] shows that $\mathrm {C}10.1$ does not admit an arc-transitive group, which also normalizes a Sylow 5-subgroup, and $\mathrm {C}14.1$ admits a subgroup of order 21 having trivial center. Finally, we consider the case when $K \leqslant {{\,\mathrm{\mathsf{Aut}}\,}}{{\,\mathrm{\Gamma }\,}}(H,X)$, where H and X are given in (1) of Table 1 with $n=p_2 \geqslant 13$.

Then $H_{\mathrm {right}}\lhd K$ and as ${{\,\mathrm{\Gamma }\,}}(H,X)$ is (K, 1)-transitive, $K_1$ (the stabilizer of the identity element $1\in H$) contains the automorphism $\alpha \in {{\,\mathrm{\mathsf{Aut}}\,}}(H)$ defined by $u^\alpha =u^\lambda$ and $y^\alpha =uy$. It is easy to show that $\alpha$ fixes no nonidentity element in $\langle u \rangle$. Let $L=\langle \rho _u, \alpha \rangle$. Then $|L|=3p_2=3 \cdot |N/N_{(p_2)^\prime }|$ and $Z(L)=1$, as required.

- $|{{\,\mathrm{\mathsf{V}}\,}}\Gamma _{N_{(p_2)^\prime }}|=2p_2^2$. In view of the facts that $N/N_{(p_2)^\prime }$ is not cyclic and of order $p_2^2 \geqslant 25$, Proposition 2.1 shows that $\Gamma _{N_{(p_2)^\prime }}={{\,\mathrm{\Gamma }\,}}(H,X)$, where H and X are given in (2) of Table 1 with $k=p_2$. We may write $K \leqslant {{\,\mathrm{\mathsf{Aut}}\,}}{{\,\mathrm{\Gamma }\,}}(H,X)$. Then K contains the automorphism $\alpha \in {{\,\mathrm{\mathsf{Aut}}\,}}(H)$ defined by $u^\alpha =v$, $v^\alpha =u^{-1}v^{-1}$ and $y^\alpha =uy$. It is easy to show that $\alpha$ fixes no nonidentity element of $\langle u, v\rangle$. Then we are done by letting $L=\langle \rho _u, \rho _v, \alpha \rangle$.

- $|{{\,\mathrm{\mathsf{V}}\,}}\Gamma _{N_{(p_2)^\prime }}|=6p_2$. According to Proposition 2.1, $\Gamma _{N_{(p_2)^\prime }}={{\,\mathrm{\Gamma }\,}}(H,X)$, where H and X are given in (1) of Table 1 with $n=3p_2$, or

$$\begin{aligned} \Gamma _{N_{(p_2)^\prime }} \cong \mathrm {C}18.1,~\mathrm {C}30.1~\text {or}~\mathrm {C}102.1. \end{aligned}$$

In the former case one can proceed in the same way as by the “$2p_2$ case”. The graph $\mathrm {C}18.1$ is impossible because $p_2 \geqslant 5$, and to exclude the graphs $\mathrm {C}30.1$ and $\mathrm {C}102.1$, one can check that these graphs do not admit an arc-transitive group, which also normalizes a Sylow 5- and 17-subgroup, respectively.

- $|{{\,\mathrm{\mathsf{V}}\,}}\Gamma _{N_{(p_2)^\prime }}|=6p_2^2$. Since $N/N_{(p_2)^\prime }$ is not cyclic, Proposition 2.1 shows that $\Gamma _{N_{(p_2)^\prime }}={{\,\mathrm{\Gamma }\,}}(H,X)$, where H and X are given in (3) of Table 1 with $k=p_2$. Then one can copy the argument used by the “$2p_2^2$” case. This completes the proof of the claim.

Let $Z=C_L(N_1)$. Since $C_N(N_1)=N_1$ it follows that $|Z| \leqslant 3$. If $|Z|=3$, then as $N_1$ is normal in G, Z is normal in L, and hence $L/C_L(Z)$ is isomorphic to a subgroup of ${{\,\mathrm{\mathsf{Aut}}\,}}(Z) \cong \mathbb {Z}_2$. Using also that $|L|=3p_2$ or $3p_2^2$, this shows that $Z \leqslant Z(L)$, contradicting that $Z(L)=1$. We obtain that L is isomorphic to a subgroup of $\mathrm {GL}(2,p_1)$. Moreover, as Z(L) is trivial and |L| is odd, L is isomorphic to a subgroup of $\mathrm {PSL}(2,p_1)$. On the other hand, it follows from Dickson’s classification of the subgroups of $\mathrm {PSL}(2,p_1)$ (cf. [24, Theorem 3.6.25]) that any nonabelian subgroup of order not divisible by $p_1$ has even order, and therefore, it cannot be isomorphic to L, a contradiction. $\square$

In the rest of the section, we discuss some consequences of the previous lemma. First, it follows that $P_i \cong \mathbb {Z}_{p_i}$ or $\mathbb {Z}_{p_i}^2$ for each $1 \leqslant i \leqslant m$.

Lemma 3.6

Assuming Hypothesis 3.1, suppose that $\Gamma _N \cong K_{3,3}$ and $P_i \cong \mathbb {Z}_{p_i}^2$ for some $1 \leqslant i \leqslant m$. Then 3 divides $|C_C(P_i)|$. 6

Proof

Let $M=N_{(p_i)^\prime }$. As N is abelian, $M \lhd G$. For $x \in G$, denote by $\bar{x}$ the image of x under the canonical epimorphism from G to G/M, and by $\bar{P}_i$ the image of $P_i$.

We claim that, for each $x \in G$,

$$\begin{aligned} o(x)=3~\text {and}~C_{P_i}(x) \ne 1 \implies x \in C_G(P_i). \end{aligned}$$

(4)

Choose $x \in G$ such that $o(x)=3$ and $C_{P_i}(x) \ne 1$. It is clear that M has more than 2-orbits. Thus $\Gamma$ is a regular M-cover of $\Gamma _M$, the kernel of the action of G on ${{\,\mathrm{\mathsf{V}}\,}}\Gamma _M$ is equal to M, and the image is isomorphic to $\bar{G}=G/M$. The group G/M has a faithful action on the graph $\Gamma _M$. As $x \notin M$, $\bar{x}$ is an automorphism of $\Gamma _M$ of order 3. The latter graph is cubic arc-transitive of order $6p_i^2$, hence by Proposition 2.5, it is isomorphic to ${{\,\mathrm{\Gamma }\,}}(H_3(k),X)$, where X is given in (3) in Table 1 and $k=p_i$.

Note that $C_{\bar{P}_i}(\bar{x}) \ne 1$ because $C_{P_i}(x) \ne 1$. We show next that $\bar{x} \in C_{\bar{G}}(\bar{P}_i)$. Recall that $H_3(p_i)=\langle u, v, y \rangle$ and $X=\{y,uy,uvy\}$, see Table 1. Consider the automorphism $\alpha$ of $H_3(p_i)$ defined by

$$\begin{aligned} u^\alpha =v,~v^\alpha =u^{-1}v^{-1}~\text {and}~y^\alpha =uy. \end{aligned}$$

The set X is mapped to itself by $\alpha$. This implies in turn that $\alpha \in {{\,\mathrm{\mathsf{Aut}}\,}}{{\,\mathrm{\Gamma }\,}}(H_3(p_i),X)$, $\alpha \in \bar{G}$, and $\bar{G}$ has a Sylow 3-subgroup of the form $E:=\langle \rho _{u^{p_i}}, \alpha \rangle \cong \mathbb {Z}_3^2$. By the Sylow Theorem, $\bar{x}$ is conjugate to an element of E in $\bar{G}$, and as $\bar{P_i} \lhd \bar{G}$, we may assume that $\bar{x} \in E$. It is not hard to show that $\alpha$ fixes no nonidentity element of $\langle u^3, v^3 \rangle$. Using also that $\bar{P_i}=\langle \rho _{u^3}, \rho _{v^3} \rangle$ and that $C_{\bar{P}_i}(\bar{x}) \ne 1$, we conclude that $\bar{x} \in C_{\bar{G}}(\bar{P}_i)$, as claimed.

Equivalently, $[x,y] \in M$ for each $y \in P_i$. As $P_i \lhd G$, $[x,y] \in P_i$ also holds, and so $[x,y] \in P_i \cap M$. The implication in (4) follows because $P_i \cap M=1$.

Since $\Gamma _N \cong K_{3,3}$, $|{{\,\mathrm{\mathsf{V}}\,}}\Gamma |$ is divisible by 3, and therefore so is |C|. Choose $x \in C$ such that $o(x)=3$. It is clear that $P_i \cap C \ne 1$, and therefore, $C_{P_i}(x) \ne 1$. It follows from (4) that $x \in C_C(P_i)$, and the lemma is proved. $\square$

Lemma 3.7

Assuming Hypothesis 3.1, if |N| is not a square, then $\Gamma$ is (G, 1)-regular.

Proof

By Proposition 2.11, $\Gamma$ is (G, s)-regular for $s \leqslant 2$. Assume on the contrary that $\Gamma$ is (G, 2)-regular. Since |N| is not a square, $P_i \cong \mathbb {Z}_{p_i}$ for some $1 \leqslant i \leqslant m$. By Lemma 3.5, N is abelian, hence $N_{(p_i)^\prime } \lhd G$. We obtain that $\Gamma _{N_{(p_i)^\prime }}$ is $(G/N_{(p_i)^\prime },2)$-regular, and it has order $2p_i$ or $6p_i$. It follows from Propositions 2.1 and 2.4 that $\Gamma _{N_{(p_i)^\prime }} \cong \mathrm {C}10.1,\, \mathrm {C}14.1, \mathrm {C}30.1$ or $\mathrm {C}102.1$. However, a computation with Magma [2] shows that ${{\,\mathrm{\mathsf{Aut}}\,}}\mathrm {C}14.1$ contains no subgroup of order 84, and the remaining graphs admit no arc-transitive group, which also normalizes a Sylow $p_i$-subgroup. $\square$

Lemma 3.8

Assuming Hypothesis 3.1, suppose that $\Gamma$ is bipartite, $\Gamma \ncong \mathrm {C}14.1$ and there exists $T < G$ such that T is abelian, semiregular and has two orbits, which are the biparts of ${{\,\mathrm{\mathsf{V}}\,}}\Gamma$. Then

• (i) ${{\,\mathrm{\mathsf{Aut}}\,}}\Gamma$ contains a regular normal subgroup D isomorphic to a generalized dihedral group. In particular, ${{\,\mathrm{\mathsf{Aut}}\,}}\Gamma$ is a soluble group.

• (ii) |N| is either squarefree or a square.

Proof

(i): We show first that $\Gamma$ has girth 6. Fix an edge $\{\mathbf {u},\mathbf {v}\}$ of $\Gamma$. Clearly, $\mathbf {u}$ and $\mathbf {v}$ belong to distinct T-orbits, and the remaining two neighbors of $\mathbf {u}$ are of the form $\mathbf {v}^x$ and $\mathbf {v}^y$ for distinct nonidentity elements $x, y \in T$. Then we have the closed walk of length 6 in $\Gamma$ given as

$$\begin{aligned} \mathbf {u}\sim \mathbf {v}^x \sim \mathbf {u}^x \sim \mathbf {v}^{yx}=\mathbf {v}^{xy} \sim \mathbf {u}^y \sim \mathbf {v}^y \sim \mathbf {u}. \end{aligned}$$

It is a folklore result that the only cubic arc-transitive graphs having girth 4 are $K_{3,3}$ and the cube graph $Q_3$, and as none of these can occur, we conclude that $\Gamma$ has girth 6.

Proposition 2.6 can be applied to $\Gamma$. It is clear that $\Gamma \ncong \mathrm {C}16.1$ nor $\mathrm {C}18.1$. Also, $\Gamma \ncong \mathrm {C}20.2$ because the latter graph admits no semiregular cyclic group of order 4. Finally, since $\Gamma \ncong \mathrm {C}14.1$, we find that ${{\,\mathrm{\mathsf{Aut}}\,}}\Gamma$ contains a normal subgroup D such that D is regular and D is a generalized dihedral group.

(ii): Assume on the contrary that |N| is neither squarefree nor a square. By (i), ${{\,\mathrm{\mathsf{Aut}}\,}}\Gamma$ is a soluble group, and hence $\Gamma$ is $({{\,\mathrm{\mathsf{Aut}}\,}}\Gamma , 1)$-regular by Lemma 3.7. It follows that ${{\,\mathrm{\mathsf{Aut}}\,}}\Gamma$ has a cyclic Sylow 2-subgroup, hence all involutions of ${{\,\mathrm{\mathsf{Aut}}\,}}\Gamma$ are conjugate to each other. Let $t_c$ be the unique involution contained in C. As $D \lhd {{\,\mathrm{\mathsf{Aut}}\,}}\Gamma$, $t_c \in D$. On the other hand, since |N| is not squarefree, $|P_i|=p_i^2$ for some $1\leqslant i \leqslant m$. The fact that N is abelian implies that $P_i \lhd G$, and thus $|C \cap P_i|=p_i$ follows by Proposition 2.9(ii). Therefore, $[t_c,x]=1$ for some $x \in P_i$, $o(x)=p$, which contradicts that D is a generalized dihedral group. $\square$

We would like to remark that Lemma 3.8(i) follows also from [7, Proposition 5.2].

4 Proof of Theorem 1.2

4.1 Necessity

Let $\Gamma ={{\,\mathrm{\Gamma }\,}}(H,X)$ with H and X are given as in Table 1. We show below that $\Gamma$ is connected, s-arc-regular with s given as in the table, and there exists a cyclic subgroup $C < {{\,\mathrm{\mathsf{Aut}}\,}}\Gamma$ such that C is semiregular and has k orbits. For this purpose, the following equivalence, which can be checked in two lines, will be useful: If $x, y \in H$ and $\sigma \in {{\,\mathrm{\mathsf{Aut}}\,}}(H)$, then

$$\begin{aligned} (x^y)^\sigma =x \iff [\rho _x, \rho _y \sigma ]=1~\text {holds in}~{{\,\mathrm{Sym}\,}}(H). \end{aligned}$$

(5)

Case 1. $\Gamma ={{\,\mathrm{\Gamma }\,}}(H,X)$ with H and X are in (1) of Table 1.

In this case $H=H_1(n)=\langle u, y \rangle$. Clearly, $\langle X \rangle =H$, and so $\Gamma$ is connected. Consider the automorphism $\alpha$ of H for which

$$\begin{aligned} u^\alpha =u^\lambda ~\text {and}~y^\alpha =uy. \end{aligned}$$

Then $X=y^{\langle \alpha \rangle }$, the orbit of y under $\langle \alpha \rangle$, hence $\alpha \in {{\,\mathrm{\mathsf{Aut}}\,}}\Gamma$, and we obtain that $\Gamma$ is $\langle H_{\mathrm {right}}, \alpha \rangle$-arc-transitive.

If $n=k$, then the group $C:=\langle \rho _y \rangle$ is clearly semiregular and has k orbits.

Let $n=3k$. Since $1+\lambda +\lambda ^2 \equiv 0\pmod n$, it follows that $\lambda \equiv 1\pmod 3$, and also that $\lambda -1$ is coprime to k. Let $x=u^k$. Then $x^\alpha =x$, and hence $[\rho _x, \alpha ]=1$, see (5). Also, $k \equiv 1\pmod 3$, see Remark 1.3, and hence there exists an integer $\mu$ such that

$$\begin{aligned} \mu (\lambda -1)\equiv -2k-1\pmod n. \end{aligned}$$

By this, $((u^\mu y)^{x^2})^\alpha =u^{(2k+\mu )\lambda +1}y=u^\mu y$, hence by (5), $[\rho _{u^\mu y}, \rho _{x^2} \alpha ]=1$. Thus $C:=\langle \rho _{u^\mu y}, \rho _{x^2} \alpha \rangle$ is cyclic and $|C|=6$.

We show next that C is also semiregular. This is equivalent to the condition that $\rho _{x^2} \alpha$ fixes no element of H. If $u^i$ is fixed by $\rho _{x^2} \alpha$, then $(i-k)\lambda \equiv i\pmod n$. This is impossible as $\lambda \equiv k \equiv 1\pmod 3$ and $n \equiv 0\pmod 3$. Finally, if $u^iy$ is fixed by $\rho _{x^2} \alpha$, then $(i+k)\lambda +1 \equiv i\pmod n$, which is also impossible by the same reason.

We have proved that $(\Gamma , \langle H_{\mathrm {right}}, \alpha \rangle ,C,k) \in \mathcal {Q}$. If $n=7$, then $\Gamma \cong \mathrm {C}14.1$, which is 4-arc-regular. If $n > 7$, then Lemma 3.8 can be applied to $(\Gamma , \langle H_{\mathrm {right}}, \alpha \rangle ,C,k)$ with $T=\langle \rho _a \rangle$ and this shows that ${{\,\mathrm{\mathsf{Aut}}\,}}\Gamma$ is a soluble group. By Lemma 3.7, $\Gamma$ is 1-arc-regular, i.e., $s=1$, as required.

Case 2. $\Gamma ={{\,\mathrm{\Gamma }\,}}(H,X)$ with H and X are in (2) of Table 1.

In this case $H=H_2(k)=\langle u, v, y \rangle$. It is easy to see that $\Gamma$ is connected. Consider the automorphisms $\alpha$ and $\beta$ of H for which

• $u^\alpha =v,~v^\alpha =u^{-1}v^{-1}~\text {and}~ y^\alpha =uy$,

• $u^\beta =uv,~v^\beta =v^{-1}~\text {and}~y^\beta =y$.

Both $\alpha$ and $\beta$ map X to itself, hence $\alpha , \beta \in {{\,\mathrm{\mathsf{Aut}}\,}}\Gamma$, and we obtain that $\Gamma$ is $\langle H_{\mathrm {right}}, \alpha , \beta \rangle$-arc-transitive.

Consider the product $\rho _{y} \beta$. As $y^\beta =y$, $[\rho _{y}, \beta ]=1$, see (5), hence $\rho _{y} \beta$ has order 2. Since it swaps the $\langle u, v \rangle$-orbits, the group $\langle \rho _{y} \beta \rangle$ is semiregular. Let $C=\langle \rho _{v},\rho _{y} \beta \rangle$. Then $(v^y)^\beta =v$, hence by (5), C is a cyclic group of order 2k. As both $\langle \rho _{v} \rangle$ and $\langle \rho _{y} \beta \rangle$ are semiregular, so is C, and it has k orbits.

We have proved that $(\Gamma , \langle H_{\mathrm {right}}, \alpha , \beta \rangle ,C,k) \in \mathcal {Q}$ and applying Lemma 3.8 with $T=\langle \rho _u, \rho _v \rangle$ shows that ${{\,\mathrm{\mathsf{Aut}}\,}}\Gamma$ is a soluble group. Since $({{\,\mathrm{\mathsf{Aut}}\,}}\Gamma ,\Gamma ,C,k) \in \mathcal {Q}$, Proposition 2.11 shows that $\Gamma$ is 2-arc-regular, i.e., $s=2$, as required.

Case 3. $\Gamma ={{\,\mathrm{\Gamma }\,}}(H,X)$ with H and X are in (3) of Table 1.

In this case $H=H_3(k)=\langle u, v, y \rangle$. Notice that, one can copy the proof used in Case 2. The graph $\Gamma$ is connected. Define the automorphisms $\alpha$ and $\beta$ of H by

• $u^\alpha =v,~v^\alpha =u^{-1}v^{-1}~\text {and}~ y^\alpha =uy$,

• $u^\beta =uv,~v^\beta =v^{-1}~\text {and}~y^\beta =y$.

We obtain that $\Gamma$ is $\langle H_{\mathrm {right}}, \alpha , \beta \rangle$-arc-transitive. The group $C:=\langle \rho _{v}, \rho _{y} \beta \rangle$ is a cyclic group of order 6k, which is also semiregular and hence has k orbits. Finally, one can derive that $\Gamma$ is 2-arc-regular in the same way as in Case 2.

Case 4. $\Gamma ={{\,\mathrm{\Gamma }\,}}(H,X)$ with H and X are in (4) of Table 1.

In this case $H=H_4(k,d)=\langle u, v, w, x, y \rangle$. Consider the automorphism $\alpha$ of H for which

• $u^\alpha =u^x=v,~v^\alpha =v^x=u^{-1}v^{-1},~ w^\alpha =w^\lambda ,~x^\alpha =x~\text {and}~y^\alpha =y^x= yx^{2}$.

A straightforward computation yields that $X=(uwy)^{\langle \alpha \rangle }$, hence $\alpha \in {{\,\mathrm{\mathsf{Aut}}\,}}\Gamma$, and we obtain that $\Gamma$ is $\langle H_{\mathrm {right}}, \alpha \rangle$-arc-transitive.

We consider below the product $\rho _{x} \alpha ^{-1}$. As $x^\alpha =x$, $[\rho _{x}, \alpha ]=1$, see (5), hence $\rho _{x} \alpha ^{-1}$ has order 3. Let $A=\langle u, v, w \rangle$. As permutations of the A-cosets in H, we have

$$\begin{aligned} \rho _{x}=(A,\, Ax,\, Ax^2)(Ay,\, Ayx,\, Ayx^2)~\mathrm {and}~ \alpha =(Ay,\, Ayx^2,\, Ayx). \end{aligned}$$

This shows that $\langle \rho _{x} \alpha ^{-1} \rangle$ is semiregular on the latter cosets, implying that it is also semiregular on ${{\,\mathrm{\mathsf{V}}\,}}\Gamma$.

Let $C=\langle \, \rho _{uv^2}, \rho _{y}, \rho _{x} \alpha ^{-1}\, \rangle$. Then $[\rho _{uv^2}, \rho _{y}]=1$, and as $(uv^2)^x=(uv^2)^\alpha$ and $y^x=y^\alpha$, both $\rho _{uv^2}$ and $\rho _{y}$ commute with $\rho _{x} \alpha ^{-1}$, see (5), so C is a cyclic group of order 6k/d. As each of $\langle \rho _{uv^2} \rangle$, $\langle \rho _{y} \rangle$ and $\langle \rho _{x} \alpha ^{-1} \rangle$ is semiregular, so is C, and it has k orbits.

We show next that $\langle (uwy)^{\langle \alpha \rangle } \rangle =H$. Put $t=uwy$ and let $\tilde{H}=\langle t^{\langle \alpha \rangle } \rangle$. Then

$$\begin{aligned} t t^\alpha =u^2vw^{1-\lambda }x^2~\text {and}~ t t^{\alpha ^2}=uv^{-1}w^{1-\lambda ^2}x. \end{aligned}$$

These and $1+\lambda +\lambda ^2\equiv 0\pmod d$ yield $t t^\alpha t t^{\alpha ^2}=u^3v^3 w^3$. Using that $\langle u,v \rangle \cong (\mathbb {Z}_{k/d})^2$, $o(w)=d$, k/d is coprime to d and that 3 does not divide k, we obtain that $uv, w \in \tilde{H}$. Thus, $v=(uv)^y=(uv)^t \in \tilde{H}$ as well, and then it is easy to deduce from this that $\tilde{H}=H$.

It remains to show that ${{\,\mathrm{\mathsf{Aut}}\,}}\Gamma =\langle H_{\mathrm {right}}, \alpha \rangle$. Assume on the contrary that ${{\,\mathrm{\mathsf{Aut}}\,}}\Gamma > \langle H_{\mathrm {right}}, \alpha \rangle$ and choose $\Gamma$ so that |H| is the smallest possible. Write $A=\langle H_{\mathrm {right}}, \alpha \rangle$ and let $M \leqslant {{\,\mathrm{\mathsf{Aut}}\,}}\Gamma$ such that $A < M$ and A is maximal in M. Let $g \in M \setminus A$. Then $\langle A, g \rangle =M$, and

$$\begin{aligned} |A : A \cap A^g| \cdot |A|=|A A^g| \leqslant |M| \leqslant 16 \cdot |A|. \end{aligned}$$

This shows that there is a prime divisor $p \geqslant 5$ of |H| such that the the Sylow p-subgroup P of H satisfies $P_{\mathrm {right}}\leqslant A \cap A^g$. Since $P_{\mathrm {right}}$ is normal in A, we conclude that $(P_{\mathrm {right}})^g=P_{\mathrm {right}}$, and hence $P_{\mathrm {right}}\lhd M$. Notice that,

$$\begin{aligned} (\Gamma _{P_{\mathrm {right}}},M/P_r,CP_{\mathrm {right}}/P_{\mathrm {right}},k^\prime ) \in \mathcal {Q}\end{aligned}$$

for some $k^\prime$. Also, as ${{\,\mathrm{\mathsf{Aut}}\,}}\Gamma _{P_{\mathrm {right}}} \geqslant M/P_{\mathrm {right}}> A/P_{\mathrm {right}}$, the minimality of |H| implies that $P=\langle u, v \rangle$ or $P=\langle w \rangle$.

Suppose first that $P=\langle u, v \rangle$. Then $H/P \cong D_{3d}$ and $\Gamma _{P_{\mathrm {right}}}$ is in (1) of Table 1. In this case, however, $\Gamma _{P_{\mathrm {right}}}$ is 1-arc-regular, a contradiction. Notice that, this argument also shows that we may assume that $|M : A| > 2$, and therefore, $\Gamma _{P_{\mathrm {right}}}$ is s-arc-regular for some $s \geqslant 3$.

Suppose now that $P=\langle w \rangle$ and let $T=\langle \rho _u, \rho _v, \rho _x\alpha ^{-1}\rangle$. We have seen above that T is abelian and semiregular. Moreover, $\Gamma _{P_{\mathrm {right}}}$ is bipartite and the corresponding biparts of ${{\,\mathrm{\mathsf{V}}\,}}\Gamma _{P_{\mathrm {right}}}$ are the orbits under $TP_{\mathrm {right}}/P_{\mathrm {right}}$. By Lemma 3.8, ${{\,\mathrm{\mathsf{Aut}}\,}}\Gamma _{P_{\mathrm {right}}}$ is a soluble group. Thus Proposition 2.11 can be applied to $(\Gamma _{P_{\mathrm {right}}},M/P_r,CP_{\mathrm {right}}/P_{\mathrm {right}},k^\prime ) \in \mathcal {Q}$ and this shows that $\Gamma _{P_{\mathrm {right}}}$ cannot be s-arc-regular for $s \geqslant 3$. This completes the proof of Case 4 as well as the necessity part of Theorem 1.2.

4.2 Sufficiency

Let $(\Gamma ,G,C,k) \in \mathcal {Q}$ such that G is soluble and k is squarefree and coprime to 6. We have to show that $\Gamma \cong {{\,\mathrm{\Gamma }\,}}(H,X)$, where H and X are in Table 1. Recall that, $G=N \rtimes A$, where N is a Hall $\{2,3\}^\prime$-subgroup and A is a Hall $\{2,3\}$-subgroup of G. It follows from Lemma 3.5 that N is abelian. This will be used a couple of times below.

Case 1. |N| is squarefree.

By Lemma 3.4, N is a cyclic group and G contains a regular normal subgroup H such that $H \cong D_n$ and $n=k$ or 3k. Clearly, $N < H$ and $|H : N|=2$ or 6. It follows that $\Gamma \cong {{\,\mathrm{\Gamma }\,}}(H,X)$, where X is an orbit of some subgroup $\langle \alpha \rangle \leqslant {{\,\mathrm{\mathsf{Aut}}\,}}(H)$ of order 3. Fix $y \in X$. Then $y \notin N$, $y^\alpha =uy$ for some $u \in N$, and $u^\alpha =u^\lambda$ for some integer $\lambda$ coprime to n. Since $X=\{y,\, uy,\, u^{1+\lambda }y\}$, which generates H, it follows that u is a generator of N. We conclude that $H=H_1(n)$ and ${{\,\mathrm{\Gamma }\,}}(H,X)$ is the graph in given (1) of Table 1. The condition $1+\lambda +\lambda ^2 \equiv 0\pmod n$ follows from the equality $y=y^{\alpha ^3}$.

Case 2. |N| is not squarefree and $\Gamma _N \cong K_2$.

In this case we apply Lemma 3.8 to $\Gamma$. This is possible because $\Gamma$ is clearly bipartite, $\Gamma \ncong \mathrm {C}14.1$, N is abelian, semiregular and its orbits coincide with the biparts of $\Gamma$, so letting $T=N$, all conditions of Lemma 3.8 hold. By Lemma 3.8(i), ${{\,\mathrm{\mathsf{Aut}}\,}}\Gamma$ contains a regular normal subgroup H isomorphic to a generalized dihedral group. Since N is an abelian normal Hall $\{2,3\}^\prime$-subgroup of G and H is regular, it follows from the Sylow Theorem that $N < H$, and as $\Gamma _N \cong K_2$, $|H : N|=2$. By Lemma 3.8(ii), |N| is a square. By Corollary 2.10, a Sylow p-subgroup of N is isomorphic to $\mathbb {Z}_p^2$ and clearly intersects C nontrivially. These yield that $N \cong \mathbb {Z}_k^2$. It follows that $\Gamma \cong {{\,\mathrm{\Gamma }\,}}(H, X)$, where X is an orbit of some subgroup $\langle \alpha \rangle \leqslant {{\,\mathrm{\mathsf{Aut}}\,}}(H)$ of order 3. Fix $y \in X$. Then $y \notin N$, $y^\alpha =uy$, and $u^\alpha =v$ for some $u, v \in N$. Since X generates H, it follows that $\langle u, v \rangle =N$, hence $o(u)=o(v)=k$. To sum up, $H=H_2(k)$ and ${{\,\mathrm{\Gamma }\,}}(H,X)$ is exactly the graph in (2) of Table 1.

Case 3. |N| is a square and $\Gamma _N \cong K_{3,3}$.

Let $M=C_G(N)$ and $C_3$ be the Sylow 3-subgroup of C. Since $\Gamma _N \cong K_{3,3}$, $|C_3|=3$. By Lemma 3.3, M is semiregular. On the other hand, as |N| is a square, a Sylow p-subgroup P of N is isomorphic to $\mathbb {Z}_p^2$, hence $[C_3,P]=1$ by Lemma 3.6. We deduce that $C_3 \leqslant M$, implying that $M=N \times C_3$. Now letting $T=M$, Lemma 3.8 can be applied to $\Gamma$. One can copy the argument used in Case 2 to deduce that $\Gamma \cong {{\,\mathrm{\Gamma }\,}}(H,X)$, where H and X are in (3) of Table 1.

Case 4. |N| is neither squarefree nor a square and $\Gamma _N \cong K_{3,3}$.

In this case $N=U \times W$, where $U \cong \mathbb {Z}_l^2$ and $W \cong \mathbb {Z}_d$ for coprime and squarefree numbers l and d. Both U and W are characteristic in N, hence also normal in G. By Proposition 2.8, $(\Gamma _U,G/U,CU/U,k^\prime ) \in \mathcal {Q}$, where $k=k^\prime \cdot |U : C \cap U|=k^\prime l$.

On the other hand, N/U is squarefree, and the argument used in Case 1 can be applied to $(\Gamma _U,G/U,CU/U,k^\prime )$ and this yields that $k^\prime =d$, hence $l=k/d$, and $U \cong (\mathbb {Z}_{k/d})^2$.

Furthermore, G/U contains a normal subgroup D such that $D \cong D_{3d}$ and D is regular on ${{\,\mathrm{\mathsf{V}}\,}}\Gamma _U$. Let $H < G$ be the unique subgroup that satisfies $U < H$ and $H/U=D$. Then $H \lhd G$, and it can be easily seen that H is regular on ${{\,\mathrm{\mathsf{V}}\,}}\Gamma$. Furthermore, by the Schur-Zassenhaus Theorem, there exists a subgroup $S \leqslant H$ such that $S \cong {{\,\mathrm{Sym}\,}}(3)$, $W \rtimes S \cong D_{3d}$ and H can be written in the form:

$$\begin{aligned} H = (U \times W) \rtimes S. \end{aligned}$$

(6)

Let $S_3$ denote the Sylow 3-subgroup of S.

Fix an arbitrary Sylow p-subgroup of U, denote this by P. Then $P \cong \mathbb {Z}_p^2$. Assume for the moment that $[S_3, P]=1$. By Proposition 2.8, $(\Gamma _{U_{p^\prime }},G/U_{p^\prime },CU_{p^\prime }/U_{p^\prime },k^{\prime \prime }) \in \mathcal {Q}$, where $k^{\prime \prime }$ is a divisor of k. As $[S_3, P]=1$, $S_3N/U_{p^\prime }$ is an abelian normal subgroup of ${{\,\mathrm{\mathsf{Aut}}\,}}\Gamma _{U_{p^\prime }}$ having 2 orbits on ${{\,\mathrm{\mathsf{V}}\,}}\Gamma _{U_{p^\prime }}$. But, then it follows from Lemma 3.8(ii) that $|N : U_{p^\prime }|$ is a square, which is not the case. Thus, $[S_3, P] \ne 1$, implying that S acts faithfully on P by conjugation because $S_3$ is the unique nontrivial proper normal subgroup of S.

We prove next that $C_P(S_3)=1$. Let x be a generator of $S_3$. Assume on the contrary that $a^x=a$ for some $a\in P$, $a \ne 1$. Recall that $S \cong {{\,\mathrm{Sym}\,}}(3)$. Then for $y \in S \setminus S_3$, $(a^y)^x=a^y$, so $[x, \langle a, a^y\rangle ]=1$. As $[S_3, P] \ne 1$, $|\langle a, a^y\rangle |=p$. Since $p \geqslant 5$, we have by the Maschke Theorem that $P=\langle a, b\rangle$ such that both $\langle a\rangle$ and $\langle b\rangle$ are normalized by S. This implies $[S_3, P]=1$, a contradiction. Thus $C_P(S_3)=1$. Since P is an arbitrary Sylow p-subgroup of U, it follows that $C_U(S_3)=1$ also holds.

Since H is a regular normal subgroup of G, it follows that $\Gamma \cong {{\,\mathrm{\Gamma }\,}}(H,X)$, where X is an orbit of a group $\langle \alpha \rangle \leqslant {{\,\mathrm{\mathsf{Aut}}\,}}(H)$ of order 3. Recall that, A is a Hall $\{2,3\}$-subgroup of G. We may assume w. l. o. g. that $S < A$. The group A cannot be semiregular and by this condition, we may also assume w. l. o. g. that $\alpha \in A$.

Notice that, $\alpha$ acts on $W \cong \mathbb {Z}_d$ as follows:

$$\begin{aligned} w^\alpha =w^\lambda ,~w \in W, \end{aligned}$$

where $\lambda$ is an integer satisfying $1+\lambda +\lambda ^2 \equiv 0\pmod d$.

The group A contains a subgroup L, which is conjugate to the subgroup of C of order 6. Let $L_3$ be the Sylow 3-subgroup of L. Thus $\langle L_3, \alpha \rangle \cong \mathbb {Z}_3^2$, and as $|C_A(L_3)|=18=|A|$, we also obtain that $A=S \times L_3$.

It follows from Lemma 3.6 and the normality of U in G that $[L_3, U]=1$. Thus $[\alpha , U] \ne 1$, for otherwise, $[\langle L_3, \alpha \rangle , U]=1$, contradicting that $S_3$ acts nontrivially on U by conjugation. Similarly, $[S_3,U] \ne 1$, and therefore, there exists $x \in S_3$ such that

$$\begin{aligned} u^\alpha =u^x~\text {for all}~u \in U. \end{aligned}$$

Fix $t \in X$. Note that $o(t)=2$. Then $t=uwy$ for some $u\in U$, $w\in W$ and $y \in S$. Note that,

$$\begin{aligned} S=\langle x, y \mid x^3=y^2=1, x^y=x^{-1} \rangle . \end{aligned}$$

(7)

Since $A=S \times L_3$, $y \in S$ and $\alpha \in A$, it follows that $y^\alpha =y$ or $y^x$ or $y^{x^2}$.

The set X generates H. This implies that the orbit $u^{\langle \alpha \rangle }$ generates U, w is a generator of W and $y^\alpha \ne y$. Set $v=u^x$ and $z=v^x$. Then $(uvz)^x=uvz$, and as $C_U(S_3)=1$, it follows that $uvz=1$, i.e., $v^x=z=u^{-1}v^{-1}$. To sum up,

$$\begin{aligned} \langle u,v \rangle =U \cong (\mathbb {Z}_{k/d})^2,~\langle w \rangle =W \cong \mathbb {Z}_d,~ u^x=v,~v^x=u^{-1}v^{-1}~\text {and}~w^x=w. \end{aligned}$$

(8)

Then $1=t^2=(uwy)^2=uu^yww^y$, by which $u^y=u^{-1}$ and $w^y=w^{-1}$. Also, $v^y=u^{xy}=u^{yx^2}= (u^{-1})^{x^2}=uv$, and we have

$$\begin{aligned} u^y=u^{-1},~v^y=uv~\text {and}~ w^y=w^{-1}. \end{aligned}$$

(9)

Recall that $y^\alpha =y^x$ or $y^{x^2}$. If $y^\alpha =y^{x^2}$, then $t^\alpha =u^\alpha w^\alpha y^\alpha =v w^\lambda yx$, and so $1=(t^\alpha )^2=v w^\lambda x^2 uv w^{-\lambda } x= vu^{-1}$, a contradiction. Thus, $y^\alpha =y^x=y x^2$, and so $t^\alpha =v w^\lambda y x^2$ and $t^{\alpha ^2}=u^{-1}v^{-1}w^{\lambda ^2}yx$. Finally, Eqs. (7)–(9) yield the presentation:

$$\begin{aligned} H= & {} \big \langle u, v, w, x, y \mid u^{k/d}=v^{k/d}=w^d=x^3=y^2=1,\\&[u,v]=[u,w]=[v,w]=1, u^x=v, \\&v^x=u^{-1}v^{-1}, w^x=w, u^y=u^{-1}, v^y=uv, w^y=w^{-1}, x^y=x^{-1} \big \rangle , \end{aligned}$$

so $H=H_4(k,d)$, and therefore, ${{\,\mathrm{\Gamma }\,}}(H,X)$ is exactly the graph in (4) of Table 1. This completes the proof of Theorem 1.2.

5 Proof of Theorem 1.5

Let $p \geqslant 3$ be a prime and $\Gamma$ be a cubic arc-transitive p-multicirculant.

If $p=3$, then according to [17, Theorem 1.1], $\Gamma \cong \mathrm {C}6.1$, $\mathrm {C}18.1$, $\mathrm {C}30.1$ or $\mathrm {C}54.1$. These are, respectively, the graphs in (6), (7), (12) and (4) of Table 2, where in the last case $p=3$.

If $p=5$, then according to [13, Theorem 1.2], $\Gamma \cong \mathrm {C}50.1$ or $\mathrm {C}150.1$. These are, respectively, the graphs in (3) and (4) of Table 2, both with $p=5$.

From now on let $p \geqslant 7$ and put $G={{\,\mathrm{\mathsf{Aut}}\,}}\Gamma$. If G is soluble, then Theorem 1.2 shows that $\Gamma$ is one of the graphs in (1)–(4) in Table 2.

Let G be insoluble and S be the soluble radical of G. For the sake of simplicity, write $\bar{G}=G/S$ and $\bar{C}=SC/S$. By Proposition 2.12(i), S is semiregular. Assume that $\Gamma$ is (G, s)-arc-regular. Then $|G|=|C| \cdot p \cdot 3 \cdot 2^{s-1}$. On the other hand, $|G|=|\bar{G}| \cdot |S|$ and $|C|=|\bar{C}| \cdot |S \cap C|$, and therefore,

$$\begin{aligned} |\bar{G} : \bar{C}| \cdot |S : S \cap C|=p \cdot 3 \cdot 2^{s-1}. \end{aligned}$$

(10)

We go through all the possibilities for $\bar{G}$ and $|\bar{C}|$ given in Table 3.

Using [8], we observe that $|\bar{G} : \bar{C}|$ is divisible by 5 if $\bar{G}$ is one of the groups in (1)–(4) in Table 3, and $|\bar{G} : \bar{C}|$ has more than 3 prime divisors for the groups in (5)–(8). As either possibility is in contradiction with Eq. (10), none of these groups is possible.

Suppose next that $\bar{G}$ is from (11) or (12). Since $|\bar{C}|=2r$, $r \geqslant 5$, it follows that r divides $|\bar{G} : \bar{C}|$, and we obtain from this that $p=r$, see Eq. (10). Using $s \leqslant 5$, we further compute that

$$\begin{aligned} 312 \leqslant \frac{(r^2-1)(r^2+1)}{2} \leqslant \frac{|\bar{G}|}{r \cdot |\bar{C}|} = \frac{|\bar{G}|}{p \cdot |\bar{C}|} \leqslant 48. \end{aligned}$$

This excludes the groups in (11) and (12).

It remains to consider the case when $\bar{G}$ is as in (9) or (10). We prove first that the possible pairs (s, p) belong to a restrictive list.

Since |C| is even and $|S \cap C|$ is odd, see Proposition 2.12(iii), it follows that $|\bar{C}|$ is even, and so $|\bar{C}|$ is a divisor of $(r \pm 1)/2$ if $\bar{G} \cong \mathrm {PSL}(2,r)$ and a divisor of $r \pm 1$ if $\bar{G} \cong \mathrm {PGL}(2,r)$ (see [24, Theorem 6.25]). Consequently, $r(r \pm 1)$ divides $|\bar{G} : \bar{C}|$. This and Eq. (10) yield that $r(r \pm 1)$ divides $p \cdot 3 \cdot 2^{s-1}$. It follows in turn that $r=p$, $(p \pm 1)$ divides $3 \cdot 2^{s-1}$ with $s \leqslant 4$, and so

$$\begin{aligned} (s,p) \in \big \{ (2,7),~(3,7),~(3,11),~(3,13),~(4,7),~(4,11),~(4,13),~(4,23) \big \}. \end{aligned}$$

(11)

Notice that, then Theorem 1.1 implies $|{{\,\mathrm{\mathsf{V}}\,}}\Gamma | \leqslant 6r^2=6p^2 \leqslant 3174$. We derive further restriction on the order $|{{\,\mathrm{\mathsf{V}}\,}}\Gamma |$.

Consider the quotient $\Gamma _S$. For sake of simplicity, write $\bar{v}$ for its vertex $v^S$, where $v \in {{\,\mathrm{\mathsf{V}}\,}}\Gamma$. Let $G_{\{v^S\}}$ denote the setwise stabilizer of $v^S$ in G. Note that $G_{\{v^S\}}=S G_v$. Since S is semiregular, $|G_v \cap S|=1$, and we compute

$$\begin{aligned} |\bar{G}_{\bar{v}}|=|G_{\{v^S\}}S/S|=|G_vS/S|=|G_v : G_v \cap S|= |G_v|=3 \cdot 2^{s-1}. \end{aligned}$$

Thus, $|\bar{G}|=3\cdot 2^{s-1} \cdot |{{\,\mathrm{\mathsf{V}}\,}}\Gamma _S|$, and hence

$$\begin{aligned} |{{\,\mathrm{\mathsf{V}}\,}}\Gamma |=\frac{|\bar{G}| \cdot |S|}{3 \cdot 2^{s-1}}=|S| \cdot \frac{p(p^2-1)}{3 \cdot 2^{\delta (s)}}, \end{aligned}$$

where $\delta (s)=s$ if $\bar{G}=\mathrm {PSL}(2,p)$, and $\delta (s)=s-1$ if $\bar{G}=\mathrm {PGL}(2,p)$. We conclude that there is an integer $\ell$ satisfying

$$\begin{aligned} |{{\,\mathrm{\mathsf{V}}\,}}\Gamma |=\ell \cdot \frac{p(p^2-1)}{3 \cdot 2^s} \leqslant 6 p^2. \end{aligned}$$

(12)

Using Magma [2], we carry out an exhaustive search for graphs $\Gamma$ in [5], whose orders satisfy Eq. (12) with (s, p) given in Eq. (11), and whose groups ${{\,\mathrm{\mathsf{Aut}}\,}}\Gamma$ are insoluble and contain a cyclic semiregular subgroup having p orbits. This results in the remaining graphs in (5) and (8)–(11) of Table 2, and by this the proof of Theorem 1.5 is completed.