On 2-distance-transitive circulants

Abstract

A complete classification is given of 2-distance-transitive circulants, which shows that a 2-distance-transitive circulant is a cycle, a Paley graph of prime order, a regular complete multipartite graph, or a regular complete bipartite graph of order twice an odd integer minus a 1-factor.

1 Introduction

In this paper, all graphs are finite, simple, and undirected. An ordered pair of adjacent vertices is called an arc. A graph ${\varGamma }$ is called arc-transitive if all arcs are equivalent under automorphisms of the graph. For a graph ${\varGamma }$ and two vertices u and v, the distance between u and v in ${\varGamma }$ is denoted by d(u, v), which is the smallest length of paths between u and v. The diameter$\mathrm{diam}({\varGamma })$ of ${\varGamma }$ is the maximum distance occurring over all pairs of vertices. An arc-transitive graph ${\varGamma }$ is said to be 2-distance-transitive if ${\varGamma }$ is not complete, and any two vertex pairs of vertices $(u_1,v_1)$ and $(u_2,v_2)$ with $d(u_1,v_1)=d(u_2,v_2)=2$ are equivalent under automorphisms.

A 2-arc is a triple of distinct vertices (u, v, w) such that v is adjacent to both u and w. A regular graph is called 2-arc-transitive if all 2-arcs are equivalent under automorphisms. A 2-arc-transitive graph is obviously 2-distance-transitive.

The concept of 2-distance-transitive graph generalizes the concepts of distance-transitive graph and 2-arc-transitive graph. Both distance-transitive graphs and 2-arc-transitive graphs have been extensively studied, see [3, 15]. The investigation of 2-distance-transitive graphs was initiated recently, see [4,5,6].

A vertex-transitive graph with n vertices is called a circulant if it has an automorphism of order n which acts freely on the set of vertices. Alspach et al. [1] classified 2-arc-transitive circulants; Miklavič and Potočnik [13] classified distance-regular circulants; Kovács [10] and Li [11] gave a characterization of arc-transitive circulants, see Theorem 2.1. The purpose of this paper is to give a complete classification of 2-distance-transitive circulants, stated in the following main theorem.

Theorem 1.1

The class of 2-distance-transitive graphs consists of cycles, Paley graphs of prime order, regular complete multipartite graphs, and regular complete bipartite graphs of order twice an odd integer minus a 1-factor.

By definition, a 2-distance-transitive circulant is an arc-transitive circulant. Thus, to prove Theorem 1.1, we only need to determine which of the arc-transitive circulants described in [10, 11] are 2-distance-transitive. However, this is unexpectedly nontrivial (see Lemmas 2.3–2.10), which motivates some interesting problems that we explain below.

For a finite group G and a subset S of G such that $1\notin S$ and $S=S^{-1}$, the Cayley graph$\mathrm{Cay}(G,S)$ of G with respect to S is the graph with vertex set G and edge set $\{\{g,sg\} \,|\,g\in G,s\in S\}$. It is known that a graph ${\varGamma }$ is a Cayley graph of G if and only if ${\varGamma }$ has an automorphism group which is isomorphic to G and regular on the vertex set, see [2, Lemma 16.3] and [17]. For a Cayley graph ${\varGamma }=\mathrm{Cay}(G,S)$, if G is a normal subgroup of $\mathrm{Aut}{\varGamma }$, then ${\varGamma }$ is called a normal Cayley graph. The study of normal Cayley graphs was initiated by Xu [18] and has been done under various additional conditions, see [7, 16]. A circulant is thus a Cayley graph of a cyclic group, and if further it is a normal Cayley graph of a cyclic group, then it is called a normal circulant. Many interesting examples of arc-transitive graphs and 2-arc-transitive graphs are constructed as normal Cayley graphs, due to the fact that the arc-transitivity and 2-arc-transitivity of a graph are equivalent to the local-transitivity and local-2-transitivity, respectively. The status for 2-distance-transitive graphs is, however, different. To our best knowledge, the known examples of 2-distance-transitive graphs are either distance-transitive or 2-arc-transitive. For instance, 2-distance-transitive normal circulants are cycles and Paley graphs by Theorem 1.1, which are distance-transitive. The following is a curious question.

Question 1.2

Is there a normal Cayley graph which is 2-distance-transitive, but neither distance-transitive nor 2-arc-transitive?

Another interesting problem is to characterize 2-distance-transitive Cayley graphs of certain special classes of groups, such as abelian groups and metacyclic groups.

2 Proof

We first introduce the classification of arc-transitive circulants given in [10, 11], which we need to prove Theorem 1.1.

Let ${\varGamma }=(V,E)$ be a connected graph with vertex set V and edge set E. Its complement graph$\overline{{\varGamma }}$ is the graph with vertex V such that two vertices are adjacent in $\overline{{\varGamma }}$ if and only if they are not adjacent in ${\varGamma }$. Let ${\varGamma }_1=(V_1,E_1)$ and ${\varGamma }_2=(V_2,E_2)$ be two graphs. Then ${\varGamma }={\varGamma }_1[{\varGamma }_2]$ denotes the lexicographic product of ${\varGamma }_1$ and ${\varGamma }_2$, where the vertex set of ${\varGamma }$ is $V_1\times V_2$, and two vertices $(u_1,u_2)$ and $(v_1,v_2)$ are adjacent in ${\varGamma }$ if either $u_1$ and $v_1$ are adjacent in ${\varGamma }_1$, or $u_1=v_1$ and $u_2,v_2$ are adjacent in ${\varGamma }_2$.

For a positive integer b, let $\mathrm{K}_b$ be the complete graph with b vertices. For a graph ${\varGamma }$, the graph consisting of b vertex disjoint copies of ${\varGamma }$ is denoted by $b{\varGamma }$, and ${\varGamma }[\overline{\mathrm{K}_b}]-b{\varGamma }$ is the graph whose vertex set is the same as ${\varGamma }[\overline{\mathrm{K}_b}]$ and edge set equals the edge set of ${\varGamma }[\overline{\mathrm{K}_b}]$ minus the edge set of $b{\varGamma }$.

Theorem 2.1

[11, Theorem 1.3] Let ${\varGamma }$ be a connected arc-transitive circulant of order n which is not a complete graph. Then either

• (1) ${\varGamma }$ is a normal circulant, or

• (2) there exists an arc-transitive circulant ${\varSigma }$ of order m such that $mb=n$ with $b,m>1$ and

  $$\begin{aligned} {\varGamma }=\left\{ \begin{array}{ll} {\varSigma }\big [\overline{\mathrm{K}_b}\big ], \ \ or \\ {\varSigma }\big [\overline{\mathrm{K}_b}\big ]-b{\varSigma }, \text{ where } (m,b)=1. \end{array}\right. \end{aligned}$$

Our proof of Theorem 1.1 is to analyze which graphs satisfying Theorem 2.1 are 2-distance-transitive. We now describe all examples of 2-distance-transitive circulants, which consist of cycles and the following three families:

Example 2.2

• (1) Let ${\varGamma }=\mathrm{K}_{m[b]}$ be a complete multipartite graph which has m parts of size b. Clearly, $\mathrm{K}_{m[b]}=\mathrm{K}_m[\overline{\mathrm{K}_b}]$. Then $\mathrm{Aut}{\varGamma }=\mathrm{S}_b\wr \mathrm{S}_m$ is 2-distance-transitive on ${\varGamma }$ and has a cyclic subgroup which is regular on the vertex set. Thus ${\varGamma }$ is a 2-distance-transitive circulant.

• (2) Let ${\varGamma }=\mathrm{K}_{b,b}-b\mathrm{K}_2$ where b is an odd integer, namely, a complete bipartite graph minus a 1-factor. Then ${\varGamma }$ is of valency $b-1$ and of diameter 3, and $\mathrm{Aut}{\varGamma }=\mathrm{S}_b\times \mathrm{S}_2$ is distance-transitive and 2-arc-transitive. It follows that ${\varGamma }$ is a 2-distance-transitive circulant.

• (3) Let $q=p^e$ be a prime power such that $q\equiv 1 \pmod {4}$. Let ${\mathbb {F}}_q$ be the finite field of order q. Then the Paley graph$\mathsf{P}(q)$ is the graph with vertex set ${\mathbb {F}}_q$, and two distinct vertices u, v are adjacent if and only if $u-v$ is a nonzero square in ${\mathbb {F}}_q$. The congruence condition on q implies that $-1$ is a square in ${\mathbb {F}}_q$, and hence $\mathsf{P}(q)$ is an undirected graph. This family of graphs was first defined by Paley in 1933, see [14]. Note that the field ${\mathbb {F}}_q$ has $(q-1)/2$ elements which are nonzero squares, so $\mathsf{P}(q)$ has valency $(q-1)/2$. Moreover, $\mathsf{P}(q)$ is a Cayley graph for the additive group $G={\mathbb {F}}_{q}^+\cong \mathbb {Z}_{p}^e$. Let w be a primitive element of ${\mathbb {F}}_q$. Then $S=\{w^2,w^4,\ldots ,w^{q-1}=1\}$ is the set of nonzero squares of ${\mathbb {F}}_q$, and $\mathsf{P}(q)=\mathrm{Cay}(G,S)$. If $q=p$ is a prime, then $\mathsf{P}(p)$ is a circulant of prime order. By [12], $\mathrm{Aut}\mathsf{P}(p) ={\mathbb {F}}_p^+:\langle w^2\rangle$. This implies that $\mathsf{P}(p)$ is a 2-distance-transitive normal circulant.

Proof of Theorem 1.1

By Example 2.2, cycles, regular complete multipartite graphs, regular complete bipartite graphs of order twice an odd integer minus a 1-factor, and Paley graphs of prime order are all distance-transitive and so 2-distance-transitive.

Conversely, let ${\varGamma }=\mathrm{Cay}(G,S)$ be a connected 2-distance-transitive circulant, and $G\cong \mathbb {Z}_n$. Then ${\varGamma }$ is arc-transitive, and so ${\varGamma }$ satisfies Theorem 2.1. If ${\varGamma }$ is of valency 2, then ${\varGamma }\cong \mathrm{C}_n$. Assume that ${\varGamma }$ has valency at least 3. Since ${\varGamma }$ is arc-transitive, either ${\varGamma }$ is a normal circulant, or ${\varGamma }$ satisfies part (2) of Theorem 2.1. We shall treat these two cases in two subsections, respectively. In Sect. 2.1, we prove that if ${\varGamma }$ is a normal circulant, then ${\varGamma }$ is a Paley graph of prime order, stated in Lemma 2.8. In Sect. 2.2, we deal with nonnormal circulants and show that, if ${\varGamma }$ is not a normal circulant, then ${\varGamma }$ is a regular complete multipartite graph, or a regular complete bipartite graph of order twice an odd integer minus a 1-factor. $\square$

We now introduce a few notations which we will use later. For a graph ${\varGamma }$ and a vertex v, we denote by ${\varGamma }_i(v)$ the i-th neighborhood of v in ${\varGamma }$, that is, the set of vertices which are at distance i from v. A sequence of vertices $v_0,v_1,\dots ,v_s$ is called an s-geodesic if $\{v_i,v_{i+1}\}$ is an edge for all $0\leqslant i\leqslant s-1$ and $d(v_0,v_s)=s$. We sometimes need to consider distances of the same pair of vertices in different graphs, so let $d_{\varGamma }(u,v)$ denote the distance of u and v in the graph ${\varGamma }$.

2.1 2-Distance-transitive normal circulants

Consider the cyclic group

$$\begin{aligned} G=\langle g\rangle \cong {\mathbb {Z}}_n. \end{aligned}$$

Let ${\varGamma }=\mathrm{Cay}(G,S)$ be a connected 2-distance-transitive circulant of valency $k\geqslant 3$. Then $G=\langle S\rangle$. Assume further that ${\varGamma }$ is a normal circulant. Let $A=\mathrm{Aut}{\varGamma }$, and let u be the vertex of ${\varGamma }$ corresponding to the identity of the group G. Then ${\varGamma }(u)=S$ and $A=G{:}A_u$. By [9, Lemma 2.1], we have

$$\begin{aligned} A_u=\mathrm{Aut}(G,S)=\{\sigma \in \mathrm{Aut}(G)\mid S^\sigma =S\}. \end{aligned}$$

Moreover, as $\mathrm{Aut}(G)$ is abelian and $A_u=\mathrm{Aut}(G,S)$ is transitive and faithful on S, it implies that $A_u$ is regular on S, and

$$\begin{aligned} |A_u|=|S|=k. \end{aligned}$$

We establish a series of lemmas to prove that ${\varGamma }$ is a Paley graph.

Lemma 2.3

${\varGamma }$ has girth 3.

Proof

Suppose that the girth of ${\varGamma }$ is greater than 4. Then, for each pair of vertices $\theta ,\theta '$ with distance $d(\theta ,\theta ')=2$, there is a unique 2-arc between $\theta$ and $\theta '$. Hence ${\varGamma }$ being 2-distance-transitive implies that it is 2-arc-transitive. By the classification given in [1, Theorem 1.1], the graph ${\varGamma }$ has girth 4, which is a contradiction.

Assume that ${\varGamma }$ has girth 4. Then, for any vertex $v\in {\varGamma }(u)=S$, we have $|{\varGamma }_2(u)\cap {\varGamma }(v)|=k-1$. Thus there are $k(k-1)$ edges between ${\varGamma }(u)$ and ${\varGamma }_2(u)$. Since $|A_u|=k$ and A is 2-distance-transitive on ${\varGamma }$, we conclude that $A_u$ acts transitively on ${\varGamma }_2(u)$, and the size $|{\varGamma }_2(u)|$ is a divisor of $|A_u|=k$. Let $w\in {\varGamma }_2(u)\cap {\varGamma }(v)$. Then $|S\cap {\varGamma }(w)|\leqslant |{\varGamma }(w)|=k$. Note that $|S\cap {\varGamma }(w)|\cdot |{\varGamma }_2(u)|=k(k-1)$. Hence $|S\cap {\varGamma }(w)|=k-1$ and $k=|{\varGamma }_2(u)|=|S|$. It follows that $|{\varGamma }_3(u)\cap {\varGamma }(w)|+|{\varGamma }_2(u)\cap {\varGamma }(w)|=1$. Set ${\varGamma }(u)=\{v=v_1,\ldots ,v_k\}$ and ${\varGamma }_2(u)=\{w=w_1,\ldots ,w_k\}$. Let ${\varGamma }_2(u)\cap {\varGamma }(v)=\{w_1,\ldots ,w_{k-1}\}$ and ${\varGamma }(u)\cap {\varGamma }(w)=\{v_1,\ldots ,v_{k-1}\}$.

Assume that $|{\varGamma }_3(u)\cap {\varGamma }(w)|=0$. Then $|{\varGamma }_2(u)\cap {\varGamma }(w)|=1$. Since ${\varGamma }$ is 2-distance-transitive, the stabilizer $A_u$ is transitive on ${\varGamma }_2(u)$, so the induced subgraph $[{\varGamma }_2(u)]\cong (k/2)\mathrm{K}_2$. As ${\varGamma }$ has girth 4, it follows that $[{\varGamma }_2(u)\cap {\varGamma }(v)]$ is an empty graph, and so $w_1$ is adjacent to $w_k$. Thus $[\{w_2,\ldots ,w_{k-1}\}]\cong (k/2-1)\mathrm{K}_2$. However, $\{w_2,\ldots ,w_{k-1}\}\subset {\varGamma }_2(u)\cap {\varGamma }(v)$, contradicting the fact that $[{\varGamma }_2(u)\cap {\varGamma }(v)]$ is an empty graph. Therefore, $|{\varGamma }_3(u)\cap {\varGamma }(w)|=1$, say ${\varGamma }_3(u)\cap {\varGamma }(w)=\{z\}$.

By the transitivity of $A_u$ on the set ${\varGamma }(u)$, we have $|{\varGamma }_2(u)\cap {\varGamma }(v_k)|=|{\varGamma }_2(u)\cap {\varGamma }(v_1)|=k-1$. Then ${\varGamma }_2(u)\cap {\varGamma }(v_k)=\{w_2,\dots ,w_k\}$ since $v_k$ and $w_1$ are not adjacent in ${\varGamma }$. Note that $(u,w),(v,z),(v_k,z)$ are all vertex pairs of distance 2 and ${\varGamma }$ is 2-distance-transitive. We have $|{\varGamma }(u)\cap {\varGamma }(w)|$, $|{\varGamma }(v)\cap {\varGamma }(z)|$ and $|{\varGamma }(v_k)\cap {\varGamma }(z)|$ are all equal to $k-1$. Hence ${\varGamma }(v)\cap {\varGamma }(z)={\varGamma }(v)\cap {\varGamma }_2(u)$ and ${\varGamma }(v_k)\cap {\varGamma }(z)={\varGamma }(v_k)\cap {\varGamma }_2(u)$. Thus ${\varGamma }(z)={\varGamma }_2(u)$ and ${\varGamma }_3(u)=\{z\}$, so ${\varGamma }$ has diameter 3 and is distance-transitive. Therefore, $\{u\}\cup {\varGamma }_3(u)$ is a block of imprimitivity of $\mathrm{Aut}{\varGamma }$ on the vertex set V, and ${\varGamma }\cong \mathrm{K}_{k+1,k+1}-(k+1)\mathrm{K}_2$. It is clear that $\mathrm{K}_{k+1,k+1}-(k+1)\mathrm{K}_2$ is not a normal circulant, which is a contradiction. Hence the girth of ${\varGamma }$ is 3. $\square$

Lemma 2.4

The order $|G|=n$ is odd.

Proof

Suppose that $|G|=n$ is an even integer. Since $G=\langle S\rangle$ and all elements of S have the same order, it follows that S consists of generators of G. Without loss of generality, we assume that $g\in S$. By Lemma 2.3, ${\varGamma }$ has girth 3. Then there is $g^i\in S$ such that $(g,g^i)$ is an arc of ${\varGamma }$. This implies that $g^{i-1}=g^ig^{-1}$ belongs to set S. Since ${\varGamma }$ is a normal circulant, each element in S is a generator of G. This means that $1,i-1,i$ are all relatively prime to n. This contradicts the fact that n is an even number. $\square$

Lemma 2.5

The second neighborhood ${\varGamma }_2(u)$ consists of generators of G.

Proof

Suppose that ${\varGamma }_2(u)$ contains an element which is not a generator of G. Then |G| is not a prime. Since $A_u=\mathrm{Aut}(G,S)$ is transitive on ${\varGamma }_2(u)$, none of the elements in ${\varGamma }_2(u)$ is a generator.

As $\langle S\rangle =G$ and all elements of S are conjugate in $\mathrm{Aut}(G)$, we may assume that $g\in S$. By Lemma 2.4, the order |G| is an odd integer. Thus $g^2$ is a generator, and $g^2\notin {\varGamma }_2(u)$. Since $g^2\in {\varGamma }(u)\cup {\varGamma }_2(u)$, $g^2\in S={\varGamma }(u)$. Assume that $g^r\in S$ where $r\leqslant n-3$. Then $g^{r+1}=g^rg\in {\varGamma }(u)\cup {\varGamma }_2(u)$. Thus either $g^{r+1}$ is a generator of G and $g^{r+1}\in S$, or $g^{r+1}$ is not a generator and $g^{r+1}\in {\varGamma }_2(u)$. Similarly, we have $g^{r+2}=g^rg^2\in {\varGamma }(u)\cup {\varGamma }_2(u)$. Therefore, either $g^{r+2}$ is a generator of G and $g^{r+2}\in S$, or $g^{r+2}$ is not a generator and $g^{r+2}\in {\varGamma }_2(u)$.

Let p be the smallest prime divisor of $n=|G|$. Then $g,g^2,\dots ,g^{p-1}\in S$, and $g^p\in {\varGamma }_2(u)$. Suppose that G is not a p-group. Let q be the second smallest prime divisor of n. By the deduction above, we have $g^\lambda \in S$ for any $1\leqslant \lambda \leqslant q-1$ with $(\lambda ,p)=1$. Noting that p is coprime to at least one of $q-2$ and $q-1$. If p and $q-2$ are coprime, then $g^{q-2}\in S$, so $g^q=g^{q-2}g^2\in {\varGamma }_2(u)$, as $g^2\in S$; if p and $q-1$ are coprime, then $g^{q-1}\in S$, so $g^q=g^{q-1}g\in {\varGamma }_2(u)$, as $g\in S$. Thus, $g^{q}$ is in ${\varGamma }_2(u)$. However ${\varGamma }$ is 2-distance-transitive and normal, which means that all elements of ${\varGamma }_2(u)$ have the same order. This contradicts the fact that $o(g^{p})\ne o(g^{q})$ and $g^{p}, g^{q}\in {\varGamma }_2(u)$. Thus G is a p-group.

Suppose that $|G|=p^r$. If $r\geqslant 3$, then by a similar argument as the previous paragraph, we have $g^\lambda \in S$ for any $1\leqslant \lambda \leqslant p^{r}-1$ with $(\lambda ,p)=1$. Hence $g^{p},g^{p^2}\in {\varGamma }_2(u)$. This is impossible since $o(g^{p})\ne o(g^{p^2})$ and all elements of ${\varGamma }_2(u)$ have the same order.

Therefore, we get $n=p^2$. Furthermore, $S=\{g^\lambda |1\leqslant \lambda \leqslant p^2-1,(p,\lambda )=1\}$ and ${\varGamma }_2(u)=\{g^{\mu p}|1\leqslant \mu \leqslant p-1\}$. Thus ${\varGamma }\cong \mathrm{K}_{p[p]}$.

Note that $\mathrm{K}_{p[p]}$ is not normal. We have that ${\varGamma }_2(u)$ has no nongenerators of G. This means ${\varGamma }_2(u)$ consists of generators of G. $\square$

Let

$$\begin{aligned} R={\varGamma }_2(u), \end{aligned}$$

the second neighborhood of the vertex u (corresponding to the identity of G).

Lemma 2.6

The stabilizer $A_u$ is regular on R, and $|R|=|S|$ divides $p-1$ for each prime divisor p of |G|.

Proof

Since A is 2-distance-transitive on ${\varGamma }$, $A_u$ is transitive on $R={\varGamma }_2(u)$. As $A_u=\mathrm{Aut}(G,S)$ is abelian and R consists of generators of G, $A_u$ is faithful on R. Thus $A_u$ is regular on $R={\varGamma }_2(u)$, and so $|R|=|A_u|=|S|$.

Let $n=p_1^{t_1}p_2^{t_2}\dots p_{\ell }^{t_{\ell }}$, where $p_1<p_2<\dots <p_\ell$ are distinct primes. Let $G=\langle x_1\rangle \times \dots \times \langle x_\ell \rangle$, where $o(x_i)=p_i^{t_i}$ for $1\leqslant i\leqslant \ell$ and $g=x_1\dots x_\ell$. Then

$$\begin{aligned} A_u\leqslant \mathrm{Aut}(G)=\mathrm{Aut}(\langle x_1\rangle )\times \dots \times \mathrm{Aut}(\langle x_\ell \rangle ). \end{aligned}$$

Set

$$\begin{aligned} B_j=\mathrm{Aut}(\langle x_1\rangle )\times \dots \times \mathrm{Aut}(\langle x_{j-1}\rangle )\times \mathrm{Aut}(\langle x_{j+1}\rangle )\times \dots \times \mathrm{Aut}(\langle x_\ell \rangle ), \end{aligned}$$

where $1\leqslant j\leqslant \ell$. We claim that $A_u\cap B_j=\{1\}$ and $A_u\cong A_uB_j/B_j\lesssim \mathbb {Z}_{p_j-1}$.

Assume that $A_u\cap B_j\ne \{1\}$. Then there exists $\sigma \in A_u\cap B_j$ such that $\sigma \ne 1$. Hence $g^\sigma =(x_1\dots x_\ell )^\sigma =(x_1\dots x_{j-1})^\sigma x_j (x_{j+1} \dots x_\ell )^\sigma$, and $g^\sigma g^{-1} \ne 1$ is not a generator of G. Observing that g and $g^\sigma$ are in S, we have $g^\sigma g^{-1}\in S\cup R$, which contradicts the fact that all elements in S and R are generators of G. Hence $A_u\cap B_j=\{1\}$ and

$$\begin{aligned} A_u\cong A_uB_j/B_j\lesssim \mathrm{Aut}(\langle x_j\rangle )\cong \mathrm{Aut}(\mathbb {Z}_{p_j^{t_j}})\cong \mathbb {Z}_{(p_j-1)p_j^{t_j-1}}. \end{aligned}$$

If $p_j$ divides $|A_u|$, then there exists $\sigma \in A_u$ such that $o(\sigma )=p_j$. Furthermore, $x_j^\sigma =x_j^{\lambda p_j+1}\ne x_j$ for some integer $\lambda$. Thus $g^\sigma g^{-1}\in \langle x_1\rangle \times \dots \times \langle x_{j-1}\rangle \times \langle x_j^{p_j}\rangle \times \langle x_{j+1}\rangle \times \dots \langle x_{\ell }\rangle$ is not a generator of G. This contradicts the fact that $g^\sigma g^{-1}\in S\cup R$. Therefore, $A_u\lesssim \mathbb {Z}_{p_j-1}$ for $1\leqslant j\leqslant \ell$. Noting that $A_u$ acts regularly on S, $|S|=|A_u|$. Hence |S| divides $p_i-1$ for $1\leqslant i\leqslant \ell$. $\square$

By virtue of Lemma 2.6, we can assume that

$$\begin{aligned} A_u=\langle \sigma \rangle \text{ and } g^\sigma =g^\lambda , \end{aligned}$$

where $\lambda$ is coprime to n. Let $g^\mu$ be an element of R and $\tau$ be an automorphism of G such that $g^\tau =g^\mu$. Then $(g^\mu )^\tau =(g^\tau )^\mu =g^{\mu ^2}$. Let $k=|S|=|A_u|$. We have

$$\begin{aligned} \begin{array}{rcl} S&{}=&{}g^{\langle \sigma \rangle }=\{g,g^\lambda ,g^{\lambda ^2},\dots ,g^{\lambda ^{k-1}}\},\\ R&{}=&{}(g^\mu )^{\langle \sigma \rangle }=\{g^\mu ,g^{\mu \lambda },g^{\mu \lambda ^2},\dots ,g^{\mu \lambda ^{k-1}}\}=S^\tau ,\\ R^\tau &{}=&{}(g^{\mu })^{\langle \sigma \rangle \tau }=(g^\mu )^{\tau \langle \sigma \rangle }=(g^{\mu ^2})^{\langle \sigma \rangle }=\{g^{\mu ^2},g^{\mu ^2\lambda },g^{\mu ^2\lambda ^2},\dots ,g^{\mu ^2\lambda ^{k-1}}\}. \end{array} \end{aligned}$$

Let

$$\begin{aligned} {\varSigma }={\varGamma }^\tau =\mathrm{Cay}(G,S)^\tau =\mathrm{Cay}(G,S^\tau )=\mathrm{Cay}(G,R). \end{aligned}$$

Then ${\varSigma }$ and ${\varGamma }$ are isomorphic. (Two graphs are isomorphic if there exists a bijection between their vertex sets which preserves the adjacency and the nonadjacency.)

Lemma 2.7

Let $x,y\in G$. Then $\mathrm{d }_{\varGamma }(x,y)=2$ if and only if $\mathrm{d }_{\varGamma }(xy^{-1},u)=2$, and the following conditions are equivalent:

• (i) $xy^{-1}\in R$;

• (ii) $\mathrm{d }_{\varSigma }(x,y)=1$;

• (iii) $\mathrm{d }_{\varGamma }(x,y)=\mathrm{d }_{\varGamma }(xy^{-1},u)=2$.

Proof

For any $y\in G$, let $\sigma _{{y}^{-1}}$ be the right translation by $y^{-1}$. Then $\sigma _{{y}^{-1}}$ is an automorphism of ${\varGamma }$ since ${\varGamma }$ is a Cayley graph of G. Thus for any $x\in G$, we have

$$\begin{aligned} d_{\varGamma }(x,y)=d_{\varGamma }(x^{\sigma _{{y}^{-1}}},y^{\sigma _{{y}^{-1}}})=d_{\varGamma }(xy^{-1},u). \end{aligned}$$

Noting that $R={\varGamma }_2(u)={\varSigma }(u)$, we have $xy^{-1}\in R$ if and only if $d_{\varGamma }(xy^{-1},u)=d_{\varGamma }(x,y)=2$. By the same argument, we also have $xy^{-1}\in R$ if and only if $d_{\varSigma }(xy^{-1},u)=d_{\varSigma }(x,y)=1$. $\square$

For two sets $B_1,B_2$, we use $B_1\sqcup B_2$ to denote $B_1\cup B_2$ when $B_1\cap B_2=\emptyset$. We denote ${\varGamma }_{\geqslant i}(u)={\varGamma }_{i}(u)\cup {\varGamma }_{ i+1}(u)\cup \cdots \cup {\varGamma }_{\mathrm{diam}({\varGamma })}(u)$.

Lemma 2.8

The graph ${\varGamma }$ is a Paley graph $\mathsf{P}(p)$, where $p\equiv 1$$(\mathsf{mod\ }4)$ is prime.

Proof

By Lemma 2.5, ${\varGamma }_2(u)=R$ contains generators of G. If the diameter of ${\varGamma }$ is 2, then all the elements in $G\setminus \{u\}$ are generators of G, and so n is an odd prime p. By Lemma 2.6, $|S|=|R|$, so $|S|=|R|=(p-1)/2$. Thus S is either the set of square elements or the set of nonsquare elements of $G\setminus \{u\}$, and ${\varGamma }$ is the Paley graph $\mathsf{P}(p)$, see also [8, Lemma 2.2].

In the remainder, we suppose that ${\varGamma }$ has diameter at least 3. Let (u, z, v, w) be a 3-geodesic of ${\varGamma }$. We set $k=|S|=|R|$, $a_1=|{\varGamma }(z)\cap S|$, $b_1=|{\varGamma }(z)\cap R|$ and $c_2=|{\varGamma }(v)\cap S|$. Let N be the number of edges in ${\varGamma }$ with one end in S and the other end in R. Then

$$\begin{aligned} N=b_1|z^{A_u}|=b_1k=c_2|v^{A_u}|=c_2k. \end{aligned}$$

Hence $b_1=c_2$.

Note that all S, R, and $R^\tau$ are orbits of $A_u$. We will argue in two cases.

Case 1$R^\tau \ne S$.

Fig. 1

Case 1

Since $\Sigma _2(u)=R^\tau$ and $R^\tau \ne S$, it follows that $S\subseteq \Sigma _{\geqslant 3}(u)$. Thus, for each $y\in R$ and $x\in S$, $d_\Sigma (x,y)\ne 1$, and it follows from Lemma 2.7 that $d_{\varGamma }(x,y)\ne 2$.

Let $w\in R^\tau$. Then there exist vertices $z\in S$ and $v\in R$, such that (u, v, w) and (u, z, v) are 2-geodesics in $\Sigma$ and ${\varGamma }$, respectively. Hence, by Lemma 2.7, $d_{\varGamma }(w,v)=2$ (Fig. 1).

Since ${\varGamma }$ is 2-distance-transitive, there exists $\eta \in A_v$ such that $u^\eta =w$. Thus

$$\begin{aligned} b_1=c_2=|{\varGamma }(v)\cap {\varGamma }(u)|=|{\varGamma }(v^\eta )\cap {\varGamma }(u^\eta )|=|{\varGamma }(v)\cap {\varGamma }(w)|. \end{aligned}$$

Since ${\varGamma }(v)\cap {\varGamma }(w)\subseteq R\cup {\varGamma }_3(u)$, we have

$$\begin{aligned} k=|{\varGamma }(v)|=|{\varGamma }(v)\cap S|+|{\varGamma }(v)\cap (R\cup {\varGamma }_3(u))| \geqslant 2b_1. \end{aligned}$$

(1)

For any $x\in S\cap {\varGamma }(z)$, (v, z, x) is a 2-arc. Since $d_{\varGamma }(v,x)\ne 2$, $d_{\varGamma }(v,x)=1$ and $x\in {\varGamma }(v)$. Thus

$$\begin{aligned} \{z\}\sqcup (S\cap {\varGamma }(z))\subseteq {\varGamma }(v)\cap S \end{aligned}$$

(2)

and $a_1+1\leqslant b_1$. Consider the valency of the vertex z, we have

$$\begin{aligned} k=|{\varGamma }(z)|=1+a_1+b_1\leqslant 2b_1. \end{aligned}$$

(3)

By inequalities (1) and (3), $k=2b_1$, so (2) can be modified into

$$\begin{aligned} \{z\}\sqcup ({\varGamma }(z)\cap S)={\varGamma }(v)\cap S. \end{aligned}$$

(4)

By the same deduction, for any $y\in {\varGamma }(z)\cap R$, we have

$$\begin{aligned} \{z\}\sqcup ({\varGamma }(z)\cap S)= {\varGamma }(y)\cap S. \end{aligned}$$

(5)

Similarly, for each $x\in {\varGamma }(z)\cap S\subset {\varGamma }(v)\cap S$,

$$\begin{aligned} \{x\}\sqcup ({\varGamma }(x)\cap S)= {\varGamma }(v)\cap S=\{z\}\sqcup (S\cap {\varGamma }(z)). \end{aligned}$$

(6)

Equalities (5) and (6) indicate that for each $x\in {\varGamma }(z)\setminus \{u\}$, ${\varGamma }(x)\cap S\subset \{z\}\sqcup ({\varGamma }(z)\cap S)$.

Let $z'\in S\setminus (\{z\}\cup {\varGamma }(z))$. Then $d_{\varGamma }(z,z')=2$. There is an automorphism $\eta \in A$ such that $z^\eta =u$ and $(z')^\eta =v$. Let $(z,x,z')$ be a 2-arc in ${\varGamma }$. Then $z'\in {\varGamma }(x)\cap S$ and $x\in {\varGamma }(z)$. Thus $x=u$ and

$$\begin{aligned} 1=|{\varGamma }(z)\cap {\varGamma }(z')|=|{\varGamma }(u)\cap {\varGamma }(v)|=b_1. \end{aligned}$$

Hence $k=2$. This contradicts the assumption that ${\varGamma }$ has valency at least 3.

Case 2 $R^\tau =S.$

Since $R^\tau =S$, it follows that ${\varGamma }_{\geqslant 3}(u)=\Sigma _{\geqslant 3}(u)$. For each $x\in R$ and $y\in \Sigma _{\geqslant 3}(u)$, we have $d_\Sigma (x,y)\ne 1$. This implies $d_{\varGamma }(x,y)\ne 2$. Thus ${\varGamma }_{\geqslant 3}(u)={\varGamma }_{3}(u)$ and $\mathrm{diam}({\varGamma })=3$ (Fig. 2).

Fig. 2

Case 2

Let $w\in {\varGamma }_3(u)$. Then there exist $z\in S, v\in R$ such that (u, z, v, w) is a 3-geodesic. Let

$$\begin{aligned} b_2= & {} |{\varGamma }(v)\cap {\varGamma }_3(u)|,\quad c_3=|{\varGamma }(w)\cap R|,\\ a_2= & {} |{\varGamma }(v)\cap R|,\quad a_3=|{\varGamma }(w)\cap {\varGamma }_3(u)|. \end{aligned}$$

Then $k=1+a_1+b_1=a_2+b_2+c_2=a_3+c_3$.

Let p be the smallest prime factor of n, and let $N'$ be the number of edges in ${\varGamma }$ with one end in R and the other end in ${\varGamma }_3(u)$. Then

$$\begin{aligned} k(k-1)\geqslant N'=kb_2\geqslant |{\varGamma }_3(u)|=n-2k-1. \end{aligned}$$

Hence $n\leqslant k^2+k+1$. By Lemma 2.6, k is a divisor of $p-1$, and thus $k+1<p+1$. Then $n\leqslant k^2+k+1=k(k+1)+1<(p-1)(p+1)+1=p^2$, and so n is a prime. This implies that w is also a generator of G and $|w^{A_u}|=|A_u|=k$. Furthermore,

$$\begin{aligned} N'=|v^{A_u}|b_2=kb_2\geqslant |w^{A_u}|c_3=kc_3. \end{aligned}$$

This means

$$\begin{aligned} c_3\leqslant b_2. \end{aligned}$$

(7)

For any $x\in {\varGamma }(w)\cap {\varGamma }_3(u)$, (v, w, x) is a 2-arc. Since $d_{\varGamma }(v,x)\ne 2$, we have $d_{\varGamma }(v,x)=1$ and $x\in {\varGamma }(v)$. Thus

$$\begin{aligned} \{w\}\sqcup ({\varGamma }(w)\cap {\varGamma }_3(u))\subseteq {\varGamma }(v)\cap {\varGamma }_3(u), \end{aligned}$$

and

$$\begin{aligned} a_3+1\leqslant b_2. \end{aligned}$$

(8)

By inequalities (7) and (8), we have

$$\begin{aligned} k=c_3+a_3\leqslant b_2+b_2-1. \end{aligned}$$

(9)

For any $x\in S\setminus \{z\}$, (z, u, x) is a 2-arc in ${\varGamma }$, and $d_{\varGamma }(z,x)\leqslant 2$. Thus

$$\begin{aligned} S=\{z\}\sqcup ({\varGamma }(z)\cap S)\sqcup ({\varGamma }_2(z)\cap S). \end{aligned}$$

Note that ${\varGamma }_2(z)\cap S=\Sigma (z)\cap S$ and

$$\begin{aligned} |\Sigma (z)\cap S|=|\Sigma ^{\tau ^{-1}}(z^{\tau ^{-1}})\cap S^{\tau ^{-1}}|=|{\varGamma }(v')\cap R|=a_2 \end{aligned}$$

for some $v'\in R$ where the graph isomorphism $\tau$ is defined in the paragraph before Lemma 2.7. We have

$$\begin{aligned} k=1+|{\varGamma }(z)\cap S|+|{\varGamma }_2(z)\cap S|=1+|{\varGamma }(z)\cap S|+|\Sigma (z)\cap S|=1+a_1+a_2. \end{aligned}$$

Hence $k=1+a_1+b_1=1+a_1+a_2$, and

$$\begin{aligned} b_1=a_2. \end{aligned}$$

(10)

For any $x\in {\varGamma }(z)\cap S$, (v, z, x) is a 2-arc in ${\varGamma }$. Thus we have $d_{\varGamma }(x,v)\leqslant 2$. Then

$$\begin{aligned} \begin{array}{rcl} \{z\}\sqcup ({\varGamma }(z)\cap S)&{}\subseteq &{} ({\varGamma }(v)\cap S)\sqcup ({\varGamma }_2(v)\cap S)\\ &{}=&{}({\varGamma }(v)\cap S)\sqcup (\Sigma (v)\cap S) \end{array} \end{aligned}$$

and

$$\begin{aligned} 1+a_1\leqslant b_1+b_1. \end{aligned}$$

Thus $b_1=k-(1+a_1)\geqslant k-2b_1=b_2$. By inequality (9),

$$\begin{aligned} k=2b_1+b_2>2b_2-1\geqslant k. \end{aligned}$$

This is a contradiction.

Therefore, ${\varGamma }$ is of diameter 2, and is a Paley graph as observed above. $\square$

2.2 2-Distance-transitive nonnormal circulants

Let ${\varGamma }=(V,E)$ be an arc-transitive circulant which is not a normal circulant. By Theorem 2.1, there exists an arc-transitive circulant ${\varSigma }$ of order m such that $mb=n$, and

$$\begin{aligned} {\varGamma }=\left\{ \begin{array}{ll} {\varSigma }\big [\overline{\mathrm{K}_b}\big ], \ \ or \\ {\varSigma }\big [\overline{\mathrm{K}_b}\big ]-b{\varSigma }, \text{ where } (m,b)=1. \end{array}\right. \end{aligned}$$

We next determine which of these graphs are 2-distance-transitive.

The vertex set V of ${\varGamma }$ is partitioned into m parts of size b, and thus we may label the vertices as

$$\begin{aligned} V=\{v_{i,j} \mid 1\leqslant i\leqslant m,\ 1\leqslant j\leqslant b\} \end{aligned}$$

such that

$$\begin{aligned} B_i:=\{v_{i,1},v_{i,2},\ldots ,v_{i,b}\},\text { where } 1\leqslant i\leqslant m \end{aligned}$$

are blocks for $\mathrm{Aut}{\varGamma }$. Let ${\mathcal {B}}=\{B_1,B_2,\dots ,B_m\}$, the corresponding block system for $\mathrm{Aut}{\varGamma }$ acting on V. The quotient graph${\varGamma }_{\mathcal {B}}$ is the graph with vertex set $V_{\mathcal {B}}={\mathcal {B}}$ such that two vertices $B_1,B_2\in {\mathcal {B}}$ are adjacent if and only if there exist $u_1\in B_1$ and $u_2\in B_2$ which are adjacent in ${\varGamma }$. Then ${\varGamma }_{\mathcal {B}}\cong {\varSigma }$, and each element $g\in \mathrm{Aut}{\varGamma }$ naturally induces a permutation $\overline{g}$ on set ${\mathcal {B}}$ which is an automorphism of the graph ${\varGamma }_{\mathcal {B}}$.

Lemma 2.9

Let u be an arbitrary vertex in ${\varGamma }$. Then, except for the case ${\varGamma }=\Sigma [\overline{\mathrm{K}_2}]-2\Sigma$, the subset $\{u\}\cup {\varGamma }_2(u)\subset V$ is a block of size b.

Proof

It is clear that $B_1$ is a block of size b for $\mathrm{Aut}{\varGamma }$. Without loss of generality, set $u=v_{1,1}$, since ${\varGamma }$ is vertex-transitive. Thus, we only need to show that $B_1=\{u\}\cup {\varGamma }_2(u)$.

Let $w=v_{1,2}\in B_1$. We have $d_{\varGamma }(u,w)\geqslant 2$ since the induced subgraph $[B_1]\cong \overline{\mathrm{K}_b}$. Suppose that $B_1,B_2$ are two vertices of ${\varGamma }_{{\mathcal {B}}}$ which are adjacent. If ${\varGamma }={\varSigma }[\overline{\mathrm{K}_b}]$, then $(u ,v_{2,1},w)$ is a 2-geodesic of ${\varGamma }$. If ${\varGamma }={\varSigma }[\overline{\mathrm{K}_b}]-b{\varSigma }$, where $b\geqslant 3$, then $(u,v_{2,3},w)$ is a 2-geodesic of ${\varGamma }$. Thus in either case, $w\in {\varGamma }_2(u)$. By the same deduction, for any $w'\in B_1\setminus \{u\}$, we have $w'\in {\varGamma }_2(u)$. Hence $B_1\subseteq \{u\}\cup {\varGamma }_2(u)$.

Let $A=\mathrm{Aut}{\varGamma }$ and $A_u$ be the stabilizer of vertex u. Since ${\varGamma }$ is 2-distance-transitive and $B_1$ is a block of V for A, we have $w^{A_u}={\varGamma }_2(u)\subseteq B_1$. Thus $\{u\}\cup {\varGamma }_2(u)= B_1$, and it is a block of size b on V for A. $\square$

Lemma 2.10

Let ${\varGamma }$ be a 2-distance-transitive circulant which is not a normal circulant. Then ${\varGamma }=\mathrm{K}_{m[b]}$ or $\mathrm{K}_{b,b}-b\mathrm{K}_2$.

Proof

Assume first that $m=2$. Since ${\varGamma }$ is of valency at least 3 by our assumption, either $b\geqslant 3$ and ${\varGamma }= \mathrm{K}_{b,b}$, or $b\geqslant 5$ and ${\varGamma }=\mathrm{K}_2[\overline{\mathrm{K}_b}]-b\mathrm{K}_2= \mathrm{K}_{b,b}-b\mathrm{K}_2$. We next consider the case where $m\geqslant 3$.

Assume that ${\varGamma }={\varSigma }[\overline{\mathrm{K}}_b]$ with $m\geqslant 3$. Let $u=v_{1,1}\in B_1$. By Lemma 2.9, $\{u\}\cup {\varGamma }_2(u)= B_1$ is a block for $A=\mathrm{Aut}{\varGamma }$. Thus there is no vertex $w\in V\setminus B_1$ at distance 2 with u in ${\varGamma }$. It follows that ${\varSigma }$ is a complete graph, and so ${\varGamma }\cong {\varSigma }[\overline{\mathrm{K}_b}]\cong \mathrm{K}_{m[b]}$.

Now, let ${\varGamma }= {\varSigma }[\overline{\mathrm{K}_b}]-b{\varSigma }$ with $m\geqslant 3$ and $b\geqslant 3$. Again, by Lemma 2.9, $\{u\}\cup {\varGamma }_2(u)= B_1$ is a block for $A=\mathrm{Aut}{\varGamma }$. Similarly, there is no vertex $w\in V\setminus B_1$ at distance 2 with u in ${\varGamma }$, and ${\varSigma }$ is a complete graph. Therefore, $(u,v_{2,2},v_{3,1})$ is a 2-geodesic in ${\varGamma }$. This contradicts the fact that $v_{3,1}\notin B_1=\{u\}\cup {\varGamma }_2(u)$.

Finally, assume that ${\varGamma }= {\varSigma }[\overline{\mathrm{K}_2}]-2{\varSigma }$ with $m\geqslant 3$. According to Lemma 2.1, m is relatively prime to b. Hence m is an odd integer. If ${\varSigma }$ is a complete graph, then ${\varGamma }=\mathrm{K}_{m[2]}-2\mathrm{K}_m$. Note that $\mathrm{K}_{m[2]}-2\mathrm{K}_m$ is isomorphic to $\mathrm{K}_{2[m]}-m\mathrm{K}_2$. We have ${\varGamma }\cong \mathrm{K}_{2[m]}-m\mathrm{K}_2$. If ${\varSigma }$ is not a complete graph, then it is clear that the quotient graph ${\varGamma }_{{\mathcal {B}}}$ is also 2-distance-transitive. By the argument above we have ${\varGamma }_{{\mathcal {B}}}$ is isomorphic to $\mathrm{C}_m$, $\mathsf{P}(p)$, or $\mathrm{K}_{m'[b']}$. When $p=5$, the Paley graph $\mathsf{P}(5)$ is isomorphic to $\mathrm{C}_5$. If ${\varGamma }_{{\mathcal {B}}}\cong \mathrm{C}_m$ then $\Gamma \cong \mathrm{C}_{2m}$ is normal, a contradiction. If ${\varGamma }_{{\mathcal {B}}}\cong \mathsf{P}(p)$ for $p>5$, there is a triangle in ${\varGamma }_{{\mathcal {B}}}$. If ${\varGamma }_{{\mathcal {B}}} \cong \mathrm{K}_{m'[b']}$, then there exists a triangle in ${\varGamma }_{{\mathcal {B}}}$ too. In either case, let $(B_1,B_2,B_3)$ be a triangle in ${\varGamma }_{{\mathcal {B}}}$ and $(B_1,B_2,B_4)$ be a 2-geodesic in ${\varGamma }_{{\mathcal {B}}}$. Since ${\varGamma }= {\varSigma }[\overline{\mathrm{K}_2}]-2{\varSigma }$, it follows that the vertex $v_{2,2}$ is adjacent to $v_{1,1}$, $v_{3,1}$, and $v_{4,1}$, and the vertex $v_{1,1}$ is not adjacent to $v_{3,1}$ and $v_{4,1}$. Hence $(v_{1,1},v_{2,2},v_{3,1})$ and $(v_{1,1},v_{2,2},v_{4,1})$ are two 2-geodesics in ${\varGamma }$. Thus there exists an automorphism $\sigma \in \mathrm{Aut}{\varGamma }$ such that $v_{1,1}^\sigma =v_{1,1}$ and $v_{3,1}^\sigma =v_{4,1}$. This is impossible since $\sigma$ induces an automorphism of ${\varGamma }_{{\mathcal {B}}}$ and $1=d_{{\varGamma }_{{\mathcal {B}}}}(B_1,B_3)\ne d_{{\varGamma }_{{\mathcal {B}}}}(B_1,B_4)=2$. $\square$