Automorphism Groups of Cubic Cayley Graphs of Dihedral Groups of Order 2ⁿp^m (n ≥ 2 and p Odd Prime)

Xianfen Kong

Department of Foundational Mathematics, Xi’an Jiaotong-Liverpool University, Suzhou, China.

DOI: 10.4236/jamp.2020.812226                

Abstract

Keywords

Automorphism Group, Dihedral Group, Cayley Graph

1. Introduction

An automorphism of a graph X is a permutation $\sigma$ of vertex set of X with the property that, for any vertices u and v, we have $\mathrm{<mo>\; \{\; </mo>}{u}^{\sigma }\mathrm{,}{v}^{\sigma }\mathrm{<mo>\; \}\; </mo>}$ is an edge of X if and only if $\mathrm{<mo>\; \{\; </mo>}u\mathrm{,}v\mathrm{<mo>\; \}\; </mo>}$ is the edge of X. As usual, we use $u^{\sigma }$ to denote the image of the vertex u under the permutation $\sigma$ and $\mathrm{<mo>\; \{\; </mo>}u\mathrm{,}v\mathrm{<mo>\; \}\; </mo>}$ to denote the edge joining vertices u and v. All automorphisms of graph X form a group under the composite operation of mapping. This group is called the full automorphism group of graph X, denoted by A in this paper.

For a graph X, we denote vertex set and edge set of X by $V\mathrm{<mo\; stretchy="false">\; (\; </mo>}X\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ and $E\mathrm{<mo\; stretchy="false">\; (\; </mo>}X\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ . $A_v$ is the stabilizer of vertex $v$ in the automorphism group of X. $X_k\mathrm{<mo\; stretchy="false">\; (\; </mo>}v\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ denotes the set of vertices at distance k from vertex v. $D_{2n}$ means the dihedral group of order 2n. A graph is called vertex-transitive if its automorphism group A is transitive on the vertex set $V\mathrm{<mo\; stretchy="false">\; (\; </mo>}X\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ . An s-arc in a graph is an ordered $\mathrm{<mo\; stretchy="false">\; (\; </mo>}s+\mathrm{1\; <mo\; stretchy="false">\; )\; </mo>}$ -tuple $\mathrm{<mo\; stretchy="false">\; (\; </mo>}{v}_0\mathrm{,}{v}_1\mathrm{,...,}{v}_{s-1}\mathrm{,}{v}_s\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ of vertices of the graph such that $v_{i-1}$ is adjacent to $v_i$ for $1\le i\le s$ and $v_{i-1}\ne v_{i+1}$ for $1\le i\le s$ . A graph is said to be s-arc-transitive if the automorphism group A acts transitively on the set of all s-arcs in X. When $s=1$ , 1-arc called arc and 1-arc transitive is called arc-transitive or symmetric.

Throughout this paper, graphs are finite, simple and undirected.

Let G be a finite group and S be a subset of G such that $1\notin S$ . The Cayley graph $X=Cay(G,S)$ on G with respect to S is defined to have vertex set $V(X)=G$ and edge set $E(X)=\left\{\right\{g,sg\left\}\right|g\in G\text{and}s\in S\}$ . Let set $S^{-1}=\left\{s^{-1}\right|s\in S\}$ . If $S^{-1}=S$ , $Cay\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{,}S\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ is undirected. If S is a generating system of G, $Cay\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{,}S\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ is connected. Two subsets S and T of group G are called equivalent if there exists a group automorphism of group G mapping S to T: $S^{\alpha }=T$ for some $\alpha \in Aut\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ . Denote by $S\equiv T$ . If S and T are equivalent, Cayley graphs $Cay\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{,}S\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ and $Cay\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{,}S\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ are isomorphic.

The right regular representation $R\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ of group G is a subgroup of the the automorphism group A of the Cayley graph X. In particular by [1], if $R\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ is the full automorphism group of X then $X=Cay(G,S)$ is called a GRR (for graphical regular representation) of G. A Cayley graph is normal if $R\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ is a normal subgroup of A. $R\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ is transitive on G hence Cayley graph is vertex-transitive. Denote $Aut(G,S)=\{\alpha \in Aut(G)|S^{\alpha }=S\}$ , the set of all automorphism of group G preserving S. $Aut\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{,}S\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ is also a subgroup of the automorphism group of Cayley graph. In particular, $Aut\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{,}S\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ is a subgroup of stabilizer of vertex identity $A_1$ . By [2] the normalizer of $R\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ in A is the semi-direct product of $R\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ and $Aut\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{,}S\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ : $N_A\mathrm{<mo\; stretchy="false">\; (\; </mo>}R\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{<mo\; stretchy="false">\; )\; </mo>\; <mo\; stretchy="false">\; )\; </mo>}=R\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{<mo\; stretchy="false">\; )\; </mo>}⋊Aut\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{,}S\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ . By [3] Proposition 1.5 X is normal if and only if $A_1=Aut\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{,}S\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ . Cayley graph X is normal if and only if the automorphism group of X is $A=R\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{<mo\; stretchy="false">\; )\; </mo>}⋊Aut\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{,}S\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ . Normality provides an approach to find automorphism groups of Cayley graphs.

In [4] the automorphism group of connected cubic Cayley graphs of order 4p is given. In [5] the automorphism group of connected cubic Cayley graphs of order 32p is given. In this paper, the automorphism group of connected cubic Cayley graphs of dihedral groups of order $2^n{p}^m$ where $n\ge 2$ and p is odd is given.

Summarising theorem 4.1, 4.2, 4.3 in Part 4 gives the main results.

Theorem 1.1. Let $G=D_{2^n{p}^m}$ be a dihedral group where $n\ge 2$ and p is an odd prime number. S is an inverse-closed generating system of three elements without identity element. Then Cayley graph $Cay\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{,}S\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ is GRR except the following cases:

1) $S\equiv \mathrm{<mo>\; \{\; </mo>}b\mathrm{,}ab\mathrm{,}{a}^{k}b\mathrm{<mo>\; \}\; </mo>}$ where $k^2\equiv 1\left(\mathrm{mod}{2}^{n-1}{p}^m\right)$ and $\gcd \mathrm{<mo\; stretchy="false">\; (\; </mo>}k{\mathrm{,2}}^{n-1}{p}^m\mathrm{<mo\; stretchy="false">\; )\; </mo>}=1$ , $Aut\mathrm{<mo\; stretchy="false">\; (\; </mo>}X\mathrm{<mo\; stretchy="false">\; )\; </mo>}\cong G\mathrm{<mo>\; :\; </mo>}{\mathbb{Z}}_2$ .

2) $S\equiv \mathrm{<mo>\; \{\; </mo>}b\mathrm{,}ab\mathrm{,}{a}^{2^{n-1}{p}^m}b\mathrm{<mo>\; \}\; </mo>}$ , $Aut\mathrm{<mo\; stretchy="false">\; (\; </mo>}X\mathrm{<mo\; stretchy="false">\; )\; </mo>}\cong {\mathbb{Z}}_2^{2^{n-2}{p}^m}⋊D_{2^{n-1}{p}^m}$ .

3) $S\equiv \mathrm{<mo>\; \{\; </mo>}a\mathrm{,}{a}^{-1}\mathrm{,}b\mathrm{<mo>\; \}\; </mo>}$ , $Aut\mathrm{<mo\; stretchy="false">\; (\; </mo>}X\mathrm{<mo\; stretchy="false">\; )\; </mo>}=G\mathrm{<mo>\; :\; </mo>}{\mathbb{Z}}_2$ .

4) $S\equiv \mathrm{<mo>\; \{\; </mo>}b\mathrm{,}ab\mathrm{,}{a}^{2^{n-2}{p}^m}\mathrm{<mo>\; \}\; </mo>}$ , $Aut\mathrm{<mo\; stretchy="false">\; (\; </mo>}X\mathrm{<mo\; stretchy="false">\; )\; </mo>}=G\mathrm{<mo>\; :\; </mo>}{\mathbb{Z}}_2$ .

2. Preliminary

Results used to prove main theorem are listed here.

Proposition 2.1. Suppose that $G=<a,b|a^n=b^2=1,b^{-1}ab=a^{-1}>$ is a dihedral group, then the automorphism group $Aut\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ of G has the following properties.

1) Any automorphism of G can be defined as $a\mapsto a^i$ and $b\mapsto a^{j}b$ where $i\in {\mathbb{Z}}_n^{\mathrm{<mo>\; *\; </mo>}}$ and $j\in {\mathbb{Z}}_n$ .

2) $Aut(G)=<\alpha >⋊<\beta >\cong {\mathbb{Z}}_n⋊{\mathbb{Z}}_n^{*}$ where $\alpha \mathrm{<mo>\; :\; </mo>}a\mapsto a\mathrm{,}b\mapsto ab\mathrm{<mo>\; ;\; </mo>}\beta \mathrm{<mo>\; :\; </mo>}a\mapsto a^i\mathrm{,}b\mapsto b\mathrm{,}i\in {\mathbb{Z}}_n^{\mathrm{<mo>\; *\; </mo>}}$ .

Proposition 2.2. Suppose G is a finite group and subsets $S\equiv T$ , then $Cay\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{,}S\mathrm{<mo\; stretchy="false">\; )\; </mo>}\cong Cay\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{,}T\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ .

Proposition 2.3. Let $G=<a,b|a^n=b^2=1,b^{-1}ab=a^{-1}>$ be the dihedral group of order 2n. Subsets $\mathrm{<mo>\; \{\; </mo>}b\mathrm{,}ab\mathrm{,}{a}^{k}b\mathrm{<mo>\; \}\; </mo>}\equiv \mathrm{<mo>\; \{\; </mo>}b\mathrm{,}ab\mathrm{,}{a}^{1-k}b\mathrm{<mo>\; \}\; </mo>}$ .

Proof Let $\sigma \in Aut\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{<mo\; stretchy="false">\; )\; </mo>\; <mo>\; :\; </mo>}a\mapsto a^{-1}\mathrm{,}b\mapsto ab$ then ${\{b,ab,a^{k}b\}}^{\sigma }=\{b,ab,a^{1-k}b\}$ .

The following sufficient and necessary condition of normality of Cayley graph is from paper [6].

Proposition 2.4. Let $X=Cay(G,S)$ be connected. Then X is a normal Cayley graph of G if and only if the following conditions are satisfied:

1) For each $\phi \in A_1$ there exists $\sigma \in Aut\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ such that $\phi {|}_{X_1(1)}=\sigma {|}_{X_1(1)}$ ;

2) For each $\phi \in A_1$ , $\phi {|}_{X_1(1)}=1_{X_1(1)}$ implies $\phi {|}_{X_2(1)}=1_{X_2(1)}$ .

A classification of locally primitive Cayley graphs of dihedral groups from paper [7] will be used.

Proposition 2.5. Let X be a locally-primitive Cayley graph of a dihedral group of order 2n. Then one of the following statements is true, where q is a prime power.

1) X is 2-arc-transitive, and one of the following holds:

a) $X=K_{2n},K_{n,n}$ or $K_{n\mathrm{,}n}-n{K}_2$ ;

b) $X=\mathcal{H}\mathcal{D}\mathrm{<mo\; stretchy="false">\; (\; </mo>11,5,2\; <mo\; stretchy="false">\; )\; </mo>}$ or $\mathcal{H}\mathcal{D}\mathrm{<mo\; stretchy="false">\; (\; </mo>11,6,2\; <mo\; stretchy="false">\; )\; </mo>}$ , the incidence or non-incidence graph of the Hadamard design on 11 points;

c) $X=\mathcal{P}\mathcal{H}\mathrm{<mo\; stretchy="false">\; (\; </mo>}d\mathrm{,}q\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ or $\mathcal{P}{\mathcal{H}}^{\prime }\mathrm{<mo\; stretchy="false">\; (\; </mo>}d\mathrm{,}q\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ , the point-hyperplane incidence or non-incidence graph of $\mathrm{<mo\; stretchy="false">\; (\; </mo>}d-\mathrm{1\; <mo\; stretchy="false">\; )\; </mo>}$ -dimension projective geometry $PG\mathrm{<mo\; stretchy="false">\; (\; </mo>}d-\mathrm{1,}q\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ , where $d\ge 3$ ;

d) $X=K_{q+1}^{2d}$ , where d is a divisor of $\frac{q-1}{2}$ if $q\equiv 1$ (mod 4), and a divisor of $q-1$ if $q\equiv 3$ (mod 4) respectively.

2) $X=\mathcal{N}{\mathcal{D}}_{2n\mathrm{,}r\mathrm{,}k}$ is a normal Cayley graph and is not 2-arc-transitive, where $n=r^t{p}_1^{e_1}{p}_2^{e_2}\cdots p_s^{e_s}\ge 13$ with $r\mathrm{,}{p}_1\mathrm{,}{p}_2\mathrm{,}\cdots \mathrm{,}{p}_s$ distinct odd primes, $t\le 1$ , $s\ge 1$ and $r\mathrm{<mo>\; |\; </mo>\; <mo\; stretchy="false">\; (\; </mo>}{p}_i-\mathrm{1\; <mo\; stretchy="false">\; )\; </mo>}$ for each i. There are exactly ${\mathrm{<mo\; stretchy="false">\; (\; </mo>}r-\mathrm{1\; <mo\; stretchy="false">\; )\; </mo>}}^{s-1}$ non-isomorphism such graphs for a given order 2n.

3. Lemmas and Propositions

In the following, group G means that $G=<a,b|a^{2^{n-1}{p}^m}=b^2=1,b^{-1}ab=a^{-1}>$ be dihedral group of order $2^n{p}^m$ where $n\ge 2$ and p is an odd prime number.

Proposition 3.1. If $S=\mathrm{<mo>\; \{\; </mo>}{a}^{i}b\mathrm{,}{a}^{j}b\mathrm{,}{a}^{r}b\mathrm{<mo>\; \}\; </mo>}$ is a generating system of G of three elements, then $S\equiv \mathrm{<mo>\; \{\; </mo>}b\mathrm{,}ab\mathrm{,}{a}^{k}b\mathrm{<mo>\; \}\; </mo>}$ for some $2\le k\le 2^{n-1}{p}^m-1$ .

There are two types of S classified by the number of subsets of two elements generating G.

Type 1: S has only one subset of two elements generating G.

Type 2: S has exactly two subsets of two elements generating G. In this case, $S\equiv \mathrm{<mo>\; \{\; </mo>}b\mathrm{,}ab\mathrm{,}{a}^{k}b\mathrm{<mo>\; \}\; </mo>}$ where $\gcd \mathrm{<mo\; stretchy="false">\; (\; </mo>}k{\mathrm{,2}}^{n-1}{p}^m\mathrm{<mo\; stretchy="false">\; )\; </mo>}=1$ .

The proof of Proposition 3.1 will be done by the following three lemmas.

Lemma 3.1. If $S=\mathrm{<mo>\; \{\; </mo>}{a}^{i}b\mathrm{,}{a}^{j}b\mathrm{,}{a}^{r}b\mathrm{<mo>\; \}\; </mo>}$ is a generating system of G of three elements, then S is equivalent to a subset of type $\mathrm{<mo>\; \{\; </mo>}b\mathrm{,}ab\mathrm{,}{a}^{k}b\mathrm{<mo>\; \}\; </mo>}$ for some $2\le k\le 2^{n-1}{p}^m-1$ .

Proof By proposition 2.1 in preliminary, automorphism group Aut(G) of dihedral group G is transitive on the set of involutions $\mathrm{<mo>\; \{\; </mo>}{a}^{i}b\mathrm{<mo>\; |\; </mo>0}\le i\le 2^{n-1}{p}^m-\mathrm{1\; <mo>\; \}\; </mo>}$ . One may assume that $b\in S$ and $S=\{b,a^{i}b,a^{j}b\}$ be a generating system of G of three elements. S has three subsets of two elements: $\mathrm{<mo>\; \{\; </mo>}b\mathrm{,}{a}^{i}b\mathrm{<mo>\; \}\; </mo>,\; <mo>\; \{\; </mo>}b\mathrm{,}{a}^{j}b\mathrm{<mo>\; \}\; </mo>}$ and $\mathrm{<mo>\; \{\; </mo>}{a}^{i}b\mathrm{,}{a}^{j}b\mathrm{<mo>\; \}\; </mo>}$ ..

Note that, subset $T\subset G$ is a generating system of G if and only if $T^{\alpha }$ is a generating system of G for any $\alpha \in Aut\left(G\right)$ .

Suppose that subset $\mathrm{<mo>\; \{\; </mo>}b\mathrm{,}{a}^{x}b\mathrm{<mo>\; \}\; </mo>}$ ( $x=i$ or $j$ ) generates G. Let $\alpha \in Aut\left(G\right)$ : $a\mapsto a^x\mathrm{,}b\mapsto b$ , then ${\{b,ab,a^{k}b\}}^{\alpha }=\{b,a^{i}b,a^{j}b\}$ for some $k\ne \mathrm{0,1}$ . Hence $S\equiv \mathrm{<mo>\; \{\; </mo>}b\mathrm{,}ab\mathrm{,}{a}^{k}b\mathrm{<mo>\; \}\; </mo>}$ .

Assume that both subset $\mathrm{<mo>\; \{\; </mo>}b\mathrm{,}{a}^{i}b\mathrm{<mo>\; \}\; </mo>}$ and $\mathrm{<mo>\; \{\; </mo>}b\mathrm{,}{a}^{j}b\mathrm{<mo>\; \}\; </mo>}$ do not generate G. Next will show that $\mathrm{<mo>\; \{\; </mo>}{a}^{i}b\mathrm{,}{a}^{j}b\mathrm{<mo>\; \}\; </mo>}$ must be able to generate G.

$G=<S>=<b,a^{i}b,a^{j}b>=<a^i,a^j><b>=<a^{\gcd (i,j)}><b>$ . Hence $\gcd (i,j)$ and $2^{n-1}{p}^m$ are mutually prime.

$G\ne <b,a^{i}b>=<a^i><b>$ . Hence $i$ and $2^{n-1}{p}^m$ are not mutually prime.

Similarly, $G\ne <b,a^{j}b>$ implies that $j$ and $2^{n-1}{p}^m$ are also not mutually prime.

$(\gcd (i,j),2^{n-1}{p}^m)=1$ , $\mathrm{<mo\; stretchy="false">\; (\; </mo>}i{\mathrm{,2}}^{n-1}{p}^m\mathrm{<mo\; stretchy="false">\; )\; </mo>}\ne 1$ and $\mathrm{<mo\; stretchy="false">\; (\; </mo>}j{\mathrm{,2}}^{n-1}{p}^m\mathrm{<mo\; stretchy="false">\; )\; </mo>}\ne 1$ imply that, for $i$ and $j$ , one number is power of 2 and the other one is power of p. Thus $i-j$ and $2^{n-1}{p}^m$ are mutually prime.

Hence, $\mathrm{<mo>\; \{\; </mo>}{a}^{i}b\mathrm{,}{a}^{j}b\mathrm{<mo>\; \}\; </mo>}$ is a generating system of G since $<a^{i}b,a^{j}b>=<a^{i-j}><a^{i}b>=G$ .

Let $\alpha \in Aut\left(G\right)$ : $a\mapsto a^{i-j}\mathrm{,}b\mapsto a^{j}b$ . Then ${\{b,ab,a^{k}b\}}^{\alpha }=\{b,a^{i}b,a^{j}b\}$ for some k. $S\equiv \mathrm{<mo>\; \{\; </mo>}b\mathrm{,}ab\mathrm{,}{a}^{k}b\mathrm{<mo>\; \}\; </mo>}$ .

Corollary 3.1. If $S=\{a^{i}b,a^{j}b,a^{r}b\}$ is a generating system of G of three elements, there exists at least one subset of two elements generating G.

Lemma 3.2. If $S=\{a^{i}b,a^{j}b,a^{r}b\}$ is a generating system of G of three elements, there are only one or two subsets of two elements of S generating G.

Proof By Lemma 3.1, we assume that $S=\{b,ab,a^{k}b\}$ where $k\ne \mathrm{0,1}$ . S has three subsets of two elements: $\mathrm{<mo>\; \{\; </mo>}b\mathrm{,}ab\mathrm{<mo>\; \}\; </mo>,\; <mo>\; \{\; </mo>}b\mathrm{,}{a}^{k}b\mathrm{<mo>\; \}\; </mo>}$ and $\mathrm{<mo>\; \{\; </mo>}ab\mathrm{,}{a}^{k}b\mathrm{<mo>\; \}\; </mo>}$ . Next we will show that it is impossible that all three subsets of two elements generating G.

$<b,a^{k}b>=<a^k><b>$ is a dihedral subgroup of G. $<ab,a^{k}b>=<a^{k-1}><ab>$ is also a dihedral subgroup of G.

For $k$ and $k-1$ , one is an even number and the other one is an odd number. The orders of elements $a^k$ and $a^{k-1}$ are different: $\circ \mathrm{<mo\; stretchy="false">\; (\; </mo>}{a}^k\mathrm{<mo\; stretchy="false">\; )\; </mo>}\ne \circ \mathrm{<mo\; stretchy="false">\; (\; </mo>}{a}^{k-1}\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ . This implies that at least one subset of $\mathrm{<mo>\; \{\; </mo>}b\mathrm{,}{a}^{k}b\mathrm{<mo>\; \}\; </mo>}$ and $\mathrm{<mo>\; \{\; </mo>}ab\mathrm{,}{a}^{k}b\mathrm{<mo>\; \}\; </mo>}$ does not generate G.

Hence there are only one or two subsets of two elements of S generating G.

Lemma 3.3. Let $S=\mathrm{<mo>\; \{\; </mo>}{a}^{i}b\mathrm{,}{a}^{j}b\mathrm{,}{a}^{r}b\mathrm{<mo>\; \}\; </mo>}$ be a generating system of G of three elements and S has two subsets of two elements generating G. If $S\equiv \mathrm{<mo>\; \{\; </mo>}b\mathrm{,}ab\mathrm{,}{a}^{k}b\mathrm{<mo>\; \}\; </mo>}$ , either $\gcd (k{\mathrm{,2}}^{n-1}{p}^m)=1$ or $\gcd (1-k{\mathrm{,2}}^{n-1}{p}^m)=1$ .

Proposition 3.2. Suppose that $S=\mathrm{<mo>\; \{\; </mo>}{a}^{i}b\mathrm{,}{a}^{j}b\mathrm{,}{a}^{r}b\mathrm{<mo>\; \}\; </mo>}$ is a generating system of G of three elements and $S\equiv \mathrm{<mo>\; \{\; </mo>}b\mathrm{,}ab\mathrm{,}{a}^{k}b\mathrm{<mo>\; \}\; </mo>}$ .

(1) If S has only one subset of two elements generating G, then $Aut\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{,}S\mathrm{<mo\; stretchy="false">\; )\; </mo>}=1$ .

(2) If S has two subsets of two elements generating G, then $Aut\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{,}S\mathrm{<mo\; stretchy="false">\; )\; </mo>}=1$ except the following two cases. $Aut\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{,}S\mathrm{<mo\; stretchy="false">\; )\; </mo>}\cong {\mathbb{Z}}_2$ if $k^2\equiv 1(\mathrm{mod}{2}^{n-1}{p}^m)$ and $\gcd \mathrm{<mo\; stretchy="false">\; (\; </mo>}k{\mathrm{,2}}^{n-1}{p}^m\mathrm{<mo\; stretchy="false">\; )\; </mo>}=1$ ; $Aut\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{,}S\mathrm{<mo\; stretchy="false">\; )\; </mo>}\cong {\mathbb{Z}}_2$ if ${\mathrm{<mo\; stretchy="false">\; (\; </mo>1}-k\mathrm{<mo\; stretchy="false">\; )\; </mo>}}^2\equiv 1(\mathrm{mod}{2}^{n-1}{p}^m)$ and $\gcd \mathrm{<mo\; stretchy="false">\; (\; </mo>1}-k{\mathrm{,2}}^{n-1}{p}^m\mathrm{<mo\; stretchy="false">\; )\; </mo>}=1$ .

Proof (1) If there is only one subset of two elements in $S=\mathrm{<mo>\; \{\; </mo>}b\mathrm{,}ab\mathrm{,}{a}^{k}b\mathrm{<mo>\; \}\; </mo>}$ generating G, then $G\ne \mathrm{<mo>\; <\; </mo>}b\mathrm{,}{a}^{k}b\mathrm{<mo>\; >\; </mo>}$ , $G\ne \mathrm{<mo>\; <\; </mo>}ab\mathrm{,}{a}^{k}b\mathrm{<mo>\; >\; </mo>}$ and $G=<b,ab>$ . For any $\sigma \in Aut(G,S)$ , ${\mathrm{<mo>\; \{\; </mo>}b\mathrm{,}ab\mathrm{<mo>\; \}\; </mo>}}^{\sigma }$ is also a generating system of G. ${\mathrm{<mo>\; \{\; </mo>}b\mathrm{,}ab\mathrm{<mo>\; \}\; </mo>}}^{\sigma }=\mathrm{<mo>\; \{\; </mo>}b\mathrm{,}ab\mathrm{<mo>\; \}\; </mo>}$ . Since $S^{\sigma }=S$ . Hence $a^{k}b=S-\mathrm{<mo>\; \{\; </mo>}b\mathrm{,}ab\mathrm{<mo>\; \}\; </mo>}$ is fixed by $\sigma$ . ${\mathrm{<mo\; stretchy="false">\; (\; </mo>}{a}^{k}b\mathrm{<mo\; stretchy="false">\; )\; </mo>}}^{\sigma }=a^{k}b$ .

If $b^{\sigma }=b$ and ${\mathrm{<mo\; stretchy="false">\; (\; </mo>}ab\mathrm{<mo\; stretchy="false">\; )\; </mo>}}^{\sigma }=ab$ then $a^{\sigma }={(abb)}^{\sigma }={(ab)}^{\sigma }{b}^{\sigma }=abb=a$ , hence $\sigma =1$ .

If $b^{\sigma }=ab$ and ${\mathrm{<mo\; stretchy="false">\; (\; </mo>}ab\mathrm{<mo\; stretchy="false">\; )\; </mo>}}^{\sigma }=b$ , then $a^{\sigma }={(abb)}^{\sigma }={(ab)}^{\sigma }{b}^{\sigma }=bab=a^{-1}$ . This implies that $a^{k}b={(a^{k}b)}^{\sigma }={(a^k)}^{\sigma }{b}^{\sigma }=a^{-k}ab=a^{1-k}b$ . Thus $a^k=a^{1-k}$ . This is a contradiction. For k and $1-k$ , one is an even number and the other one is an odd number. This implies that the orders of the element $a^k$ and $a^{1-k}$ are not equal: $\circ \mathrm{<mo\; stretchy="false">\; (\; </mo>}{a}^k\mathrm{<mo\; stretchy="false">\; )\; </mo>}\ne \circ \mathrm{<mo\; stretchy="false">\; (\; </mo>}{a}^{1-k}\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ .

Hence $Aut\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{,}S\mathrm{<mo\; stretchy="false">\; )\; </mo>}=1$ .

(2) If there are two subsets of two elements of S generating G, we assume that $\gcd (k,2^{n-1}{p}^m)=1$ . $G=<b,ab>=<b,a^{k}b>$ and $G\ne <ab,a^{k}b>$ .

Since subset $\mathrm{<mo>\; \{\; </mo>}ab\mathrm{,}{a}^{k}b\mathrm{<mo>\; \}\; </mo>}$ is the only subset of two elements not generating G, ${\{ab,a^{k}b\}}^{\sigma }=\{ab,a^{k}b\}$ for any $\sigma \in Aut\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{,}S\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ . $b=S-\{ab,a^{k}b\}$ is fixed by $\sigma$ . ${(ab)}^{\sigma }=ab$ or $a^{k}b$ .

If ${(ab)}^{\sigma }=ab$ , then $\sigma =1$ .

If ${(ab)}^{\sigma }=a^{k}b$ , then $a^{\sigma }={(abb)}^{\sigma }={(ab)}^{\sigma }{b}^{\sigma }=a^{k}bb=a^k$ . ${(a^{k}b)}^{\sigma }={(a^k)}^{\sigma }{b}^{\sigma }={(a^k)}^{k}b=a^{k^2}b=ab$ . So $k^2\equiv 1(\mathrm{mod}{2}^{n-1}{p}^m)$ .

Hence $Aut\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{,}S\mathrm{<mo\; stretchy="false">\; )\; </mo>}=1$ if $k^2\cancel{\equiv }1(\mathrm{mod}{2}^{n-1}{p}^m)$ . $Aut\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{,}S\mathrm{<mo\; stretchy="false">\; )\; </mo>}\cong {\mathbb{Z}}_2$ if $k^2\equiv 1(\mathrm{mod}{2}^{n-1}{p}^m)$ .

Similarly, when $\gcd (1-k,2^{n-1}{p}^m)=1$ , $Aut\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{,}S\mathrm{<mo\; stretchy="false">\; )\; </mo>}=1$ if ${\mathrm{<mo\; stretchy="false">\; (\; </mo>1}-k\mathrm{<mo\; stretchy="false">\; )\; </mo>}}^2\cancel{\equiv }1(\mathrm{mod}{2}^{n-1}{p}^m)$ . $Aut\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{,}S\mathrm{<mo\; stretchy="false">\; )\; </mo>}\cong {\mathbb{Z}}_2$ if ${\mathrm{<mo\; stretchy="false">\; (\; </mo>1}-k\mathrm{<mo\; stretchy="false">\; )\; </mo>}}^2\equiv 1(\mathrm{mod}{2}^{n-1}{p}^m)$ .

Proposition 3.3. Suppose that S is inverse-closed generating system of three elements of G, then $S\equiv \mathrm{<mo>\; \{\; </mo>}a\mathrm{,}{a}^{-1}\mathrm{,}b\mathrm{<mo>\; \}\; </mo>}$ , $\mathrm{<mo>\; \{\; </mo>}b\mathrm{,}ab\mathrm{,}{a}^{2^{n-2}{p}^m}\mathrm{<mo>\; \}\; </mo>}$ or $\mathrm{<mo>\; \{\; </mo>}b\mathrm{,}ab\mathrm{,}{a}^{k}b\mathrm{<mo>\; \}\; </mo>\; <mo\; stretchy="false">\; (\; </mo>}k\ne \mathrm{0,1\; <mo\; stretchy="false">\; )\; </mo>}$ .

Proof Since S contains three elements and inverse-closed, there must be an involution in S. There are two orbits of involutions in G under the action of group automorphism $Aut\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ : $\mathrm{<mo>\; \{\; </mo>}{a}^{2^{n-2}{p}^m}\mathrm{<mo>\; \}\; </mo>}$ and $\mathrm{<mo>\; \{\; </mo>}{a}^{i}b\mathrm{<mo>\; |\; </mo>0}\le i\le 2^{n-1}{p}^m-\mathrm{1\; <mo>\; \}\; </mo>}$ .

Suppose that $a^{2^{n-2}{p}^m}\in S$ . $S-\mathrm{<mo>\; \{\; </mo>}{a}^{2^{n-2}{p}^m}\mathrm{<mo>\; \}\; </mo>}$ is also inverse-closed hence it is a set of two involutions from orbit $\mathrm{<mo>\; \{\; </mo>}{a}^{i}b\mathrm{<mo>\; |\; </mo>0}\le i\le 2^{n-1}{p}^m-\mathrm{1\; <mo>\; \}\; </mo>}$ . S generating G implies that $S-\mathrm{<mo>\; \{\; </mo>}{a}^{2^{n-2}{p}^m}\mathrm{<mo>\; \}\; </mo>}$ also generates G. We get $S\equiv \mathrm{<mo>\; \{\; </mo>}b\mathrm{,}ab\mathrm{,}{a}^{2^{n-2}{p}^m}\mathrm{<mo>\; \}\; </mo>}$ .

Suppose that S contains an involution from $\mathrm{<mo>\; \{\; </mo>}{a}^{i}b\mathrm{<mo>\; |\; </mo>0}\le i\le 2^{n-1}{p}^m-\mathrm{1\; <mo>\; \}\; </mo>}$ . $Aut\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ is transitive on this orbit, we can assume that $b\in S$ . If $S-\mathrm{<mo>\; \{\; </mo>}b\mathrm{<mo>\; \}\; </mo>}$ contains an involution, $S\equiv \mathrm{<mo>\; \{\; </mo>}b\mathrm{,}ab\mathrm{,}{a}^{k}b\mathrm{<mo>\; \}\; </mo>\; <mo\; stretchy="false">\; (\; </mo>}k\ne \mathrm{0,1\; <mo\; stretchy="false">\; )\; </mo>}$ by Proposition 3.1 and 2.1. If $S-\mathrm{<mo>\; \{\; </mo>}b\mathrm{<mo>\; \}\; </mo>}$ contains no involutions, $S\equiv \mathrm{<mo>\; \{\; </mo>}b\mathrm{,}a\mathrm{,}{a}^{-1}\mathrm{<mo>\; \}\; </mo>}$ by Proposition 2.1.

4. Results

By Proposition 3.3, we only need to discuss $X=Cay\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{,}S\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ for $S=\mathrm{<mo>\; \{\; </mo>}a\mathrm{,}{a}^{-1}\mathrm{,}b\mathrm{<mo>\; \}\; </mo>,\; <mo>\; \{\; </mo>}b\mathrm{,}ab\mathrm{,}{a}^{2^{n-2}{p}^m}\mathrm{<mo>\; \}\; </mo>}$ and $\mathrm{<mo>\; \{\; </mo>}b\mathrm{,}ab\mathrm{,}{a}^{k}b\mathrm{<mo>\; \}\; </mo>\; <mo\; stretchy="false">\; (\; </mo>}k\ne \mathrm{0,1\; <mo\; stretchy="false">\; )\; </mo>}$ .

Firstly, we discuss $X=Cay\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{,\; <mo>\; \{\; </mo>}b\mathrm{,}ab\mathrm{,}{a}^{k}b\mathrm{<mo>\; \}\; </mo>\; <mo\; stretchy="false">\; )\; </mo>\; <mo\; stretchy="false">\; (\; </mo>}k\ne \mathrm{0,1\; <mo\; stretchy="false">\; )\; </mo>}$ .

Theorem 4.1. Suppose that $S=\mathrm{<mo>\; \{\; </mo>}{a}^{i}b\mathrm{,}{a}^{j}b\mathrm{,}{a}^{m}b\mathrm{<mo>\; \}\; </mo>}$ is a generating system of three involutions of G and $S\equiv \mathrm{<mo>\; \{\; </mo>}b\mathrm{,}ab\mathrm{,}{a}^{k}b\mathrm{<mo>\; \}\; </mo>}$ .

X is GRR except the following cases.

(1) When $\gcd (k{\mathrm{,2}}^{n-2}{p}^m)=1$ , $k^2\equiv 1(\mathrm{mod}{2}^{n-1}{p}^m)$ and $k\ne 2^{n-2}{p}^m+1$ then $Aut\mathrm{<mo\; stretchy="false">\; (\; </mo>}X\mathrm{<mo\; stretchy="false">\; )\; </mo>}\cong R\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{<mo\; stretchy="false">\; )\; </mo>\; <mo>\; :\; </mo>}{\mathbb{Z}}_2$ .

(2) When $\gcd (1-k{\mathrm{,2}}^{n-2}{p}^m)=1$ , ${\mathrm{<mo\; stretchy="false">\; (\; </mo>1}-k\mathrm{<mo\; stretchy="false">\; )\; </mo>}}^2\equiv 1(\mathrm{mod}{2}^{n-1}{p}^m)$ and $k\ne 2^{n-2}{p}^m$ then $Aut\mathrm{<mo\; stretchy="false">\; (\; </mo>}X\mathrm{<mo\; stretchy="false">\; )\; </mo>}\cong R\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{<mo\; stretchy="false">\; )\; </mo>\; <mo>\; :\; </mo>}{\mathbb{Z}}_2$ .

(3) If $k=2^{n-2}{p}^m+1$ or $k=2^{n-2}{p}^m$ , then $Aut\mathrm{<mo\; stretchy="false">\; (\; </mo>}X\mathrm{<mo\; stretchy="false">\; )\; </mo>}\cong {\mathbb{Z}}_2^{2^{n-2}{p}^m}⋊D_{2^{n-1}{p}^m}$ .

Proof Let $S=\{b,ab,a^{k}b\}$ where $2\le k\le 2^{n-1}{p}^m-1$ and $X=Cay(G,S)$ . Classify X in two cases: there are 4-cycles in X and there is no 4-cycle in X.

(1) Note that $X_2(1)=\{a,a^k,a^{-1},a^{k-1},a^{-k},a^{1-k}\}$ is the set of vertices at distance 2 from vertex 1.

If there are 4-cycles in X, some vertices in $X_2\mathrm{<mo\; stretchy="false">\; (\; </mo>1\; <mo\; stretchy="false">\; )\; </mo>}$ are coincident. Solving $a=a^{k-1}$ and $a^{-1}=a^{1-k}$ we get $k=2$ . Solving $a=a^{-k}$ and $a^k=a^{-1}$ we get $k=-1$ . Solving $a^k=a^{-k}$ we get $k=2^{n-2}{p}^m$ . Solving $a^{k-1}=a^{1-k}$ we get $k=2^{n-2}{p}^m+1$ . There is no solution for other equations. Note that −1 and $2^{n-2}{p}^m+1$ are two solutions of equation $k^2\equiv 1(\mathrm{mod}{2}^{n-1}{p}^m)$ . 2 and $2^{n-2}{p}^m$ are two solutions of equation ${\mathrm{<mo\; stretchy="false">\; (\; </mo>1}-k\mathrm{<mo\; stretchy="false">\; )\; </mo>}}^2\equiv 1(\mathrm{mod}{2}^{n-1}{p}^m)$ . Since $\mathrm{<mo>\; \{\; </mo>}b\mathrm{,}ab\mathrm{,}{a}^{2}b\mathrm{<mo>\; \}\; </mo>}\equiv \mathrm{<mo>\; \{\; </mo>}b\mathrm{,}ab\mathrm{,}{a}^{-1}b\mathrm{<mo>\; \}\; </mo>}$ and $\mathrm{<mo>\; \{\; </mo>}b\mathrm{,}ab\mathrm{,}{a}^{2^{n-2}{p}^m}b\mathrm{<mo>\; \}\; </mo>}\equiv \mathrm{<mo>\; \{\; </mo>}b\mathrm{,}ab\mathrm{,}{a}^{2^{n-2}{p}^m+1}b\mathrm{<mo>\; \}\; </mo>}$ we only discuss $k=2$ and $k=2^{n-2}{p}^m$ .

(1.1) When $k=2$ , $X=C_{2^{n-1}{p}^m}\times K_2$ is a cylinder as Figure 1. Hence $A\cong D_{2^n{p}^m}\times {\mathbb{Z}}_2$ .

(1.2) When $k=2^{n-2}{p}^m$ , X is a thickened 2-cover of the cycle graph $C_{2^{n-1}{p}^m}$ as Figure 2. All 4-cycles in X form an imprimitive block system of A and the

Figure 1. $X=Cay(G,\{b,ab,a^{2}b\})$ .

Figure 2. $X=Cay(G,\{b,ab,a^{2^{n-2}{p}^m}b\})$ .

kernel of the action of A on the imprimitive block system is isomorphic to ${\mathbb{Z}}_2^{2^{n-2}{p}^m}$ . Thus $A\cong Z_2^{2^{n-2}{p}^m}⋊D_{2^{n-1}{p}^m}$ .

(2) Suppose that there is no 4-cycle in X. We will count 6-cycles passing through vertex 1.

$X_3(1)=\{a^{-k}b,a^{-1}b,a^{1-k}b,a^{2-k}b,a^{k-1}b,a^{k+1}b,a^{2}b,a^{2k-1}b,a^{2k}b\}$ is the set of

vertices at distance 3 from vertex 1.

a) Solving $a^{2k}b=a^{2-k}b$ and $a^{2k-1}b=a^{1-k}b$ , we get $3k\equiv 2(\mathrm{mod}{2}^{n-1}{p}^m)$ . Solving $a^{2k}b=a^{1-k}b$ and $a^{2k-1}b=a^{-k}b$ , we get $3k\equiv 1(\mathrm{mod}{2}^{n-1}{p}^m)$ . Solving $a^{k-1}b=a^{2}b$ and $a^{-1}b=a^{2-k}b$ we get $k=3$ . Solving $a^{-k}=a^2$ and $a^{k+1}=a^{-1}$ we get $k=-2$ . There is no solution for other equations.

The induced subgraph of the set of vertices at distance less than or equal to 3 from vertex 1 in X are isomorphic in these four cases. The following uses $Cay\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{,\; <mo>\; \{\; </mo>}b\mathrm{,}ab\mathrm{,}{a}^{3}b\mathrm{<mo>\; \}\; </mo>\; <mo\; stretchy="false">\; )\; </mo>}$ as representative to discuss. See Figure 3.

We count the number of 6-cycles passing through vertex 1. There are four 6-cycles through edge $\mathrm{<mo>\; \{\; </mo>1,}b\mathrm{<mo>\; \}\; </mo>}$ . There are five 6-cycles through edge $\mathrm{<mo>\; \{\; </mo>1,}ab\mathrm{<mo>\; \}\; </mo>}$ . There are three 6-cycles through edge $\mathrm{<mo>\; \{\; </mo>1,}{a}^{3}b\mathrm{<mo>\; \}\; </mo>}$ . For any $\sigma \in A_1$ , $A_1$ fixes edged $\mathrm{<mo>\; \{\; </mo>1,}b\mathrm{<mo>\; \}\; </mo>,\; <mo>\; \{\; </mo>1,}ab\mathrm{<mo>\; \}\; </mo>,\; <mo>\; \{\; </mo>1,}{a}^{3}b\mathrm{<mo>\; \}\; </mo>}$ and hence $\sigma$ fixes vertices set $X_1(1)=\{b,ab,a^{3}b\}$ pointwise. $\sigma$ fixes all vertices on X by the connectivity of X and the transitivity of A on $V\mathrm{<mo\; stretchy="false">\; (\; </mo>}X\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ . Hence $A_1=1$ . X is GRR.

b) Suppose that $k\cancel{\equiv }3$ , $k\cancel{\equiv }-2$ , $3k\cancel{\equiv }2$ , $3k\cancel{\equiv }1$ (mod $2^{n-1}{p}^m$ ). Then the induced subgraph of the set of vertices at distance less than or equal to 3 from vertex 1 in X is the as Figure 4.

Firstly, show that the action of $A_1$ on $X_1\mathrm{<mo\; stretchy="false">\; (\; </mo>1\; <mo\; stretchy="false">\; )\; </mo>}$ is faithful.

Let $\sigma \in A_1$ and $\sigma$ fixes $X_1\mathrm{<mo\; stretchy="false">\; (\; </mo>1\; <mo\; stretchy="false">\; )\; </mo>}$ pointwise. Passing through vertices $\mathrm{<mo>\; \{\; </mo>1,}b\mathrm{,}ab\mathrm{<mo>\; \}\; </mo>}$ ,

Figure 3. $X=\text{inducedsubgraphof}Cay(G,\{b,ab,a^{3}b\})$ .

Figure 4. $X=\text{inducedsubgraph}Cay(G,\{b,ab,a^{k}b\})$ .

there is a unique 6-cycle $[1,b,a^k,a^{1-k}b,a^{k-1},ab]\triangleq C_1$ . Passing through vertices $\mathrm{<mo>\; \{\; </mo>1,}b\mathrm{,}{a}^{k}b\mathrm{<mo>\; \}\; </mo>}$ , there is a unique 6-cycle $[1,b,a,a^{k-1}b,a^{1-k},a^{k}b]\triangleq C_2$ . Passing through vertices $\mathrm{<mo>\; \{\; </mo>1,}ab\mathrm{,}{a}^{k}b\mathrm{<mo>\; \}\; </mo>}$ , there is a unique 6-cycle $[1,ab,a^{-1},a^{k+1}b,a^{-k},a^{k}b]\triangleq C_3$ . For any $\alpha \in A$ , the image of a cycle of length l under $\alpha$ is also a cycle of length l. Note that $\sigma \in A_1$ fixes $\mathrm{<mo>\; \{\; </mo>1,}b\mathrm{,}ab\mathrm{,}{a}^{k}b\mathrm{<mo>\; \}\; </mo>}$ pointwise, hence $C_1^{\sigma }$ is also a 6-cycle passing through vertices $\mathrm{1,}b\mathrm{,}ab$ . Hence $C_1^{\sigma }=C_1$ . Follow the same argument, $C_2^{\sigma }=C_2,C_3^{\sigma }=C_3$ . So $\sigma$ fixes all vertices on cycles $C_1\mathrm{,}{C}_2\mathrm{,}{C}_3$ . In particular, $\sigma$ fixes $X_2\mathrm{<mo\; stretchy="false">\; (\; </mo>1\; <mo\; stretchy="false">\; )\; </mo>}$ pointwise. By the connectivity of X and the transitivity of A on $V\mathrm{<mo\; stretchy="false">\; (\; </mo>}X\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ , we get $A_1$ acts on $X_1(1)=S$ faithfully.

Next, show that X is normal.

$A_1$ acting on $X_1\mathrm{<mo\; stretchy="false">\; (\; </mo>1\; <mo\; stretchy="false">\; )\; </mo>}$ faithfully implies that $A_1$ is isomorphic to a subgroup of symmetric group of degree 3. $A_1\lesssim S_3$ .

If $A_1\cong A_3$ or $S_3$ , then $A_1$ is transitive on $X_1\mathrm{<mo\; stretchy="false">\; (\; </mo>1\; <mo\; stretchy="false">\; )\; </mo>}$ . Since $\mathrm{<mo>\; |\; </mo>}{X}_1\mathrm{<mo\; stretchy="false">\; (\; </mo>1\; <mo\; stretchy="false">\; )\; </mo>\; <mo>\; |\; </mo>}=3$ is prime, X is a locally-primitive Cayley graph. Theorem 1.5 in [7] gives a classification of locally primitive Cayley graphs of dihedral groups which has been listed as Proposition 2.5 in this paper.

Since the order of G is $2^n{p}^m$ where $n\ge 2$ and p is odd, $Cay\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{,}S\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ is not on the list of locally-primitive Cayley graphs. Thus, $A_1$ is not transitive on $X_1\mathrm{<mo\; stretchy="false">\; (\; </mo>1\; <mo\; stretchy="false">\; )\; </mo>}$ . $A_1\cong {\mathbb{Z}}_1$ or ${\mathbb{Z}}_2$ . $|A:R(G)|=\left|A_1\right|=1$ or 2, $R(G)⊴A$ . X is normal. $A=R\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{<mo\; stretchy="false">\; )\; </mo>}⋊Aut\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{,}S\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ .

By Proposition 3.2 and part(1) of this proof, $A=R\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{<mo\; stretchy="false">\; )\; </mo>\; <mo>\; :\; </mo>}{\mathbb{Z}}_2$ if $k^2\equiv 1(\mathrm{mod}{2}^{n-1}{p}^m)$ , $k\ne 2^{n-2}{p}^m+1$ and $\gcd \mathrm{<mo\; stretchy="false">\; (\; </mo>}k{\mathrm{,2}}^{n-1}{p}^m\mathrm{<mo\; stretchy="false">\; )\; </mo>}=1$ or ${\mathrm{<mo\; stretchy="false">\; (\; </mo>1}-k\mathrm{<mo\; stretchy="false">\; )\; </mo>}}^2\equiv 1(\mathrm{mod}{2}^{n-1}{p}^m)$ , $k\ne 2^{n-2}{p}^m$ and $\gcd \mathrm{<mo\; stretchy="false">\; (\; </mo>1}-k{\mathrm{,2}}^{n-1}{p}^m\mathrm{<mo\; stretchy="false">\; )\; </mo>}=1$ .

Theorem 4.2. Suppose that $S\equiv \mathrm{<mo>\; \{\; </mo>}a\mathrm{,}{a}^{-1}\mathrm{,}b\mathrm{<mo>\; \}\; </mo>}$ , then X is normal and $A=G\mathrm{<mo>\; :\; </mo>}{\mathbb{Z}}_2$ .

Proof Suppose that $S\equiv \mathrm{<mo>\; \{\; </mo>}a\mathrm{,}{a}^{-1}\mathrm{,}b\mathrm{<mo>\; \}\; </mo>}$ and $X=Cay\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{,}S\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ . Cayley graph X is also a cylinder as Figure 5. Hence $A=D_{2^n{p}^m}\times {\mathbb{Z}}_2$ .

Theorem 4.3. Suppose that $S\equiv \mathrm{<mo>\; \{\; </mo>}b\mathrm{,}ab\mathrm{,}{a}^{2^{n-2}{p}^m}\mathrm{<mo>\; \}\; </mo>}$ , then X is normal and $A=G\mathrm{<mo>\; :\; </mo>}{\mathbb{Z}}_2$ .

Proof Suppose that $S\equiv \mathrm{<mo>\; \{\; </mo>}b\mathrm{,}ab\mathrm{,}{a}^{2^{n-2}{p}^m}\mathrm{<mo>\; \}\; </mo>}$ and $X=Cay\mathrm{<mo\; stretchy="false">\; (\; </mo>}G\mathrm{,}S\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ . The Cayley graph is an Möbius ladder as Figure 6. Hence, $A\text{=}{D}_{2^n{p}^m}⋊{\mathbb{Z}}_2$ .

Figure 5. $X=Cay(G,\{a,a^{-1},b\})$

Figure 6. $X=Cay(G,\{b,a{b}^{-1},a^{2^{n-2}}{p}^m\})$

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.

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