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

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Kong, X. (2020) Automorphism Groups of Cubic Cayley Graphs of Dihedral Groups of Order 2npm (n ≥ 2 and p Odd Prime). Journal of Applied Mathematics and Physics, 8, 3075-3084. doi: 10.4236/jamp.2020.812226.

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 $\left\{{u}^{\sigma },{v}^{\sigma }\right\}$ is an edge of X if and only if $\left\{u,v\right\}$ is the edge of X. As usual, we use ${u}^{\sigma }$ to denote the image of the vertex u under the permutation $\sigma$ and $\left\{u,v\right\}$ 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 $V\left(X\right)$ and $E\left(X\right)$ . ${A}_{v}$ is the stabilizer of vertex $v$ in the automorphism group of X. ${X}_{k}\left(v\right)$ 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 $V\left(X\right)$ . An s-arc in a graph is an ordered $\left(s+1\right)$ -tuple $\left({v}_{0},{v}_{1},...,{v}_{s-1},{v}_{s}\right)$ 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 $X=Cay\left(G,S\right)$ on G with respect to S is defined to have vertex set $V\left(X\right)=G$ and edge set $E\left(X\right)=\left\{\left\{g,sg\right\}|g\in G\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}s\in S\right\}$ . Let set ${S}^{-1}=\left\{{s}^{-1}|s\in S\right\}$ . If ${S}^{-1}=S$ , $Cay\left(G,S\right)$ is undirected. If S is a generating system of G, $Cay\left(G,S\right)$ 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 $\alpha \in Aut\left(G\right)$ . Denote by $S\equiv T$ . If S and T are equivalent, Cayley graphs $Cay\left(G,S\right)$ and $Cay\left(G,S\right)$ are isomorphic.

The right regular representation $R\left(G\right)$ of group G is a subgroup of the the automorphism group A of the Cayley graph X. In particular by [1], if $R\left(G\right)$ is the full automorphism group of X then $X=Cay\left(G,S\right)$ is called a GRR (for graphical regular representation) of G. A Cayley graph is normal if $R\left(G\right)$ is a normal subgroup of A. $R\left(G\right)$ is transitive on G hence Cayley graph is vertex-transitive. Denote $Aut\left(G,S\right)=\left\{\alpha \in Aut\left(G\right)|{S}^{\alpha }=S\right\}$ , the set of all automorphism of group G preserving S. $Aut\left(G,S\right)$ is also a subgroup of the automorphism group of Cayley graph. In particular, $Aut\left(G,S\right)$ is a subgroup of stabilizer of vertex identity ${A}_{1}$ . By [2] the normalizer of $R\left(G\right)$ in A is the semi-direct product of $R\left(G\right)$ and $Aut\left(G,S\right)$ : ${N}_{A}\left(R\left(G\right)\right)=R\left(G\right)⋊Aut\left(G,S\right)$ . By [3] Proposition 1.5 X is normal if and only if ${A}_{1}=Aut\left(G,S\right)$ . Cayley graph X is normal if and only if the automorphism group of X is $A=R\left(G\right)⋊Aut\left(G,S\right)$ . 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 $Cay\left(G,S\right)$ is GRR except the following cases:

1) $S\equiv \left\{b,ab,{a}^{k}b\right\}$ where ${k}^{2}\equiv 1\left(\mathrm{mod}{2}^{n-1}{p}^{m}\right)$ and $\mathrm{gcd}\left(k{,2}^{n-1}{p}^{m}\right)=1$ , $Aut\left(X\right)\cong G:{ℤ}_{2}$ .

2) $S\equiv \left\{b,ab,{a}^{{2}^{n-1}{p}^{m}}b\right\}$ , $Aut\left(X\right)\cong {ℤ}_{2}^{{2}^{n-2}{p}^{m}}⋊{D}_{{2}^{n-1}{p}^{m}}$ .

3) $S\equiv \left\{a,{a}^{-1},b\right\}$ , $Aut\left(X\right)=G:{ℤ}_{2}$ .

4) $S\equiv \left\{b,ab,{a}^{{2}^{n-2}{p}^{m}}\right\}$ , $Aut\left(X\right)=G:{ℤ}_{2}$ .

2. Preliminary

Results used to prove main theorem are listed here.

Proposition 2.1. Suppose that $G=$ is a dihedral group, then the automorphism group $Aut\left(G\right)$ of G has the following properties.

1) Any automorphism of G can be defined as $a↦{a}^{i}$ and $b↦{a}^{j}b$ where $i\in {ℤ}_{n}^{*}$ and $j\in {ℤ}_{n}$ .

2) $Aut\left(G\right)=<\alpha >⋊<\beta >\cong {ℤ}_{n}⋊{ℤ}_{n}^{*}$ where $\alpha :a↦a,b↦ab;\beta :a↦{a}^{i},b↦b,i\in {ℤ}_{n}^{*}$ .

Proposition 2.2. Suppose G is a finite group and subsets $S\equiv T$ , then $Cay\left(G,S\right)\cong Cay\left(G,T\right)$ .

Proposition 2.3. Let $G=$ be the dihedral group of order 2n. Subsets $\left\{b,ab,{a}^{k}b\right\}\equiv \left\{b,ab,{a}^{1-k}b\right\}$ .

Proof Let $\sigma \in Aut\left(G\right):a↦{a}^{-1},b↦ab$ then ${\left\{b,ab,{a}^{k}b\right\}}^{\sigma }=\left\{b,ab,{a}^{1-k}b\right\}$ .

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

Proposition 2.4. Let $X=Cay\left(G,S\right)$ 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 $\sigma \in Aut\left(G\right)$ such that $\phi {|}_{{X}_{1}\left(1\right)}=\sigma {|}_{{X}_{1}\left(1\right)}$ ;

2) For each $\phi \in {A}_{1}$ , $\phi {|}_{{X}_{1}\left(1\right)}={1}_{{X}_{1}\left(1\right)}$ implies $\phi {|}_{{X}_{2}\left(1\right)}={1}_{{X}_{2}\left(1\right)}$ .

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 ${K}_{n,n}-n{K}_{2}$ ;

b) $X=\mathcal{H}\mathcal{D}\left(11,5,2\right)$ or $\mathcal{H}\mathcal{D}\left(11,6,2\right)$ , the incidence or non-incidence graph of the Hadamard design on 11 points;

c) $X=\mathcal{P}\mathcal{H}\left(d,q\right)$ or $\mathcal{P}{\mathcal{H}}^{\prime }\left(d,q\right)$ , the point-hyperplane incidence or non-incidence graph of $\left(d-1\right)$ -dimension projective geometry $PG\left(d-1,q\right)$ , 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) $X=\mathcal{N}{\mathcal{D}}_{2n,r,k}$ 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 $r,{p}_{1},{p}_{2},\cdots ,{p}_{s}$ distinct odd primes, $t\le 1$ , $s\ge 1$ and $r|\left({p}_{i}-1\right)$ for each i. There are exactly ${\left(r-1\right)}^{s-1}$ non-isomorphism such graphs for a given order 2n.

3. Lemmas and Propositions

In the following, group G means that $G=$ be dihedral group of order ${2}^{n}{p}^{m}$ where $n\ge 2$ and p is an odd prime number.

Proposition 3.1. If $S=\left\{{a}^{i}b,{a}^{j}b,{a}^{r}b\right\}$ is a generating system of G of three elements, then $S\equiv \left\{b,ab,{a}^{k}b\right\}$ 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, $S\equiv \left\{b,ab,{a}^{k}b\right\}$ where $\mathrm{gcd}\left(k{,2}^{n-1}{p}^{m}\right)=1$ .

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

Lemma 3.1. If $S=\left\{{a}^{i}b,{a}^{j}b,{a}^{r}b\right\}$ is a generating system of G of three elements, then S is equivalent to a subset of type $\left\{b,ab,{a}^{k}b\right\}$ 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 $\left\{{a}^{i}b|0\le i\le {2}^{n-1}{p}^{m}-1\right\}$ . One may assume that $b\in S$ and $S=\left\{b,{a}^{i}b,{a}^{j}b\right\}$ be a generating system of G of three elements. S has three subsets of two elements: $\left\{b,{a}^{i}b\right\},\left\{b,{a}^{j}b\right\}$ and $\left\{{a}^{i}b,{a}^{j}b\right\}$ ..

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 $\left\{b,{a}^{x}b\right\}$ ( $x=i$ or $j$ ) generates G. Let $\alpha \in Aut\left(G\right)$ : $a↦{a}^{x},b↦b$ , then ${\left\{b,ab,{a}^{k}b\right\}}^{\alpha }=\left\{b,{a}^{i}b,{a}^{j}b\right\}$ for some $k\ne 0,1$ . Hence $S\equiv \left\{b,ab,{a}^{k}b\right\}$ .

Assume that both subset $\left\{b,{a}^{i}b\right\}$ and $\left\{b,{a}^{j}b\right\}$ do not generate G. Next will show that $\left\{{a}^{i}b,{a}^{j}b\right\}$ must be able to generate G.

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

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

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

$\left(\mathrm{gcd}\left(i,j\right),{2}^{n-1}{p}^{m}\right)=1$ , $\left(i{,2}^{n-1}{p}^{m}\right)\ne 1$ and $\left(j{,2}^{n-1}{p}^{m}\right)\ne 1$ 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, $\left\{{a}^{i}b,{a}^{j}b\right\}$ is a generating system of G since $<{a}^{i}b,{a}^{j}b>=<{a}^{i-j}><{a}^{i}b>=G$ .

Let $\alpha \in Aut\left(G\right)$ : $a↦{a}^{i-j},b↦{a}^{j}b$ . Then ${\left\{b,ab,{a}^{k}b\right\}}^{\alpha }=\left\{b,{a}^{i}b,{a}^{j}b\right\}$ for some k. $S\equiv \left\{b,ab,{a}^{k}b\right\}$ .

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

Lemma 3.2. If $S=\left\{{a}^{i}b,{a}^{j}b,{a}^{r}b\right\}$ 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 $S=\left\{b,ab,{a}^{k}b\right\}$ where $k\ne 0,1$ . S has three subsets of two elements: $\left\{b,ab\right\},\left\{b,{a}^{k}b\right\}$ and $\left\{ab,{a}^{k}b\right\}$ . Next we will show that it is impossible that all three subsets of two elements generating G.

$=<{a}^{k}>$ is a dihedral subgroup of G. $=<{a}^{k-1}>$ 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: $\circ \left({a}^{k}\right)\ne \circ \left({a}^{k-1}\right)$ . This implies that at least one subset of $\left\{b,{a}^{k}b\right\}$ and $\left\{ab,{a}^{k}b\right\}$ does not generate G.

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

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

Proposition 3.2. Suppose that $S=\left\{{a}^{i}b,{a}^{j}b,{a}^{r}b\right\}$ is a generating system of G of three elements and $S\equiv \left\{b,ab,{a}^{k}b\right\}$ .

(1) If S has only one subset of two elements generating G, then $Aut\left(G,S\right)=1$ .

(2) If S has two subsets of two elements generating G, then $Aut\left(G,S\right)=1$ except the following two cases. $Aut\left(G,S\right)\cong {ℤ}_{2}$ if ${k}^{2}\equiv 1\left(\mathrm{mod}{2}^{n-1}{p}^{m}\right)$ and $\mathrm{gcd}\left(k{,2}^{n-1}{p}^{m}\right)=1$ ; $Aut\left(G,S\right)\cong {ℤ}_{2}$ if ${\left(1-k\right)}^{2}\equiv 1\left(\mathrm{mod}{2}^{n-1}{p}^{m}\right)$ and $\mathrm{gcd}\left(1-k{,2}^{n-1}{p}^{m}\right)=1$ .

Proof (1) If there is only one subset of two elements in $S=\left\{b,ab,{a}^{k}b\right\}$ generating G, then $G\ne $ , $G\ne $ and $G=$ . For any $\sigma \in Aut\left(G,S\right)$ , ${\left\{b,ab\right\}}^{\sigma }$ is also a generating system of G. ${\left\{b,ab\right\}}^{\sigma }=\left\{b,ab\right\}$ . Since ${S}^{\sigma }=S$ . Hence ${a}^{k}b=S-\left\{b,ab\right\}$ is fixed by $\sigma$ . ${\left({a}^{k}b\right)}^{\sigma }={a}^{k}b$ .

If ${b}^{\sigma }=b$ and ${\left(ab\right)}^{\sigma }=ab$ then ${a}^{\sigma }={\left(abb\right)}^{\sigma }={\left(ab\right)}^{\sigma }{b}^{\sigma }=abb=a$ , hence $\sigma =1$ .

If ${b}^{\sigma }=ab$ and ${\left(ab\right)}^{\sigma }=b$ , then ${a}^{\sigma }={\left(abb\right)}^{\sigma }={\left(ab\right)}^{\sigma }{b}^{\sigma }=bab={a}^{-1}$ . This implies that ${a}^{k}b={\left({a}^{k}b\right)}^{\sigma }={\left({a}^{k}\right)}^{\sigma }{b}^{\sigma }={a}^{-k}ab={a}^{1-k}b$ . 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: $\circ \left({a}^{k}\right)\ne \circ \left({a}^{1-k}\right)$ .

Hence $Aut\left(G,S\right)=1$ .

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

Since subset $\left\{ab,{a}^{k}b\right\}$ is the only subset of two elements not generating G, ${\left\{ab,{a}^{k}b\right\}}^{\sigma }=\left\{ab,{a}^{k}b\right\}$ for any $\sigma \in Aut\left(G,S\right)$ . $b=S-\left\{ab,{a}^{k}b\right\}$ is fixed by $\sigma$ . ${\left(ab\right)}^{\sigma }=ab$ or ${a}^{k}b$ .

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

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

Hence $Aut\left(G,S\right)=1$ if ${k}^{2}\overline{)\equiv }1\left(\mathrm{mod}{2}^{n-1}{p}^{m}\right)$ . $Aut\left(G,S\right)\cong {ℤ}_{2}$ if ${k}^{2}\equiv 1\left(\mathrm{mod}{2}^{n-1}{p}^{m}\right)$ .

Similarly, when $\mathrm{gcd}\left(1-k,{2}^{n-1}{p}^{m}\right)=1$ , $Aut\left(G,S\right)=1$ if ${\left(1-k\right)}^{2}\overline{)\equiv }1\left(\mathrm{mod}{2}^{n-1}{p}^{m}\right)$ . $Aut\left(G,S\right)\cong {ℤ}_{2}$ if ${\left(1-k\right)}^{2}\equiv 1\left(\mathrm{mod}{2}^{n-1}{p}^{m}\right)$ .

Proposition 3.3. Suppose that S is inverse-closed generating system of three elements of G, then $S\equiv \left\{a,{a}^{-1},b\right\}$ , $\left\{b,ab,{a}^{{2}^{n-2}{p}^{m}}\right\}$ or $\left\{b,ab,{a}^{k}b\right\}\left(k\ne 0,1\right)$ .

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 $Aut\left(G\right)$ : $\left\{{a}^{{2}^{n-2}{p}^{m}}\right\}$ and $\left\{{a}^{i}b|0\le i\le {2}^{n-1}{p}^{m}-1\right\}$ .

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

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

4. Results

By Proposition 3.3, we only need to discuss $X=Cay\left(G,S\right)$ for $S=\left\{a,{a}^{-1},b\right\},\left\{b,ab,{a}^{{2}^{n-2}{p}^{m}}\right\}$ and $\left\{b,ab,{a}^{k}b\right\}\left(k\ne 0,1\right)$ .

Firstly, we discuss $X=Cay\left(G,\left\{b,ab,{a}^{k}b\right\}\right)\left(k\ne 0,1\right)$ .

Theorem 4.1. Suppose that $S=\left\{{a}^{i}b,{a}^{j}b,{a}^{m}b\right\}$ is a generating system of three involutions of G and $S\equiv \left\{b,ab,{a}^{k}b\right\}$ .

X is GRR except the following cases.

(1) When $\mathrm{gcd}\left(k{,2}^{n-2}{p}^{m}\right)=1$ , ${k}^{2}\equiv 1\left(\mathrm{mod}{2}^{n-1}{p}^{m}\right)$ and $k\ne {2}^{n-2}{p}^{m}+1$ then $Aut\left(X\right)\cong R\left(G\right):{ℤ}_{2}$ .

(2) When $\mathrm{gcd}\left(1-k{,2}^{n-2}{p}^{m}\right)=1$ , ${\left(1-k\right)}^{2}\equiv 1\left(\mathrm{mod}{2}^{n-1}{p}^{m}\right)$ and $k\ne {2}^{n-2}{p}^{m}$ then $Aut\left(X\right)\cong R\left(G\right):{ℤ}_{2}$ .

(3) If $k={2}^{n-2}{p}^{m}+1$ or $k={2}^{n-2}{p}^{m}$ , then $Aut\left(X\right)\cong {ℤ}_{2}^{{2}^{n-2}{p}^{m}}⋊{D}_{{2}^{n-1}{p}^{m}}$ .

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

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

If there are 4-cycles in X, some vertices in ${X}_{2}\left(1\right)$ 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 ${k}^{2}\equiv 1\left(\mathrm{mod}{2}^{n-1}{p}^{m}\right)$ . 2 and ${2}^{n-2}{p}^{m}$ are two solutions of equation ${\left(1-k\right)}^{2}\equiv 1\left(\mathrm{mod}{2}^{n-1}{p}^{m}\right)$ . Since $\left\{b,ab,{a}^{2}b\right\}\equiv \left\{b,ab,{a}^{-1}b\right\}$ and $\left\{b,ab,{a}^{{2}^{n-2}{p}^{m}}b\right\}\equiv \left\{b,ab,{a}^{{2}^{n-2}{p}^{m}+1}b\right\}$ we only discuss $k=2$ and $k={2}^{n-2}{p}^{m}$ .

(1.1) When $k=2$ , $X={C}_{{2}^{n-1}{p}^{m}}×{K}_{2}$ is a cylinder as Figure 1. Hence $A\cong {D}_{{2}^{n}{p}^{m}}×{ℤ}_{2}$ .

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

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

kernel of the action of A on the imprimitive block system is isomorphic to ${ℤ}_{2}^{{2}^{n-2}{p}^{m}}$ . 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.

${X}_{3}\left(1\right)=\left\{{a}^{-k}b,{a}^{-1}b,{a}^{1-k}b,{a}^{2-k}b,{a}^{k-1}b,{a}^{k+1}b,{a}^{2}b,{a}^{2k-1}b,{a}^{2k}b\right\}$ 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 $3k\equiv 2\left(\mathrm{mod}{2}^{n-1}{p}^{m}\right)$ . Solving ${a}^{2k}b={a}^{1-k}b$ and ${a}^{2k-1}b={a}^{-k}b$ , we get $3k\equiv 1\left(\mathrm{mod}{2}^{n-1}{p}^{m}\right)$ . 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 $Cay\left(G,\left\{b,ab,{a}^{3}b\right\}\right)$ 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 $\left\{1,b\right\}$ . There are five 6-cycles through edge $\left\{1,ab\right\}$ . There are three 6-cycles through edge $\left\{1,{a}^{3}b\right\}$ . For any $\sigma \in {A}_{1}$ , ${A}_{1}$ fixes edged $\left\{1,b\right\},\left\{1,ab\right\},\left\{1,{a}^{3}b\right\}$ and hence $\sigma$ fixes vertices set ${X}_{1}\left(1\right)=\left\{b,ab,{a}^{3}b\right\}$ pointwise. $\sigma$ fixes all vertices on X by the connectivity of X and the transitivity of A on $V\left(X\right)$ . Hence ${A}_{1}=1$ . X is GRR.

b) Suppose that $k\overline{)\equiv }3$ , $k\overline{)\equiv }-2$ , $3k\overline{)\equiv }2$ , $3k\overline{)\equiv }1$ (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 ${X}_{1}\left(1\right)$ is faithful.

Let $\sigma \in {A}_{1}$ and $\sigma$ fixes ${X}_{1}\left(1\right)$ pointwise. Passing through vertices $\left\{1,b,ab\right\}$ ,

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

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

there is a unique 6-cycle $\left[1,b,{a}^{k},{a}^{1-k}b,{a}^{k-1},ab\right]\triangleq {C}_{1}$ . Passing through vertices $\left\{1,b,{a}^{k}b\right\}$ , there is a unique 6-cycle $\left[1,b,a,{a}^{k-1}b,{a}^{1-k},{a}^{k}b\right]\triangleq {C}_{2}$ . Passing through vertices $\left\{1,ab,{a}^{k}b\right\}$ , there is a unique 6-cycle $\left[1,ab,{a}^{-1},{a}^{k+1}b,{a}^{-k},{a}^{k}b\right]\triangleq {C}_{3}$ . 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 $\left\{1,b,ab,{a}^{k}b\right\}$ pointwise, hence ${C}_{1}^{\sigma }$ is also a 6-cycle passing through vertices $1,b,ab$ . 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 ${C}_{1},{C}_{2},{C}_{3}$ . In particular, $\sigma$ fixes ${X}_{2}\left(1\right)$ pointwise. By the connectivity of X and the transitivity of A on $V\left(X\right)$ , we get ${A}_{1}$ acts on ${X}_{1}\left(1\right)=S$ faithfully.

Next, show that X is normal.

${A}_{1}$ acting on ${X}_{1}\left(1\right)$ 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 ${X}_{1}\left(1\right)$ . Since $|{X}_{1}\left(1\right)|=3$ 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, $Cay\left(G,S\right)$ is not on the list of locally-primitive Cayley graphs. Thus, ${A}_{1}$ is not transitive on ${X}_{1}\left(1\right)$ . ${A}_{1}\cong {ℤ}_{1}$ or ${ℤ}_{2}$ . $|A:R\left(G\right)|=|{A}_{1}|=1$ or 2, $R\left(G\right)⊴A$ . X is normal. $A=R\left(G\right)⋊Aut\left(G,S\right)$ .

By Proposition 3.2 and part(1) of this proof, $A=R\left(G\right):{ℤ}_{2}$ if ${k}^{2}\equiv 1\left(\mathrm{mod}{2}^{n-1}{p}^{m}\right)$ , $k\ne {2}^{n-2}{p}^{m}+1$ and $\mathrm{gcd}\left(k{,2}^{n-1}{p}^{m}\right)=1$ or ${\left(1-k\right)}^{2}\equiv 1\left(\mathrm{mod}{2}^{n-1}{p}^{m}\right)$ , $k\ne {2}^{n-2}{p}^{m}$ and $\mathrm{gcd}\left(1-k{,2}^{n-1}{p}^{m}\right)=1$ .

Theorem 4.2. Suppose that $S\equiv \left\{a,{a}^{-1},b\right\}$ , then X is normal and $A=G:{ℤ}_{2}$ .

Proof Suppose that $S\equiv \left\{a,{a}^{-1},b\right\}$ and $X=Cay\left(G,S\right)$ . Cayley graph X is also a cylinder as Figure 5. Hence $A={D}_{{2}^{n}{p}^{m}}×{ℤ}_{2}$ .

Theorem 4.3. Suppose that $S\equiv \left\{b,ab,{a}^{{2}^{n-2}{p}^{m}}\right\}$ , then X is normal and $A=G:{ℤ}_{2}$ .

Proof Suppose that $S\equiv \left\{b,ab,{a}^{{2}^{n-2}{p}^{m}}\right\}$ and $X=Cay\left(G,S\right)$ . The Cayley graph is an Möbius ladder as Figure 6. Hence, $A\text{=}{D}_{{2}^{n}{p}^{m}}⋊{ℤ}_{2}$ .

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

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

Conflicts of Interest

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

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