The cubic power graph of finite abelian groups

Abstract

Let G be a finite abelian group with identity 0. For an integer $n\ge 2,$ the additive power graph ${\Gamma }_{\mathit{apq}}(G)$ of G is the simple undirected graph with vertex set G in which two distinct vertices x and y are adjacent if and only if x + y = nt for some $t\in G$ with $nt\ne 0.$ When $n=2,$ the additive power graph has been studied in the name of square graph of finite abelian groups. In this paper, we study the additive power graph of G with n = 3 and name the graph as the cubic power graph. The cubic power graph of G is denoted by ${\Gamma }_{\mathit{cpg}}(G).$ More specifically, we obtain the diameter and the girth of the graph ${\Gamma }_{\mathit{cpg}}(G)$ and its complement ${\overline{\Gamma }}_{\mathit{cpg}}(G).$ Using these, we obtain a condition for ${\Gamma }_{\mathit{cpg}}(G)$ and its complement ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ to be self-centered. Also, we obtain the independence number, the clique number and the chromatic number of ${\Gamma }_{\mathit{cpg}}(G)$ and its complement ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ and hence we prove that ${\Gamma }_{\mathit{cpg}}(G)$ and its complement ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ are weakly perfect. Also, we discuss about the perfectness of ${\Gamma }_{\mathit{cpg}}(G).$ At last, we obtain a condition for ${\Gamma }_{\mathit{cpg}}(G)$ and its complement ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ to be vertex pancyclic.

Keywords:

• Abelian group
• additive power graph
• cubic power graph
• complement
• perfect
• chromatic number

2000 Mathematics Subject Classification:

• 05C25
• 05C40
• 05C69

1. Introduction

The definition of Cayley graph was introduced by Arthur Cayley in 1878 to explain the concept of abstract groups which are described by a set of generators. Cayley graphs from finite groups are well studied in several aspects. Especially, Cayley graphs are used as communication grid in the routing problem due to its varied properties like regular and vertex transitive [6]. Several aspects of Cayley graphs are studied in [4, 7, 8, 13–17]. There are several other graph constructions from finite groups and rings [1, 5]. For a finite commutative ring R, the set of squares of R (elements of the form x² for some $x\in R$) is a very interesting subset from algebraic point of view. Sen Gupta and Sen [2, 10] introduced and studied about the square element graph $\mathbb{S}q(R)$ of R. In fact, the square element graph $\mathbb{S}q(R)$ of R is the simple undirected graph with vertex set $R^{*}=R\setminus \left\{0\right\}$ and two vertices a and b are adjacent if and only if $a+b=x^2$ for some $x\in R^{*}.$ Subsequently Sen Gupta and Sen [10, 11] studied about the square element graph of arbitrary rings (not necessarily commutative) and they also studied about the square element graph of semigroups in [3]. Snowden [12] studied about square roots in finite full transformation semigroups. For a finite abelian group G with identity 0, Raveendra Prathap and Tamizh Chelvam [9] introduced and studied about the square graph of G. The square graph ${\Gamma }_{sq}(G)$ of G is the simple undirected graph with vertex set G and two distinct vertices x and y are adjacent if and only if $x+y=2t$ for some $t\in G$ and $2t\ne 0.$ Having introduced the square graph ${\Gamma }_{sq}(G),$ Raveendra Prathap and Tamizh Chelvam [9] studied several properties of its complement. Now, the concept of the square graph is generalized as the additive power graph ${\Gamma }_{\mathit{apq}}(G)$ of G as the simple undirected graph with vertex set G in which two distinct vertices x and y are adjacent if and only if x + y = nt for some $t\in G,$ positive integer n and $nt\ne 0.$ Note that when $n=2,$ the additive power graph ${\Gamma }_{\mathit{apq}}(G)$ becomes the square graph of ${\Gamma }_{sq}(G).$ Actually the square graph and cubic power graph of a finite abelian group G are generalizations of the Cayley graph on G [6].

In this paper, we study about the additive power graph ${\Gamma }_{\mathit{apq}}(G)$ of a finite abelian group G where $n=3.$ That is we study about the cubic power ${\Gamma }_{\mathit{cpg}}(G)$ graph of finite abelian groups. More specifically, we obtain the diameter and the girth of ${\Gamma }_{\mathit{cpg}}(G)$ and its complement ${\overline{\Gamma }}_{\mathit{cpg}}(G).$ Further we obtain a necessary and sufficient condition for ${\Gamma }_{\mathit{cpg}}(G)$ and its complement ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ to be self-centered. Also, we obtain the independence number, the clique number and the chromatic number of ${\Gamma }_{\mathit{cpg}}(G)$ and its complement ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ and hence we prove that ${\Gamma }_{\mathit{cpg}}(G)$ and its complement ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ are weakly perfect. Also, we discuss about the perfectness of ${\Gamma }_{\mathit{cpg}}(G).$ At last, we obtain a condition for ${\Gamma }_{\mathit{cpg}}(G)$ and its complement ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ to be vertex pancyclic.

2. Preliminaries

First let us recollect some basic definitions of graph theory which are essential for this paper. By a graph $\Gamma =(V,E),$ we mean Γ is a finite graph with vertex set V and edge set E. A graph Γ is said to be complete if each pair of distinct vertices is joined by an edge. We use K_n to denote the complete graph with n vertices. A graph Γ is said to be connected if there exists a path between every pair of distinct vertices in $\Gamma .$ The distance d(u, v) between the vertices u and v in Γ is the length of the shortest path between u and v. If no path exists between u and v in $\Gamma ,$ then d(a, b) = $\infty .$ For a vertex $v\in V(\Gamma ),$ the eccentricity of v is the maximum distance from v to any vertex in $\Gamma .$ That is, $e(v)=\max \{d(v,w):w\in V(\Gamma )\}.$ The radius of Γ is the minimum eccentricity among the vertices of $\Gamma .$ i.e., radius$(\Gamma )=\min \{e(v):v\in V(G)\}.$ The diameter of Γ is the maximum eccentricity among the vertices of $\Gamma .$ i.e., diam$(\Gamma )=\max \{e(v):v\in V(G)\}.$ The girth of Γ is the length of a shortest cycle in Γ and is denoted by $gr(\Gamma ).$ A graph Γ is said to be self-centered if the eccentricity of every vertex of Γ is the same. In other words, a graph is a self-centered graph if radius and diameter of the graph are equal. A graph Γ with n vertices is called a refinement of the star graph if $K_{1,n-1}$ is a subgraph of Γ and Γ is not complete. For a vertex $v\in V(G),$ N(x) is the set of all vertices in G which are adjacent to v and $N[x]=N(x)\cup \left\{x\right\}.$

The degree of a vertex v is the number of the edges in Γ which are incident with v. A clique of Γ is a maximal complete subgraph of Γ and the number of vertices in the largest clique of Γ is called the clique number of Γ and is denoted by $\omega (\Gamma ).$ A graph Γ with n vertices is called pancyclic if it contains a cycle of length k for every $3\le k\le n,$ and it is called vertex pancyclic if every vertex is contained in a cycle of length k for every $3\le k\le n.$ An independent set is a set of vertices in a graph, in which no two vertices are adjacent and cardinality of a maximal independent set is called the independent number. A vertex coloring of Γ is a mapping $c:V(\Gamma )\to S.$ The elements of S are called colors. If $\left|S\right|=k,$ we say that c is a k-coloring. A coloring is said to be proper if adjacent vertices have different colors. A graph is k-colorable if it has a proper k-coloring. The chromatic number is the least k such that Γ is k-colorable [18, 19].

Let G be a finite abelian group and $\left|G\right|=3^{\alpha }\times p_1^{{\alpha }_1}\times \cdots \times p_r^{{\alpha }_r}$ where 3 and $p_i^{\prime }s$ are distinct primes and $\alpha ,{\alpha }_i\in {\mathbb{Z}}^{+}\cup \left\{0\right\}$ for $1\le i\le r.$ Then G is the direct product of finite cyclic groups. i.e., $G\cong {\mathbb{Z}}_{3^{\alpha }}\times {\prod }_{i=1}^r{\mathbb{Z}}_{p_i^{{\alpha }_i}}.$ A partition $P(\alpha )$ of a positive integer $\alpha ,$ also called an integer partition, is a way of writing $\alpha \ge 1$ as the sum of positive integers. We write a partition of $\alpha$ as $P(\alpha )=m_1+m_2+\dots +m_k$ with $m_i\ge 1.$ If $\left|G\right|$ is divisible by 3, then we have $G\cong {\prod }_{i=1}^k{\mathbb{Z}}_{3^{m_i}}\times H$ where ${\left\{0\right\}}^k\times H$ is a subgroup of G of order ${\prod }_{i=1}^r{p}_i^{{\alpha }_i}$ and $H\cong {\prod }_{i=1}^r{\mathbb{Z}}_{p_i^{{\alpha }_i}}.$

Remark 2.1.

For a finite abelian group G, let $G_1=\left\{3t\right|t\in G\}\subseteq G$ and $G_2=G\setminus G_1.$ Suppose $G\cong {\mathbb{Z}}_{3^{\alpha }}\times H.$ Then $G_1(G)=G_1({\mathbb{Z}}_{3^{\alpha }})\times H$ and $G_2(G)=G_2({\mathbb{Z}}_{3^{\alpha }})\times H.$ Note that two distinct vertices x and y are adjacent in ${\Gamma }_{\mathit{cpg}}(G)$ if and only if $x+y\in G_1\setminus \left\{0\right\}.$

Remark 2.2.

1. If $\left|G\right|$ is not divisible by 3, then G = G₁ and $G_2=\varphi .$

2. Assume that $\left|G\right|$ is divisible by 3 and so $\alpha \ge 1.$ Consider a partition $P(\alpha )=m_1+\cdots +m_k$ of $\alpha .$ Here $k\ge 1.$ Now, with respect to a partition of $\alpha ,$ one can associate $2^k,$ k-tuples of 0’s and 1’s. For a k-tuple, $\ell =(a_1,a_2,\dots ,a_k),$ let $$X_{\ell }=\prod _{i=1}^k{H}_i\times H\text{where}{H}_i=\{\begin{array}{cc}{G}_1({\mathbb{Z}}_{3^{m_i}})\hfill & \text{if}{a}_i=0;\hfill \\ G_2({\mathbb{Z}}_{3^{m_i}})\hfill & \text{if}{a}_i=1.\hfill \end{array}$$

3. One can check that in either case, $G_1=X_{(0,0,\dots ,0)}=X_1,G_2={\cup }_{i=2}^{2^k}{X}_i.$

4. One can verify that $\left|X_{(0,0,\dots ,0)}\right|=\frac{\left|G\right|}{3^k},\left|X_{(1,0,\dots ,0)}\right|=2\frac{\left|G\right|}{3^k},\left|X_{(1,1,\dots ,0)}\right|=2^2\frac{\left|G\right|}{3^k},\dots ,\left|X_{(1,1,\dots ,1)}\right|=2^k\frac{\left|G\right|}{3^k}.$

3. Properties of ${\Gamma }_{\mathrm{cpg}}(G)$

In this section, we list out some basic properties of the cubic power graph ${\Gamma }_{\mathit{cpg}}(G)$ of a finite abelian group G. In particular, we prove that ${\Gamma }_{\mathit{cpg}}(G)$ is connected and obtain the diameter, the girth, the independence number, the chromatic number and the clique number of ${\Gamma }_{\mathit{cpg}}(G).$ Using these parameters, we conclude that ${\Gamma }_{\mathit{cpg}}(G)$ is weakly perfect. At the end of the section, we prove that ${\Gamma }_{\mathit{cpg}}(G)$ is vertex pancyclic and not Eulerian.

Lemma 3.1.

Let G be a finite abelian group, $\left|G\right|=3^{\alpha }.t,$ $\alpha \ge 1,$ $\alpha =m_1+\cdots +m_k,$ and t be relatively prime to 3. If $x\in X_i$ and $y\in X_j$ for $i\ne j,$ then x is not adjacent to y in ${\Gamma }_{\mathit{cpg}}(G).$

Proof.

Let $x\in X_i$ and $y\in X_j$ for $i\ne j$ and $1\le i,j\le 2^k.$

Then $i=(a_1,\dots ,a_{\xi },\dots ,a_k)$ and $j=(b_1,\dots ,b_{\xi },\dots ,b_k)$ with $a_{\xi }\ne b_{\xi }$ for some $\xi ,1\le \xi \le k.$ Let $x=(x_1,\dots ,x_{\xi },\dots ,x_r)$ and $y=(y_1,\dots ,y_{\xi },\dots ,y_r).$

If $a_{\xi }=0,$ then $b_{\xi }=1.$ As per the definition of $X_i,$ we get that $x_{\xi }\in G_1({\mathbb{Z}}_{3^{m_{\xi }}})$ and $y_{\xi }\in G_2({\mathbb{Z}}_{3^{m_{\xi }}})$ and so $x_{\xi }+y_{\xi }\notin G_1({\mathbb{Z}}_{3^{m_{\xi }}})$ which implies that $x+y\notin G_1.$ Hence x is not adjacent to y in ${\Gamma }_{\mathit{cpg}}(G).$ □

In this following, we obtain the diameter and the girth of ${\Gamma }_{\mathit{cpg}}(G)$ where $\left|G\right|$ is not divisible by 3.

Theorem 3.2.

Let G be a finite abelian group and $\left|G\right|$ is not divisible by 3. Then

1. $de{g}_{{\Gamma }_{\mathit{cpg}}(G)}(x)=\{\begin{array}{cc}\left|G\right|-1\hfill & \mathit{\text{if}}x=-x\mathit{\text{for}}x\in G;\hfill \\ \left|G\right|-2\hfill & \mathit{\text{otherwise}}.\hfill \end{array}$

2. ${\Gamma }_{\mathit{cpg}}(G)$ is connected.

3. $\mathit{diam}({\Gamma }_{\mathit{cpg}}(G))=\{\begin{array}{cc}1\hfill & \mathit{\text{if}}\mathrm{G}\cong {\mathbb{Z}}_2^{\mathrm{k}}\mathit{\text{for}}\mathrm{}\mathit{\text{some}}\mathrm{}\mathit{\text{integer}}k\ge 1;\hfill \\ 2\hfill & \mathit{\text{otherwise}}.\hfill \end{array}$

4. $gr({\Gamma }_{\mathit{cpg}}(G))=\{\begin{array}{cc}\infty \hfill & \mathit{\text{if}}\mathrm{G}\cong {\mathbb{Z}}_2;\hfill \\ 3\hfill & \mathit{\text{otherwise}}.\hfill \end{array}$

5. ${\Gamma }_{\mathit{cpg}}(G)$ is a refinement of the star graph$K_{1,\left|G\right|-1}.$

Proof.

Assume that $\left|G\right|$ is not divisible by 3. As observed in Remark 2.2(1), we have G = G₁.

1. Assume that $x=-x$ for $x\in G.$ For every $y\in G,$ we have $x+y\in G_1\setminus \left\{0\right\}$ and so x is adjacent to all y in ${\Gamma }_{\mathit{cpg}}(G).$ Thus deg${}_{{\Gamma }_{\mathit{cpg}}(G)}(x)=\left|G\right|-1.$ Let $x\in G$ be such that $x\ne -x.$ Then $x+y\in G_1\setminus \left\{0\right\}$ for all $y\in G.$ Thus x is adjacent to all $y\in G\setminus \{-x\}$ in ${\Gamma }_{\mathit{cpg}}(G).$ From this, we have deg${}_{{\Gamma }_{\mathit{cpg}}(G)}(x)=\left|G\right|-2.$ This is also true for $-x\in G.$

2. By (1), ${\Gamma }_{\mathit{cpg}}(G)$ is connected.

3. Let $G\cong {\mathbb{Z}}_2^k.$ Then $x=-x$ for every $x\in G.$ Thus ${\Gamma }_{\mathit{cpg}}(G)\cong K_{\left|G\right|}$ and so diam$({\Gamma }_{\mathit{apq}}(G))=1.$ Now, assume that $G\cancel{\cong }{\mathbb{Z}}_2^k$ and $0\ne x\in G$ is an element in G. Then there exists an element $y\in G$ such that $x\ne y$ and $x+y=0.$ Thus x is not adjacent to y but there exists a path $x-0-y$ of length two in G. Thus diam$({\Gamma }_{\mathit{cpg}}(G))=2.$

4. If $G\cong {\mathbb{Z}}_2,$ then ${\Gamma }_{\mathit{cpg}}(G)\cong K_2$ and so gr$({\Gamma }_{\mathit{cpg}}(G))=\infty .$ If $G\cancel{\cong }{\mathbb{Z}}_2,$ then $\left|G\right|\ge 4$ as $\left|G\right|$ is not divisible by 3. Let $0,x,y\in G$ with $x+y\ne 0.$ Then $0-x-y-0$ is a cycle of length 3 in G and so gr$({\Gamma }_{\mathit{apq}}(G))=3.$

5. Follow from the proof (1) and (3), if $G\cong {\mathbb{Z}}_2^k,$ then ${\Gamma }_{\mathit{cpg}}(G)$ is complete and if $G\cancel{\cong }{\mathbb{Z}}_2^k,$ then ${\Gamma }_{\mathit{cpg}}(G)$ is a refinement of the star graph $K_{1,\left|G\right|-1}$ with center either 0 or x in G of order is 2. □

In the following theorem, we obtain the diameter and girth of ${\Gamma }_{\mathit{cpg}}(G)$ where $\left|G\right|$ is divisible by 3.

Theorem 3.3.

Let G be a finite abelian group. Assume that $\left|G\right|=3^{\alpha }.t,\alpha \ge 1,$ t is relatively prime to 3 and ${\sum }_{i=1}^k{m}_i=\alpha .$ Then

1. $de{g}_{{\Gamma }_{\mathit{cpg}}(G)}(x)=\{\begin{array}{cc}\frac{\left|G\right|}{3^k}-1\hfill & \mathit{\text{if}}x=-x\mathit{\text{and}}x\in G_1=\{3t:t\in G\};\hfill \\ \frac{\left|G\right|}{3^k}-2\hfill & \mathit{\text{if}}x\ne -x\mathit{\text{and}}x\in G_1;\hfill \\ \frac{\left|G\right|}{3^k}-1\hfill & \mathit{\text{if}}x\in G_2=G\setminus G_1.\hfill \end{array}$

2. If $G\cong {\mathbb{Z}}_3^k\times {\mathbb{Z}}_2^m,$ then $〈G_1〉\cong K_{\left|G_1\right|}.$ Otherwise $〈G_1〉$ is a refinement of the star graph $K_{1,\frac{\left|G\right|}{3^k}-1}.$

3. $〈G_2〉$ is a regular graph of degree $\frac{\left|G\right|}{3^k}-1.$

4. ${\Gamma }_{\mathit{cpg}}(G)$ is isomorphic to disjoint union of $〈G_1〉$ and $〈G_2〉.$

Proof.

Note that $\left|G_1\right|=\frac{\left|G\right|}{3^k}.$

1. Let $x=-x$ for $x\in G_1.$ Then, for every $y\in G_1,$ we have $x+y\in G_1\setminus \left\{0\right\}$ and so x is adjacent to all y in ${\Gamma }_{\mathit{cpg}}(G).$ Thus deg$(x)=\left|G_1\right|-1=\frac{\left|G\right|}{3^k}-1.$ Let $x\ne -x$ for $x\in G_1.$ In this case $x\in G_1$ is adjacent to all $y\in G_1\setminus \{-x\}.$ Thus deg$(x)=\left|G_1\right|-2=\frac{\left|G\right|}{3^k}-2.$

   Let $z\in G_2.$ Then $z+y\cancel{=}3t$ for all $y\in G_1.$ This implies that, when $z\in G_2$ and $z+y=3t\in G_1\setminus \left\{0\right\},$ then one must have $y\in G_2.$ From this, for $z\in G_2,$ we have $N(z)=\{x_2-z,x_3-z,\dots ,x_{\frac{\left|G\right|}{3^k}}-z\}$ where $G_1=\{0,x_2,\dots ,x_{\frac{\left|G\right|}{3^k}}\}.$ Hence deg$(z)=|N(z)|=\frac{\left|G\right|}{3^k}-1.$

2. As observed in the proof of (1), $〈G_1〉$ is either $K_{\left|G_1\right|}$ or a refinement of the star graph with center 0 or x in G of order 2.

3. Follows from the Case (iii) of (1).

4. Note that $G_1\cap G_2=\varphi$ and $G_1\cup G_2=G.$ By Lemma 3.1, ${\Gamma }_{\mathit{cpg}}(G)$ is isomorphic to disjoint union of $〈G_1〉$ and $〈G_2〉.$ □

From Lemma 3.1 and Theorem 3.3 we have the following corollary.

Corollary 3.4.

Let G be a finite abelian group. Assume that $\left|G\right|=3^{\alpha }.t,\alpha \ge 1,$ t is relatively prime to 3 and ${\sum }_{i=1}^k{m}_i=\alpha .$ Then

1. $\mathrm{deg}(x)=\{\begin{array}{cc}\frac{\left|G\right|}{3^k}-1\hfill & \mathit{\text{if}}x=-x\mathit{\text{and}}x\in G_1=\{3t:t\in G\};\hfill \\ \frac{\left|G\right|}{3^k}-2\hfill & \mathit{\text{if}}x\ne -x\mathit{\text{and}}x\in G_1;\hfill \\ \frac{\left|G\right|}{3^k}-1\hfill & \mathit{\text{if}}x\in G_2=G\setminus G_1.\hfill \end{array}$

2. If $G\cong {\mathbb{Z}}_3^k\times {\mathbb{Z}}_2^m,$ then $〈G_1〉\cong K_{\left|G_1\right|}.$ Otherwise $〈G_1〉$ is a refinement of the star graph $K_{1,\frac{\left|G\right|}{3^k}-1}$ with center0.

3. The induced subgraph $<X_{\ell }>(2\le \ell \le 2^k)$ is a regular graph of degree $\frac{\left|G\right|}{3^k}-1.$ Further $〈G_2〉$ is a regular graph of degree $\frac{\left|G\right|}{3^k}-1.$

4. ${\Gamma }_{\mathit{cpg}}(G)$ is the disjoint union of $〈X_{\ell }〉$ for $1\le \ell \le 2^k.$

By substituting t = 1 in the statement of Corollary 3.4, we have the following corollary.

Corollary 3.5.

Let G be a finite abelian group of order $3^{\alpha }$ with $\alpha \ge 1$ and ${\sum }_{i=1}^k{m}_i=\alpha .$ Then

1. $\mathrm{deg}(x)=\{\begin{array}{cc}\frac{\left|G\right|}{3^k}-1\hfill & \mathit{\text{if}}x=-x\mathit{\text{and}}x\in G_1=\{3t:t\in G\};\hfill \\ \frac{\left|G\right|}{3^k}-2\hfill & \mathit{\text{if}}x\ne -x\mathit{\text{and}}x\in G_1;\hfill \\ \frac{\left|G\right|}{3^k}-1\hfill & \mathit{\text{if}}x\in G_2=G\setminus G_1.\hfill \end{array}$

2. If $G\cong {\mathbb{Z}}_3^k,$ then $〈G_1〉\cong K_{\left|G_1\right|}.$ Otherwise $〈G_1〉$ is a refinement of the star graph $K_{1,\frac{\left|G\right|}{3^k}-1}$ with center 0.

3. The induced subgraph $<X_{\ell }>(2\le \ell \le 2^k)$ is a regular graph of degree $\frac{\left|G\right|}{3^k}-1.$ Further $〈G_2〉$ is a regular graph of degree $\frac{\left|G\right|}{3^k}-1.$

4. ${\Gamma }_{\mathit{cpg}}(G)$ is the disjoint union of $2^k$ induced subgraphs $〈X_{\ell }〉.$

Recall that a graph Γ is self-centered if the eccentricity of every vertex in Γ is same. In the following theorem, we obtain a condition for ${\Gamma }_{\mathit{cpg}}(G)$ to be self-centered.

Theorem 3.6.

Let G be a finite abelian group with $\left|G\right|=3^{\alpha }.t$ and t is relatively prime to 3. Then ${\Gamma }_{\mathit{cpg}}(G)$ is self-centered if and and only if $\alpha =0$ and $G\cong {\mathbb{Z}}_2^n$ for some $n\ge 1.$

Proof.

If $G\cong {\mathbb{Z}}_2^n$ then $\alpha =0,$ G = G₁ and $x=-x$ for every $x\in G.$ Thus ${\Gamma }_{\mathit{cpg}}(G)\cong K_{\left|G\right|}$ and ${\Gamma }_{\mathit{cpg}}(G)$ is self-centered.

Conversely assume that ${\Gamma }_{\mathit{cpg}}(G)$ is self-centered. If $\left|G\right|$ is divisible by 3, then by Theorem 3.3(4), ${\Gamma }_{\mathit{cpg}}(G)$ is disconnected, a contradiction to ${\Gamma }_{\mathit{cpg}}(G)$ is self-centered. Hence $\left|G\right|$ is not divisible by 3 and so $\alpha =0.$ Assume that $G\cancel{\cong }{\mathbb{Z}}_2^n.$ Then G contains elements x, y such that $x=-x$ and $y\ne -y.$ Now we have, $e_{{\Gamma }_{\mathit{cpg}}(G)}(x)=1$ and $e_{{\Gamma }_{\mathit{cpg}}(G)}(y)=2,$ which is contradiction to ${\Gamma }_{\mathit{cpg}}(G)$ is self-centered. Hence $G\cong {\mathbb{Z}}_2^n.$ □

In the following theorem, we obtain the independence number of ${\Gamma }_{\mathit{cpg}}(G).$

Theorem 3.7.

Let G be a finite abelian group with $\left|G\right|=3^{\alpha }.t$ and t is relatively prime to 3. Then the independence number $$\beta ({\Gamma }_{\mathit{cpg}}(G))=\{\begin{array}{cc}1\mathit{\text{or}}2\hfill & \mathit{\text{if}}\alpha =0;\hfill \\ 1+\frac{\left|G\right|}{3}\mathit{\text{or}}2+\frac{\left|G\right|}{3}\hfill & \mathit{\text{if}}\alpha \ge 1.\hfill \end{array}$$

Proof.

Case 1. Assume that $\alpha =0.$ i.e., $\left|G\right|$ is not divisible by 3 and hence $G=G_1.$

If $G\cong {\mathbb{Z}}_2^n,$ then ${\Gamma }_{\mathit{cpg}}(G)\cong K_{\left|G\right|}$ and so $\beta ({\Gamma }_{\mathit{cpg}}(G))=1.$ If $G\cancel{\cong }{\mathbb{Z}}_2^n,$ then $x\in G$ is adjacent to all vertices except $-x\in G.$ Hence $S=\{x,-x\}$ is a maximal independent set and so $\beta ({\Gamma }_{\mathit{cpg}}(G))=2.$

Case 2. Assume that $\alpha \ne 0.$ i.e., $\left|G\right|$ is divisible by 3 and so by Theorem 3.3(4), $〈G_1〉$ and $〈G_2〉$ are disjoint induced subgraphs of ${\Gamma }_{\mathit{cpg}}(G).$ Hence $\beta ({\Gamma }_{\mathit{cpg}}(G))=\beta (〈G_1〉)+\beta (〈G_2〉).$

Assume that $p(\alpha )=m_1+m_2+\dots +m_k.$

Case 2.1. Assume that t = 1 and $k=1.$ This implies that $\left|G\right|=3^{\alpha }$ and $G\cong {\mathbb{Z}}_{3^{\alpha }}.$

Case 2.1.1. If $\alpha =1,$ then $G_1=\left\{0\right\}$ and so ${\Gamma }_{\mathit{cpg}}(G)=\underset{\left|G\right|}{\cup }{K}_1.$ This gives that $\beta (〈G〉)=\left|G\right|.$

Case 2.1.2. If $\alpha >1,$ then there exists $x\in G_1$ such that $x\ne -x.$ Hence $\{x,-x\}$ is a maximal independent set of $〈G_1〉$ and so $\beta (〈G_1〉)=2.$ Further in this case $S(G_2({\mathbb{Z}}_{3^{\alpha }}))=\{1+3j|0\le j\le 3^{\alpha -1}-1\}\text{and}\{2+3\mathrm{j}|0\le \mathrm{j}\le 3^{\alpha -1}-1\}$ is a maximal independent set of $〈G_2({\mathbb{Z}}_{3^{\alpha }})〉$ and so $\beta (〈G_2({\mathbb{Z}}_{3^{\alpha }})〉)=\frac{3-1}{2}.\frac{\left|G\right|}{3}.$ Thus ${\Gamma }_{\mathit{cpg}}(G)=\frac{\left|G\right|}{3}+2.$

Case 2.2. Assume that t > 1 or $k>1.$

Case 2.2.1. Suppose $G\cong {\mathbb{Z}}_3^k.$ In this case $\alpha =k=1+1+\dots +1.$ Also ${\Gamma }_{\mathit{cpg}}(G)\cong \underset{\left|G\right|}{\cup }{K}_1$ and so $\beta ({\Gamma }_{\mathit{cpg}}(G))=\left|G\right|.$

Case 2.2.2. In the remaining cases $G\cong {\prod }_{i=1}^k{\mathbb{Z}}_{3^{m_i}}\times H$ where $\left|H\right|=t$ and t is relatively prime to 3. In this case also $〈G_1〉$ is a proper refinement of a star graph and there exists $x\in G_1$ such that $x\ne -x.$ Hence $\{x,-x\}$ is a maximal independent set of $〈G_1〉$ and so $\beta (〈G_1〉)=2.$

If $G\cong {\mathbb{Z}}_{3^{m_i}}\times H,$ then $\{1+3j|0\le j\le 3^{m_i-1}-1\}\times H$ is a maximal independent set in $〈G_2({\mathbb{Z}}_{3^{m_i}}\times H)〉$ and so $\beta (〈G_2({\mathbb{Z}}_{3^{m_i}}\times H)〉)=\frac{\left|{\mathbb{Z}}_{3^{m_i}}\right|}{3}.\left|H\right|$

In general, if $G\cong {\prod }_{i=1}^k{\mathbb{Z}}_{3^{m_i}}\times H$ and $S(G_2({\mathbb{Z}}_{3^{m_i}}))$ is a maximal independent set in $G_2({\mathbb{Z}}_{3^{m_i}}),$ then $S(G_2)={\mathbb{Z}}_{3^{m_1}}\times {\mathbb{Z}}_{3^{m_2}}\times \cdots \times S(G_2({\mathbb{Z}}_{3^{m_i}}))\times \cdots \times {\mathbb{Z}}_{3^{m_k}}\times H$ is a maximal independent set of $〈G_2〉$ in ${\Gamma }_{\mathit{cpg}}(G).$ Thus $\beta (〈G_2〉)=\left|{\mathbb{Z}}_{3^{m_1}}\right|\times \left|{\mathbb{Z}}_{3^{m_2}}\right|\times \cdots \times |S(G_2({\mathbb{Z}}_{3^{m_i}}))|\times \cdots \times \left|{\mathbb{Z}}_{3^{m_k}}\right|\times \left|H\right|=\frac{\left|G\right|}{3}$ in either of the cases.

Hence according to Cases 2.1 and 2.2, we have $\beta ({\Gamma }_{\mathit{cpg}}(G))=1+\frac{\left|G\right|}{3}$ or $2+\frac{\left|G\right|}{3}.$ □

In the following theorem, we obtain the clique number of ${\Gamma }_{\mathit{cpg}}(G)$.

Theorem 3.8.

Let G be a finite abelian group. Assume that $\left|G\right|=3^{\alpha }.t,$ t is relatively prime to 3 and $T=\{x\in G:x=-x\}.$ Then the clique number $$\omega ({\Gamma }_{\mathit{cpg}}(G))=\{\begin{array}{cc}\left|T\right|+\lceil \frac{G|-|T|}{2}\rceil \hfill & \mathit{\text{if}}\alpha =0;\hfill \\ \left|T\right|+\lceil \frac{\frac{\left|G\right|}{3^k}-\left|T\right|}{2}\rceil \hfill & \mathit{\text{if}}\alpha \ge 1\mathit{\text{and}}\alpha =\sum _{i=1}^k{m}_i.\hfill \end{array}$$

Proof.

Case 1. Let $\alpha =0.$ Then $\left|G\right|$ is not divisible by 3. By Theorem 3.2, any $x\in G$ with $x=-x,$ is adjacent to other all vertices in ${\Gamma }_{\mathit{cpg}}(G).$ On the other hand, x with $x\ne -x,$ is adjacent with all but $-x.$ Then the subgraph induced by $S=\{x:x=-x,x\in G\}\cup \{x\text{or}-x:x\ne -x,x\in G\}$ is a maximal complete subgraph of ${\Gamma }_{\mathit{cpg}}(G)$ and $\left|S\right|=\left|T\right|+\lceil \frac{\left|G\right|-\left|T\right|}{2}\rceil .$ Hence $\omega ({\Gamma }_{\mathit{cpg}}(G))=\left|T\right|+\lceil \frac{\left|G\right|-\left|T\right|}{2}\rceil .$

Case 2. Assume that $\alpha \ne 0.$ Then $\left|G\right|$ is divisible by 3. By Theorem 3.3, $〈G_1〉$ and $〈G_2〉$ are disjoint induced subgraphs of ${\Gamma }_{\mathit{cpg}}(G).$ From this, we get that $\omega ({\Gamma }_{\mathit{cpg}}(G))=\max \{\omega (〈G_1〉),\omega (〈G_2〉)\}.$

If $G\cong {\mathbb{Z}}_{3^{m_i}}\times H,$ then $\left|G_1\right|=\frac{\left|G\right|}{3}.$ Applying Case 1 for $G_1,$ we get that $\omega (〈G_1({\mathbb{Z}}_{3^{m_i}}\times H)〉)=\left|T\right|+\lceil \frac{\left|G_1\right|-\left|T\right|}{2}\rceil .=\left|T\right|+\lceil \frac{\frac{\left|G\right|}{3}-\left|T\right|}{2}\rceil .$

In general, if $G\cong {\prod }_{i=1}^k{\mathbb{Z}}_{3^{m_i}}\times H,$ then applying successively the above fact, we get that the clique number $\omega (〈G_1〉)=\left|T\right|+\lceil \frac{\left|G_1\right|-\left|T\right|}{2}\rceil =\left|T\right|+\lceil \frac{\frac{\left|G\right|}{3^k}-\left|T\right|}{2}\rceil .$

Let $G\cong {\mathbb{Z}}_{3^{m_i}}.$ For $x\in G_2({\mathbb{Z}}_{3^{m_i}}),$ by Theorem 3.3, we have deg$(x)>1.$ We claim that K₃ is not an induced subgraph of $〈G_2({\mathbb{Z}}_{3^{m_i}})〉.$ Suppose not, let $K_3=〈S〉=〈\{a,b,c\}〉\subseteq 〈G_2〉.$ From this, we have that $a+b=3t_1,b+c=3t_2$ and $c+a=3t_3$ and $a+a,$ b + b and c + c are not 3 multiplies. At the same time, we have $b=3t_1-a,c=3t_2-b=3(t_2-t_1)+a$ and so $2a=3(t_3-t_2+t_1)$ which is a contradiction to a + a is not a 3 multiple. Thus $\omega (〈G_2({\mathbb{Z}}_{3^{m_i}})〉)=2$ and so $\omega (〈G_2({\mathbb{Z}}_{3^{m_i}}\times H)〉)=2.$ In general, if $G\cong {\prod }_{i=1}^k{\mathbb{Z}}_{3^{m_i}}\times H,$ then $\omega (〈G_2〉)=2.$

Hence $\omega ({\Gamma }_{\mathit{cpg}}(G))=\max \left\{\left|T\right|+\lceil \frac{\frac{\left|G\right|}{3^k}-\left|T\right|}{2}\rceil ,2\right\}=\left|T\right|+\lceil \frac{\frac{\left|G\right|}{3^k}-\left|T\right|}{2}\rceil .$ □

In the following theorem, we obtain the chromatic number of ${\Gamma }_{\mathit{cpg}}(G).$

Theorem 3.9.

Let G be a finite abelian group. Assume that $\left|G\right|=3^{\alpha }.t,$ t is relatively prime to 3 and $T=\{x\in G:x=-x\}.$ Then the chromatic number $$\chi ({\Gamma }_{\mathit{cpg}}(G))=\{\begin{array}{cc}\left|T\right|+\lceil \frac{G|-|T|}{2}\rceil \hfill & \mathit{\text{if}}\alpha =0;\hfill \\ \left|T\right|+\lceil \frac{\frac{\left|G\right|}{3^k}-\left|T\right|}{2}\rceil \hfill & \mathit{\text{if}}\alpha \ge 1\mathit{\text{and}}\alpha ={\sum }_{i=1}^k{m}_i.\hfill \end{array}$$

Proof.

In general we have $\chi ({\Gamma }_{\mathit{cpg}}(G))\ge \omega ({\Gamma }_{\mathit{cpg}}(G)).$

Case 1. Assume that $\left|G\right|$ is not divisible by 3 and $T=\{x\in G:x=-x\}.$ In this case $\alpha =0.$ By Theorem 3.2, any $x\in G$ with $x=-x,$ is adjacent to other all vertices in ${\Gamma }_{\mathit{cpg}}(G).$ Assigning different colors to all vertices in T and assign a color to each x and – x for those with $x\ne -x.$ This assignment gives a proper coloring for ${\Gamma }_{\mathit{cpg}}(G)$ and so $\left|T\right|+\lceil \frac{\left|G\right|-\left|T\right|}{2}\rceil \ge \chi ({\Gamma }_{\mathit{cpg}}(G))\ge \omega ({\Gamma }_{\mathit{cpg}}(G)).$ Thus, by Theorem 3.8, we get that $\chi ({\Gamma }_{\mathit{cpg}}(G))=\omega ({\Gamma }_{\mathit{cpg}}(G))=\left|T\right|+\lceil \frac{\left|G\right|-\left|T\right|}{2}\rceil .$

Case 2. Assume that $\left|G\right|$ is divisible by 3. By Theorem 3.3, $〈G_1〉$ and $〈G_2〉$ are disjoint induced subgraphs of ${\Gamma }_{\mathit{cpg}}(G)$ and so $\chi ({\Gamma }_{\mathit{cpg}}(G))=\max \{\chi (〈G_1〉),\chi (〈G_2〉)\}.$

If $G\cong {\mathbb{Z}}_{3^{m_i}},$ as observed in the proof of Case 1 of Theorem 3.3, 0 is adjacent to other all vertices in $〈G_1({\mathbb{Z}}_{3^{m_i}})〉$ and so one has to assign separate colors to 0 and all vertices. Any $x\in 〈G_1({\mathbb{Z}}_{3^{m_i}})\setminus \left\{0\right\}〉$ is not adjacent to $-x.$ Hence the color of x can be assigned to –x. This assignment shall give a proper coloring for $〈G_1({\mathbb{Z}}_{3^{m_i}})〉.$ Hence $\left|T\right|+\lceil \frac{\frac{\left|G\right|}{3}-\left|T\right|}{2}\rceil =\omega (〈G_1({\mathbb{Z}}_{3^{m_i}})〉)\le \chi (〈G_1({\mathbb{Z}}_{3^{m_i}})〉)\le \left|T\right|+\lceil \frac{\frac{\left|G\right|}{3}-\left|T\right|}{2}\rceil .$

In general if $G\cong {\prod }_{i=1}^k{\mathbb{Z}}_{3^{m_i}}\times H,$ then $\left|T\right|+\lceil \frac{\frac{\left|G\right|}{3^k}-\left|T\right|}{2}\rceil =\omega (〈G_1〉)\le \chi (〈G_1〉)\le \left|T\right|+\lceil \frac{\frac{\left|G\right|}{3^k}-\left|T\right|}{2}\rceil .$ Thus $\omega (〈G_1〉)=\chi (〈G_1〉)=\left|T\right|+\lceil \frac{\frac{\left|G\right|}{3^k}-\left|T\right|}{2}\rceil .$

If $G\cong {\mathbb{Z}}_{3^{m_i}},$ as observed in the proof of Case 2.2.2 of Theorem 3.7, $\{1+3j|0\le j\le 3^{m_i-1}-1\}$ and $\{2+3j|0\le j\le 3^{m_i-1}-1\}$ are independent sets in $〈G_2({\mathbb{Z}}_{3^{m_i}})〉$ and hence same color can be assigned to each of these vertices. Further, the remaining vertices can be colored by another color and so $\chi (〈G_2({\mathbb{Z}}_{3^{m_i}})〉)=2.$

In general if $G\cong {\prod }_{i=1}^k{\mathbb{Z}}_{3^{m_i}}\times H,$ then $2=\omega (〈G_2〉)\le \chi (〈G_2〉)\le 2$ and so $\omega (〈G_2〉)=\chi (〈G_2〉)=2.$

Hence $\chi ({\Gamma }_{\mathit{cpg}}(G))=\max \left\{\left|T\right|+\lceil \frac{\frac{\left|G\right|}{3^k}-\left|T\right|}{2}\rceil ,2\right\}=\left|T\right|+\lceil \frac{\frac{\left|G\right|}{3^k}-\left|T\right|}{2}\rceil .$ □

Note that, a graph Γ is said to be weakly perfect if clique and chromatic numbers of Γ are same. From Theorems 3.8 and 3.9, one can conclude that ${\Gamma }_{\mathit{cpg}}(G)$ is weakly perfect and the same is stated below.

Theorem 3.10.

Let G be a finite abelian group. Then the cubic power graph ${\Gamma }_{\mathit{cpg}}(G)$ is weakly perfect.

A graph Γ with n vertices is called pancyclic if it contains a cycle of length k for every $i(3\le i\le n).$ Further Γ is said to be vertex pancyclic if every vertex of Γ is contained in a cycle of length k for every $i(3\le i\le n).$ In the following theorem, we prove that ${\Gamma }_{\mathit{cpg}}(G)$ is vertex pancyclic where G is a finite abelian group and $\left|G\right|$ is not divisible by 3.

Theorem 3.11.

Let G be a finite abelian group. If $\left|G\right|$ is not divisible by 3, then the cubic power graph ${\Gamma }_{\mathit{cpg}}(G)$ is vertex pancyclic.

Proof.

By Theorem 3.2, $G=G_1.$ Let $T=\{x\in G:x=-x\}=\{x_1,x_2,\dots ,x_n\}$ and $G\setminus T=\{y_1,-y_1,y_2,-y_2,\dots ,y_m,-y_m\}.$ Note that $〈T〉$ is a complete subgraph of ${\Gamma }_{\mathit{cpg}}(G)$ and every $y_i\in G\setminus T$ is adjacent to all vertices in $G\setminus T$ except $-y_i.$ Hence the cycle $C:x_1-x_2-\dots -x_n-y_1-y_2-\dots -y_m-(-y_1)-(-y_2)-\dots -(-y_m)-x_1$ contains a cycle a length i for $3\le i\le |V({\Gamma }_{\mathit{cpg}}(G))|.$ Hence ${\Gamma }_{\mathit{cpg}}(G)$ is vertex pancyclic. □

A famous characterization for Eulerian graph is as follows: Any connected graph Γ is an Eulerian graph if and only if all its vertices are of even degree. Now, we observe that ${\Gamma }_{\mathit{cpg}}(G)$ is not Eulerian.

Theorem 3.12.

For any finite abelian group G, ${\Gamma }_{\mathit{cpg}}(G)$ is not Eulerian.

Proof.

If $\left|G\right|$ is divisible by 3, by Theorem 3.3, ${\Gamma }_{\mathit{cpg}}(G)$ is disconnected and hence ${\Gamma }_{\mathit{cpg}}(G)$ is not Eulerian. Assume that $\left|G\right|$ is not divisible by 3. Then $G=G_1.$ By Theorem 3.2, there exists vertices of odd degree and so ${\Gamma }_{\mathit{cpg}}(G)$ is not Eulerian. □

4. Properties of the complement ${\overline{\Gamma }}_{\mathrm{cpg}}(G)$

In this section, we study about the complement ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ of the cubic power graph ${\Gamma }_{\mathit{cpg}}(G)$ of finite abelian groups.

Theorem 4.1.

Let G be a finite abelian group and $\left|G\right|$ is not divisible by 3 and $T=\{x\in G:x=-x\}.$ Then ${\overline{\Gamma }}_{\mathit{cpg}}(G)\cong \underset{\left|T\right|}{\cup }{K}_1\underset{\frac{\left|G\right|-\left|T\right|}{2}}{\cup }{K}_2.$

Proof.

If $\left|G\right|$ is not divisible by 3, then $G=G_1.$ For every $x,y\in G,$ there exists $t\in G$ such that $x+y=3t.$ Hence x and y are adjacent in ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ if and only if $x+y=0$ if and only if $y=-x.$

When $x=y,$ we get K₁ and when $x\ne y,$ we get $K_2.$ Hence ${\overline{\Gamma }}_{\mathit{cpg}}(G)\cong \underset{\left|T\right|}{\cup }{K}_1\underset{\frac{\left|G\right|-\left|T\right|}{2}}{\cup }{K}_2,$ where $T=\{x\in G:x=-x\}.$ □

Theorem 4.2.

Let G be a finite abelian group. Then ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ is connected if and only if $\left|G\right|$ is divisible by 3.

Proof.

If ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ is connected, by Theorem 4.1, $\left|G\right|$ is divisible by 3.

Conversely, assume that $\left|G\right|$ is divisible by 3. By Theorem 3.3, ${\Gamma }_{\mathit{cpg}}(G)$ is disconnected and so ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ is connected. □

Now, let us observe some properties of the induced subgraphs of ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ induced by the subsets $X_{\ell }$ of G.

Theorem 4.3.

Let G be a finite abelian group. Assume that $\left|G\right|=3^{\alpha }.t$, t is relatively prime to 3 and ${\sum }_{i=1}^k{m}_i=\alpha .$ Then the following hold.

1. The induced subgraph $〈G_1〉$ of ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ is either K₁ or $\cup K_1$ or $(\cup K_1)\cup (\cup K_2);$

2. For $1\le i,j\le 2^k$ and $i\ne j,$ let $x\in 〈X_i〉$ and $y\in 〈X_j〉.$ Then x and y are adjacent in ${\overline{\Gamma }}_{\mathit{cpg}}(G);$

3. For $2\le \ell \le 2^k,$ the induced subgraph $〈X_{\ell }〉$ is a regular subgraph of ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ degree $\left|X_{\ell }\right|-\left|G_1\right|.$ Further $〈G_2〉$ is a regular subgraph of ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ of degree $\left|G_2\right|-\left|G_1\right|.$

4. diam $({\overline{\Gamma }}_{\mathit{cpg}}(G))=1\mathit{\text{or}}2;$

5. gr$({\overline{\Gamma }}_{\mathit{cpg}}(G))=3;$

6. ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ is self-centered.

Proof.

1. First let us consider the case that $\left|G\right|$ is divisible by 3. Note that $G_1=X_{(0,\dots ,0)}=X_1.$ Let $x,y\in X_1.$ Then x adjacent to y in $〈X_1〉\subset {\overline{\Gamma }}_{\mathit{cpg}}(G)$ if and only if $x+y=0.$

   By Theorem 3.3(1.1), if $G\cong {\mathbb{Z}}_3^k,$ then $X_1=\left\{0\right\}.$ Hence $〈X_1〉\cong K_1$ in ${\overline{\Gamma }}_{\mathit{cpg}}(G).$

   By Theorem 3.3(1.1), if $G\cong {\mathbb{Z}}_3^k\times {\mathbb{Z}}_2^r$ and $x\in X_1,$ then $x=-x.$ Hence $〈X_1〉\cong \cup K_1$ in ${\overline{\Gamma }}_{\mathit{cpg}}(G).$

   In the remaining possibilities, by Theorem 3.3(1.1), for $x\in X_1$ with $x=-x,$ we have $〈x〉\cong K_1$ and $〈y,-y〉\cong K_2$ for all $y\in X_1\setminus \left\{x\right\}$ with $y\ne -y.$ Hence $〈X_1〉\cong (\cup K_1)\cup (\cup K_2)$ in ${\overline{\Gamma }}_{\mathit{cpg}}(G).$

2. By Lemma 3.1, $x\in X_i$ and $y\in X_j$ are not adjacent in ${\Gamma }_{\mathit{cpg}}(G)$ and so they are adjacent in ${\overline{\Gamma }}_{\mathit{cpg}}(G).$

3. Note that $〈G_2〉={\cup }_{i=2}^{2^k}{X}_i.$ For every $x\in G_2,$ there exists a $\ell$ such that $x\in X_{\ell }.$ By Theorem 3.3((1.3) and (1.4)), deg${}_{{\Gamma }_{\mathit{cpg}}(G)}(x)=\frac{\left|G\right|}{3^k}-1=\left|G_1\right|-1$ of the induced subgraph $〈X_{\ell }〉\subseteq {\Gamma }_{\mathit{cpg}}(G).$ This gives that deg${}_{{\overline{\Gamma }}_{\mathit{cpg}}(G)}(x)=\left|X_{\ell }\right|-(|G_1|-1)-1=\left|X_{\ell }\right|-\left|G_1\right|$ of the induced subgraph $〈X_{\ell }〉\subseteq {\overline{\Gamma }}_{\mathit{cpg}}(G).$ Further with regard to $〈G_2〉$ in ${\overline{\Gamma }}_{\mathit{cpg}}(G),$ we have deg$(x)=\left|X_i\right|-\left|G_1\right|+{\sum }_{j=1,j\ne i}^{2^k}\left|X_j\right|=\left|G_2\right|-\left|G_1\right|.$

4. Assume that $G\cong {\mathbb{Z}}_3^k.$ Then ${\overline{\Gamma }}_{\mathit{cpg}}(G)\cong K_{\left|G\right|}$ and so diam $({\overline{\Gamma }}_{\mathit{cpg}}(G))=1.$

   In the remaining possibilities, $\left|X_{\ell }\right|\ge 2$ for $1\le \ell \le 2^k.$ From (2), $x\in X_i$ and $y\in X_j$ are adjacent in ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ and so diam $({\overline{\Gamma }}_{\mathit{cpg}}(G))=2.$

5. Consider $0\in G_1$ and $x\in G_2.$ Then $〈\{0,x,-x\}〉\cong K_3\subseteq {\overline{\Gamma }}_{\mathit{cpg}}(G).$ Hence gr${\overline{\Gamma }}_{\mathit{cpg}}(G)=3.$

6. Assume that $G\cong {\mathbb{Z}}_3^k.$ Then ${\overline{\Gamma }}_{\mathit{cpg}}(G)\cong K_{\left|G\right|}$ and so $e_{{\overline{\Gamma }}_{\mathit{cpg}}(G)}(x)=1$ for all $x\in G.$

In the remaining possibilities, $x\in G_1$ is not adjacent to 0 in ${\overline{\Gamma }}_{\mathit{cpg}}(G).$ From (2), $e_{{\overline{\Gamma }}_{\mathit{cpg}}(G)}(x)=2$ for all $x\in G_1.$ For $x\in G_2,$ from (3), there exists at least one $y\in G_2$ which is not adjacent to x. Hence $e_{{\overline{\Gamma }}_{\mathit{cpg}}(G)}(x)=2$ for all $x\in G_2.$

Thus ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ is self-centered in both the cases. □

In the following theorem, we obtain the independence number of ${\overline{\Gamma }}_{\mathit{cpg}}(G).$

Theorem 4.4.

Let G be a finite abelian group. Assume that $\left|G\right|=3^{\alpha }.t$, t is relatively prime to 3 and $T=\{x\in G:x=-x\}.$ Then the independence number $$\beta ({\overline{\Gamma }}_{\mathit{cpg}}(G))=\{\begin{array}{cc}\frac{\left|G\right|+\left|T\right|}{2}\hfill & \mathit{\text{if}}\alpha =0;\hfill \\ \left|T\right|or\frac{\frac{\left|G\right|}{3^k}+\left|T\right|}{2}\hfill & \mathit{\text{if}}\alpha \ge 1\mathit{\text{and}}\alpha =\sum _{i=1}^k{m}_i.\hfill \end{array}$$

Proof.

Case 1. Let $\alpha =0.$ Then $\left|G\right|$ is not divisible by 3. By Theorem 4.1, ${\overline{\Gamma }}_{\mathit{cpg}}(G)\cong \underset{\left|T\right|}{\cup }{K}_1\underset{\frac{\left|G\right|-\left|T\right|}{2}}{\cup }{K}_2.$ Hence $\beta ({\overline{\Gamma }}_{\mathit{cpg}}(G))=\frac{\left|G\right|+\left|T\right|}{2}.$

Case 2. Let $\alpha \ge 1$ and $\alpha ={\sum }_{i=1}^k{m}_i.$ Then $\left|G\right|$ is divisible by 3.

If $G\cong {\mathbb{Z}}_3^k,$ then ${\overline{\Gamma }}_{\mathit{cpg}}(G)\cong K_{\left|G\right|}$ and so $\beta ({\overline{\Gamma }}_{\mathit{cpg}}(G))=1=\left|T\right|.$

In the remaining possibilities, $\left|X_{\ell }\right|\ge 2$ for $1\le \ell \le 2^k.$ By Theorem 4.3(2), any independent set S is contained in one $X_{\ell }$ for some $\ell ,$ $1\le \ell \le 2^k.$

1. Suppose S is an independent set in ${\overline{\Gamma }}_{\mathit{cpg}}(G),S\subset X_{\ell }$ for some $\ell ,$ $2\le \ell \le 2^k$ and $\left|S\right|\ge 3.$ Let $\{x_1,x_2,x_3\}\subseteq S$ Then $x_1+x_2,x_1+x_3\in G_1\setminus \left\{0\right\}$ and so $x_1+x_2=3t_1$ and $x_1+x_3=3t_2$ for some $t_1,t_2\in G$ which implies $x_2+x_3=3(t_1+t_2)-2x_1\in G_2.$ Thus x₂ is adjacent to x₃ in ${\overline{\Gamma }}_{\mathit{cpg}}(G),$ which is contradiction to our assumption that S is independent. Hence $S\subset X_1$ or $\left|S\right|\le 2.$

2. Assume that $S\subset X_1$ is an independent set. By Theorem 4.3(1), $〈X_1〉\cong \cup K_1\text{or}(\cup K_1)\cup (\cup K_2).$ Thus $\left|S\right|\ge 2.$

Combining the possibilities of (a) and (b), we get that the maximal independent set S of ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ is a subset of $X_1.$ From this $\left|S\right|\ge 2$ in ${\overline{\Gamma }}_{\mathit{cpg}}(G).$ Hence $\beta ({\overline{\Gamma }}_{\mathit{cpg}}(G))=\left|T\right|\text{or}\frac{\left|G_1\right|+\left|T\right|}{2}.$ □

In the following theorem, we obtain the clique number of the ${\overline{\Gamma }}_{\mathit{cpg}}(G).$

Theorem 4.5.

Let G be a finite abelian group. Assume that $\left|G\right|=3^{\alpha }.t$, t is relatively prime to 3 and $\alpha ={\sum }_{i=1}^k{m}_i$ Then the clique number $$\omega ({\overline{\Gamma }}_{\mathit{cpg}}(G))=\{\begin{array}{cc}1\mathit{\text{or}}2\hfill & \mathit{\text{if}}\alpha =0;\hfill \\ \left|G\right|\mathit{\text{or}}1+\frac{\left|G_2\right|}{2}\mathit{\text{or}}2+\frac{\left|G_2\right|}{2}\hfill & \mathit{\text{if}}\alpha \ge 1.\hfill \end{array}$$

Proof.

Case 1. Assume that $\left|G\right|$ is not divisible by 3.

Case 1.1. If $G\cong {\mathbb{Z}}_2^n,$ then ${\overline{\Gamma }}_{\mathit{cpg}}(G)\cong \cup K_1$ and so $\omega ({\overline{\Gamma }}_{\mathit{cpg}}(G))=1.$

Case 1.2. If $G\cancel{\cong }{\mathbb{Z}}_2^n,$ by Theorem 4.1, ${\overline{\Gamma }}_{\mathit{cpg}}(G)\cong \underset{\left|T\right|}{\cup }{K}_1\underset{\frac{\left|G\right|-\left|T\right|}{2}}{\cup }{K}_2$ and so $\omega ({\overline{\Gamma }}_{\mathit{cpg}}(G))=2.$

Case 2. Assume that $\left|G\right|$ is divisible by 3.

Case 2.1. If $G\cong {\mathbb{Z}}_3^k,$ then ${\overline{\Gamma }}_{\mathit{cpg}}(G)\cong K_{\left|G\right|}$ and so $\omega ({\overline{\Gamma }}_{\mathit{cpg}}(G))=\left|G\right|.$

Case 2.2. In the remaining possibilities $G\cong {\prod }_{i=1}^k{\mathbb{Z}}_{3^{m_i}}\times H.$

Case 2.2.1. By Theorem 4.3(1), we get $〈G_1〉\cong \cup K_1(\text{or})(\cup {\mathrm{K}}_1)\cup (\cup {\mathrm{K}}_2).$ Thus $\omega (〈G_1〉)=1\text{or}2.$

Case 2.2.2. As observed the proof of Case 2.2.2 in Theorem 3.7, if $G\cong {\mathbb{Z}}_{3^{m_i}}\times H,$ then $S(G_2({\mathbb{Z}}_{3^{m_i}}\times H))=\{1+3j|0\le j\le 3^{m_i-1}-1\}\times H$ is a maximal clique of $〈G_2({\mathbb{Z}}_{3^{m_i}}\times H)〉$ and so $\omega (〈G_2({\mathbb{Z}}_{3^{m_i}}\times H)〉)=\frac{|G_2({\mathbb{Z}}_{3^{m_i}}\times H)|}{2}.$

Let $\ell =(a_1,a_2,\dots ,a_k).$ Let $\beta \in \{1,2,\dots ,k\}$ be the first index such that $a_{\beta }=1$ and $a_i=0\mathit{foralli}\in \{1,\dots ,\beta -1\}.$ Then $S(X_{\ell })=H_1\times H_2\times \cdots \times H_k\times H$ where $H_i=\{\begin{array}{cc}S(G_2({\mathbb{Z}}_{3^{m_{\beta }}}))\hfill & \text{if}i=\beta ;\hfill \\ G_2({\mathbb{Z}}_{3^{m_i}})\hfill & \text{if}i\ne \beta \text{and}{a}_i=1;\hfill \\ G_1({\mathbb{Z}}_{3^{m_i}})\hfill & \text{if}i\ne \beta \text{and}{a}_i=0.\hfill \end{array}$ is a maximal clique of $〈X_{\ell }〉$ and so $|S(X_{\ell })|=\frac{\left|X_{\ell }\right|}{2}.$

By Remark 2.2 (4), $|S(G_2)|={\sum }_{\ell =2}^{2^k}|S(X_{\ell })|={\sum }_{\ell =2}^{2^k}\frac{\left|X_{\ell }\right|}{2}=\frac{\left|G_2\right|}{2}.$

Hence from the Cases 2.1, 2.2.1 and 2.2.2, we have $\omega ({\Gamma }_{\mathit{cpg}}(G))=\left|G\right|$ or $1+\frac{\left|G_2\right|}{2}$ or $2+\frac{\left|G_2\right|}{2}.$ □

In the following theorem, we obtain the chromatic number of ${\overline{\Gamma }}_{\mathit{cpg}}(G).$

Theorem 4.6.

Let G be a finite abelian group. Assume that $\left|G\right|=3^{\alpha }.t$, t is relatively prime to 3 and $\alpha ={\sum }_{i=1}^k{m}_i$ Then the chromatic number $$\chi ({\overline{\Gamma }}_{\mathit{cpg}}(G))=\{\begin{array}{cc}1\mathit{\text{or}}2\hfill & \mathit{\text{if}}\alpha =0;\hfill \\ \left|G\right|\mathit{\text{or}}1+\frac{\left|G_2\right|}{2}\mathit{\text{or}}2+\frac{\left|G_2\right|}{2}\hfill & \mathit{\text{if}}\alpha \ge 1.\hfill \end{array}$$

Proof.

Case 1. Assume that $\left|G\right|$ is not divisible by 3.

Case 1.1. If $G\cong {\mathbb{Z}}_2^n,$ then ${\overline{\Gamma }}_{\mathit{cpg}}(G)\cong \cup K_1$ and so $\chi ({\overline{\Gamma }}_{\mathit{cpg}}(G))=1.$

Case 1.2. If $G\cancel{\cong }{\mathbb{Z}}_2^n,$ then ${\overline{\Gamma }}_{\mathit{cpg}}(G)\cong \underset{\left|T\right|}{\cup }{K}_1\underset{\frac{\left|G\right|-\left|T\right|}{2}}{\cup }{K}_2$ and so $\chi ({\overline{\Gamma }}_{\mathit{cpg}}(G))=2.$

Case 2. Assume that $\left|G\right|$ is divisible by 3.

Case 2.1. If $G\cong {\mathbb{Z}}_3^k,$ then ${\overline{\Gamma }}_{\mathit{cpg}}(G)\cong K_{\left|G\right|}$ and so $\chi ({\overline{\Gamma }}_{\mathit{cpg}}(G))=\left|G\right|.$

Case 2.2. In the remaining possibilities, $G\cong {\prod }_{i=1}^k{\mathbb{Z}}_{3^{m_i}}\times H.$

By Theorem 4.3 (2), ${\Gamma }_{\mathit{cpg}}(G)$ is a complete bipartite graph with vertex partitions $〈G_1〉$ and $〈G_2〉.$ Then $\chi ({\Gamma }_{\mathit{cpg}}(G))=\chi (〈G_1〉)+\chi (〈G_2〉).$

Case 2.2.1. By Theorem 4.3(1), we get $〈G_1〉\cong \cup K_1\text{or}(\cup {\mathrm{K}}_1)\cup (\cup {\mathrm{K}}_2).$ Thus $\chi (〈G_1〉)=1\text{or}2.$

Case 2.2.2. Assume that $G\cong {\mathbb{Z}}_{3^{m_i}}.$ If $x\in G_2({\mathbb{Z}}_{3^{m_i}})$ and $x=1+3j,$ then there exists $y=2+3j\in G_2({\mathbb{Z}}_{3^{m_i}})$ such that x and y are not adjacent in $〈G_2({\mathbb{Z}}_{3^{m_i}})〉.$ Thus the same color can be assigned to both x and y in $G_2({\mathbb{Z}}_{3^{m_i}}).$ Hence number of colors needed to color $G_2({\mathbb{Z}}_{3^{m_i}})$ is $\frac{|G_2({\mathbb{Z}}_{3^{m_i}})|}{2}.$ Thus we get that $\frac{|G_2({\mathbb{Z}}_{3^{m_i}})|}{2}=\omega (〈G_2({\mathbb{Z}}_{3^{m_i}})〉)\le \chi (〈G_2({\mathbb{Z}}_{3^{m_i}})〉)\le \frac{|G_2({\mathbb{Z}}_{3^{m_i}})|}{2}$ and so $\chi (〈G_2({\mathbb{Z}}_{3^{m_i}})〉)=\frac{|G_2({\mathbb{Z}}_{3^{m_i}})|}{2}.$

In general, let $G\cong {\prod }_{i=1}^k{\mathbb{Z}}_{3^{m_i}}\times H$ and $x=(x_1,\dots ,x_i,\dots ,x_k,h).$ Applying selection of y_i as above, for each x_i for each $i(1\le i\le k),$ we get that $y\in G_2$ such that x and y are not adjacent in ${\overline{\Gamma }}_{\mathit{cpg}}(G).$ Hence the same color can be assigned to both x and y and so number of colors needed to color G₂ is $|c(G_2)|.$ Thus $\frac{\left|G_2\right|}{2}=\omega (〈G_2〉)\le \chi (〈G_2〉)\le \frac{\left|G_2\right|}{2}$ and so $\omega (〈G_2〉)=\chi (〈G_2〉)=\frac{\left|G_2\right|}{2}.$ From the Cases 2.1, 2.2.1 and 2.2.2, we have $$\chi ({\Gamma }_{\mathit{cpg}}(G))=\left|G\right|\text{or}1+\frac{\left|G_2\right|}{2}\text{or}2+\frac{\left|G_2\right|}{2}.$$

From Theorems 4.5 and 4.6, one can conclude that ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ is weakly perfect and the same is stated below.

Theorem 4.7.

For any finite abelian group G, the graph ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ is weakly perfect.

In the following theorem, we give a necessary and sufficient condition for ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ to be Eulerian.

Theorem 4.8.

Let G be a finite abelian group. Then ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ is Eulerian if and only if $G\cong {\mathbb{Z}}_3^n\text{or}{\mathbb{Z}}_3^{\mathrm{n}}\times {\mathbb{Z}}_2^{\mathrm{r}}.$

Proof.

Assume that $G\cong {\mathbb{Z}}_3^n\text{or}{\mathbb{Z}}_3^{\mathrm{n}}\times {\mathbb{Z}}_2^{\mathrm{r}}.$ By Theorem 4.2, ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ is connected. To prove that ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ is Eulerian, it is enough to show that each of the vertex in ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ is of even degree.

If $G\cong {\mathbb{Z}}_3^n,$ then $\left|G\right|$ is odd and ${\overline{\Gamma }}_{\mathit{cpg}}(G)\cong K_{\left|G\right|}.$ Hence each vertex of ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ is of even degree.

If $G\cong {\mathbb{Z}}_3^n\times {\mathbb{Z}}_2^r,$ then both $\left|G_1\right|$ and $\left|G_2\right|$ are even. By Theorem 4.3(1), $〈G_1〉\cong \cup K_1.$ By Theorem 4.3(3), the induced subgraph $〈G_2〉$ of ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ is a regular graph of degree $\left|G_2\right|-\left|G_1\right|.$ Then each vertex of the induced subgraph $〈G_2〉$ is of even degree. By Theorem 4.3(2), each vertex in G₁ is adjacent to all the vertices in $G_2.$ Thus each vertex of ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ is of even degree.

Conversely, assume that ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ is Eulerian. Then each vertex of ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ is of even degree. Suppose $G\cancel{\cong }{\mathbb{Z}}_3^n$ and $G\cancel{\cong }{\mathbb{Z}}_3^n\times {\mathbb{Z}}_2^r.$ Then deg$(0)=\left|G_2\right|$ and their exists an element $x\in G_1$ such that deg$(x)=\left|G_2\right|+1,$ an odd number. Hence it is a contradiction. From this $G\cong {\mathbb{Z}}_3^n\text{or}{\mathbb{Z}}_3^{\mathrm{n}}\times {\mathbb{Z}}_2^{\mathrm{r}}.$ □

In the following theorem, we give a necessary and sufficient condition for ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ to be vertex pancyclic.

Theorem 4.9.

Let G be a finite abelian group whose order is divisible by 3. Then ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ is vertex pancyclic.

Proof.

Assume that $\left|G\right|$ is divisible by 3. By Theorem 4.2, ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ is connected.

For $1\le \ell \le 2^k,$ let $X_{\ell }=\{x_{\ell ,1},\dots ,x_{\ell ,\left|X_{\ell }\right|}\}.$ One can arrange the sets $X_{\ell }$ such that $\left|X_1\right|<\left|X_2\right|\le \left|X_3\right|\le \dots \le \left|X_{2^k-2}\right|\le \left|X_{2^k-1}\right|<\left|X_{2^k}\right|.$

By Theorem 4.5, $S(X_{\ell })$ is a maximal clique of $〈X_{\ell }〉$ for $1\le \ell \le 2^k$ and $X={\cup }_{\ell =2}^{2^k}S(X_{\ell })$ is a maximal clique of $〈G_2〉$ and so $〈X〉\cong K_{\frac{\left|G_2\right|}{2}}.$ Further, by Theorem 4.3(2), for $1\le i,j\le 2^k$ and $i\ne j,$ $x\in 〈X_i〉$ and $y\in 〈X_j〉$ are adjacent in ${\overline{\Gamma }}_{\mathit{cpg}}(G).$

Let $X_1^{*}=X_1,$ $X_i^{*}=X_i\setminus S(X_i)$ for $2\le i\le 2^k.$ By Remark 2.2(4), $\left|X_{2^k-1}\right|=2^{k-1}.\frac{\left|G\right|}{3^k}$ and by Theorem 4.13, $|S(X_{\ell })|=\frac{\left|X_{\ell }\right|}{2}.$ These together imply that $\left|X_{2^k-1}^{*}\right|=2^{k-2}.\frac{\left|G\right|}{3^k},$ $\left|X_{2^k}^{*}\right|=2^{k-1}.\frac{\left|G\right|}{3^k}$ and so $\left|X_{2^k}^{*}\right|-\left|X_{2^k-1}^{*}\right|=2^{k-2}\frac{\left|G\right|}{3^k}.$

Let $s={\sum }_{\ell =2}^{2^k-1}\frac{\left|X_{\ell }\right|}{2}.$ Note that $s>2^{k-1}.\frac{\left|G\right|}{3^k}$ and the elements of the set X can be arranged such that $X=\{a_1,a_2,\dots ,a_s\}\cup S(X_{2^k}^{*})$ and $S(X_{2^k}^{*})=\{b_{s+1},b_{s+2},\dots ,b_{s+(2^{k-1})\frac{\left|G\right|}{3^k}=\frac{\left|G_2\right|}{2}}\}.$ Let $X_{\ell }^{*}=\{x_{\ell ,1},\dots ,x_{\ell ,\left|X_{\ell }^{*}\right|}\}$ for $1\le \ell \le 2^k.$

Now we have the spanning cycle $C:x_{1,1}-x_{2,1}-\cdots -x_{2^k,1}-x_{1,2}-x_{2,2}-\cdots -x_{2^k,2}-x_{1,3}-\cdots -x_{1,\left|X_1^{*}\right|}-x_{2,\left|X_1^{*}\right|}-\cdots -x_{2^k,\left|X_1^{*}\right|}-x_{2,\left|X_1^{*}\right|+1}-x_{3,\left|X_1^{*}\right|+1}-\cdots -x_{2^k,\left|X_1^{*}\right|+1}-x_{2,\left|X_1^{*}\right|+2}-x_{3,\left|X_1^{*}\right|+2}-\cdots -x_{2^k,\left|X_1^{*}\right|+2}-x_{2,\left|X_1^{*}\right|+3}-\cdots -x_{2,\left|X_2^{*}\right|}-x_{3,\left|X_2^{*}\right|}-\cdots -x_{2^k,\left|X_2^{*}\right|}-x_{3,\left|X_2^{*}\right|+1}-x_{4,\left|X_2^{*}\right|+1}-\cdots -x_{2^k,\left|X_2^{*}\right|+1}-x_{3,\left|X_2^{*}\right|+2}-x_{4,\left|X_2^{*}\right|+2}-\cdots -x_{2^k,\left|X_2^{*}\right|+2}-x_{3,\left|X_2^{*}\right|+3}-\cdots -x_{3,\left|X_3^{*}\right|}-x_{4,\left|X_3^{*}\right|}-\cdots -x_{2^k,\left|X_3^{*}\right|}-x_{4,\left|X_3^{*}\right|+1}-\cdots -x_{2^k-1,\left|X_{2^k-2}^{*}\right|+1}-x_{2^k,\left|X_{2^k-2}^{*}\right|+1}-\cdots -x_{2^k-1,\left|X_{2^k-1}^{*}\right|}-x_{2^k,\left|X_{2^k-1}^{*}\right|}-a_1-x_{2^k,\left|X_{2^k-1}^{*}\right|+1}-a_2-x_{2^k,\left|X_{2^k-1}^{*}\right|+2}-\cdots -x_{2^k,\left|X_{2^k}^{*}\right|}-a_{(2^{k-1}).\frac{\left|G\right|}{3^k}}-\cdots -a_s-b_{s+1}-\cdots -b_{\frac{\left|G_2\right|}{2}}-x_{1,1}.$

Let m be an integer such that $3\le m\le \left|G\right|.$ Select m vertices starting from $x_{1,1}$ in C. If the mth vertex $x_{i,j}\in X_1,$ then choose a different $x_{i,j}\in G\setminus (X_1\cup X_{2^k}).$ The first vertex is in X₁ and the mth vertex $x_{i,j}$ is in a different $X_{\ell }$ and so the first and last vertices are adjacent. Suppose $x_{i,j}\notin X_1,$ obviously the first vertex $x_{1,1}$ and the last vertex $x_{i,j}$ are in two different $X_{\ell }$ and so they are adjacent. This gives a cycle of length m in ${\overline{\Gamma }}_{\mathit{cpg}}(G).$ Hence ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ is pancyclic. □

We shall conclude this section by proving that ${\Gamma }_{\mathit{cpg}}(G)$ is perfect. Note that a graph Γ is perfect if neither Γ nor $\overline{\Gamma }$ contain an induced odd cycle of degree at least five. So to conclude ${\Gamma }_{\mathit{cpg}}(G)$ is perfect, one has to prove that neither ${\Gamma }_{\mathit{cpg}}(G)$ nor ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ contains an induced cycle $C_5.$ Assume that $\left|G\right|$ is divisible by 3. By Theorem 4.3(1), $〈G_1〉=〈X_1〉$ contains no cycle. Hence we consider the induced subgraphs $〈X_{\ell }〉$ for $2\le \ell \le 2^k$ in the following theorem.

Lemma 4.10.

Let G be a finite abelian group such that $\left|G\right|$ is divisible by 3. For $\ell ,2\le \ell \le 2^k,$ the induced subgraphs $〈X_{\ell }〉$ of ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ contain no induced cycle of odd length greater than 3.

Proof.

Let G be a finite abelian group and $\left|G\right|$ be divisible by 3. Assume that $〈X_{\ell }〉$ contains an induced cycle of odd length greater than 3, for some $\ell (2\le \ell \le 2^k).$

Case 1. Assume that $G\cong {\mathbb{Z}}_3^k.$ Then ${\overline{\Gamma }}_{\mathit{cpg}}(G)\cong K_{\left|G\right|},$ which implies that K₃ is a subgraph of any cycle C in $〈X_{\ell }〉$ of length grader than 3, which is a contradiction. Hence $〈X_{\ell }〉$ contains no induced cycle of odd length greater than 3 for any $\ell ,2\le \ell \le 2^k.$

Case 2. In the remaining possibilities, $G\cong {\prod }_{i=1}^k{\mathbb{Z}}_{3^{m_i}}\times H.$

(a). Let $G\cong {\mathbb{Z}}_{3^{m_i}}.$ As observed in the proof of Case 2.2.2 in Theorem 4.5, $A=\{1+3j|0\le j\le 3^{m_i-1}-1\}$ and $B=\{2+3j|0\le j\le 3^{m_i-1}-1\}$ are maximal cliques of $〈G_2({\mathbb{Z}}_{3^{m_i}})〉.$ Let $C:a-b-c-d-e-a$ be a cycle of length 5 in $G_2({\mathbb{Z}}_{3^{m_i}}).$ Then either $\left|\right\{a,b,c,d,e\}\cap A|\ge 3$ or $\left|\right\{a,b,c,d,e\}\cap B|\ge 3$ and so K₃ is a subgraph of C in $〈G_2({\mathbb{Z}}_{3^{m_i}})〉,$ which is a contradiction. Hence $〈G_2({\mathbb{Z}}_{3^{m_i}})〉$ of ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ contains no induced cycle of odd length greater than 3.

One can conclude the proof of general case as in the proof of later part of Case 2.2.2 in Theorem 4.5. Hence $〈X_{\ell }〉$ of ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ contains no induced cycle of odd length greater than 3 for $2\le \ell \le 2^k.$ □

In the following theorem, we prove that ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ is perfect.

Theorem 4.11.

Let G be a finite abelian group. Then the cubic power graph ${\Gamma }_{\mathit{cpg}}(G)$ is perfect.

Proof.

Case 1. Assume that $\left|G\right|$ is divisible by 3.

Claim 1. ${\Gamma }_{\mathit{cpg}}(G)$ contains no induced cycle of odd length greater than 3. Suppose ${\Gamma }_{\mathit{cpg}}(G)$ contains an induced cycle C of odd length greater than 3. By Corollary 3.4(4), ${\Gamma }_{\mathit{cpg}}(G)$ is the disjoint union of $〈X_{\ell }〉$ for $1\le \ell \le 2^k.$ Thus $C\subseteq X_{\ell }$ for some $\ell (1\le \ell \le 2^k).$

Suppose $C\subseteq 〈X_1〉.$ By Theorem 3.3(2) and Corollary 3.5(2), $〈G_1〉$ is either complete or refinement of a star graph with center 0 in ${\Gamma }_{\mathit{cpg}}(G)$ and so $K_3\subset C$ in $〈G_1〉,$ which is a contradiction.

Suppose $C\subseteq X_{\ell },$ for some $\ell (2\le \ell \le 2^k).$ Let $C:a-b-c-d-e-a$ and so $V(C)=\{a,b,c,d,e\}.$ Then $a+b=3t_1,b+c=3t_2,c+d=3t_3,d+e=3t_4$ for some $3t_1,3t_2,3t_3,3t_4\in G_1\setminus \left\{0\right\}.$ From this, we get that $b=3t_1-a,$ $c=[3(t_2-t_1)+a],$ $d=[3(t_3-t_2+t_1)-a]$ and $e=[3(t_4-t_3+t_2-t_1)+a].$ This in turn implies that a is adjacent to d and b is adjacent to e. Thus $K_3\subset C$ in $〈X_{\ell }〉$ in ${\Gamma }_{\mathit{cpg}}(G)$ for all $\ell (2\le \ell \le 2^k),$ which is a contradiction. Hence ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ contains no induced cycle of length odd order greater than 3.

Claim 2. ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ contains no induced cycle of odd length greater than 3.

By Theorem 4.3(1), $〈G_1〉=〈X_1〉$ contains no cycle. By Lemma 4.10, $〈X_{\ell }〉$ contains no induced cycle of length odd order greater than 3 for $2\le \ell \le 2^k.$ Hence ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ contains no induced cycle of length odd order greater than 3.

Case 2. Assume that $\left|G\right|$ is not divisible by 3.

By Theorem 3.2(5), ${\Gamma }_{\mathit{cpg}}(G)$ is either complete or a refinement of a star graph with center 0 and so ${\Gamma }_{\mathit{cpg}}(G)$ contains no induced cycle of odd length greater than 3.

By Theorem 4.1, ${\overline{\Gamma }}_{\mathit{cpg}}(G)\cong \underset{\left|T\right|}{\cup }{K}_1\underset{\frac{\left|G\right|-\left|T\right|}{2}}{\cup }{K}_2.$ Hence ${\overline{\Gamma }}_{\mathit{cpg}}(G)$ contains no induced cycle of odd length greater than 3.

Hence ${\Gamma }_{\mathit{cpg}}(G)$ is a perfect graph. □

Additional information

Funding

This research work is supported by CSIR Emeritus Scientist Scheme (No. 21 (1123)/20/EMR-II) of Council of Scientific and Industrial Research, Government of India through the second author.