$C_4$ and $C_6$ decomposition of the tensor product of complete graphs

TL;DR

This paper establishes necessary and sufficient conditions for decomposing tensor products of complete graphs into cycles of length 4 and 6, and applies these results to even regular complete multipartite graphs.

Contribution

It provides the first complete characterization of $C_4$ and $C_6$ decompositions of tensor products of complete graphs, extending graph decomposition theory.

Findings

Conditions for $C_4$-decomposition of $K_m \times K_n$

Conditions for $C_6$-decomposition of $K_m \times K_n$

Every even regular complete multipartite graph with edges divisible by 4 or 6 is decomposable into $C_4$ or $C_6$

Abstract

Let $G$ be a simple and finite graph. A graph is said to be \textit{decomposed} into subgraphs $H_1$ and $H_2$ which is denoted by $G= H_1 \oplus H_2$, if $G$ is the edge disjoint union of $H_1$ and $H_2$. If $G= H_1 \oplus H_2 \oplus H_3 \oplus \cdots \oplus H_k$, where\ $H_1$,$H_2$,$H_3$, ..., $H_k$ are all isomorphic to $H$, then $G$ is said to be $H$-decomposable. Futhermore, if $H$ is a cycle of length $m$ then we say that $G$ is $C_m$-decomposable and this can be written as $C_m|G$. Where $ G\times H$ denotes the tensor product of graphs $G$ and $H$, in this paper, we prove the necessary and sufficient conditions for the existence of $C_4$-decomposition (respectively, $C_6$-decomposition ) of $K_m \times K_n$. Using these conditions it can be shown that every even regular complete multipartite graph $G$ is $C_4$-decomposable (respectively, $C_6$-decomposable) if the number of edges of $G$ is divisible by $4$ (respectively, $6$).

Full text

$H_{1}$,$H_{2}$,$H_{3}$, …, $H_{k}$ are all isomorphic to $H$, then $G$ is said to be $H$-decomposable. Futhermore, if $H$ is a cycle of length $m$ then we say that $G$ is $C_{m}$-decomposable and this can be written as $C_{m}|G$. Where $G\times H$ denotes the tensor product of graphs $G$ and $H$, in this paper, we prove the necessary and sufficient conditions for the existence of $C_{4}$-decomposition (respectively, $C_{6}$-decomposition ) of $K_{m}\times K_{n}$. Using these conditions it can be shown that every even regular complete multipartite graph $G$ is $C_{4}$-decomposable (respectively, $C_{6}$-decomposable) if the number of edges of $G$ is divisible by $4$ (respectively, $6$).

1Present Address: Air Force Institute of Technology, Kaduna, Nigeria

2Present Address: Federal University of Agriculture, Makurdi, Nigeria

1. Introduction

Let $C_{m}$, $K_{m}$ and $K_{m}-I$ denote cycle of length $m$, complete graph on $m$ vertices and complete graph on $m$ vertices minus a $1$-factor respectively. By an $m$-cycle we mean a cycle of length $m$. Let $K_{n,n}$ denote the complete bipartite graph with $n$ vertices in each bipartition set and $K_{n,n}-I$ denote $K_{n,n}$, with a $1$-factor removed. All graphs considered in this paper are simple and finite. A graph is said to be decomposed into subgraphs $H_{1}$ and $H_{2}$ which is denoted by $G=H_{1}\oplus H_{2}$, if $G$ is the edge disjoint union of $H_{1}$ and $H_{2}$. If $G=H_{1}\oplus H_{2}\oplus\cdots\oplus H_{k}$, where $H_{1}$,$H_{2}$, …, $H_{k}$ are all isomorphic to $H$, then $G$ is said to be $H$-decomposable. Futhermore, if $H$ is a cycle of length $m$ then we say that $G$ is $C_{m}$-decomposable and this can be written as $C_{m}|G$. A $k$-factor of $G$ is a $k$-regular spanning subgraph. A $k$-factorization of a graph $G$ is a partition of the edge set of $G$ into $k$-factors. A $C_{k}$-factor of a graph is a $2$-factor in which each component is a cycle of length $k$. A resolvable $k$-cycle decomposition (for short $k$-RCD) of $G$ denoted by $C_{k}||G$, is a $2$-factorization of $G$ in which each $2$-factor is a $C_{k}$-factor.

For two graphs $G$ and $H$ their tensor product $G\times H$ has vertex set $V(G)\times V(H)$ in which two vertices $(g_{1},h_{1})$ and $(g_{2},h_{2})$ are adjacent whenever $g_{1}g_{2}\in E(G)$ and $h_{1}h_{2}\in E(H)$. From this, note that the tensor product of graphs is distributive over edge disjoint union of graphs, that is if $G=H_{1}\oplus H_{2}\oplus\cdots\oplus H_{k}$, then $G\times H=(H_{1}\times H)\oplus(H_{2}\times H)\oplus\cdots\oplus(H_{k}\times H)$. Now, for $h\in V(H)$, $V(G)\times h=\{(v,h)|v\in V(G)\}$ is called a column of vertices of $G\times H$ corresponding to $h$. Further, for $y\in V(G)$, $y\times V(H)=\{(y,v)|v\in V(H)\}$ is called a layer of vertices of $G\times H$ corresponding to $y$. It is true that $K_{m}\times K_{2}$ is isomorphic to the complete bipartite graph $K_{m,m}$ with the edges of a perfect matching removed, i.e. $K_{m}\times K_{2}\cong K_{m,m}-I$, where $I$ is a $1$-factor of $K_{m,m}$.

The Lexicographic product $G\ast H$ of two graphs $G$ and $H$ is the graph having the vertex set $V(G)\times V(H)$, in which two vertices $(g_{1},h_{1})$ and $(g_{2},h_{2})$ are adjacent if either $g_{1},g_{2}\in E(G)$; or $g_{1}=g_{2}$ and $h_{1},h_{2}\in E(H)$.

The problem of finding $C_{k}$-decomposition of $K_{2n+1}$ or $K_{2n}-I$ where $I$ is a $1$-factor of $K_{2n}$, is completely settled by Alspach, Gavlas and Šajna in two different papers (see [2, 17]). A generalization to the above complete graph decomposition problem is to find a $C_{k}$-decomposition of $K_{m}\ast\overline{K}_{n}$, which is the complete $m$-partite graph in which each partite set has $n$ vertices. The study of cycle decompositions of $K_{m}\ast\overline{K}_{n}$ was initiated by Hoffman et al. [6]. In the case when $p$ is a prime, the necessary and sufficient conditions for the existence of $C_{p}$-decomposition of $K_{m}\ast\overline{K}_{n}$, $p\geq 5$ is obtained by Manikandan and Paulraja in [10, 11, 13]. Billington [3] has studied the decomposition of complete tripartite graphs into cycles of length 3 and 4. Furthermore, Cavenagh and Billington [5] have studied $4$-cycle, $6$-cycle and $8$-cycle decomposition of complete multipartite graphs. Billington et al. [4] have solved the problem of decomposing $(K_{m}\ast\overline{K}_{n})$ into $5$-cycles. Similarly, when $p\geq 3$ is a prime, the necessary and sufficient conditions for the existence of $C_{2p}$-decomposition of $K_{m}\ast\overline{K}_{n}$ is obtained by Smith (see [19]). For a prime $p\geq 3$, it was proved in [20] that $C_{3p}$-decomposition of $K_{m}\ast\overline{K}_{n}$ exists if the obvious necessary conditions are satisfied. As the graph $K_{m}\times K_{n}\cong K_{m}\ast\overline{K}_{n}-E(nK_{m})$ is a proper regular spanning subgraph of $K_{m}\ast\overline{K}_{n}$. It is therefore natural to think about the cycle decomposition of $K_{m}\times K_{n}$.

The results in [10, 11, 13] also gives the necessary and sufficient conditions for the existence of a $p$-cycle decomposition, (where $p\geq 5$ is a prime number) of the graph $K_{m}\times K_{n}$. In [12] it was shown that the tensor product of two regular complete multipartite graph is Hamilton cycle decomposable. Muthusamy and Paulraja in [14] proved the existence of $C_{kn}$-factorization of the graph $C_{k}\times K_{mn}$, where $mn\neq 2(\textrm{mod}\ 4)$ and $k$ is odd. Paulraja and Kumar [16] showed that the necessary conditions for the existence of a resolvable $k$-cycle decomposition of tensor product of complete graphs are sufficient when $k$ is even. In this paper, we prove the necessary and sufficient conditions for $K_{m}\times K_{n}$, where $m,n\geq 2$, to have a $C_{4}$-decomposition (respectively, $C_{6}$-decomposition). Among other results, here we prove the following main results. It is not surprising that the conditions in Theorem 1.1 and Theorem 1.2 are ”symmetric” with respect to $m$ and $n$ since $K_{m}\times K_{n}\cong K_{n}\times K_{m}$.

Theorem 1.1**.**

For $m,n\geq 2$, $C_{4}|K_{m}\times K_{n}$ if and only if either

1. (1)

$n\equiv 0\ (\textrm{mod}\ 4)$* and $m$ is odd,* 2. (2)

$m\equiv 0\ (\textrm{mod}\ 4)$* and $n$ is odd or* 3. (3)

$m\ or\ n\equiv 1\ (\textrm{mod}\ 4)$**

Theorem 1.2**.**

For $2\leq m,n,C_{6}|K_{m}\times K_{n}$ if and only if $m\equiv 1\ or\ 3\ (\textrm{mod}\ 6)$ or $n\equiv 1\ or\ 3\ (\textrm{mod}\ 6)$.

Theorem 1.3**.**

Let $m$ be an even integer and $m\geq 6$, then $C_{6}|K_{m}-I\times K_{n}$ if and only if $m\equiv 0\ or\ 2\ (\textrm{mod}\ 6)$

2. Preliminaries

2.1. Definition

Let $\rho$ be a permutation of the vertex set $V$ of a graph $G$. For any subset $U$ of $V$, $\rho$ acts as a function from $U$ to $V$ by considering the restriction of $\rho$ to $U$. If $H$ is a subgraph of $G$ with vertex set $U$, then $\rho(H)$ is a subgraph of $G$ provided that for each edge $xy\in E(H)$, $\rho(x)\rho(y)\in E(G)$. In this case, $\rho(H)$ has vertex set $\rho(U)$ and edge set $\{\rho(x)\rho(y):xy\in E(H)\}$.

 

Next, we give some existing results on cycle decomposition of complete graphs.

Theorem 2.1**.**

[7]** Let $m$ be an odd integer and $m\geq 3$. If $m\equiv 1\ or\ 3\ (\textrm{mod}\ 6)$ then $C_{3}|K_{m}$.

Theorem 2.2**.**

[17]** Let $n$ be an odd integer and $m$ be an even integer with $3\leq m\leq n$. The graph $K_{n}$ can be decomposed into cycles of length $m$ whenever $m$ divides the number of edges in $K_{n}$.

Now we have the following lemma, this lemma gives the cycle decomposition of the complete graph $K_{m}$ into cycles of length $3$ and $4$.

Lemma 2.3**.**

For $m\equiv 5(\textrm{mod}\ 6)$, there exist positive integers $p$ and $q$ such that $K_{m}$ is decomposable into $p$ $3$-cycles and $q$ $4$-cycles.

Proof.

Let the vertices of $K_{m}$ be $0,1,...,m-1$. The $4$-cycles are $(i,i+1+2s,i-1,i+2+2s),s=0,1,...,(m-i)/2-2,i=1,3,...,m-4$. The $3$-cycles are $(m-1,i-1,i),i=1,3,...,m-2$. Hence the proof. ∎

The following theorem is on the complete bipartite graph minus a $1$-factor, it was obtained by Ma et. al [9].

Theorem 2.4**.**

[9]** Let $m$ and $n$ be positive integers. Then there exist an $m$ cycle system of $K_{n,n}-I$ if and only if $n\equiv 1\ (\textrm{mod}\ 2)$, $m\equiv 0\ (\textrm{mod}\ 2)$, $4\leq m\leq 2n$ and $n(n-1)\equiv 0\ (\textrm{mod}\ m)$.

From the theorem above we have the following corollary.

Corollary 2.5**.**

The graph $K_{n,n}-I$, where $I$ is a $1$-factor of $K_{n,n}-I$ admits a $C_{4}$ decomposition if and only if $n\equiv 1\ (\textrm{mod}\ 4)$.

The following result is on the complete bipartite graphs.

Theorem 2.6**.**

[18]** The complete bipartite graph $K_{a,b}$ can be decomposed into cycles of length $2k$ if and only if $a$ and $b$ are even, $a\geq k$, $b\geq k$ and $2k$ divides $ab$.

The next result is on cycle decomposition of complete graph minus a $1$-factor.

Theorem 2.7**.**

[17]** Let $n$ be an even integer and $m$ be an odd integer with $3\leq m\leq n$. The graph $K_{n}-I$ can be decomposed into cycles of length $m$ whenever $m$ divides the number of edges in $K_{n}-I$.

3. $C_{4}$** Decomposition of $C_{m}\times K_{n}$**

We begin this section with the following lemma.

Lemma 3.1**.**

$C_{4}|C_{3}\times K_{4}$.

Proof.

Following from the definition of the tensor product of graphs, let $U^{1}=\{u_{1},v_{1},w_{1}\}$, $U^{2}=\{u_{2},v_{2},w_{2}\}$,$...$, $U^{4}=\{u_{4},v_{4},w_{4}\}$ form the partite sets of vertices in the product $C_{3}\times K_{4}$. For $1\leq i,j\leq 4$, surely $U^{i}\cup U^{j}$, $i\neq j$ induces a $K_{3,3}-I$, where $I$ is a $1$-factor of $K_{3,3}$.

A $C_{4}$ decomposition of $C_{3}\times K_{4}$ is given below:

$\{u_{1},v_{4},u_{2},w_{3}\}$,$\{u_{1},v_{3},u_{4},w_{2}\}$,$\{u_{1},v_{2},u_{3},w_{4}\}$,$\{u_{2},v_{3},w_{2},v_{1}\}$,$\{u_{3},v_{1},w_{3},v_{4}\}$,

$\{u_{2},w_{1},v_{3},w_{4}\}$,$\{u_{3},w_{2},v_{4},w_{1}\}$,$\{u_{4},v_{1},w_{4},v_{2}\}$ and $\{u_{4},w_{1},v_{2},w_{3}\}$ ∎

Next, we have the following lemma which follows from Lemma 3.1.

Lemma 3.2**.**

$C_{4}|C_{3}\times K_{5}$.

Proof.

Suppose we fix the $4$-cycles already given in Lemma 3.1, clearly the graph which remains after removing the edges of $C_{3}\times K_{4}$ from $C_{3}\times K_{5}$ can be decomposed into $3$ copies of $K_{2,4}$. Now, by Theorem 2.6 the graph $K_{2,4}$ can be decomposed into cycles of length $4$. Hence $C_{4}|C_{3}\times K_{5}$. ∎

The following theorem is an extension of Lemma 3.1 and Lemma 3.2.

Theorem 3.3**.**

$C_{4}|C_{3}\times K_{n}$* if and only if $n\equiv 0\ or\ 1\ (\textrm{mod}\ 4)$.*

Proof.

Suppose that $C_{4}|C_{3}\times K_{n}$. The graph $C_{3}\times K_{n}$ has $3n(n-1)$ edges. For $C_{4}|C_{3}\times K_{n}$ it implies that $n(n-1)\equiv 0\ (\textrm{mod}\ 4)$. Hence $n\equiv 0\ or\ 1\ (\textrm{mod}\ 4)$.

Following the definition of tensor product of graphs, let $U^{1}=\{u_{1},v_{1},w_{1}\}$, $U^{2}=\{u_{2},v_{2},w_{2}\}$,…, $U^{n}=\{u_{n},v_{n},w_{n}\}$ form the partite sets of vertices in the product $C_{3}\times K_{n}$. For $1\leq i,j\leq n$, surely $U^{i}\cup U^{j}$, $i\neq j$ induces a $K_{3,3}-I$, where $I$ is a $1$-factor of $K_{3,3}$.

Next, we prove the sufficiency in two cases.

Case 1. Whenever $n\equiv 0(\textrm{mod}\ 4)$. Let $n=4t$ where $t\geq 1$.

Next we note that $C_{3}\times K_{n}\cong(C_{3}\times K_{4})\ +\ (C_{3}\times K_{4})\ +\ (C_{3}\times K_{4})\ +\ \cdots+\ (C_{3}\times K_{4})\ +\ H^{*}$, $H^{*}$ is the graph containing the edges of $C_{3}\times K_{n}$ which are not covered by these $t$ copies of $C_{3}\times K_{4}$. By Lemma 3.1 the product $C_{3}\times K_{4}$ admits a $C_{4}$-decomposition.

Furthermore, we define the set $U=\{u^{1},u^{2},...,u^{p}\}$,$V=\{v^{1},v^{2},...,v^{p}\}$ and $W=\{w^{1},w^{2},...,w^{p}\}$ where $p=n/4$ and for $j=1,2,...,p$, $u^{j}=\{u_{i}|4j-3\leq i\leq 4j\}$, $v^{j}=\{v_{i}|4j-3\leq i\leq 4j\}$ and $w^{j}=\{w_{i}|4j-3\leq i\leq 4j\}$.

Now, $H^{*}$ is decomposable into graphs isomorphic to $K_{4,4n-4}$. Indeed, the $K_{4,4n-4}$ graphs in the decomposition of $H^{*}$ are induced by $(u^{i}\cup v^{1}\cup v^{2}\cup\cdots\cup v^{p})\setminus v^{i}$, $(u^{i}\cup w^{1}\cup w^{2}\cup\cdots\cup w^{p})\setminus w^{i}$ and $(v^{i}\cup w^{1}\cup w^{2}\cup\cdots\cup w^{p})\setminus w^{i}$, $i=1,2,...,p$. By Theorem 2.6 $C_{4}|K_{4,4n-4}$. Therefore we have decomposed $C_{3}\times K_{n}$ into $4$-cycles when $n\equiv 0(\textrm{mod}\ 4)$.

Case 2. Whenever $n\equiv 1(\textrm{mod}\ 4)$. Let $n=4t+1$ where $t\geq 1$.

By removing $U^{1}$, we obtain a copy of $C_{3}\times K_{n-1}$, so we may apply Case 1. The remaining structure can be decomposed into $3K_{2,4t}$ and by Theorem 2.6 $C_{4}|K_{2,4t}$. Therefore $C_{4}|C_{3}\times K_{n}$ when $n\equiv 1(\textrm{mod}\ 4)$. ∎

4. $C_{6}$** Decomposition of $C_{3}\times K_{n}$**

Theorem 4.1**.**

For all $n$, $C_{6}|C_{3}\times K_{n}$.

Proof.

Following from the definition of tensor product of graphs, let $U^{1}=\{u_{1},v_{1},w_{1}\}$, $U^{2}=\{u_{2},v_{2},w_{2}\}$,$...$, $U^{n}=\{u_{n},v_{n},w_{n}\}$ form the partite set of vertices in $C_{3}\times K_{n}$. Also, $U^{i}$ and $U^{j}$ has an edge in $C_{3}\times K_{n}$ for $1\leq i,j\leq n$ and $i\neq j$ if the subgraph induce $K_{3,3}-I$, where $I$ is a $1$-factor of $K_{3,3}$. Now, each subgraph $U^{i}\cup U^{j}$ is isomorphic to $K_{3,3}-I$. But $K_{3,3}-I$ is a cycle of length six. Hence the proof. ∎

Example 4.2**.**

*The graph $C_{3}\times K_{7}$ can be decomposed into cycles of length $6$.

Solution. Let the partite sets (layers) of the tripartite graph $C_{3}\times K_{7}$ be $U=\{u_{1},u_{2},...,u_{7}\}$, $V=\{v_{1},v_{2},...,v_{7}\}$ and $W=\{w_{1},w_{2},...,w_{7}\}$. We assume that the vertices of $U,V$ and $W$ having same subscripts are the corresponding vertices of the partite sets. A $6$-cycle decomposition of $C_{3}\times K_{7}$ is given below:

$\{u_{1},v_{2},w_{1},u_{2},v_{1},w_{2}\}$,$\{u_{1},v_{3},w_{1},u_{3},v_{1},w_{3}\}$,$\{u_{2},v_{3},w_{2},u_{3},v_{2},w_{3}\}$,

$\{u_{1},v_{4},w_{1},u_{4},v_{1},w_{4}\}$,$\{u_{2},v_{4},w_{2},u_{4},v_{2},w_{4}\}$,$\{u_{3},v_{4},w_{3},u_{4},v_{3},w_{4}\}$,

$\{u_{1},v_{5},w_{1},u_{5},v_{1},w_{5}\}$,$\{u_{2},v_{5},w_{2},u_{5},v_{2},w_{5}\}$,$\{u_{3},v_{5},w_{3},u_{5},v_{3},w_{5}\}$,

$\{u_{4},v_{5},w_{4},u_{5},v_{4},w_{5}\}$,$\{u_{1},v_{6},w_{1},u_{6},v_{1},w_{6}\}$,$\{u_{2},v_{6},w_{2},u_{6},v_{2},w_{6}\}$,

$\{u_{3},v_{6},w_{3},u_{6},v_{3},w_{6}\}$,$\{u_{4},v_{6},w_{4},u_{6},v_{4},w_{6}\}$,$\{u_{5},v_{6},w_{5},u_{6},v_{5},w_{6}\}$,

$\{u_{1},v_{7},w_{1},u_{7},v_{1},w_{7}\}$,$\{u_{2},v_{7},w_{2},u_{7},v_{2},w_{7}\}$,$\{u_{3},v_{7},w_{3},u_{7},v_{3},w_{7}\}$,

$\{u_{4},v_{7},w_{4},u_{7},v_{4},w_{7}\}$,$\{u_{5},v_{7},w_{5},u_{7},v_{5},w_{7}\}$,$\{u_{6},v_{7},w_{6},u_{7},v_{6},w_{7}\}$.*

5. $C_{m}$** Decomposition of $C_{m}\times K_{n}$**

Next, we establish the following lemma.

Lemma 5.1**.**

For all $n\geq 3$, $C_{4}|C_{4}\times K_{n}$.

Proof.

From the definition of tensor product of graphs, let $U^{1}=\{u_{1},v_{1},w_{1},x_{1}\}$, $U^{2}=\{u_{2},v_{2},w_{2},x_{2}\}$,$...$, $U^{n}=\{u_{n},v_{n},w_{n},x_{n}\}$ form the partite sets of vertices in the product $C_{4}\times K_{n}$. Also, for $1\leq i,j\leq n$ and $i\neq j$, $U^{i}\cup U^{j}$ induces $K_{4,4}-2I$, where $I$ is a $1$-factor of $K_{4,4}$. Now, each set $U^{i}\cup U^{j}$ is isomorphic to $K_{4,4}-2I$. But $K_{4,4}-2I$ admits a $4$-cycle decomposition. Hence the proof. ∎

Lemma 5.2**.**

$C_{6}|C_{6}\times K_{2}$**

Proof.

Let the partite set of the bipartite graph $C_{6}\times K_{2}$ be $\{u_{1},u_{2},...,u_{6}\}$,

$\{v_{1},v_{2},...,$ $v_{6}\}$. We assume that the vertices having the same subscripts are the corresponding vertices of the partite sets. Now $C_{6}\times K_{2}$ can be decomposed into $6$-cycles which are $\{u_{1},v_{2},u_{3},v_{4},u_{5},v_{6}\}$ and $\{v_{1},u_{2},v_{3},u_{4},v_{5},u_{6}\}$. ∎

Theorem 5.3**.**

For all $n$, $C_{6}|C_{6}\times K_{n}$.

Proof.

Let the partite set of the $6$-partite graph $C_{6}\times K_{n}$ be $U=\{u_{1},u_{2},...,$

$u_{n}\}$, $V=\{v_{1},v_{2},...,v_{n}\}$, $W=\{w_{1},w_{2},...,w_{n}\}$, $X=\{x_{1},x_{2},...,x_{n}\}$, $Y=\{y_{1},y_{2},...,y_{n}\}$ and $Z=\{z_{1},z_{2},...,z_{n}\}$: we assume that the vertices of $U,V,W,X,Y$ and $Z$ having the same subscripts are the corresponding vertices of the partite sets. Let $U^{1}=\{u_{1},v_{1},w_{1},x_{1},y_{1},z_{1}\}$, $U^{2}=\{u_{2},v_{2},w_{2},x_{2},y_{2},z_{2}\}$, …, $U^{n}=\{u_{n},v_{n},w_{n},x_{n},y_{n},z_{n}\}$ be the sets of these vertices having the same subscripts. By the definition of the tensor product, each $U^{i}$, $1\leq i\leq n$ is an independent set and the subgraph induced by each $U^{i}\cup U^{j}$, $1\leq i,j\leq n$ and $i\neq j$ is isomorphic to $C_{6}\times K_{2}$. Now by Lemma 5.2 the graph $C_{6}\times K_{2}$ admits a $6$-cycle decomposition. This completes the proof. ∎

Furthermore, we quote the following result on decomposition of the tensor product of graphs into cycles of odd length.

Lemma 5.4**.**

[11]** For $k\geq 1$ and $m\geq 3$, $C_{2k+1}|C_{2k+1}\times K_{m}$

The next theorem is an extension of Lemma 5.4.

Theorem 5.5**.**

For $m\geq 3$ and $n\geq 2$, $C_{m}|C_{m}\times K_{n}$

Proof.

We shall split the proof of this lemma into two cases.

**Case 1.**When $m=2k+1$, $k\geq 1$

The proof of this case is immediate from Lemma 5.4.

**Case 2.**When $m=2k$, $k\geq 2$

Following from the definition of tensor product of graphs, let $U_{1}=\{u^{1}_{1},u^{2}_{1},u^{3}_{1},...,u^{m}_{1}\}$, $U_{2}=\{u^{1}_{2},u^{2}_{2},u^{3}_{2},...,u^{m}_{2}\}$,$...$, $U_{n}=\{u^{1}_{n},u^{2}_{n},u^{3}_{n},...,u^{m}_{n}\}$ form the partite sets of vertices in the product $C_{m}\times K_{n}$. Now, for $1\leq i,j\leq n$ and $i\neq j$, the subgraph induced by $U_{i}\cup U_{j}$ is isomorphic to $K_{m,m}-(m-2)I$, where $I$ is a $1$-factor of $K_{m,m}$. But $K_{m,m}-(m-2)I$ admits an $m$-cycle decomposition. Hence the proof. ∎

6. Proof of the Main Theorem

Proof of Theorem 1.1. Assume that $C_{4}|K_{m}\times K_{n}$, for some $m$ and $n$ with $2\leq m,n$. Then every vertex of $K_{m}\times K_{n}$ has even degree and $4$ divides the number of edges of $K_{m}\times K_{n}$. These two conditions translates to $(m-1)(n-1)$ being even and $8|mn(m-1)(n-1)$ respectively. Hence by the first fact, $m$ or $n$ has to be odd, i.e. has to be congruent to $1\ or\ 3\ or\ 5\ (\textrm{mod}\ 6)$. The second condition is satisfied precisely when one of the following holds.

1. (1)

$n\equiv 0\ (\textrm{mod}\ 4)$ and $m$ is odd, 2. (2)

$m\equiv 0\ (\textrm{mod}\ 4)$ and $n$ is odd, or 3. (3)

$m\ or\ n\equiv 1\ (\textrm{mod}\ 4)$.

Next we proceed to prove the sufficiency in two cases.

Case 1. Since the tensor product is commutative, we may assume that $m$ is odd and so $m\equiv\ 1\ or\ 3\ or\ 5\ (\textrm{mod}\ 6)$. Suppose that $n\equiv 0\ (\textrm{mod}\ 4)$.

Subcase 1. Let $m\equiv 1\ or\ 3\ (\textrm{mod}\ 6)$

Now since $m\equiv 1\ or\ 3\ (\textrm{mod}\ 6)$ it implies that by Theorem 2.1 $C_{3}|K_{m}$. Therefore, the graph $K_{m}\times K_{n}=((C_{3}\times K_{n})\oplus\cdots\oplus(C_{3}\times K_{n}))$. But $n\equiv 0\ (\textrm{mod}\ 4)$ therefore by Theorem 3.3 we have that $C_{4}|C_{3}\times K_{n}$. Hence $C_{4}|K_{m}\times K_{n}$.

Subcase 2. Let $m\equiv 5(\textrm{mod}\ 6)$

By Lemma 2.3, there exist positive integers $p$ and $q$ such that $K_{m}$ is decomposable into $p$ $3$-cycles and $q$ $4$-cycles. Hence $K_{m}\times K_{n}$ has a decomposition into $p$ copies of $C_{3}\times K_{n}$ and $q$ copies of $C_{4}\times K_{n}$. By Theorem 3.3 $C_{4}|C_{3}\times K_{n}$ and also Lemma 5.1 shows that $C_{4}|C_{4}\times K_{n}$. Hence $C_{4}|K_{m}\times K_{n}$.

Case 2. By commutativity of the tensor product we assume that $m\equiv 1\ (\textrm{mod}\ 4)$. The graph $K_{m}\times K_{n}=((K_{m}\times K_{2})\oplus\cdots\oplus(K_{m}\times K_{2}))$. Since $m\equiv 1\ (\textrm{mod}\ 4)$, by Corollary 2.5, $C_{4}|K_{m,m}-I$, and $K_{m}\times K_{2}\cong K_{m,m}-I$. Hence $C_{4}|K_{m}\times K_{n}$. This completes the proof.                                                                                    $\Box$

Proof of Theorem 1.2. Assume that $C_{6}|K_{m}\times K_{n}$ for some $m$ and $n$ with $2\leq m,n$. Then every vertex of $K_{m}\times K_{n}$ has even degree and $6$ divides in the number of edges of $K_{m}\times K_{n}$. These two conditions translate to $(m-1)(n-1)$ being even and $6|m(m-1)n(n-1)$ respectively. Hence, by the first fact $m$ or $n$ has to be odd, i.e., has to be congruent to $1\ or\ 3\ or\ 5\ (\textrm{mod}\ 6)$. The second fact can now be used to show that they cannot both be congruent to $5\ (\textrm{mod}\ 6)$. It now follows that $m\equiv 1\ or\ 3\ (\textrm{mod}\ 6)$ or $n\equiv 1\ or\ 3\ (\textrm{mod}\ 6)$.

Conversely, let $m\equiv 1\ or\ 3\ (\textrm{mod}\ 6)$. By Theorem 2.1, $C_{3}|K_{m}$ and hence $K_{m}\times K_{n}=((C_{3}\times K_{n})\oplus\cdots\oplus(C_{3}\times K_{n}))$. Since $C_{6}|C_{3}\times K_{n}$ by Theorem 4.1.

Finally, if $n\equiv 1\ or\ 3\ (\textrm{mod}\ 6)$, the above argument can be repeated with the roles of $m$ and $n$ interchanged to show again that $C_{6}|K_{m}\times K_{n}$. This completes the proof.

 

Proof of Theorem 1.3. Assume that $C_{6}|K_{m}-I\times K_{n},m\geq 6$. Certainly, $6|mn(m-2)(n-1)$. But we know that if $6|m(m-2)$ then $6|mn(m-2)(n-1)$. But $m$ is even therefore $m\equiv 0\ or\ 2\ (\textrm{mod}\ 6)$.

Conversely, let $m\equiv 0\ or\ 2\ (\textrm{mod}\ 6)$. Notice that for each $m$, $\frac{m(m-2)}{2}$ is a multiple of $3$. Thus by Theorem 2.7 $C_{3}|K_{m}-I$ and hence $K_{m}-I\times K_{n}=((C_{3}\times K_{n})\oplus\cdots\oplus(C_{3}\times K_{n}))$. From Theorem 4.1 $C_{6}|C_{3}\times K_{n}$. The proof is complete.

7. CONCLUSION

In view of the results obtained in this paper we draw our conclusion by the following remark and corollary.

Remark 7.1**.**

The product $K_{m}\times K_{n}$ can also be viewed as an even regular complete multipartite graph. So by the conditions given in Theorem 1.1 we have that every even regular complete multipartite graph $G$ is $C_{4}$-decomposable if the number of edges of $G$ is divisible by $4$.

Corollary 7.2**.**

*For any simple graph $G$. If

1. (1)

$C_{3}|G$* then $C_{6}|G\times K_{n}$, whenever $n\geq 2$.* 2. (2)

$C_{6}|G$* then $C_{6}|G\times K_{n}$, whenever $n\geq 2$.*

Proof.

We only need to show that $C_{3}|G$. Applying Theorem 4.1 gives the result. ∎

**Comments

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The results presented in this work are split in two different papers [1, 15].