On a graph of monogenic semigroups

The history of studying zero-divisor graphs began with commutative rings in the paper [1], and then it continued with commutative and noncommutative rings in some of the joint papers written by Anderson (see, for instance, [2–4]) and some other authors (see, for instance, [5, 6]). After that DeMeyer et al. and some other authors studied these special graphs of commutative and noncommutative semigroups [7–9]. Since then a very large number of studies have been added in the literature about zero-divisor graphs. It is obvious that the reason for studying this subject is to give a great opportunity to characterize over the algebraic structure that studied on it.

In [7–9], by considering the (commutative) semigroup S with zero, the zero-divisor graph $\mathrm{\Gamma }(S)$ is defined as an undirected graph with vertices $Z{(S)}^{\ast }=Z(S)\mathrm{\setminus }\{0\}$ and for the set of nonzero zero-divisors of S, where for distinct $x,y\in Z{(S)}^{\ast }$, the vertices x and y are adjacent if and only if $xy=0$. In the light of this definition, one can always get some new varieties of graphs by changing the rule of adjacency of vertices.

Then we define an undirected graph (actually, a type of zero-divisor graph) $\mathrm{\Gamma }({\mathcal{S}}_M)=(V,E)$ associated to ${\mathcal{S}}_M$ as follows. The vertices are the nonzero zero-divisors (in other words, all nonzero elements) of ${\mathcal{S}}_M$, and any two distinct vertices $x^i$ and $x^j$ ($1\le i,j\le n$) are adjacent in case of $x^i\cdot x^j=0$ with the rule $x^i\cdot x^j=x^{i+j}=0$ if and only if $i+j>n$ (see Figures 1 and 2 below). We write $x^i{x}^j\in E(\mathrm{\Gamma }({\mathcal{S}}_M))$ if vertices $x^i$ and $x^j$ are adjacent. In this paper, we mainly obtain some certain values for the diameter, girth, maximum and minimum degree, chromatic number, clique number, degree sequence, irregularity index and domination number of the graph $\mathrm{\Gamma }({\mathcal{S}}_M)$. Moreover, the number of triangles for this special graph has been calculated. We also note that by studying this new graph $\mathrm{\Gamma }({\mathcal{S}}_M)$, we have reached some strict equalities for the bounds given in [8] for some of the properties of commutative semigroups. For example, in here, we get $diam(\mathrm{\Gamma }({\mathcal{S}}_M))=2$ and $girth(\mathrm{\Gamma }({\mathcal{S}}_M))=3$ while they were presented by ≤3 and ≤4, respectively, in [8].

In this section, by considering the graph $\mathrm{\Gamma }({\mathcal{S}}_M)$ defined as in the first section, we will mainly deal with the graph properties, namely diameter, girth, maximum and minimum degrees, domination number and finally irregularity index of it. In fact, it is quite well known that most of these properties can be obtained by checking the distance or the total number of vertices in any graph G. So, the methods in the proofs of the results in this section will be followed by this idea.

We thus obtain the following result.

Theorem 1 For any monogenic semigroup ${\mathcal{S}}_M$ as given in (1), the diameter of the graph $\mathrm{\Gamma }({\mathcal{S}}_M)$ is 2.

Proof It is clear that the vertex x of $\mathrm{\Gamma }({\mathcal{S}}_M)$ is pendant, and so the diameter can be figured out by considering the distance between this vertex and one of the other vertices in the vertex set. Therefore, x is only connected with the vertex $x^n$, and since $x^n$ is adjacent to all other vertices (i.e., $x^n\cdot x^i=0$, $1\le i\le n$), we finally get $diam(\mathrm{\Gamma }({\mathcal{S}}_M))=2$, as required. □

It is known that the girth of a simple graph G is the length of the shortest cycle contained in G. However, if G does not contain any cycle, then the girth of it is assumed to be infinity.

Theorem 2 For any monogenic semigroup ${\mathcal{S}}_M$ as given in (1), the girth of the graph $\mathrm{\Gamma }({\mathcal{S}}_M)$ is 3.

Proof By the definition of $\mathrm{\Gamma }({\mathcal{S}}_M)$, since $x^n\cdot x^{n-1}=0$, $x^{n-1}\cdot x^2=0$ and $x^n\cdot x^2=0$, we then have $x^n-x^{n-1}-x^2-x^n$, which implies the result, as desired. □

The degree ${deg}_G(v)$ of a vertex v of G is the number of vertices adjacent to v. Among all degrees, the maximum degree $\mathrm{\Delta }(G)$ (or the minimum degree $\delta (G)$) of G is the number of the largest (or the smallest) degrees in G (see [10]). By considering maximum or minimum degrees, another result can be presented as follows.

Proof Let us consider the vertex $x^n$ of $\mathrm{\Gamma }({\mathcal{S}}_M)$. It is clear that $x^n\cdot x^i=0$ for any $1\le i\le n$, as $n+i>n$. By the definition of $\mathrm{\Gamma }({\mathcal{S}}_M)$, we get $x^n{x}^i\in E(\mathrm{\Gamma }({\mathcal{S}}_M))$, $1\le i\le n$. Thus, we have $\mathrm{\Delta }(\mathrm{\Gamma }({\mathcal{S}}_M))=n-1$.

On the other hand, let us consider the vertex x of $\mathrm{\Gamma }({\mathcal{S}}_M)$. Then the equality $x\cdot x^i=0$ satisfies only if $i=n$. Nevertheless, for every $i=\{1,2,\dots ,n-1\}$, we have $x\cdot x^i\ne 0$. That means the unique vertex x is only connected to the vertex $x^n$ (i.e., $x^{n}x\in E(\mathrm{\Gamma }({\mathcal{S}}_M))$), which implies $\delta (\mathrm{\Gamma }({\mathcal{S}}_M))=1$, as required. □

Example 1 Consider the graph $\mathrm{\Gamma }({\mathcal{S}}_{M_6})$, as drawn in Figure 1, with the vertex set $V(\mathrm{\Gamma }({\mathcal{S}}_{M_6}))=\{x,x^2,x^3,x^4,x^5,x^6\}$. By Theorems 1, 2 and 3, we certainly have $diam(\mathrm{\Gamma }({\mathcal{S}}_{M_6}))=2$, $girth(\mathrm{\Gamma }({\mathcal{S}}_{M_6}))=3$, $\mathrm{\Delta }(\mathrm{\Gamma }({\mathcal{S}}_{M_6}))=5$, $\delta (\mathrm{\Gamma }({\mathcal{S}}_{M_6}))=1$.

The degree sequence, denote by $DS(G)$, is a sequence of degrees of vertices of a graph G. In [11], a new parameter for graphs, namely the irregularity index of G, has been recently defined and denoted by $\mathit{MWB}(G)$. In fact $\mathit{MWB}(G)$ is the number of distinct terms in the set $DS(G)$. (At this point, we should note that although this new index is denoted by $t(G)$ in the paper [11], we prefer to denote it by $\mathit{MWB}(G)$ not to make any confusion with the material in Section 4 of this paper.)

We recall that for a real number r, the notation $\lfloor r\rfloor$ denotes the greatest integer ≤r while $\lceil r\rceil$ denotes the least integer ≥r. This fact will be needed for some of the theorems in this paper.

and $\mathit{MWB}(\mathrm{\Gamma }({\mathcal{S}}_M))=n-1$, respectively.

Proof In $\mathrm{\Gamma }({\mathcal{S}}_M)$, since the vertex x is connected only with the vertex $x^n$, then we clearly obtain that the degree of x is 1. Secondly, let us consider the vertex $x^2\in V(\mathrm{\Gamma }({\mathcal{S}}_M))$. Then, as a similar idea, it is only connected with the vertices $x^n$ and $x^{n-1}$, which implies that the degree of $x^2$ is equal to 2. Now, if we apply the same progress to all remaining vertices, then we see that

Moreover,

Now, if we keep following the same procedure, then we get that

Hence, by the definition of degree sequence, we clearly obtain the set $DS(\mathrm{\Gamma }({\mathcal{S}}_M))$ as depicted in the theorem. Nevertheless, it is easily seen that the irregularity index $\mathit{MWB}(\mathrm{\Gamma }({\mathcal{S}}_M))=n-1$, as required. □

A subset D of the vertex set $V(G)$ of a graph G is called a dominating set if every vertex $V(G)\mathrm{\setminus }D$ is joined to at least one vertex of D by an edge. Additionally, the domination number $\gamma (G)$ is the number of vertices in the smallest dominating set for G (see [10]).

Now, in our case, by considering the definition of $\mathrm{\Gamma }({\mathcal{S}}_M)$, the vertex $x^n$ is the only element adjacent to all the other vertices. In other words, $x^n{x}^i\in E(\mathrm{\Gamma }({\mathcal{S}}_M))$ for $1\le i\le n$, and so the dominating set contains only the element $x^n$. This simple fact gives the following result about the domination number of $\mathrm{\Gamma }({\mathcal{S}}_M)$.

Example 2 As an example of Theorems 4 and 5, let us consider the graphs $\mathrm{\Gamma }({\mathcal{S}}_{M_4})$ and $\mathrm{\Gamma }({\mathcal{S}}_{M_5})$ as drawn in Figure 2. In here, $\gamma (\mathrm{\Gamma }({\mathcal{S}}_{M_4}))=1$, $DS(\mathrm{\Gamma }({\mathcal{S}}_{M_4}))=\{1,2,2,3\}$ and $\mathit{MWB}(\mathrm{\Gamma }({\mathcal{S}}_{M_4}))=3$. Moreover, $DS(\mathrm{\Gamma }({\mathcal{S}}_{M_5}))=\{1,2,2,3,4\}$ and $\mathit{MWB}({\mathcal{S}}_{M_5})=4$.

We note that for the graph $\mathrm{\Gamma }({\mathcal{S}}_{M_6})$ in Example 1, $\mathit{MWB}(\mathrm{\Gamma }({\mathcal{S}}_{M_6}))=5$.

Since perfect graphs are directly related to the terms coloring and clique numbers, let us start this section by reminding the definitions of them.

Basically, the coloring of G is to be an assignment of colors (elements of some set) to the vertices of G, one color to each vertex, so that adjacent vertices are assigned distinct colors. If n colors are used, then the coloring is referred to as n-coloring. If there exists an n-coloring of G, then G is called n-colorable. The minimum number n for which G is n-colorable is called the chromatic number of G and is denoted by $\chi (G)$. In addition, there exists another graph parameter, namely the clique of a graph G. In fact, depending on the vertices, each of the maximal complete subgraphs of G is called a clique. Moreover, the largest number of vertices in any clique of G is called the clique number and denoted by $\omega (G)$. In general, it is well known that $\chi (G)\ge \omega (G)$ for any graph G (see, for instance, [10]). For every induced subgraph $H\subseteq G$ of G, if $\chi (H)=\omega (H)$, then G is called a perfect graph [12].

In here, we will state and prove the chromatic and clique numbers for the graph $\mathrm{\Gamma }({\mathcal{S}}_M)$ separately, and so the perfectness of this special graph will be obtained.

Proof As usual, let us consider the graph $\mathrm{\Gamma }({\mathcal{S}}_M)$ associated with ${\mathcal{S}}_M$ as defined in (1). Now, if we first take account of the vertex $x^n$, then it is easy to see that $x^n$ is adjacent to all the other vertices. That means the color used for $x^n$ cannot be used for the remaining vertices. So, let us suppose that the color for $x^n$ is labeled ${\mathcal{C}}_1$.

Secondly, let us consider the vertex $x^{n-1}$. Since $x^{n-1}$ is adjacent to all vertices except the vertex x, the color for $x^{n-1}$, say ${\mathcal{C}}_2$, can also be used only for x. Similarly, the vertex $x^{n-2}$ is adjacent to all vertices except the vertices x and $x^2$. Thus the color, say ${\mathcal{C}}_3$, for $x^{n-2}$ can also be used only for the vertex $x^2$. (The color ${\mathcal{C}}_2$ has already been used for x in the previous step.)

By applying same progress, one can see that to handle the number of coloring for all vertices in the set $V(\mathrm{\Gamma }({\mathcal{S}}_M))$, we must add 1 to the least integer $\ge \frac{n-1}{2}$. In other words, a total $1+\lceil \frac{n-1}{2}\rceil$ colors should be needed, which gives the required chromatic number in the theorem. □

The following lemma (proof can be seen directly by mathematical induction) plays a central role in the proof of Theorem 7 below.

Lemma 1 For any $n\in {\mathbb{N}}^{+}$, there always exists $n-\lceil \frac{n}{2}\rceil =\lceil \frac{n-1}{2}\rceil$.

It is easy to see that the number of elements in $V(A)$ is $n-\lceil \frac{n}{2}\rceil +1$, which equals $\lceil \frac{n-1}{2}\rceil +1$ by Lemma 1.

By contradiction, we will show that the set $V(A)$ is maximal. Suppose to the contrary that $|V(A)|>\lceil \frac{n-1}{2}\rceil +1$. If any two vertices $x^i$ and $x^j$ ($x^i,x^j\in V(\mathrm{\Gamma }({\mathcal{S}}_M))$, $1\le i,j\le \lceil \frac{n}{2}\rceil -1$) is in $V(A)$, then we arrive at a contradiction, as $i+j\le n-1$. Otherwise, exactly any one vertex $x^i$ ($x^i\in V(\mathrm{\Gamma }({\mathcal{S}}_M))$, $1\le i\le \lceil \frac{n}{2}\rceil -1$) is in $V(A)$ as $|V(A)|>\lceil \frac{n-1}{2}\rceil +1$. Again, we arrive at a contradiction as $i+\lceil \frac{n}{2}\rceil \le n$, $1\le i\le \lceil \frac{n}{2}\rceil -1$. Thus, the set $V(A)$ is maximal. Hence, $\omega (\mathrm{\Gamma }({\mathcal{S}}_M))=1+\lceil \frac{n-1}{2}\rceil$, as required. □

Now, by keeping in our mind the definition of perfect graphs [12] as depicted in the beginning of this section and considering Theorems 6, 7, we can obtain the perfectness of the graph $\mathrm{\Gamma }({\mathcal{S}}_M)$ as in the following corollary.

Remark 1 Since $\chi (\mathrm{\Gamma }({\mathcal{S}}_M))=\omega (\mathrm{\Gamma }({\mathcal{S}}_M))=1+\lceil \frac{n-1}{2}\rceil$, the graph $\mathrm{\Gamma }({\mathcal{S}}_M)$ is perfect.

Notice that for $\mathrm{\Gamma }({\mathcal{S}}_{M_5})$ and $\mathrm{\Gamma }({\mathcal{S}}_{M_6})$ as drawn in Figures 1 and 2(ii), respectively, we have $\chi (\mathrm{\Gamma }({\mathcal{S}}_{M_5}))=3=\omega (\mathrm{\Gamma }({\mathcal{S}}_{M_5}))$ and $\chi (\mathrm{\Gamma }({\mathcal{S}}_{M_6}))=4=\omega (\mathrm{\Gamma }({\mathcal{S}}_{M_6}))$, respectively.

We recall that any graph G is called Berge if no induced subgraph of G is an odd cycle of length of at least five or the complement of one (see [13]). The following lemma proved by Chudnovsky et al. in [14] figures out the relationship between perfect and Berge graphs. (This lemma is named strong perfect conjecture in some studies.)

Lemma 2 ([14])

A graph is perfect if and only if it is Berge.

By using this relationship depicted in Lemma 2, one can also prove the perfectness of $\mathrm{\Gamma }({\mathcal{S}}_M)$ as in the following result.

Theorem 8 Let ${\mathcal{S}}_M$ be a monogenic semigroup as in (1). Then the graph of $\mathrm{\Gamma }({\mathcal{S}}_M)$ is perfect.

Proof Assume that any induced subgraph of $\mathrm{\Gamma }({\mathcal{S}}_M)$ contains an odd cycle $C_{2k+1}$ and its number of vertices are $2k+1$ ($k\ge 2$). Let us assume that these vertices are $x^{a_1},x^{a_2},\dots ,x^{a_{2k+1}}$ in $C_{2k+1}$ such that $x^{a_1}-x^{a_2}-x^{a_3}-\cdots -x^{a_{2k+1}}-x^{a_1}$, where $a_1<a_2<\cdots <a_{2k+1}$. Since $x^{a_i}{x}^{a_{i+1}}\in E(\mathrm{\Gamma }({\mathcal{S}}_M))$ ($1\le i\le 2k$) and $x^{a_1}{x}^{a_{2k+1}}\in E(\mathrm{\Gamma }({\mathcal{S}}_M))$, by the definition of $\mathrm{\Gamma }({\mathcal{S}}_M)$, we get $x^{a_1}{x}^{a_j}\in E(\mathrm{\Gamma }({\mathcal{S}}_M))$, $3\le j\le 2k$ as $a_1+a_j>a_1+a_2\ge n+1$. This implies that no odd cycle induced subgraph of length of at least 5 is in $\mathrm{\Gamma }({\mathcal{S}}_M)$, which is a contradiction. Finally, by Lemma 2, we conclude that $\mathrm{\Gamma }({\mathcal{S}}_M)$ is perfect. □

Moreover, by [12], since the complement of a perfect graph is also perfect, we also obtain that the complement $\overline{\mathrm{\Gamma }({\mathcal{S}}_M)}$ of $\mathrm{\Gamma }({\mathcal{S}}_M)$ is perfect. As a final note, we may refer to [10, 15] for some other properties of perfect graphs which are clearly satisfied for $\mathrm{\Gamma }({\mathcal{S}}_M)$.

In this section, by using the spectral graph theory, we count the number of triangles in the graph $\mathrm{\Gamma }({\mathcal{S}}_M)$ associated with ${\mathcal{S}}_M$ as defined in (1).

We recall that the notation $t(\mathrm{\Gamma }({\mathcal{S}}_M))$ denotes the number of triangles for any simple undirected graph $\mathrm{\Gamma }({\mathcal{S}}_M)$. In fact, the following lemma on $t(\mathrm{\Gamma }({\mathcal{S}}_M))$ will be needed for our main result of this section.

Lemma 3 ([16])

where $|N_i\cap N_j|$ denotes the cardinality of common neighbors.

Before presenting this result, let us consider the adjacency matrix (here the first row and column correspond to vertex $x^n$, the second row and column correspond to vertex $x^{n-1},\dots$ , etc.) of $\mathrm{\Gamma }({\mathcal{S}}_M)$ which is defined by its entries $a_{ij}=1$ if $x^i{x}^j\in E(\mathrm{\Gamma }({\mathcal{S}}_M))$ and 0 otherwise (see $A(\mathrm{\Gamma }({\mathcal{S}}_M))$ in Figure 3). Since $A(\mathrm{\Gamma }({\mathcal{S}}_M))$ is symmetric, its eigenvalues are real. Without loss of generality, we can write them as ${\lambda }_1\ge {\lambda }_2\ge \cdots \ge {\lambda }_n$ and call them the eigenvalues of $\mathrm{\Gamma }({\mathcal{S}}_M)$. Moreover, the second power $A^2(\mathrm{\Gamma }({\mathcal{S}}_M))$ of the adjacency matrix is written as the form given in Figure 4.

 □

From the above, we certainly have

and

By continuing these calculations, we finally reach

We have two possible cases for n:

Using (5) and (6) in $t(\mathrm{\Gamma }({\mathcal{S}}_M))=\frac{1}{6}{\sum }_{i=1}^n{c}_{i,i}$, we get the result, as desired.

By Figures 1 and 2, it is quite easy to see that $t(\mathrm{\Gamma }({\mathcal{S}}_{M_4}))$, $t(\mathrm{\Gamma }({\mathcal{S}}_{M_5}))$ and $t(\mathrm{\Gamma }({\mathcal{S}}_{M_6}))$ are equal to 1, 2 and 5, respectively.

For given arbitrary graphs $G_1$ and $G_2$, the Cartesian product $G_1\mathrm{\square }{G}_2$ is defined as the graph on the vertex set $V(G_1)\times V(G_2)$ with vertices $u=(u_1,u_2)$ and $v=(v_1,v_2)$ which are connected by an edge if and only if either $u_1=v_1$ and $u_2{v}_2\in E(G_2)$ or $u_2=v_2$ and $u_1{v}_1\in E(G_1)$. This subject has been studied extensively by several authors (see, for instance, [17–19]).

respectively, as given in (1). Without loss of generality, we can assume that $n\ge m$. In this section, the following results will be dealt with: the diameter, girth, chromatic number and clique number of the graph $\mathrm{\Gamma }({\mathcal{S}}_M^1)\mathrm{\square }\mathrm{\Gamma }({\mathcal{S}}_M^2)$.

For simplicity, the graph $\mathrm{\Gamma }({\mathcal{S}}_M^1)\mathrm{\square }\mathrm{\Gamma }({\mathcal{S}}_M^2)$ will be denoted by a single letter $\mathcal{T}$.

Proof Since the vertex $(x_1,x_2)$ of the graph $\mathcal{T}$ definitely has neighborhoods, the diameter can be figured out by considering the distance between $(x_1,x_2)$ and one of the other vertices in $V(\mathcal{T})$. It is easily deduced that $(x_1,x_2)$ is just adjacent to the vertices $(x_1,x_2^m)$ and $(x_1^n,x_2)$. However, for $1\le i<n$ and $1\le j<m$, since $x_1^n\cdot x_1^i=0$ and $x_2^j\cdot x_2^m=0$, we clearly obtain each of $(x_1,x_2^m)$ and $(x_1^n,x_2)$ is adjacent to the vertex $(x_1^n,x_2^m)$. These facts can be shown systematically as

Therefore, $diam(\mathcal{T})=4$, as required. □

Proof Since $x^n\cdot x^{n-1}=0$, $x^{n-1}\cdot x^2=0$ and $x^n\cdot x^i=0$, we clearly have $(x_1^n,x_2^m)-(x_1^n,x_2^{m-1})-(x_1^n,x_2^2)-(x_1^n,x_2^m)$. This gives the result. □

We note that the chromatic number of the Cartesian product of simple graphs $G_1$ and $G_2$ satisfies the equality $\chi (G_1\mathrm{\square }{G}_2)=max\{\chi (G_1),\chi (G_2)\}$ (cf. [20]). Now, we replace $G_1$ by ${\mathcal{S}}_M^1$ and $G_2$ by ${\mathcal{S}}_M^2$, and then considering Theorem 6, we clearly obtain the following result about the chromatic number for the graph $\mathcal{T}$.

Furthermore, we also get the next theorem for the clique number of $\mathcal{T}$.

i.e., for all related powers i, j, a and b, the subgraph will be complete if $(x_1^i,x_2^j)(x_1^a,x_2^b)\in E(\mathcal{T})$.

Hence, we obtain $\omega (\mathcal{T})=1+\lceil \frac{n-1}{2}\rceil$ as required. □

Remark 2 For any two graphs $G_1$ and $G_2$, it is presented in [17] that $\omega (G_1\mathrm{\square }{G}_2)\ge max\{\omega (G_1),\omega (G_2)\}$. However, by considering Theorems 11 and 13, since $\chi (\mathcal{T})=\omega (\mathcal{T})=1+\lceil \frac{n-1}{2}\rceil$, we obtain the strict equality $\omega (\mathcal{T})=max\{\omega (\mathrm{\Gamma }({\mathcal{S}}_M^1)),\omega (\mathrm{\Gamma }({\mathcal{S}}_M^2))\}$ for our special graphs studied in this paper.