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 first recall that for any simple graph G, the distance (length of the shortest path) between two vertices u, v of G is denoted by {d}_{G}(u,v). Actually, the diameter of G is defined by
diam(G)=max\{{d}_{G}(x,y):x\text{and}y\text{are vertices of}G\}.
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}^{n1}=0, {x}^{n1}\cdot {x}^{2}=0 and {x}^{n}\cdot {x}^{2}=0, we then have {x}^{n}{x}^{n1}{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.
Theorem 3 For any monogenic semigroup {\mathcal{S}}_{M} as given in (1), the maximum and minimum degrees of \mathrm{\Gamma}({\mathcal{S}}_{M}) are
\mathrm{\Delta}(\mathrm{\Gamma}({\mathcal{S}}_{M}))=n1\phantom{\rule{1em}{0ex}}\phantom{\rule{0.25em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}\delta (\mathrm{\Gamma}({\mathcal{S}}_{M}))=1.
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}))=n1.
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 ,n1\}, 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.
Theorem 4 Let {\mathcal{S}}_{M} be a monogenic semigroup as given in (1). Then the degree sequence and irregularity index of \mathrm{\Gamma}({\mathcal{S}}_{M}) are given by
DS(\mathrm{\Gamma}({\mathcal{S}}_{M}))=\{1,2,3,\dots ,\lfloor \frac{n}{2}\rfloor 1,\lfloor \frac{n}{2}\rfloor ,\lfloor \frac{n}{2}\rfloor ,\lfloor \frac{n}{2}\rfloor +1,\lfloor \frac{n}{2}\rfloor +2,\dots ,n2,n1\}
and \mathit{MWB}(\mathrm{\Gamma}({\mathcal{S}}_{M}))=n1, 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}^{n1}, 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

the degree of vertex {x}^{\lfloor \frac{n}{2}\rfloor 1} is \lfloor \frac{n}{2}\rfloor 1 and

the degree of vertex {x}^{\lfloor \frac{n}{2}\rfloor} is \lfloor \frac{n}{2}\rfloor, but

the vertex {x}^{\lfloor \frac{n}{2}\rfloor +1} has the same degree as the vertex {x}^{\lfloor \frac{n}{2}\rfloor}.
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}))=n1, 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}).
Theorem 5 Let {\mathcal{S}}_{M} be a monogenic semigroup as given in (1). Then
\gamma (\mathrm{\Gamma}({\mathcal{S}}_{M}))=1.
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.