On the Planarity of Graphs Associated with Symmetric and Pseudo Symmetric Numerical Semigroups

Abstract

Let $S(m,e)$ be a class of numerical semigroups with multiplicity m and embedding dimension e. We call a graph $G_S$ an $S(m,e)$-graph if there exists a numerical semigroup $S\in S(m,e)$ with $V\left(G_S\right)=\{x:x\in g\left(S\right)\}$ and $E\left(G_S\right)=\{xy\iff x+y\in S\}$, where $g\left(S\right)$ denotes the gap set of S. The aim of this article is to discuss the planarity of $S(m,e)$-graphs for some cases where S is an irreducible numerical semigroup.

1. Introduction and Preliminaries

In the last couple of decades, researchers have been assigning graphs to various kinds of algebraic structures, which opens new horizons to study algebraic structures with the help of graphs’ theoretic properties and vice versa. The first paper in this direction was the work by Beck [1], where he assigned a graph with the zero divisor elements of a commutative ring and called it a zero divisor graph. After that, many generalizations of this concept were provided by different researchers. Presently, assigning a graph to an algebraic object and studying the interplay between the properties of algebraic objects and with properties of the graph is an active area of research. The most studied concepts among these are the zero divisor graph [2], extended zero divisor graph [3], Cayley graph [4], nilpotent graphs [5], etc. Recently, Binyamin et al. [6] assigned a graph to the numerical semigroup and studied some properties of this graph. In a similar way, a graph is assigned to the ideal of a numerical semigroup by Binyamin et al. [7] who studied its metric dimension [8] and planarity [9].

Let $\mathbb{N}$ be the set of non-negative integers. A subset $S\subset \mathbb{N}$ is said to be numerical semigroup if $0\in S$, $x+y\in S$ for all $x,y\in S$, and $\mathbb{N}\setminus S$ is finite. The least positive integer $x\in S$ is called the multiplicity of S and the set $\mathbb{N}\setminus S$ is called the gap set of S. We use the notations $m\left(S\right)$ and $g\left(S\right)$ to denote the multiplicity and gap set of S, respectively. The number of elements of $g\left(S\right)$ is called the genus of S. The largest integer that belongs to the set $g\left(S\right)$ is called the Frobenius number and is denoted by $F\left(S\right)$ or simply F. A numerical semigroup is said to be symmetric if for any $x\in \mathbb{N}\setminus S$, then $F-x\in S$. Similarly, a numerical semigroup is said to be pseudo symmetric if for any $x\in \mathbb{N}\setminus S$, either $F-x\in S$ or $x=\frac{F}{2}$. An important property of numerical semigroup is that it is always finitely generated and there exists a minimal system of generators of S. Let $x_1,x_2,\dots ,x_n$ be a minimal system of generators of S, then we write $S=\langle x_1,x_2,\dots ,x_n\rangle$. The number of elements in the minimal system of generators of S is called the embedding dimension of S and is denoted by $e\left(S\right)$. A numerical semigroup is called irreducible if it cannot be written as an intersection of two numerical semigroups containing it properly. It is well-known that an irreducible numerical semigroup is either symmetric or pseudo-symmetric [10]. To read more about the theory of numerical semigroups, the readers can see the book by [11].

Let G be a simple graph with the vertex set and edge set denoted by V and E, respectively. The degree of a vertex $x\in V$ is the number of edges incident to it. A graph is called complete if every pair of vertices has an edge between them. A complete graph on n vertices is denoted by $K_n$. A complete subgraph of a graph G is called clique and the clique of largest possible size in G is called a maximum clique of G. The number of vertices in the maximum clique of G is called clique number and is denoted by $cl\left(G\right)$. A graph G is called bipartite if its vertex set V can be partitioned into two sets, $V_1$ and $V_2$, and the edges are from elements of $V_1$ to the elements of $V_2$. If all the vertices of $V_1$ are adjacent to all vertices of $V_2$, then G is called a complete bipartite graph. If $\mid V_1\mid =m$ and $\mid V_1\mid =n$, $K_{m,n}$ denotes the complete bipartite graph. A graph is called planar if it can be embedded in a plane. In other words, a graph is planar if it can be drawn in a plane such that its edges intersect at end points (or no edges cross each other). A graph H is called minor of G if H can be formed by deleting edges and vertices and by contracting edges. The planarity of a graph can be checked using the famous Wagner’s Theorem which states that a graph G is planar if and only if it contains neither $K_5$ nor $K_{3,3}$ as a graph minor [12].

We use the notation $S(m,e)$ to denote the class of numerical semigroups with multiplicity m and embedding dimension e. Following the idea of Binyamin et al. [6], a graph G can be assigned to any numerical semigroup S by considering the vertex set of G as the gap set $g\left(S\right)$ and any two vertices are adjacent if their sum belongs to S. In this work, we introduced the notion of $S(m,e)$ graph. We call a graph $G_S$ an $S(m,e)$-graph if there exist a numerical semigroup $S\in S(m,e)$ with $V\left(G_S\right)=\{x:x\in g\left(S\right)\}$ and $E\left(G_S\right)=\{xy\iff x+y\in S\}$, where $g\left(S\right)$ denotes the gap set of S. Now, finding a closed subset of $g\left(S\right)$ is equivalent to finding a clique of graph $G_S$ (which is very difficult to compute in general). In this article, we computed a clique of graph $G_S$ of order 5 and as a consequence, we deduced that the graph $G_S$ is non-planar. The aim of this article is to discuss the planarity of $S(m,e)$-graphs for some cases when S is an irreducible numerical semigroup.

2. Planarity of Graphs Associated with Numerical Semigroups of Embedding Dimension 2

In this section, we discuss the planarity of the graph $G_S$ associated with the numerical semigroup $S\in S(m,2)$. It is well-known that every numerical semigroup of embedding dimension 2 is symmetrical and for any $S\in S(m,2)$, we have $\mid g\left(S\right)\mid =\frac{F+1}{2}$. We prove that if $\mid G_S\mid >4$, then $G_S$ is always non-planar. The following results can be immediately obtained from Theorem 1 [6].

Proposition 1

([6]). Every $S(m,e)$-graph for $m>3$, is not a complete graph.

Proposition 2

([6]). If $S(3,2)$-graph is complete then $G_S\cong K_3$ or $K_4.$

Proposition 3

([6]). If $S(3,3)$-graph is complete then $G_S\cong K_2$ or $K_3$.

Proposition 4

([6]). Every $S(2,2)$-graph is complete.

Lemma 1.

Let $G_S$ be a graph associated with $S=\langle m,b\rangle$. If $|G_S |>4$, then one of the following conditions hold:

• If $m=3$ then $\{F,F-3,F-6,F-9,F-b\}\subseteq g(\langle m,b\rangle ).$
• If $m\ge 4$ then $\{F,F-m,F-2m,F-b,F-(m+b)\}\subseteq g(\langle m,b\rangle ).$

Proof. 

If $m=3$ then clearly $0,3,6,9,b\in S$, and therefore $F,F-3,F-6,F-9,F-b\notin S$. Furthermore, $|G_S |>4$ gives $F\ge 11$ and $b\ge 7$. Please note that

$$F-b=b-3.$$

This implies

$$F-3,F-6,F-9,F-b>0,$$

and therefore

$$F,F-3,F-6,F-9,F-b\in g(\langle m,b\rangle ).$$

Since $0,m,2m,b,b+m\in S$, therefore, $F,F-m,F-2m,F-b,F-(m+b)\notin S$. Also

$$F-m=mb-b-2m=(m-1)b-2m,$$

$$F-2m=mb-b-3m=(m-1)b-3m,$$

$$F-b=mb-m-2b=(m-2)b-m,$$

and

$$F-(m+b)=mb-2m-2b=(m-2)b-2m.$$

If $m\ge 4$, then clearly

$$F-m,F-2m,F-b,F-(m+b)>0.$$

This implies

$$F,F-m,F-2m,F-b,F-(m+b)\in g(\langle m,b\rangle ).$$

□

Theorem 1.

Let $G_S$ be an $S(m,2)$-graph, where $m\ge 2$. If $|G_S|>4$ then $cl\left(G_S\right)\ge 5$.

Proof. 

We may assume that $S=<m,b>$. If $m=2$ then from Proposition 4, we have $G_S\cong K_n$ with $n\ge 5$, as $|G_S |>4$. This gives $cl\left(G_S\right)\ge 5$ in this case.

Now if $m=3$ then from Lemma 1, we have

$$\{F,F-3,F-6,F-9,F-b\}\subseteq g(\langle m,b\rangle ).$$

Clearly $2F-3,2F-6,2F-9,2F-b\in S$. Now we need to show

$$2F-12,2F-15,2F-(3+b),2F-(6+b),2F-(9+b)\in S.$$

Since $|G_S |>4$ therefore $F\ge 11$ and $b\ge 7$. Please note that

$$F-(2F-12)=12-F\notin S,$$

$$F-(2F-15)=15-F\notin S,$$

$$F-(2F-(3+b)=(3+b)-F=3-(F-b)\notin S,$$

$$F-(2F-(6+b)=(6+b)-F=9-b\notin S,$$

$$F-(2F-(9+b)=(6+b)-F=12-b\notin S.$$

This gives

$$2F-12,2F-15,2F-(3+b),2F-(6+b),2F-(9+b)\in S,$$

and therefore $cl\left(G_S\right)\ge 5$ (see Figure 1).

If $m\ge 4$ then again from Lemma 1, we have

$$\{F,F-m,F-2m,F-b,F-(m+b)\}\subseteq g(\langle m,b\rangle ).$$

Clearly, $2F-m,2F-2m,2F-b,2F-(m+b)\in S$. Please note that

$$F-(2F-3m)=3m-F=m-(F-2m)\notin S,$$

$$F-(2F-(2m+b\left)\right)=2m+b-F=m-(F-(m+b\left)\right)\notin S,$$

$$F-(2F-(3m+b)=3m+b-F.$$

This implies $3m+b-F=4m+(2-m)b$ or $3m+b-F=(4-b)m+2b$. Since $4\le m<b$, therefore, both possibilities give $3m+b-F\notin S$. Furthermore,

$$F-(2F-(m+2b\left)\right)=m+2b-F.$$

We have either $m+2b-F=2m+(3-m)b$ or $3m+b-F=(2-b)m+3b$. Again $4\le m<b$, give $m+2b-F\notin S$. This implies

$$2F-3m,2F-(2m+b),2F-(3m+b),2F-(m+2b)\in S.$$

Consequently, we obtain $cl\left(G_S\right)\ge 5$ (see Figure 2). □

Corollary 1.

For $m\ge 2$, every $S(m,2)$-graph, whose order is greater than 4 is non-planar.

3. Planarity of Graphs Associated with Irreducible Numerical Semigroups of Maximal Embedding Dimension

A numerical semigroup S is said to have a maximal embedding dimension if its multiplicity and embedding dimension are the same. It is proved in [10] that a numerical semigroup of maximal embedding dimension is irreducible if its embedding dimension is either 2 or 3. In this section, we discuss the planarity of the graph $G_S$ in the case $S\in S(3,3)$.

Lemma 2.

Let $S=\langle 3,3+x,3+2x\rangle$, where x is not a multiple of 3. If $\left|g\right(S\left)\right|>6$, then

$$\{F,F-3,F-6,F-9,\frac{F}{2}\}\subseteq g\left(S\right).$$

Proof. 

Please note that $x>5$ and $F>10$, as $\left|g\right(S\left)\right|>6$. This implies $F-3,F-6,F-9>0.$ Since S is pseudo symmetric and $3,6,9\in S$, therefore,

$$\{F,F-3,F-6,F-9,\frac{F}{2}\}\subseteq g\left(S\right).$$

□

Theorem 2.

Let $G_S$ be an $S(m,e)$-graph, where S is an irreducible numerical semigroup of maximal embedding dimension. If $|G_S|>4$ then $cl\left(G_S\right)\ge 4$.

Proof. 

Since S is an irreducible numerical semigroup of maximal embedding dimension, then from Proposition 6 of [10], it follows that either $m=2=e$ or $m=3=e.$ If $m=2=e$, then from Proposition 4, it follows that $cl\left(G_S\right)\ge 5$. Now if $m=3=e$, then from Proposition 7 of [10], we have $S=\langle 3,3+x,3+2x\rangle$, where x is not a multiple of 3.

If $|G_S|=5$ then $x=4$. This implies $S=\langle 3,7,11\rangle$ and $g\left(S\right)=\{1,2,4,5,8\}$ such that $1+4\notin S$ and

$$2+4,2+5,2+8,4+5,4+8,5+8\in S.$$

This gives $cl\left(G_S\right)=4$. Similarly, If $|G_S|=6$ then $x=5$, therefore, $S=\langle 3,8,13\rangle$ and $g\left(S\right)=\{1,2,4,5,7,10\}$. Clearly $cl\left(G_S\right)=4$.

Now if $|G_S |>6$ then from Lemma 2, we have

$$\{F,F-3,F-6,F-9,\frac{F}{2}\}\subseteq g\left(S\right).$$

Clearly, $2F-3,2F-6,2F-9,\frac{3F}{2}\in S$. Now we show that $2F-9,2F-12,\frac{3F}{2}-3,2F-15,\frac{3F}{2}-6,\frac{3F}{2}-9\in S$. This is easy to see that none of $2F-9,2F-12,\frac{3F}{2}-3,2F-15,\frac{3F}{2}-6$ and $\frac{3F}{2}-9$ is equal to $\frac{F}{2}$. Please note that

$$F-(2F-9)=9-F=3-(F-6),$$

$$F-(2F-12)=12-F=3-(F-9),$$

$$F-(2F-15)=15-F,$$

$$F-(\frac{3F}{2}-3)=3-\frac{F}{2},$$

$$F-(\frac{3F}{2}-6)=6-\frac{F}{2}$$

and

$$F-(\frac{3F}{2}-9)=9-\frac{F}{2}.$$

Since $F-6,F-9,\frac{F}{2}\in g\left(S\right)$, therefore, $9-F,12-F,3-\frac{F}{2}\notin S$. Furthermore, since $F\ge 14$, therefore, $15-F,6-\frac{F}{2},9-\frac{F}{2}\notin S$. This implies $cl\left(G_S\right)\ge 5$ (see Figure 3). □

Corollary 2.

Let $G_S$ be an $S(m,e)$-graph, where S is an irreducible numerical semigroup of maximal embedding dimension. If $|G_S|>5$ then $G_S$ is non-planar.

Proof. 

If $m=2=e,$ then the result follows immediately from Theorem 1. If $m=3=e,$ then $S=<3,x+3,2x+3>$ and $F=2x$. Now, if $\mid G_S\mid =6$ then $x=5$ and therefore $S=<3,8,13>.$ This implies $cl\left(G_S\right)=4$ (see Figure 4). By contraction of $e_4{e}_5$ and by removing multiple edges, we obtain the minor of $G_S$ isomorphic to $K_5$ (see Figure 5). Now if $\mid G_S\mid >6$ then from Theorem 2 it follows that $cl\left(G_S\right)\ge 5$. Hence $G_S$ is non-planar. □

4. Planarity of Graphs Associated with Irreducible Numerical Semigroups of Arbitrary Embedding Dimension

In this section, we discuss the planarity of different classes of irreducible numerical semigroups of arbitrary embedding dimensions.

Lemma 3.

Let $G_S$ be an $S(m,e)$-graph, where $S=\langle m,m+1,qm+2q+2,\dots ,qm+(m-1)\rangle$ with $m\ge 2q+3$, $e=m-2q$ and $q\ge 1$. Then,

$$\{F,F-m,F-(m+1),F-(2m+1),F-(qm+2q+2)\}\subseteq g\left(S\right).$$

Proof. 

Since $0,m,m+1,2m+1,qm+2q+2\in S$, therefore, $F,F-m,F-(m+1),F-(2m+1),F-(qm+2q+2)\notin S$. From Lemma 1 of [13], it follows that S is symmetric and $F=2qm+2q+1$. Please note that

$$F=2qm+2q+1=(m+1)+(2q-1)m+2q.$$

Since $q\ge 1$, therefore, $F>m+1$. This implies $F-m,F-(m+1)>0$ and therefore $F-m,F-(m+1)\in g\left(S\right)$. Now consider

$$F-(2m+1)=2qm+2q+1-2m-1=(2q-2)m+2q>0,$$

$$F-(qm+2q+2)=2qm+2q+1-qm-2q-2=qm-1>0.$$

This gives $F-(2m+1),F-(qm+2q+2)\in g\left(S\right)$. Consequently, we obtain the required result. □

Theorem 3.

Let $G_S$ be an $S(m,e)$-graph, where $S=\langle m,m+1,qm+2q+2,\dots ,qm+(m-1)\rangle$ with $m\ge 2q+3$, $e=m-2q$ and $q\ge 1$. Then, $G_S$ is nonplanar.

Proof. 

From Lemma 3, we have

$$\{F,F-m,F-(m+1),F-(2m+1),F-(qm+2q+2)\}\subseteq g\left(S\right).$$

Clearly $2F-m,2F-(m+1),2F-(2m+1),2F-(qm+2q+2)\in S$. We need to show $2F-(3m+1),2F-(3m+2),2F-\left(\right(q+1)m+2q+2),2F-\left(\right(q+1)m+2q+3),2F-\left(\right(q+2)m+2q+3)\in S$. For this, we consider

$$F-(2F-(3m+1\left)\right)=(3m+1)-F=m-(F-(2m+1\left)\right)\notin S,$$

$$F-(2F-(3m+2\left)\right)=(3m+2)-F=(3-2q)m-(2q-1).$$

For $q=1$, we have $F-(2F-(3m+2\left)\right)=m-1$ and for $q>1$, $F-(2F-(3m+2\left)\right)<0$. Both cases give $F-(2F-(3m+2\left)\right)\notin S$.

$$F-(2F-((q+1)m+2q+2\left)\right)=\left(\right(q+1)m+2q+2))-F,$$

$$=m-(F-(qm+2q+2\left)\right)\notin S.$$

$$F-(2F-((q+1)m+2q+3\left)\right)=\left(\right(q+1)m+2q+3))-F,$$

$$=(1-q)m+2\notin S.$$

$$F-(2F-((q+2)m+2q+3\left)\right)=\left(\right(q+2)m+2q+3))-F=(2-q)m+2.$$

For $q=1$, we have $F-(2F-((q+2)m+2q+3\left)\right)=m+2$, for $q=2$, we have $F-(2F-((q+2)m+2q+3\left)\right)=2$ and for $q>2$, $F-(2F-((q+2)m+2q+3\left)\right)<0$. All three cases give $F-(2F-((q+2)m+2q+3\left)\right)\notin S$. This implies $cl\left(G_S\right)\ge 5$ and consequently $G_S$ is non-planar (see Figure 6). □

Lemma 4.

Let $G_S$ be an $S(m,e)$-graph, where $S=\langle m,m+1,(q+1)m+q+2,\dots ,(q+1)m+m-q-2\rangle$ with $m\ge 2q+4$, $e=m-2q-1$ and $q\ge 0$.

• If $q=0$ and $|G_S|>6$, then $\{F,F-m,F-(m+1),F-(m+2),F-(m+3)\}\subseteq g\left(S\right).$
• If $q>0$ then $\{F,F-m,F-2m,F-(m+1),F-(2m+1)\}\subseteq g\left(S\right).$

Proof. 

Since $0,m,m+1,m+2,m+3,2m,2m+1\in S$, therefore, $F,F-m,F-2m,F-(m+1),F-(m+2),F-(m+3),F-(2m+1)\notin S$. From Lemma 3 of [13], it follows that S is symmetric and $F=2(q+1)m-1$.

If $q=0$ and $|G_S |>6$ then $F=2m-1$ with $m>6$. Please note that $F-m=m-1>0$, $F-(m+1)=m-2>0$, $F-(m+2)=m-3>0$ and $F-(m+3)=m-4>0$. This implies

$$\{F,F-m,F-(m+1),F-(m+2),F-(m+3)\}\subseteq g\left(S\right).$$

Now if $q>0$, then $F\ge 4m-1$ with $m\ge 6$. We have $F-m\ge 3m-1>0$, $F-2m\ge 2m-1>0$, $F-(m+1)\ge 3m-2>0$ and $F-(2m+1)\ge 2m-2>0$. This gives

$$\{F,F-m,F-2m,F-(m+1),F-(2m+1)\}\subseteq g\left(S\right).$$

□

Theorem 4.

Let $G_S$ be an $S(m,e)$-graph, where $S=\langle m,m+1,(q+1)m+q+2,\dots ,(q+1)m+m-q-2\rangle$ with $m\ge 2q+4$, $e=m-2q-1$ and $q\ge 0$.

• If $q=0$ and $|G_S|>6$, then $G_S$ is non-planar.
• If $q>0$, then $G_S$ is non-planar.

Proof. 

If $q=0$ and $|G_S |>6$, then from Lemma 4, it follows that

$$\{F,F-m,F-(m+1),F-(m+2),F-(m+3)\}\subseteq g\left(S\right).$$

Clearly, $2F-m,2F-(m+1),2F-(m+2),2F-(m+3)\in S$. Also

$$F-(2F-(2m+1\left)\right)=2m+1-F=m-(F-(m+1\left)\right)\notin S.$$

$$F-(2F-(2m+2\left)\right)=2m+2-F=m-(F-(m+2\left)\right)\notin S.$$

$$F-(2F-(2m+3\left)\right)=2m+3-F=m-(F-(m+3\left)\right)\notin S.$$

$$F-(2F-(2m+4\left)\right)=2m+4-F=5\notin S.$$

$$F-(2F-(2m+5\left)\right)=2m+5-F=6\notin S.$$

Now if $q>0$, then again from Lemma 4, we have

$$\{F,F-m,F-2m,F-(m+1),F-(2m+1)\}\subseteq g\left(S\right).$$

Clearly, $2F-m,2F-2m,2F-(m+1),2F-(2m+1)\in S$. We have to show $2F-3m,2F-(3m+1),2F-(4m+1),2F-(3m+2)\in S$. For this, we consider

$$F-(2F-3m)=3m-F=m-(F-2m)\notin S.$$

$$F-(2F-(3m+1\left)\right)=3m+1-F=m-(F-(2m+1\left)\right)\notin S.$$

$$F-(2F-(4m+1\left)\right)=4m+1-F=m-2\left(\right(1-q)m+1)\notin S.$$

$$F-(2F-(3m+2\left)\right)=3m+2-F=m-2(1-q)m+3\notin S.$$

Both cases implies $cl\left(G_S\right)\ge 5$, therefore, $G_S$ is non-planar (see Figure 7 and Figure 8). □

Lemma 5.

Let $G_S$ be an $S(m,e)$-graph, where $S=\langle m,m+1,(q+1)m+q+2,\dots ,(q+1)m+m-q-3,(q+1)m+(m-1)\rangle$ with $m\ge 2q+5$, $e=m-2q-1$ and $q\ge 0$. Then,

$$\{F,\frac{F}{2},F-m,F-(m+1),F-((q+1)m+q+2)\}\subseteq g\left(S\right).$$

Proof. 

Since $m,m+1,(q+1)m+q+2\in S$, therefore, $F-m,F-(m+1),F-\left(\right(q+1)m+q+2)\notin S$. From Lemma 2 of [14], it follows that S is pseudo-symmetric and $F=2(q+1)m-2$. This implies

$$F-m=(2q+1)m-2>0.$$

$$F-(m+1)=(2q+1)m-3>0.$$

$$F-\left(\right(q+1)m+q+2)=(q+1)m-(q+4)>0,$$

since $m\ge 2q+5$. This gives

$$\{F,\frac{F}{2},F-m,F-(m+1),F-((q+1)m+q+2)\}\subseteq g\left(S\right).$$

□

Theorem 5.

Let $G_S$ be an $S(m,e)$-graph, where $S=\langle m,m+1,(q+1)m+q+2,\dots ,$$(q+1)m+m-q-3,(q+1)m+(m-1)\rangle$ with $m\ge 2q+5$, $e=m-2q-1$ and $q\ge 0$. If $m\ge 6$ then $G_S$ is non-planar.

Proof. 

From Lemma 5, we have

$$\{F,\frac{F}{2},F-m,F-(m+1),F-((q+1)m+q+2)\}\subseteq g\left(S\right).$$

Then $\frac{3F}{2},2F-m,2F-(m+1),2F-((q+1)m+q+2)\in S$. Now consider

$$F-(\frac{3F}{2}-m)=m-\frac{F}{2}\notin S.$$

$$F-(\frac{3F}{2}-(m+1))=m+1-\frac{F}{2}=2-qm\notin S.$$

$$F-(\frac{3F}{2}-((q+1)m+q+2))=(q+1)m+q+2-\frac{F}{2}=q+3\notin S,$$

since $q+3<m$.

$$F-(2F-(m+1\left)\right)=2m+1-F=m-(F-(m+1\left)\right)\notin S.$$

$$F-(2F-((q+2)m+q+2\left)\right)=(q+2)m+q+2-F=(1-m)q+4\notin S.$$

$$F-(2F-((q+2)m+q+3\left)\right)=(q+2)m+q+3-F=(1-m)q+5\notin S.$$

This implies $cl\left(G_S\right)\ge 5$ and therefore $G_S$ is non-planar (see Figure 9). □

Lemma 6.

Let $G_S$ be an $S(m,e)$-graph, where $S=\langle m,m+1,qm+2q+3,\dots ,qm+m-1,$$(q+1)m+q+2\rangle$ with $m\ge 2q+4$, $e=m-2q$ and $q\ge 1$. Then,

$$\{F,\frac{F}{2},F-m,F-(m+1),F-2m\}\subseteq g\left(S\right).$$

Proof. 

Since $m,m+1,2m\in S$, therefore, $F-m,F-(m+1),F-2m\notin S$. From Lemma 4 of [14], it follows that S is pseudo-symmetric and $F=2qm+2q+2$. This implies

$$F-m=(2q-1)m+q+2>0.$$

$$F-(m+1)=(2q-1)m+q+1>0.$$

$$F-2m=(2q-2)m+q+2>0.$$

This gives

$$\{F,\frac{F}{2},F-m,F-(m+1),F-2m\}\subseteq g\left(S\right).$$

□

Theorem 6.

Let $G_S$ be an $S(m,e)$-graph, where $S=\langle m,m+1,qm+2q+3,\dots ,$$qm+m-1,(q+1)m+q+2\rangle$ with $m\ge 2q+4$, $e=m-2q$ and $q\ge 1$. Then,

$$\{F,\frac{F}{2},F-m,F-(m+1),F-2m\}\subseteq g\left(S\right).$$

Proof. 

From Lemma 6, it follows that

$$\{F,\frac{F}{2},F-m,F-(m+1),F-2m\}\subseteq g\left(S\right).$$

Note that $\frac{3F}{2},2F-m,2F-2m,2F-(m+1)\in S$. Now consider

$$F-(\frac{3F}{2}-m)=m-\frac{F}{2}\notin S.$$

$$F-(\frac{3F}{2}-2m)=2m-\frac{F}{2}=(2-q)m-(q+1)\notin S,$$

$$F-(\frac{3F}{2}-(m+1))=m+1-\frac{F}{2}=(1-q)m-q\notin S,$$

since $m>q+1$.

$$F-(2F-3m)=3m-F=m-(F-2m)\notin S.$$

$$F-(2F-(2m+1\left)\right)=2m+1-F=m-(F-(m+1\left)\right)\notin S.$$

$$F-(2F-(3m+1\left)\right)=3m+1-F=(3-2q)m-(2q+1)\notin S,$$

since $m>2q+1$. This gives $cl\left(G_S\right)\ge 5$ and hence $G_S$ is non-planar (see Figure 10). □

5. Conclusions

Numerical semigroups have applications in many fields. One of the important applications of a numerical semigroup is in finding the non-negative solutions of linear diophantine equations. Following the idea of Binyamin et al. [6], we introduced the concept of $S(m,e)$ graph. In this work, we discussed the planarity of $S(m,e)$-graphs in the case when the numerical semigroup is either symmetric or pseudo-symmetric. To answer the planarity of any general $S(m,e)$ graph is still an open problem.