On the Semigroup Whose Elements Are Subgraphs of a Complete Graph

Let Kn be a complete graph on n vertices. Denote by SKn the set of all subgraphs of Kn. For each G, H ∈ SKn, the ring sum of G and H is a graph whose vertex set is V(G) ∪ V(H) and whose edges are that of either G or H, but not of both. Then SKn is a semigroup under the ring sum. In this paper, we study Green’s relations on SKn and characterize ideals, minimal ideals, maximal ideals, and principal ideals of SKn. Moreover, maximal subsemigroups and a class of maximal congruences are investigated. Furthermore, we prescribe the natural order on SKn and consider minimal elements, maximal elements and covering elements of SKn under this order.


Introduction and Preliminaries
One of several ways to study the algebraic structures in mathematics is to consider the relations between graph theory and semigroup theory known as algebraic graph theory.It is a branch of mathematics concerning the study of graphs in connection to semigroup theory in which algebraic methods are applied to problems about graphs.Cayley graphs of semigroups are special graphs such that many authors have widely studied, see for examples [1][2][3].Studying on characterization of those graphs is a way of considering the relations between graphs and semigroups in the sense that such graphs are constructed from semigroups.On the other hand, the construction of a semigroup from a given graph is also interesting to study.However, there are no authors considering the properties of semigroups which are constructed from graphs.Certain special types of connected graphs are also interesting to study and look for some applications, especially a complete graph which is a graph in which every two distinct vertices are adjacent.Some authors considered a complete graph for applying its structure to complete their research, see for examples [4,5].Furthermore, an algebraic formation on connected graphs has been studied in the sense of defining some binary operations among a pair of such graphs.Several authors investigated some properties of families of graphs together which graph operations, see for examples [6][7][8].
In this paper, we consequently construct a new semigroup from a complete graph and study some valuable properties of such a semigroup.We need to consider that all sets mentioned in this paper are assumed to be finite sets.Some basic preliminaries, useful notations, and valuable mathematical terminologies needed in what follows are prescribed.Note that a graph G is an order pair (V(G), E(G)) of a nonempty vertex set V(G) and an edge set E(G).For all undefined notions and notations, we refer the reader to [9,10].Now, we give the description of the semigroup focused in this paper.Let K n be a complete graph on n vertices and SK n the set of all subgraphs of K n .For each G, H ∈ SK n , the ring sum G ⊕ H of G and H is a graph whose vertex set V(G ⊕ H) = V(G) ∪ V(H) and whose edges are that of either G or H, but not of both, that is, It is easy to verify that (SK n , ⊕) is a commutative semigroup.For convenience, we write GH instead of G ⊕ H.
Throughout this paper, we shall denote the set of n vertices of K n by the set X n = {v 1 , v 2 , . . ., v n }.For each a nonempty subset A of X n , denote by φ A a graph with a vertex set V(φ A ) = A and E(φ A ) = ∅ which is called an empty graph.For convenience, if A = {v}, then we write φ v instead of φ {v} .We obviously obtain that G 2 = φ V(G) and Gφ A = G for every G ∈ SK n and A ⊆ V(G).
In this section, we consider the regularity and Green's relations on the semigroup SK n .Moreover, we also determine the rank of SK n .Proposition 1. SK n is a regular semigroup.
Proof.Let G be an element of SK n .We will show that we conclude that G 3 = G which implies that G is regular.
Moreover, we observe that the set of all empty graphs in SK n forms the set of all idempotents of SK n , denoted by E (SK n ), that is, Clearly, |E (SK n )| = 2 n − 1.Furthermore, since SK n is regular and its idempotents commute, it follows from Theorem 5.1.1 [9] (p.145) that SK n is an inverse semigroup.
Next, we will describe Green's relations on SK n .
Conversely, assume that V(H) ⊆ V(G).Define K to be the graph with the vertex set V(K) = V(G) and the edge set Therefore, G = KH.
Furthermore, we can directly obtain that L = R = H = D = J since SK n is commutative.
Given a nonempty subset A of a semigroup S, denote by A the subsemigroup of S generated by A. The rank of S, denoted by rank(S), is the minimum cardinality of a generating set for a semigroup S.
In order to consider the rank of SK n , we shall denote by H[e] an induced subgraph of H induced by e where e ∈ E(H).Let T ij denote a graph in SK n with V(T ij ) = {v i , v j } where i = j and E(T ij ) = {v i v j }.T[e].This means that T can be written as a product of some elements of Moreover, we can easily observe that both of H ∈ N and G ∈ M cannot be written as a product of other elements in SK n .Therefore, all elements in M ∪ N must belong to every generating set of SK n .Consequently,

Ideals of SK n
This section presents the characterizations of ideals, minimal ideals, maximal ideals, and principal ideals of SK n .
Let C be a nonempty subset of a power set P(X n ).The set C is said to be an upper set of P(X n Now, we present the characterization of ideals of SK n as follows.
Theorem 2. The ideals of SK n are precisely the sets where C is an upper set of P(X n ).
Proof.Let C be an upper set of P(X n ).We will show that I C is an ideal of SK n .Since C = ∅, we get Conversely, let I be any ideal of SK n and let A = {V(G) : G ∈ I}.Then A is a nonempty subset of P(X n ).We will prove that A is an upper set of

This certainly completes the proof of our assertion.
In what follows, we define a subset of SK n , which plays an essential role for characterizing minimal ideals and maximal ideals of SK n .
The following lemma shows some facts about ideals of SK n which are useful for proving the next theorem.Lemma 1.Let I be an ideal of SK n .If S(1) ⊆ I, then I = SK n .
Proof.We assume that S(1) ⊆ I. Let T ∈ SK n .For each v ∈ V(T), we obtain that T = Tφ v .Since φ v ∈ S(1) ⊆ I and I is an ideal of SK n , we have T ∈ I which certainly implies that I = SK n , as required.
An ideal M of a semigroup S is said to be minimal if every ideal I of S contained in M coincides with M. Further, M is said to be maximal if every proper ideal of S containing M coincides with M.
The following results describe the characterizations of minimal ideals and maximal ideals of SK n , respectively.Theorem 3. S(n) is the unique minimal ideal of SK n .
Proof.We first show that S(n) is an ideal of SK n .It is easy to investigate that {X n } is an upper set.By Theorem 2, we obtain that I {X n } is an ideal of SK n . Consider we can conclude that S(n) is an ideal of SK n .Next, let I be an ideal of SK n such that I ⊆ S(n).
Let H ∈ S(n).Thus V(H) = V(G) and H = Hφ V(G) ∈ I for any G ∈ I which implies that S(n) ⊆ I. Therefore, S(n) = I, that is, S(n) is a minimal ideal of SK n .We now let J be a minimal ideal of SK n and G ∈ J. Hence φ V(G) ∈ J and φ X n = φ X n φ V(G) ∈ J. Let K ∈ S(n).Then K = Kφ X n ∈ J, we obtain that S(n) ⊆ J.It follows that S(n) = J by the minimality of J. Theorem 4. Maximal ideals of SK n are precisely the sets SK n \{φ v } where v ∈ X n .
Proof.We first prove that SK n \{φ v } is an ideal of SK n .Let T denote the set SK n \{φ v } where v ∈ X n .Let G ∈ T and H ∈ SK n .Suppose to the contrary that GH Next, let I be a maximal ideal of SK n .If S(1) ⊆ I, then I = SK n by Lemma 1 which contradicts to the maximality of I. Hence S(1) I, that is, there exists φ v / ∈ I for some v ∈ X n which implies that Let S be a semigroup and a ∈ S. The smallest ideal of S containing a is S 1 aS 1 where S 1 is the monoid obtained from S by adjoining an identity 1 if necessary.We shall call it the principal ideal of S generated by a.The following theorem provides the necessary and sufficient conditions for ideals of SK n to be principal.Theorem 5. I C is a principal ideal of SK n if and only if there exists unique A ∈ C such that |A| = k where k = min{|C| : C ∈ C} and A ⊆ X for all X ∈ C.

Proof. Assume that
Conversely, assume that the conditions hold.Since A ∈ C, we have φ A ∈ I C which leads to the fact that (SK Consequently, I C is a principal ideal of SK n generated by φ A which completes the proof of our assertion.

Remark 1.
Let A be a nonempty subset of X n .Define A to be the family of all supersets of A. By Theorem 5, we can conclude that I A is a principal ideal of SK n .It is not difficult to verify that if A = B, then Therefore, the number of principal ideals of SK n equals the number of nonempty subsets of X n which equals 2 n − 1, certainly.Example 1.This example illustrates the ideal, minimal ideal, maximal ideal, and principal ideal of SK 3 .All elements of SK 3 are shown in Figure 1 where each block is an L-class of SK 3 .In addition, we observe that

Maximal Subsemigroups and a Class of Maximal Congruences of SK n
This section presents the results of maximal subsemigroups and maximal congruences of SK n .A nonempty proper subset M of a semigroup S is called a maximal subsemigroup if M is a subsemigroup of S and any proper subsemigroup of S containing M must be M. Theorem 6.A maximal subsemigroup of SK n is one of the following forms.
Proof.We have known that SK n \{φ v } is a subsemigroup of SK n for all v ∈ X n by Theorem 4. So we need to prove that SK n \{T ij } is a subsemigroup of SK n for any distinct i, j ∈ {1, 2, . . ., n}.
Then there exists only one of them containing {v i v j }.Without loss of generality, suppose that E(G) Let S be a maximal subsemigroup of SK n .We consider the following two cases.
Then φ v / ∈ S for some v ∈ X n .Thus S ⊆ SK n \{φ v } and hence S = SK n \{φ v } since S is a maximal subsemigroup of SK n .Therefore, S is of the form (i).
Let ρ be a congruence on a semigroup S. We call ρ a maximal congruence if δ is a congruence on S with ρ δ ⊆ S × S implies δ = S × S.
Next, we show that ρ is a maximal congruence on SK n .Assume that δ is a congruence on SK n such that ρ δ ⊆ SK n × SK n .Then there exists (φ v , K) ∈ δ where we obtain by the transitivity of δ that (G, H) = (φ v , H) ∈ δ.Thus δ = SK n × SK n , as required.

Natural Order on SK n
In this section, we prescribe the natural order on SK n and investigate minimal elements and maximal elements of SK n with respect to this order.Furthermore, we consider lower covers and upper covers of elements that are not minimal and maximal, respectively.We also give the necessary and sufficient conditions for the existence of an infimum and a supremum of a nonempty subset of SK n .
On an inverse semigroup S, the natural order is defined in a natural way.For given a, b ∈ S, we define a ≤ b if there exists an idempotent e ∈ S such that a = be.The following theorem characterizes the natural order on SK n .
Remark 2. In fact, the order relation ≤ is compatible with the multiplication on a semigroup S which can be seen in [9] (p.152).
Let (P, ≤) be a partially ordered set.An element a of P is called minimal if for each p ∈ P, p ≤ a implies p = a.For the definition of a maximal element, we can define in an analogous manner of a minimal element.
We now present the characterizations of minimal elements and maximal elements of SK n , respectively.
Proof.Assume that G is minimal.We have known that V(G) ⊆ X n .Suppose that there exists v ∈ X n \V(G).Let H be a graph such that V(H) = V(G) ∪ {v} and E(H) = E(G).Thus V(G) V(H) and then H < G which contradicts to the minimality of G. Hence V(G) = X n .
Conversely, assume that V(G) = X n .Let H ∈ SK n be such that H ≤ G.By Theorem 8, we have X n = V(G) ⊆ V(H) and E(G) = E(H).Hence V(H) = X n = V(G) which leads to H = G.Therefore, G is minimal.
Theorem 10.Let G ∈ SK n .Then G is maximal if and only if either G = φ v for some v ∈ X n or G contains no isolated vertices.
Proof.Assume that G is maximal.Suppose that G = φ v for all v ∈ X n and G contains an isolated vertex, say v, that is, deg(v) = 0. Let H be a graph such that V(H) = V(G)\{v} and E(H) = E(G).Then V(H) V(G) which implies that G < H by Theorem 8.This contradicts the maximality of G.
Conversely, it is easy to verify that G is maximal when G = φ v for all v ∈ X n .Now, we assume that G contains no isolated vertices.Let H ∈ SK n be such that G ≤ H. Then V(H) ⊆ V(G) and E(H) = E(G).Let v ∈ V(G).Since deg(v) > 0, there exists u ∈ V(G) such that vu ∈ E(G) = E(H), and thus v ∈ V(H).Hence V(H) = V(G) which leads to H = G.Therefore, G is maximal in SK n .
Let (P, ≤) be a partially ordered set.A lower cover of p ∈ P is an element l of P such that l < p and there is no l ∈ P in which l < l < p.An upper cover of p ∈ P is an element u ∈ P such that p < u and there is no u ∈ P in which p < u < u.
The following lemma describes the existence of lower covers and upper covers of elements in SK n .
We claim that every element of SK n can be generated by some elements of M ∪ N .Let T ∈ SK n .If T contains isolated vertices, then those isolated vertices can be generated by corresponding elements in N .So we now consider in the case where T has no isolated vertices.It is not difficult to verify that the set of all subgraphs T[e] where e ∈ E(T) is a subset of M and T = ⊕ e∈E(T)