1. Introduction

The elements of a group of order two play a very important role not only in group theory but in other branches of mathematics, they are known as involutions. We call elements of order greater than two as non-involutions in this paper. The structure called −set is constructed with the concept of inverses and reveal some properties related to involutions ^[7]. In fact, a group has only one −set if and only if it is an elementary abelian 2-group. A subset D of a group G is a −set whenever every element of G not in D has its inverse in D. This paper shows results that would lead to the comparison of the numbers of non-involutions of two arbitrary groups. We study connections of structural-preserving mappings between groups and their corresponding −set families.

We borrow concepts and notations from set theory ^[5]. Let X and Y be sets, then is the complement of in . If is a function with then called the image of in The cardinality of a set is denoted by . We denote the set of all involutions of a group together with the identity element by ; that is,

A −set of group is a minimum − set if and only if the inverse of each is not in ^[1]. Note that for a finite group this idea coincides with the minimum −sets mentioned in ^[6]. We write as the family of all −sets of a group and the subset containing all minimum −sets ^[1]. It was shown in ^[7] that is a semigroup with respect to union of sets.

We deviate a little to discuss the motivation of −set and some related literature. The definition of −set is based on dominating sets of graphs. Let be a graph and is said to dominate if for any there exists such that (see ^[2]). As mentioned in ^[1], a special type of graph constructed from a group was introduced by Kandasamy and Smarandache ^[4] in 2009. An identity graph of a nontrivial group is an undirected graph formed by adjoining every non-identity element to the identity e of and are connected whenever In view of identity graphs of finite groups, the points contained in a minimum −set form a special type of induced subgraph called stars ^[1]. Hence, we can view as a family of stars related to the group.

2. Results

We start by showing how can be generated from the corresponding .

Proposition 1 Let G be a group. Then generates as a semigroup. Moreover, if

where

Proof: Let be in . If then and we only have one −set in this case. That is, . Assume and denote Consider an nonempty subset of such that, for each We observe that can be expressed as

where and are in Thus, generates

We remark that a minimum −set cannot be written as a union of two distinct −sets. Let x be in G. Then we write

and

The following lemma in ^[7] gives a certain characterization of the involutions in G.

Lemma 1 ^[7] Let x be a non-identity element of a group G. Then x is an involution if and only if

The following proposition is a refinement of Lemma 1.

Proposition 2 Let G be a nontrivial group. A non-identity element x in G is an involution if and only if

Proof: Let x be an involution in G. Since then Suppose Let and by Proposition 1, where and are elements of. By assumption, and are both in . Hence, This means that and by Lemma 1, is an involution.

The proposition below proves that an isomorphism of families of −sets preserves the minimality property.

Proposition 3 Let G and H be groups and be a semigroup isomorphism. Then D is a minimum −set of G if and only if is a minimum −set of H.

Proof: The case is trivial. Suppose Assume is minimum while is not. Then there exists at least one pair both in As in the proof of Proposition 1, there exist and in such that with and It follows that there exist distinct and in such that and But this implies that and so This is a contradiction to a remark following Proposition 1.

For the converse, suppose is a minimum while is not. There exist distinct and in where Hence, where this is absurd.

Proposition 4 Let G and H be groups and be a semigroup isomorphism. If and then where

Proof: Suppose D is in not containing an element x of G. Then is an element of where is the only pair of inverses in this −set. As in the proof of Proposition 1,

where is also in The homomorphic property of implies that

where in by Proposition 3. Further, there must exist in where (WLOG)

Suppose there exists another element z which shares the same characteristic with y.

We may assume that and are in while and are in As a consequence of the above argument, can be expressed as

where the three factors are distinct elements of . By the surjective property of and Proposition 3, there exist and in such that

This means that

By the properties of , we have

and so

Since the three factors on the right handside of equation are distinct elements of we get at least two pairs of inverses. But we only have and from the left handside of , this is absurd. Hence, and must be the only pair of inverses in and so

Let us now state and prove the main result of this paper.

Theorem 1 Let and be groups with Then is isomorphic to if and only if there exists a bijection such that for any x in

Proof: Let be an isomorphism. We form the bijection such that for any in Firstly, we choose a fix in Let be in then either or If then the pair and is unique in By Proposition 4, there exists a unique pair in where We can now form and If then and we proceed as in the first case. Therefore, if then there exists a unique such that and

We show that is an injection by way of contradiction. Suppose that in such that Since a is mapped to and to where then Now, we form and in :

• If then

• If then

• If then

• If then

Hence, we have the following cases:

Case 1: and

•

•

Case 2: and

•

•

Case 3: and

•

•

Case 4: and

•

•

Note that in any of the cases above,

for some

Now, the only pair of inverses in is while only in Let Since thenand Hence,

Since is injective, we have

This further implies that and are both in this is a contradiction.

To show that it is surjective, assume an element Using in either or If , then and the pair is unique in Further,

where is a minimum −set of By Proposition 4 and the isomorphism we have in which contains a unique pair However,

where is in WLOG, we may have and . Thus, we take and in which

On the other hand, given that , then We proceed as above knowing that is the only pair of inverses in By following the same pattern of reasoning, we will still obtain a unique pair and from in which we can write and Hence, is surjective. Summing up, we have the required bijection.

For the converse, suppose there exists a bijection such that for any in We form a semigroup isomorphism Let be in then where We define by

where is the image of with respect to The verification that is an isomorphism is a routine.

We prove more properties involving morphisms and −sets.

Proposition 5 Let be a monomorphism of groups G and H. Then

i. If is a −et of H then D is a − set of G;

ii. If is a minimum −set of H then D is a minimum − set of G.

Proof: (i) Let be a −set of H and Since is injective, then must not be in By assumption, is in This implies that

(ii) Suppose is a minimum −set of By part (i), is a −set of If then By assumption, Thus, and this proves our claim.

We observe that if s a singleton semigroup (that is, ) then the following hold true vacuously.

Lemma 2 Let be a mapping of groups G and H where If an isomorphism has the property that for and then

Proof: Let and be an isomorphism such that with as above. From the proof of Proposition 4,

where in Now we have implying that cannot be in Otherwise, we will get which is absurd.

Theorem 2 Let be a monomorphism of groups G and H where Then there exists an isomorphism such that for every and if and only if

Proof: Suppose is an isomorphism such that for every and If then for some Thus, Assuming that would imply which means since is injective. This is a contradiction. Hence, must be in Now, if then choose a minimum −set of say not containing . By Proposition 4 and

for some By property of

As in Lemma 2, It is now evident that

For the converse, assume that We now have a bijection such that for all By Theorem 1, we have the isomorphism defined by

where with for some Let be in and Suppose where Then But we have Consequently,

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