Journal of Mathematics

Volume 2013 (2013), Article ID 915250, 7 pages

http://dx.doi.org/10.1155/2013/915250

Research Article

-Semihypergroups and Regular Relations

S. Mirvakili,¹ S. M. Anvariyeh,² and B. Davvaz²

¹Department of Mathematics, Payame Noor University, Tehran, Iran

²Department of Mathematics, Yazd University, Yazd, Iran

Received 30 October 2012; Accepted 29 December 2012

Academic Editor: Feng Feng

Copyright © 2013 S. Mirvakili et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract.

In this paper we discuss the structure of -semihypergroups. We prove some basic results and present several examples of -semihypergroups. Also, we obtain some properties of regular and strongly regular relations on a -semihypergroup and construct a -semigroup from a -semihypergroup by using the notion of fundamental relation.

1. Introduction

The algebraic hyperstructure notion was introduced in 1934 by the French mathematician Marty [1], at the 8th Congress of Scandinavian Mathematicians. He published some notes on hypergroups, using them in different contexts: algebraic functions, rational fractions, and noncommutative groups. Algebraic hyperstructures are a suitable generalization of classical algebraic structures. In a classical algebraic structure, the composition of two elements is an element, while in an algebraic hyperstructure, the composition of two elements is a set.

Around the 1940s, the general aspects of the theory, the connections with groups, and various applications in geometry were studied. The theory knew an important progress starting with the 1970s, when its research area enlarged. A recent book on hyperstructures [2] points out to their applications in cryptography, codes, automata, probability, geometry, lattices, binary relations, graphs, and hypergraphs. Many authors studied different aspects of semihypergroups, for instance, Bonansinga and Corsini [3], Corsini [4], Davvaz [5], Davvaz and Poursalavati [6], Fasino and Freni [7], Guţan [8], Hasankhani [9], Leoreanu [10], and Onipchuk [11].

In 1986, Sen and Saha [12] defined the notion of a -semigroup as a generalization of a semigroup. One can see that -semigroups are generalizations of semigroups. Many classical notions of semigroups have been extended to -semigroups and a lot of results on -semigroups are published by a lot of mathematicians, for instance, Chattopadhyay [13, 14], Hila [15], Saha [16], Sen et al. [12, 17–20], Seth [21], and Sardar et al. [22]. In [23–25], Anvariyeh et al. introduced the notion of a -semihypergroup as a generalization of a semihypergroup. Many classical notions of semigroups and semihypergroups have been extended to -semihypergroups and a lot of results on -semihypergroups are obtained.

2. Basic Definitions

In this section, we recall certain definitions and results needed for our purpose.

Let and be two nonempty sets, the set of all mapping from to , and a set of some mappings from to . The usual composition of two elements of cannot be defined. But if we take from and from , then the usual mapping composition can be defined. Also, we see that and for and . Sen and Saha [12] defined the notion of a -semigroup as a generalization of a semigroup as follows.

Definition 1. Let and be two nonempty sets. Then is called a -semigroup if there exists a mapping written as satisfying the following identity:

Let be a nonempty subset of . Then is called a sub--semigroup of if for all and .

Example 2. If is the set of matrices and is a set of some matrices over the field of real numbers, then we can define such that

where and . An algebraic system that satisfying the associativity property of the above type is a -semigroup.

Example 3. Let and . Then is a gamma-semigroup under the usual multiplication. Next, let . We have as a nonempty subset of and for all and . Then is a sub--semigroup of .

More examples of -semigroup can be found in [12, 16].

Let be a nonempty set and let be the set of all nonempty subsets of . A hyperoperation on is a map and the couple is called a hypergroupoid. If and are non-empty subsets of , then we denote , and .

Definition 4. A hypergroupoid is called a semihypergroup if for all of one has , which means that

A semihypergroup is called a hypergroup if for all , one has

3. -Semihypergroups

In this section, we introduce the notion of a -semihypergroup which is a generalization of the notion of -semigroup and semihypergroup.

Definition 5. Let and be two non-empty sets. is called a -semihypergroup if every be a hyperoperation on , that is, for every , and for every and one has .

If every is an operation, then is a -semigroup.

If is a hypergroup for every , then is called a -hypergroup.

Let and be two non-empty subsets of and , we define the following:

Also

A nonempty subset of is called a sub--semihypergroup of if .

is called a right (left) hyperideal of if (), and is called a hyperideal of if it is both a right and a left hyperideal. Clearly, every right (left) hyperideal is a sub--semihypergroup.

A -semihypergroup is called commutative if for all and we have .

Example 6. Let be a semihypergroup and be a nonempty set. We define for every and . Then, is a -semihypergroup.

Example 7. Let be a semihypergroup and be a non-empty subset of . We define for every and . Then is a -semihypergroup.

Example 8. Let and , for every and . We define

Then, is a hyperoperation. For every and we have

This means that is a -semihypergroup.

Also, if , then is not a hypergroup, because for every we have . So is not a -hypergroup.

Example 9. Let and be defined as follows:

It is not difficult to see that is a -hypergroup.

Example 10. Let and be defined as follows:

It is not difficult to see that is a -hypergroup. Also, is a hypergroup and is a group.

Example 11. Let be a group and be normal subgroups of such that . Set where for every , the hyperoperation be as follows:

For every we have

Also, and so is a -semihypergroup. It is easy to see that is a -hypergroup.

Theorem 12. Let be a -semihypergroup and exists such that be a hypergroup. Then, for every , is a hypergroup.

Proof. For every and there exists such that

Now, there exists such that and so is a hypergroup.

Remark 13. Let be a -group and for some , element be an identity of semigroup . Then, in general case is not an identity element of , where . For example, let and be as follows:

is an identity element of and is an identity element of .

Lemma 14. Let be a -semihypergroup and . If be a scalar identity of and , then .

Proof. We prove that for every , . We have the following:

Theorem 15. Let be a -semihypergroup and be a scalar identity of , that is, for every , one has . Then and is a semihypergroup.

Proof. It is obtained from Lemma 14.

Example 16. Theorem 15 is not true for identity elements of -semihypergroups. In Example 9, we see that and are identity elements and .

Theorem 17. Let be a -semihypergroup and for every , and semihypergroup has a scalar identity . Then, one has .

Proof. By Lemma 14, we have

In general, the above theorem is not true for a -semihypergroup with identity elements.

Example 18. Let and be as follows:

In -hypergroup we have as a scalar identity of , an identity of , and is an identity of and an scalar identity of , but

Definition 19. Let be a -semihypergroup and be a -semihypergroup. A map is called a homomorphism if there exists a bijective map such that for every and :

If we set , then .

Lemma 20. Let be a -semihypergroup, be a -semihypergroup, and be a homomorphism. Then one has the following.

(1)If is a sub--semihypergroup of , then is a sub--semihypergroup of S′. (2)If T′ is a sub--semihypergroup of S′ and , then is a sub--semihypergroup of .

Proof. It is straightforward.

4. Regular and Strongly Regular Relations

If is a -semihypergroup and is an equivalence relation on , then for all pairs of nonempty subsets of , we set if and only if for every there exists such that and for every there exists such that . The equivalence relation is said to be regular to the right if implies for all and . Analogously, we can define regular to the left. Moreover, is called regular if it is regular to the right and to the left.

Also, we set if and only if for all and . The equivalence relation is said to be strongly regular to the right if implies that for all and . Analogously, we can define strongly regular to the left. Moreover, is called strongly regular if it is strongly regular to the right and to the left.

Let be a -semihypergroup and be an equivalence relation on . Let be the equivalence class of with respect to and let . For every we define on as follows:

Now, we set .

Theorem 21. Let be a -semihypergroup and be an equivalence relation on . The following conditions are equivalent:

(1) is a regular relation, (2) is a -semihypergroup.

Proof. : We will prove that for every the hyperoperation is welldefined. Let and . If and , we prove that . We have ,, and is regular so and . Therefore, for every , there exists such that and for every , there exists such that . Thus, for every there exists such that and . Therefore, for every there exists such that , that is,

Also, by a similar way, we have

Now, we see that for every and ,

Let . Then there exists such that . Hence, we can suppose that and . From this it follows that , as a consequence . In a similar way, we obtain

: Let then and so for every , we have . Now, let . Then and so there exists such that . In the same way, for every there exists such that . Therefore, and so is regular.

Theorem 22. Let be a -semihypergroup and be a regular equivalence relation on . Then, the canonical projection is an epimorphism and if is a -hypergroup, then is a -hypergroup.

Proof. First, we show that is a homomorphism. Let , where . Thus, there exists such that , and so , hence .

Now, let thus there exists such that , so . Therefore, . Finally, if is a -hypergroup, then for all and , we have

This implies that is a -hypergroup.

Theorem 23. Let be a -hypergroup, be a -hypergroup and be a homomorphism. Then the equivalence relation associated with , that is,

is regular and the function is an isomorphism.

Proof. Let and . Then, for every and we have

So for every , there exists such that , this means that . Also, for every , there exists such that , this means that . Therefore, is regular to the right. In a similar way, we obtain to be regular to the left and so is regular. Finally, for every and for every we have:

Therefore, is an isomorphism.

Theorem 24. Let be a -semihypergroup and be an equivalence relation on . The following conditions are equivalent:

(1) is a strongly regular relation, (2)if and then   for every , (3) is a -semigroup.

Proof. and are clear.

: Let and . Since is strongly regular, then for every we have and so is singleton. Now, by Theorem 21, the proof is completed.

: Let , and , Since , are singleton, then for every and we have and so is strongly regular to the right and the same way implies that is strongly regular to the left. Therefore, is strongly regular.

Theorem 25. Let be a -hypergroup. Let such that a relation be a strongly regular on the semihypergroup and exists such that for every , and . Then, for every , the relation is a strongly regular on a semihypergroup , that means is a strongly regular relation on a -semihypergroup .

Proof. Let and such that . Since is a -hypergroup, then there exists such that . By hypothesis , thus, we have

Since and , then . In the same way, we obtain . Therefore, is a strongly regular relation on a -semihypergroup .

Theorem 26. Let be a -semihypergroup. Then,

(1)if is a -semigroup and is a homomorphism, then the equivalence relation associated to is strongly regular;(2)if is a -hypergroup and a strongly regular relation, then is a -group.

Proof. (1): We have . Since is a -semigroup, we obtain

for every and , and implies that

Therefore, .

(2): It is obtained from Theorems 22 and 24.

Definition 27. Let be a -semihypergroup and be a relation on . We say that is a fundamental relation on if is the smallest strongly regular equivalence relation on .

In fact, the fundamental relation is the smallest equivalence relation on a -semihypergroup such that the quotient is a -semigroup.

Definition 28. Let be a -semihypergroup and be a relation on as follows:

if and only if there exist , and such that

and let be the transitive closure of the relation .

Remark 29. Let be a semihypergroup. First time, the relation on introduced by Koskas [26] is as follows:

Therefore, the relation is a generalization of the relation .

Lemma 30. Let be a -semihypergroup. Then, for every one has .

Proof. It is  straightforward.

Lemma 31. The relation is reflexive and symmetric on and is an equivalence relation, moreover one has

Proof. Clearly, is reflexive and symmetric, so is an equivalence relation. Since for every , , then and so .

Example 32. Let be the -hypergroup in Example 11. For every , we set . Then, we have the following:

Also, it is not difficult to see that . Every is a hypergroup and so (see [27]). Therefore, we obtain and is an equivalence relation on .

Theorem 33. The relation is a strongly regular equivalence relation on the -semihypergroup .

Proof. By Lemma 31, is an equivalence relation. In order to prove that it is strongly regular, we show first that

for every , and . If , then there exist , and such that

For every and and every and , we obtain:

Thus, and so . In the same way, we can show that .

Now, if , then there exist and such that , whence, we obtain

Finally, for all and for all , taking , ,,, we have and so . Therefore,

Similarly, we can prove that . Hence, is strongly regular.

Corollary 34. Let be a -semihypergroup. Then, the quotient is a -semigroup.

Proof. By Theorems 24 and 33, the proof is completed.

Theorem 35. Let be a -semihypergroup. Then the equivalence relation is the smallest strongly regular equivalence relation on and so , where is the fundamental relation on .

Proof. Let be a strongly regular relation. Now, let so there exist , and such that

Since is a strongly regular relation, then is a -semigroup and so

where . Thus, and so we have . Finally, if , then there exist and such that . Therefore, , and transitivity of implies that . Thus, .

Theorem 36. If is a -hypergroup, then the quotient is a -group.

Proof. Let and . Then, there exists such that so and so is a -group.

Theorem 37. If is a -semihypergroup with an identity element , then and so is a semigroup.

Proof. Since is an identity element of -semigroup , then by Theorem 15, the proof is completed.

Theorem 38. If is a -hypergroup with a left (right) identity element , then and so is a semigroup.

Proof. It is easy to see that for every ,    there is a left identity element of group and so is an identity. By Theorem 15, the proof is completed.

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