Γ-Semihypergroups and Regular Relations

Abstract

In this paper we discuss the structure of Γ-semihypergroups. We prove some basic results and present several examples of Γ-semihypergroups. Also, we obtain some properties of regular and strongly regular relations on a Γ-semihypergroup and construct a Γ-semigroup from a Γ-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 Γ-semigroup as a generalization of a semigroup. One can see that Γ-semigroups are generalizations of semigroups. Many classical notions of semigroups have been extended to Γ-semigroups and a lot of results on Γ-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 Γ-semihypergroup as a generalization of a semihypergroup. Many classical notions of semigroups and semihypergroups have been extended to Γ-semihypergroups and a lot of results on Γ-semihypergroups are obtained.

2. Basic Definitions

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

Let 𝐴 and 𝐵 be two nonempty sets, 𝑀 the set of all mapping from 𝐴 to 𝐵, and Γ a set of some mappings from 𝐵 to 𝐴. The usual composition of two elements of 𝑀 cannot be defined. But if we take 𝑓,𝑔 from 𝑀 and 𝛼 from Γ, then the usual mapping composition 𝑓𝛼𝑔 can be defined. Also, we see that 𝑓𝛼𝑔∈𝑀 and (𝑓𝛼𝑔)𝛽ℎ=𝑓𝛼(𝑔𝛽ℎ) for 𝑓,𝑔,ℎ∈𝑀 and 𝛼,𝛽∈Γ. Sen and Saha [12] defined the notion of a Γ-semigroup as a generalization of a semigroup as follows.

Definition 1. Let 𝑀={𝑎,𝑏,𝑐,…} and Γ={𝛼,𝛽,𝛾,…} be two nonempty sets. Then 𝑀 is called a Γ-semigroup if there exists a mapping 𝑀×Γ×𝑀→𝑀 written as (𝑎,𝛾,𝑏)↦𝑎𝛾𝑏 satisfying the following identity: (𝑎𝛼𝑏)𝛽𝑐=𝑎𝛼𝑏𝛽𝑐∀𝑎,𝑏,𝑐∈𝑀,∀𝛼,𝛽∈Γ. (1)

Let 𝐾 be a nonempty subset of 𝑀. Then 𝐾 is called a sub-Γ-semigroup of 𝑀 if 𝑎𝛾𝑏∈𝐾 for all 𝑎,𝑏∈𝐾 and 𝛾∈Γ.

Example 2. If 𝑀 is the set of 𝑚×𝑛 matrices and Γ is a set of some 𝑛×𝑚 matrices over the field of real numbers, then we can define 𝐴_𝑚,𝑛𝛼_𝑛,𝑚𝐵_𝑚,𝑛 such that 𝐴_𝑚,𝑛𝛼_𝑛,𝑚𝐵_𝑚,𝑛𝛽_𝑛,𝑚𝐶_𝑚,𝑛=𝐴_𝑚,𝑛𝛼_𝑛,𝑚𝐵_𝑚,𝑛𝛽_𝑛,𝑚𝐶_𝑚,𝑛, (2) where 𝐴_𝑚,𝑛,𝐵_𝑚,𝑛,𝐶_𝑚,𝑛∈𝑀 and 𝛼_𝑛,𝑚,𝛽_𝑛,𝑚∈Γ. An algebraic system that satisfying the associativity property of the above type is a Γ-semigroup.

Example 3. Let 𝑀=[0,1] and Γ={1/𝑛∣𝑛i𝑠apositiveinteger}. Then 𝑀 is a gamma-semigroup under the usual multiplication. Next, let 𝐾=[0,1]. We have 𝐾 as a nonempty subset of 𝑀 and 𝑎𝛾𝑏∈𝐾 for all 𝑎,𝑏∈𝐾 and 𝛾∈Γ. Then 𝐾 is a sub-Γ-semigroup of 𝑀.

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

Let 𝐻 be a nonempty set and let ℘^∗(𝐻) be the set of all nonempty subsets of 𝐻. A hyperoperation on 𝐻 is a map ∘∶𝐻×𝐻→℘^∗(𝐻) and the couple (𝐻,∘) is called a hypergroupoid. If 𝐴 and 𝐵 are non-empty subsets of 𝐻, then we denote 𝐴∘𝐵=⋃_{𝑎∈𝐴,𝑏∈𝐵}𝑎∘𝑏, 𝑥∘𝐴={𝑥}∘𝐴 and 𝐴∘𝑥=𝐴∘{𝑥}.

Definition 4. A hypergroupoid (𝐻,∘) is called a semihypergroup if for all 𝑥,𝑦,𝑧 of 𝐻 one has (𝑥∘𝑦)∘𝑧=𝑥∘(𝑦∘𝑧), which means that _{𝑢∈𝑥∘𝑦}𝑢∘𝑧=_{𝑣∈𝑦∘𝑧}𝑥∘𝑣. (3) A semihypergroup (𝐻,∘) is called a hypergroup if for all 𝑥∈𝐻, one has 𝑥∘𝐻=𝐻∘𝑥=𝐻. (4)

3. Γ-Semihypergroups

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

Definition 5. Let 𝑆 and Γ be two non-empty sets. 𝑆 is called a Γ-semihypergroup if every 𝛾∈Γ be a hyperoperation on 𝑆, that is, 𝑥𝛾𝑦⊆𝑆 for every 𝑥,𝑦∈𝑆, and for every 𝛼,𝛽∈Γ and 𝑥,𝑦,𝑧∈𝑆 one has 𝑥𝛼(𝑦𝛽𝑧)=(𝑥𝛼𝑦)𝛽𝑧.

If every 𝛾∈Γ is an operation, then 𝑆 is a Γ-semigroup.

If (𝑆,𝛾) is a hypergroup for every 𝛾∈Γ, then 𝑆 is called a Γ-hypergroup.

Let 𝐴 and 𝐵 be two non-empty subsets of 𝑆 and 𝛾∈Γ, we define the following: 𝐴𝛾𝐵=∪𝑎𝛾𝑏∣𝑎∈𝐴,𝑏∈𝐵. (5) Also 𝐴Γ𝐵=∪𝑎𝛾𝑏∣𝑎∈𝐴,𝑏∈𝐵and𝛾∈Γ=_𝛾∈Γ𝐴𝛾𝐵. (6)

A nonempty subset 𝐴 of 𝑆 is called a sub-Γ-semihypergroup of 𝑆 if 𝐴Γ𝐴⊆𝐴.

𝐴 is called a right (left) hyperideal of 𝑆 if 𝐴Γ𝑆⊆𝐴 (𝑆Γ𝐴⊆𝐴), and is called a hyperideal of 𝑆 if it is both a right and a left hyperideal. Clearly, every right (left) hyperideal is a sub-Γ-semihypergroup.

A Γ-semihypergroup 𝑆 is called commutative if for all 𝑥,𝑦∈𝑆 and 𝛾∈Γ we have 𝑥𝛾𝑦=𝑦𝛾𝑥.

Example 6. Let (𝑆,∘) be a semihypergroup and Γ be a nonempty set. We define 𝑥𝛾𝑦=𝑥∘𝑦 for every 𝑥,𝑦∈𝑆 and 𝛾∈Γ. Then, 𝑆 is a Γ-semihypergroup.

Example 7. Let (𝑆,∘) be a semihypergroup and Γ be a non-empty subset of 𝑆. We define 𝑥𝛾𝑦=𝑥∘𝛾∘𝑦 for every 𝑥,𝑦∈𝑆 and 𝛾∈Γ. Then 𝑆 is a Γ-semihypergroup.

Example 8. Let 𝑆=[0,1] and Γ=ℕ, for every 𝑥,𝑦∈𝑆 and 𝛾∈Γ. We define 𝛾∶𝑆×𝑆→𝒫^∗(𝑆)𝑥𝛾𝑦=0,𝑥𝑦𝛾. (7) Then, 𝛾 is a hyperoperation. For every 𝑥,𝑦,𝑧∈𝑆 and 𝛼,𝛽∈Γ we have 𝑥𝛼𝑦𝛽𝑧=0,𝑥𝑦𝑧.𝛼𝛽=𝑥𝛼𝑦𝛽𝑧 (8) This means that 𝑆 is a Γ-semihypergroup.

Also, if 𝛾∈Γ, then (𝑆,𝛾) is not a hypergroup, because for every 0≤𝑡<1 we have 𝑡𝛾𝑆=[0,𝑡/𝛾]≠𝑆. So 𝑆 is not a Γ-hypergroup.

Example 9. Let 𝑆={0,1} and Γ={𝛼,𝛽} be defined as follows: 𝛼0100{}10,1{}𝛽0,1}{0,1010{}10,1}{0,1{0,1}{0,1} (9) It is not difficult to see that 𝑆 is a Γ-hypergroup.

Example 10. Let 𝑆={0,1} and Γ={𝛽,𝛾} be defined as follows: 𝛽010{}10,1}{0,1{}𝛾0,1}{0,101010110 (10) It is not difficult to see that 𝑆 is a Γ-hypergroup. Also, (𝑆,𝛽) is a hypergroup and (𝑆,𝛾) is a group.

Example 11. Let (𝐺,∘) be a group and 𝐻₁,…,𝐻_𝑘 be normal subgroups of 𝐺 such that 𝐻₁⊆𝐻₂⊆⋯𝐻_𝑘. Set Γ={𝛾₁,…,𝛾_𝑘} where for every 1≤𝑖≤𝑘, the hyperoperation 𝛾_𝑖 be as follows: 𝑥𝛾_𝑖𝑦=𝐻_𝑖∘𝑥∘𝑦,∀𝑥,𝑦∈𝐺. (11) For every 1≤𝑖,𝑗≤𝑘 we have 𝑥𝛾_𝑖𝑦𝛾_𝑗𝑧=𝑥𝛾_𝑖𝐻_𝑗∘𝑦∘𝑧=𝐻_𝑖∘𝑥∘𝐻_𝑗∘𝑦∘𝑧=𝐻_𝑖∘𝐻_𝑗∘𝑥∘𝑦∘𝑧=𝐻_{max{𝑖,𝑗}}∘𝑥∘𝑦∘𝑧. (12) Also, 𝑥𝛾_𝑖(𝑦𝛾_𝑗𝑧)=𝐻_{max{𝑖,𝑗}}∘𝑥∘𝑦∘𝑧 and so 𝐺 is a Γ-semihypergroup. It is easy to see that 𝐺 is a Γ-hypergroup.

Theorem 12. Let 𝑆 be a Γ-semihypergroup and exists 𝛼∈Γ such that (𝑆,𝛼) be a hypergroup. Then, for every 𝛾∈Γ, (𝑆,𝛾) is a hypergroup.

Proof. For every 𝑥,𝑧∈𝑆 and 𝛾∈Γ there exists 𝑦∈𝑆 such that 𝑧∈𝑥𝛾𝑥𝛼𝑦=𝑥𝛾𝑥𝛼𝑦. (13) Now, there exists 𝑢∈𝑥𝛼𝑦 such that 𝑧∈𝑥𝛾𝑢 and so (𝑆,𝛾) is a hypergroup.

Remark 13. Let 𝑆 be a Γ-group and for some 𝛾∈Γ, element 𝑒 be an identity of semigroup (𝑆,𝛾). Then, in general case 𝑒 is not an identity element of (𝑆,𝛽), where 𝛽∈Γ. For example, let 𝑆={0,1} and Γ={𝛽,𝛾} be as follows: 𝛽010110𝛾0101010110 (14)0 is an identity element of (𝑆,𝛾) and 1 is an identity element of (𝑆,𝛾).

Lemma 14. Let 𝑆 be a Γ-semihypergroup and 𝛼,𝛽∈Γ. If 𝑒 be a scalar identity of (𝑆,𝛼) and (𝑆,𝛽), then 𝛼=𝛽.

Proof. We prove that for every 𝑥,𝑦∈Γ, 𝑥𝛼𝑦=𝑥𝛽𝑦. We have the following: 𝑥𝛼𝑦=𝑥𝛼𝑒𝛽𝑦=(𝑥𝛼𝑒)𝛽𝑦=𝑥𝛽𝑦. (15)

Theorem 15. Let 𝑆 be a Γ-semihypergroup and 𝑒 be a scalar identity of 𝑆, that is, for every (𝑥,𝛾)∈𝑆×Γ, one has 𝑥𝛾𝑒=𝑥=𝑒𝛾𝑥. Then card(Γ)=1 and 𝑆 is a semihypergroup.

Proof. It is obtained from Lemma 14.

Example 16. Theorem 15 is not true for identity elements of Γ-semihypergroups. In Example 9, we see that 0 and 1 are identity elements and 𝛼≠𝛽.

Theorem 17. Let 𝑆 be a Γ-semihypergroup and for every 𝛾∈Γ, and semihypergroup (𝑆,𝛾) has a scalar identity 𝑒_𝛾∈𝑆. Then, one has card(Γ)≤card(𝑆).

Proof. By Lemma 14, we have card(Γ)=card𝑒_𝛾∣𝛾∈Γ≤card(𝑆). (16)

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

Example 18. Let 𝑆={0,1} and Γ={𝛼,𝛽,𝛾} be as follows: 𝛽010110𝛽01010{}10,1}{0,1{}𝛾0,1}{0,101010110 (17)

In Γ-hypergroup 𝑆 we have 0 as a scalar identity of (𝑆,𝛼), an identity of (𝑆,𝛽), and 1 is an identity of (𝑆,𝛽) and an scalar identity of (𝑆,𝛾), but card(Γ)=3>card(𝑆)=2. (18)

Definition 19. Let 𝑆 be a Γ-semihypergroup and 𝑆′ be a Γ′-semihypergroup. A map 𝑓∶𝑆→𝑆′ is called a homomorphism if there exists a bijective map ℎ∶Γ→Γ′ such that for every 𝛾∈Γ and 𝑥,𝑦∈𝑆: 𝑓𝑥𝛾𝑦=𝑓(𝑥)ℎ𝛾𝑓𝑦. (19) If we set ℎ(𝛾)=𝛾′, then 𝑓(𝑥𝛾𝑦)=𝑓(𝑥)𝛾′𝑓(𝑦).

Lemma 20. Let 𝑆 be a Γ-semihypergroup, 𝑆′ be a Γ′-semihypergroup, and 𝑓∶𝑆→𝑆′ be a homomorphism. Then one has the following. (1)If 𝑇 is a sub-Γ-semihypergroup of 𝑆, then 𝑓(𝑇) is a sub-Γ′-semihypergroup of S′. (2)If T′ is a sub-Γ′-semihypergroup of S′ and 𝑓⁻¹(𝑇′)≠∅, then 𝑓⁻¹(𝑇′) is a sub-Γ-semihypergroup of 𝑆.

Proof. It is straightforward.

4. Regular and Strongly Regular Relations

If 𝑆 is a Γ-semihypergroup and 𝑅⊆𝑆×𝑆 is an equivalence relation on 𝑆, then for all pairs (𝐴,𝐵) of nonempty subsets of 𝑆, we set 𝐴𝑅𝐵 if and only if for every 𝑎∈𝐴 there exists 𝑏∈𝐵 such that 𝑎𝑅𝑏 and for every 𝑏∈𝐵 there exists 𝑎∈𝐴 such that 𝑎𝑅𝑏. The equivalence relation 𝑅 is said to be regular to the right if 𝑥𝑅𝑦 implies 𝑥𝛾𝑎𝑅𝑦𝛾𝑎 for all (𝑥,𝑦,𝑎)∈𝑆³ and 𝛾∈Γ. Analogously, we can define regular to the left. Moreover, 𝑅 is called regular if it is regular to the right and to the left.

Also, we set 𝐴𝑅𝐵 if and only if 𝑎𝜌𝑏 for all 𝑎∈𝐴 and 𝑏∈𝐵. The equivalence relation 𝑅 is said to be strongly regular to the right if 𝑥𝑅𝑦 implies that 𝑥𝛾𝑎𝑅𝑦𝛾𝑎 for all (𝑥,𝑦,𝑎)∈𝑆³ and 𝛾∈Γ. Analogously, we can define strongly regular to the left. Moreover, 𝑅 is called strongly regular if it is strongly regular to the right and to the left.

Let 𝑆 be a Γ-semihypergroup and 𝑅 be an equivalence relation on 𝑆. Let 𝑅(𝑎) be the equivalence class of 𝑎 with respect to 𝑅 and let 𝑆/𝑅={𝑅(𝑎)∣𝑎∈𝑆}. For every 𝛾∈Γ we define 𝛾/𝑅 on 𝑆/𝑅 as follows: 𝑅(𝑎)𝛾𝑅𝑅(𝑏)=𝑅(𝑥)∣𝑥∈𝑎𝛾𝑏. (20) Now, we set Γ/𝑅={𝛾/𝑅∣𝛾∈Γ}.

Theorem 21. Let 𝑆 be a Γ-semihypergroup and 𝑅 be an equivalence relation on 𝑆. The following conditions are equivalent:(1)𝑅 is a regular relation, (2)𝑆/𝑅 is a Γ/𝑅-semihypergroup.

Proof. (1⇒2): We will prove that for every 𝛾∈Γ the hyperoperation 𝛾/𝑅 is welldefined. Let 𝛾∈Γ and 𝑥,𝑦∈𝑆. If 𝑥′∈𝑅(𝑥) and 𝑦′∈𝑅(𝑦), we prove that 𝑅(𝑥′)(𝛾/𝑅)𝑅(𝑦′)=𝑅(𝑥)(𝛾/𝑅)𝑅(𝑦). We have 𝑥′𝑅𝑥,𝑦′𝑅𝑦, and 𝑅 is regular so 𝑥𝛾𝑦𝑅𝑥𝛾𝑦′ and 𝑥𝛾𝑦′𝑅𝑥′𝛾𝑦′. Therefore, for every 𝑤∈𝑥𝛾𝑦, there exists 𝑢∈𝑥𝛾𝑦′ such that 𝑤𝑅𝑢 and for every 𝑣∈𝑥𝛾𝑦′, there exists 𝑧∈𝑥′𝛾𝑦′ such that 𝑧𝑅𝑣. Thus, for every 𝑤∈𝑥𝛾𝑦 there exists 𝑣∈𝑥′𝛾𝑦′ such that 𝑤𝑅𝑧 and 𝑅(𝑤)=𝑅(𝑧). Therefore, for every 𝑅(𝑤)∈𝑅(𝑥′)(𝛾/𝑅)𝑅(𝑦′) there exists 𝑧∈𝑥𝛾𝑦 such that 𝑅(𝑤)=𝑅(𝑧)∈𝑅(𝑥)(𝛾/𝑅)𝑅(𝑦), that is, 𝑅𝛾𝑥′𝑅𝑅(𝑥)𝛾𝑦′⊆𝑅𝑅𝑅𝑦. (21) Also, by a similar way, we have 𝑅(𝑥)𝛾𝑅𝑅𝑦𝛾⊆𝑅𝑥′𝑅𝑅𝑦′. (22) Now, we see that for every 𝛼/𝑅,𝛽/𝑅∈Γ/𝑅 and 𝑅(𝑥),𝑅(𝑦),𝑅(𝑧)∈𝑆/𝑅, 𝑅(𝑥)𝛼𝑅𝑅𝑦𝛽𝑅𝑅(𝑧)=𝑅(𝑥)𝛼𝑅𝑅𝑦𝛽𝑅𝑅(𝑧). (23) Let 𝑅(𝑤)∈𝑅(𝑥)(𝛼/𝑅)(𝑅(𝑦)(𝛽/𝑅)𝑅(𝑧)). Then there exists 𝑅(𝑣)∈𝑅(𝑦)(𝛽/𝑅)𝑅(𝑧) such that 𝑅(𝑤)∈𝑅(𝑥)(𝛼/𝑅)𝑅(𝑣). Hence, we can suppose that 𝑣∈𝑦𝛽𝑧 and 𝑤∈𝑥𝛼(𝑦𝛽𝑧)=(𝑥𝛼𝑦)𝛽𝑧. From this it follows that 𝑤∈𝑥𝛼𝑣, as a consequence 𝑅(𝑤)∈(𝑅(𝑥)(𝛼/𝑅)𝑅(𝑦))(𝛽/𝑅)𝑅(𝑧). In a similar way, we obtain 𝑅(𝑥)𝛼𝑅𝑅𝑦𝛽𝑅𝑅(𝑧)(𝑥)𝛼⊆𝑅𝑅𝑅𝑦𝛽𝑅𝑅(𝑧). (24)
(2⇒1): Let 𝑥𝑅𝑥′ then 𝑅(𝑥)=𝑅(𝑥′) and so for every 𝑦∈𝑆, we have 𝑅(𝑥)(𝛾/𝑅)𝑅(𝑦)=𝑅(𝑥′)(𝛾/𝑅)𝑅(𝑦). Now, let 𝑢∈𝑥𝛾𝑦. Then 𝑅(𝑢)∈𝑅(𝑥)(𝛾/𝑅)𝑅(𝑦)=𝑅(𝑥′)(𝛾/𝑅)𝑅(𝑦) and so there exists 𝑣∈𝑥′𝛾𝑦 such that 𝑅(𝑣)=𝑅(𝑢). In the same way, for every 𝑧∈𝑥′𝛾𝑦 there exists 𝑤∈𝑥𝛾𝑦 such that 𝑅(𝑤)=𝑅(𝑧). Therefore, 𝑥𝛾𝑦𝑅𝑥′𝛾𝑦 and so 𝑅 is regular.

Theorem 22. Let 𝑆 be a Γ-semihypergroup and 𝑅 be a regular equivalence relation on 𝑆. Then, the canonical projection 𝜋∶𝑆→𝑆/𝑅 is an epimorphism and if 𝑆 is a Γ-hypergroup, then 𝑆/𝑅 is a Γ/𝑅-hypergroup.

Proof. First, we show that 𝜋 is a homomorphism. Let 𝑅(𝑧)∈𝜋(𝑥𝛾𝑦), where 𝛾∈Γ. Thus, there exists 𝑧′∈𝑥𝛾𝑦 such that 𝑅(𝑧)=𝜋(𝑧′), and so 𝑅(𝑧)=𝑅(𝑧′)∈𝜋(𝑥)(𝛾/𝑅)𝜋(𝑦), hence 𝜋(𝑥𝛾𝑦)⊆𝜋(𝑥)𝛾/𝑅𝜋(𝑦).
Now, let 𝑅(𝑧)∈𝜋(𝑥)(𝛾/𝑅)𝑅(𝑦) thus there exists 𝑧′∈𝑅(𝑧) such that 𝑧′∈𝑥𝛾𝑦, so 𝑅(𝑧)=𝜋(𝑧′)∈𝜋(𝑥𝛾𝑦). Therefore, 𝜋(𝑥𝛾𝑦)=𝜋(𝑥)(𝛾/𝑅)𝜋(𝑦). Finally, if 𝑆 is a Γ-hypergroup, then for all 𝑥∈𝑆 and 𝛾∈Γ, we have 𝜋(𝑆)𝛾𝑅𝜋(𝑥)(𝑆)=𝑆=𝜋𝑆𝛾𝑥=𝜋𝑅(𝑆)(𝑥)𝛾=𝜋=𝜋𝑅𝜋(𝑆)=𝜋𝑥𝛾𝑆. (25) This implies that 𝑆/𝑅 is a Γ/𝑅-hypergroup.

Theorem 23. Let 𝑆 be a Γ-hypergroup, 𝑆^′ be a Γ^′-hypergroup and 𝑓∶𝑆→𝑆^′ be a homomorphism. Then the equivalence relation 𝑅 associated with 𝑓, that is, 𝑥𝑅𝑦⟺𝑓(𝑥)=𝑓𝑦, (26) is regular and the function 𝜓∶𝑓(𝑆)→𝑆/𝑅,𝜓(𝑓(𝑥))=𝑅(𝑥) is an isomorphism.

Proof. Let 𝑥,𝑦∈𝑆 and 𝑥𝑅𝑦. Then, for every 𝑎∈𝑆 and 𝛾∈Γ we have 𝑓𝑥𝛾𝑎=𝑓(𝑥)𝛾′𝑓(𝑎)=𝑓𝑦𝛾′𝑓(𝑎)=𝑓𝑦𝛾𝑎. (27) So for every 𝑢∈𝑥𝛾𝑎, there exists 𝑣∈𝑦𝛾𝑎 such that 𝑓(𝑢)=𝑓(𝑣), this means that 𝑢𝑅𝑣. Also, for every 𝑤∈𝑦𝛾𝑎, there exists 𝑧∈𝑥𝛾𝑎 such that 𝑓(𝑧)=𝑓(𝑤), this means that 𝑧𝑅𝑤. Therefore, 𝑅 is regular to the right. In a similar way, we obtain 𝑅 to be regular to the left and so 𝑅 is regular. Finally, for every 𝑥,𝑦∈𝑆 and for every 𝛾∈Γ we have: 𝜓𝑓(𝑥)𝑦𝑓=𝑅(𝑧)(𝑥)𝛾𝛾′𝑓=𝜓𝑥𝛾𝑦∣𝑧∈𝑥𝛾𝑦=𝑅𝑅𝑅𝑦𝑓(𝑥)𝛾=𝜓𝑅𝑓𝑦,𝜓⁻¹𝑅(𝑥)𝛾𝑅𝑅𝑦=𝜓⁻¹𝑅(𝑧)=∣𝑧∈𝑥𝛾𝑦_{𝑧∈𝑥𝛾𝑦}𝑓(𝑧)=𝑓𝑥𝛾𝑦=𝑓(𝑥)𝛾′𝑓𝑦=𝜓⁻¹(𝑅(𝑥))𝛾′𝜓⁻¹𝑅𝑦. (28)(29) Therefore, 𝜓 is an isomorphism.

Theorem 24. Let 𝑆 be a Γ-semihypergroup and 𝑅 be an equivalence relation on 𝑆. The following conditions are equivalent: (1)𝑅 is a strongly regular relation, (2)if 𝑥₁𝑅𝑦₁ and 𝑥₂𝑅𝑦₂ then 𝑥₁𝛾𝑥₂𝑅𝑦₁𝛾𝑦₂  for every 𝛾∈Γ, (3)𝑆/𝑅 is a Γ/𝑅-semigroup.

Proof. (1⇒2) and (2⇒1) are clear.
(1⇒3): Let 𝑥,𝑦∈𝑆 and 𝛾∈Γ. Since 𝑅 is strongly regular, then for every 𝑢,𝑣∈𝑥𝛾𝑦 we have 𝑅(𝑢)=𝑅(𝑣) and so 𝑅(𝑥)(𝛾/𝑅)𝑅(𝑦) is singleton. Now, by Theorem 21, the proof is completed.
(3⇒1): Let 𝑥𝑅𝑦, 𝑎∈𝑆 and 𝛾∈Γ, Since 𝑅(𝑥)(𝛾/𝑅)𝑅(𝑎)={𝑅(𝑧)∣𝑧∈𝑥𝛾𝑎}, 𝑅(𝑦)(𝛾/𝑅)𝑅(𝑎)={𝑅(𝑡)∣𝑡∈𝑦𝛾𝑎} are singleton, then for every 𝑧∈𝑥𝛾𝑎 and 𝑡∈𝑦𝛾𝑎 we have 𝑧𝑅𝑡 and so 𝑅 is strongly regular to the right and the same way implies that 𝑅 is strongly regular to the left. Therefore, 𝑅 is strongly regular.

Theorem 25. Let 𝑆 be a Γ-hypergroup. Let 𝛼∈Γ such that a relation 𝑅 be a strongly regular on the semihypergroup (𝑆,𝛼) and exists 𝑢∈𝑆 such that for every (𝑥,𝛾)∈𝑆×Γ, 𝑥𝛾𝑢𝑅𝑥𝛾𝑢 and 𝑢𝛾𝑥𝑅𝑢𝛾𝑥. Then, for every 𝛾∈Γ, the relation 𝑅 is a strongly regular on a semihypergroup (𝑆,𝛾), that means 𝑅 is a strongly regular relation on a Γ-semihypergroup 𝑆.

Proof. Let 𝛾∈Γ and 𝑥,𝑦,𝑎∈𝑆 such that 𝑥𝑅𝑦. Since 𝑆 is a Γ-hypergroup, then there exists 𝑣∈𝑆 such that 𝑎∈𝑢𝛼𝑣. By hypothesis 𝑥𝛾𝑢𝑅𝑦𝛾𝑢, thus, we have 𝑥𝛾𝑢𝛼𝑣𝑅𝑦𝛾𝑢𝛼𝑣⇒𝑥𝛾(𝑢𝛼𝑣)𝑅𝑦𝛾(𝑢𝛼𝑣). (30) Since 𝑥𝛾𝑎⊆𝑥𝛾(𝑢𝛼𝑣) and 𝑦𝛾𝑎⊆𝑦𝛾(𝑢𝛼𝑣), then 𝑥𝛾𝑎𝑅𝑥𝛾𝑎. In the same way, we obtain 𝑎𝛾𝑥𝑅𝑎𝛾𝑦. Therefore, 𝑅 is a strongly regular relation on a Γ-semihypergroup 𝑆.

Theorem 26. Let 𝑆 be a Γ-semihypergroup. Then,(1)if 𝐻 is a Γ^′-semigroup and 𝜙∶𝑆→𝐻 is a homomorphism, then the equivalence relation 𝑅 associated to 𝜙 is strongly regular;(2)if 𝑆 is a Γ-hypergroup and 𝑅 a strongly regular relation, then 𝑆/𝑅 is a Γ/𝑅-group.

Proof. (1): We have 𝑥𝑅𝑦⇔𝜙(𝑥)=𝜙(𝑦). Since 𝐻 is a Γ′-semigroup, we obtain 𝜙𝑥𝛾𝑎∈𝜙(𝑥)𝛾^′𝜙(𝑎), (31) for every 𝛾∈Γ and 𝑎∈𝑆, and 𝑥𝑅𝑦 implies that 𝜙𝑥𝛾𝑎=1=𝜙𝑦𝛾𝑎. (32) Therefore, 𝑥𝛾𝑎𝑅𝑦𝛾𝑎.
(2): It is obtained from Theorems 22 and 24.

Definition 27. Let 𝑆 be a Γ-semihypergroup and 𝜃 be a relation on 𝑆. We say that 𝜃 is a fundamental relation on 𝑆 if 𝜃 is the smallest strongly regular equivalence relation on 𝑆.
In fact, the fundamental relation 𝜃 is the smallest equivalence relation on a Γ-semihypergroup 𝑆 such that the quotient 𝑆/𝜃 is a Γ/𝜃-semigroup.

Definition 28. Let 𝑆 be a Γ-semihypergroup and 𝜌 be a relation on 𝑆 as follows:
𝑥𝜌𝑦 if and only if there exist 𝑛∈ℕ, 𝑥₁,…𝑥_𝑛∈𝑆 and 𝛾₁,…,𝛾_𝑛−1∈Γ such that 𝑥,𝑦⊆𝑥₁𝛾₁𝑥₂𝛾₂⋯𝛾_𝑛−2𝑥_𝑛−1𝛾_𝑛−1𝑥_𝑛, (33) and let 𝜌^∗ be the transitive closure of the relation 𝜌.

Remark 29. Let (𝐻,∘) be a semihypergroup. First time, the relation 𝛽_∘ on 𝐻 introduced by Koskas [26] is as follows: 𝑥𝛽_∘𝑦⟺∃𝑥₁,…,𝑥_𝑛∈𝐻,suchthat𝑥,𝑦⊆𝑥₁∘𝑥₂∘⋯∘𝑥_𝑛. (34) Therefore, the relation 𝜌 is a generalization of the relation 𝛽.

Lemma 30. Let 𝑆 be a Γ-semihypergroup. Then, for every 𝛾∈Γ one has 𝛽_𝛾⊆𝜌.

Proof. It is  straightforward.

Lemma 31. The relation 𝜌 is reflexive and symmetric on 𝑆 and 𝜌^∗ is an equivalence relation, moreover one has _𝛾∈Γ𝛽_𝛾^∗⊆𝜌^∗. (35)

Proof. Clearly, 𝜌 is reflexive and symmetric, so 𝜌^∗ is an equivalence relation. Since for every 𝛾∈Γ, 𝛽_𝛾⊆𝜌, then ⋃_𝛾∈Γ𝛽_𝛾⊆𝜌 and so (⋃_𝛾∈Γ𝛽_𝛾)^∗⊆𝜌^∗.

Example 32. Let 𝐺 be the Γ-hypergroup in Example 11. For every 1≤𝑖≤𝑛, we set 𝛽_𝛾𝑖=𝛽_𝑖. Then, we have the following: 𝛽₁⊆𝛽₂⊆⋯⊆𝛽_𝑘⊆𝜌. (36) Also, it is not difficult to see that 𝛽_𝑘=𝜌. Every (𝐺,𝛾_𝑖) is a hypergroup and so 𝛽=𝛽^∗ (see [27]). Therefore, we obtain 𝜌=𝛽_𝑘=𝛽^∗_𝑘 and 𝜌 is an equivalence relation on 𝑆.

Theorem 33. The relation 𝜌^∗ is a strongly regular equivalence relation on the Γ-semihypergroup 𝑆.

Proof. By Lemma 31, 𝜌^∗ is an equivalence relation. In order to prove that it is strongly regular, we show first that 𝑥𝜌𝑦⇒𝑥𝛾𝑎𝜌𝑦𝛾𝑎,𝑎𝛾𝑥𝜌𝑎𝛾𝑦, (37) for every 𝑎∈𝑆, and 𝛾∈Γ. If 𝑥𝜌𝑦, then there exist 𝑛∈ℕ, 𝑥₁,…𝑥_𝑛∈𝑆 and 𝛾₁,…,𝛾_𝑛−1∈Γ such that 𝑥,𝑦⊆𝑥₁𝛾₁𝑥₂𝛾₂⋯𝛾_𝑛−2𝑥_𝑛−1𝛾_𝑛−1𝑥_𝑛. (38) For every 𝑎∈𝑆 and 𝛾∈Γ and every 𝑢∈𝑥𝛾𝑎 and 𝑣∈𝑦𝛾𝑎, we obtain: 𝑢∈𝑥𝛾𝑎⊆𝑥₁𝛾₁𝑥₂𝛾₂⋯𝛾_𝑛−2𝑥_𝑛−1𝛾_𝑛−1𝑥_𝑛𝛾𝑎,𝑣∈𝑦𝛾𝑎⊆𝑥₁𝛾₁𝑥₂𝛾₂⋯𝛾_𝑛−2𝑥_𝑛−1𝛾_𝑛−1𝑥_𝑛𝛾𝑎. (39) Thus, 𝑢𝜌𝑣 and so 𝑥𝛾𝑎𝜌𝑦𝛾𝑎. In the same way, we can show that 𝑥𝜌𝑦⇒𝑎𝛾𝑥𝜌𝑎𝛾𝑦.
Now, if 𝑥𝜌^∗𝑦, then there exist 𝑚∈ℕ and 𝑥=𝑤₀,𝑤₁,…,𝑤_𝑚−1,𝑤_𝑚=𝑦∈𝑆 such that 𝑥=𝑤₀𝜌𝑤₁𝜌⋯𝜌𝑤_𝑚−1𝜌𝑤_𝑚=𝑦, whence, we obtain 𝑥𝛾𝑎=𝑤₀𝛾𝑎𝜌𝑤₁𝛾𝑎𝜌⋯𝜌𝑤_𝑚−1𝛾𝑎𝜌𝑤_𝑚𝛾𝑎=𝑦𝛾𝑎. (40)
Finally, for all 𝑣∈𝑥𝛾𝑎=𝑤₀𝛾𝑎 and for all 𝑤∈𝑤_𝑚𝛾𝑎=𝑦𝛾𝑎, taking 𝑧₁∈𝑤₁𝛾𝑎, 𝑧₂∈𝑤₂𝛾𝑎,…,𝑧_𝑚−1∈𝑤_𝑚−1𝛾𝑎, we have 𝑣𝜌𝑧₁𝜌𝑧₂𝜌⋯𝜌𝑧_𝑚−1𝜌𝑤 and so 𝑣𝜌^∗𝑤. Therefore, 𝑥𝜌^∗𝑦⇒𝑥𝛾𝑎𝜌^∗𝑦𝛾𝑎. (41) Similarly, we can prove that 𝑥𝜌^∗𝑦⇒𝑎𝛾𝑥𝜌^∗𝑎𝛾𝑦. Hence, 𝜌^∗ is strongly regular.

Corollary 34. Let 𝑆 be a Γ-semihypergroup. Then, the quotient 𝑆/𝜌^∗ is a Γ/𝜌^∗-semigroup.

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

Theorem 35. Let 𝑆 be a Γ-semihypergroup. Then the equivalence relation 𝜌^∗ is the smallest strongly regular equivalence relation on 𝑆 and so 𝜌^∗=𝜃, where 𝜃 is the fundamental relation on 𝑆.

Proof. Let 𝑅 be a strongly regular relation. Now, let 𝑥𝜌𝑦 so there exist 𝑛∈ℕ, 𝑥₁,…𝑥_𝑛∈𝑆 and 𝛾₁,…,𝛾_𝑛−1∈Γ such that 𝑥,𝑦⊆𝑥₁𝛾₁𝑥₂𝛾₂⋯𝛾_𝑛−2𝑥_𝑛−1𝛾_𝑛−1𝑥_𝑛. (42) Since 𝑅 is a strongly regular relation, then 𝑆/𝑅 is a Γ/𝑅-semigroup and so 𝑅(𝑥)=𝑅𝑥₁𝛾̈₁𝑅𝑥₂𝛾̈₂⋯𝛾̈_𝑛−2𝑅𝑥_𝑛−1𝛾̈_𝑛−1𝑅𝑥_𝑛,𝑅𝑦=𝑅𝑥₁𝛾̈₁𝑅𝑥₂𝛾̈₂⋯𝛾̈_𝑛−2𝑅𝑥_𝑛−1𝛾̈_𝑛−1𝑅𝑥_𝑛, (43) where 𝛾̈_𝑖=𝛾_𝑖/𝑅. Thus, 𝑅(𝑥)=𝑅(𝑦) and so we have 𝑥𝜌𝑦⇒𝑥𝑅𝑦. Finally, if 𝑥𝜌^∗𝑦, then there exist 𝑚∈ℕ and 𝑥=𝑤₀,𝑤₁,…,𝑤_𝑚−1,𝑤_𝑚=𝑦∈𝑆 such that 𝑥=𝑤₀𝜌𝑤₁𝜌⋯𝜌𝑤_𝑚−1𝜌𝑤_𝑚=𝑦. Therefore, 𝑥=𝑤₀𝑅𝑤₁𝑅⋯𝑅𝑤_𝑚−1𝑅𝑤_𝑚=𝑦, and transitivity of 𝑅 implies that 𝑥𝑅𝑦. Thus, 𝜌^∗⊆𝑅.

Theorem 36. If 𝑆 is a Γ-hypergroup, then the quotient 𝑆/𝜌^∗ is a Γ/𝜌^∗-group.

Proof. Let 𝜌^∗(𝑥),𝜌^∗(𝑦)∈𝑆/𝜌^∗ and 𝛾/𝜌^∗∈Γ/𝜌^∗. Then, there exists 𝑧∈𝑆 such that 𝑥∈𝑦𝛾𝑧 so 𝜌^∗(𝑥)=𝜌^∗(𝑦)𝛾/𝜌^∗𝜌^∗(𝑧) and so 𝑆/𝜌^∗ is a Γ/𝜌^∗-group.

Theorem 37. If 𝑆 is a Γ-semihypergroup with an identity element 𝑒∈𝑆, then card(Γ/𝜌^∗)=1 and so 𝑆/𝜌^∗ is a semigroup.

Proof. Since 𝜌^∗(𝑒) is an identity element of Γ/𝜌^∗-semigroup 𝑆/𝜌^∗, then by Theorem 15, the proof is completed.

Theorem 38. If 𝑆 is a Γ-hypergroup with a left (right) identity element 𝑒∈𝑆, then card(Γ/𝜌^∗)=1 and so 𝑆/𝜌^∗ is a semigroup.

Proof. It is easy to see that for every 𝛾∈Γ,  𝜌^∗(𝑒)  there is a left identity element of group (𝑆/𝜌^∗,𝛾/𝜌^∗) and so 𝜌^∗(𝑒) is an identity. By Theorem 15, the proof is completed.