# Γ-Semihypergroups and Regular Relations

Authors: S. Mirvakili; S. M. Anvariyeh; B. Davvaz
Journal: Journal of Mathematics
 (2013)

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## Abstract

In this paper we discuss the structure ofΓ-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.


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## Body

## 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.LetA 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.
LetM={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:
(1)(aαb)βc=aα(bβc)∀a,b,c∈M,∀α,β∈Γ.LetK 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.
IfM 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 Am,nαn,mBm,n such that
(2)(Am,nαn,mBm,n)βn,mCm,n=Am,nαn,m(Bm,nβn,mCm,n),
where Am,n,Bm,n,Cm,n∈M and αn,m,βn,m∈Γ. An algebraic system that satisfying the associativity property of the above type is a Γ-semigroup.Example 3.
LetM=[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 ofΓ-semigroup can be found in [12, 16].LetH 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∈Ba∘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
(3)⋃u∈x∘yu∘z=⋃v∈y∘zx∘v.
A semihypergroup (H,∘) is called a hypergroup if for all x∈H, one has
(4)x∘H=H∘x=H.


## 3.Γ-Semihypergroups

In this section, we introduce the notion of aΓ-semihypergroup which is a generalization of the notion of Γ-semigroup and semihypergroup.Definition 5.
LetS 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 everyγ∈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.LetA and B be two non-empty subsets of S and γ∈Γ, we define the following:
(5)AγB=∪{aγb∣a∈A,b∈B}.
Also
(6)AΓB=∪{aγb∣a∈A,b∈Bandγ∈Γ}=⋃γ∈ΓAγB.A nonempty subsetA 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.AΓ-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.
LetS=[0,1] and Γ=ℕ, for every x,y∈S and γ∈Γ. We define
(7)γ:S×S→𝒫*(S)xγy=[0,xyγ].
Then, γ is a hyperoperation. For every x,y,z∈S and α,β∈Γ we have
(8)(xαy)βz=[0,xyzαβ]=xα(yβz).
This means that S is a Γ-semihypergroup.Also, ifγ∈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.
LetS={0,1} and Γ={α,β} be defined as follows:
(9)α0100{0,1}1{0,1}{0,1}β010{0,1}{0,1}1{0,1}{0,1}
It is not difficult to see that S is a Γ-hypergroup.Example 10.
LetS={0,1} and Γ={β,γ} be defined as follows:
(10)β010{0,1}{0,1}1{0,1}{0,1}γ01001110
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,…,Hk be normal subgroups of G such that H1⊆H2⊆⋯Hk. Set Γ={γ1,…,γk} where for every 1≤i≤k, the hyperoperation γi be as follows:
(11)xγiy=Hi∘x∘y,∀x,y∈G.
For every 1≤i,j≤k we have
(12)xγi(yγjz)=xγi(Hj∘y∘z)=Hi∘x∘(Hj∘y∘z)=Hi∘Hj∘x∘y∘z=Hmax{i,j}∘x∘y∘z.
Also, xγi(yγjz)=Hmax{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.
LetS 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 everyx,z∈S and γ∈Γ there exists y∈S such that
(13)z∈(xγx)αy=xγ(xαy).
Now, there exists u∈xαy such that z∈xγu and so (S,γ) is a hypergroup.Remark 13.
LetS 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:
(14)β01010101γ010011100 is an identity element of (S,γ) and 1 is an identity element of (S,γ).Lemma 14.
LetS 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 everyx,y∈Γ, xαy=xβy. We have the following:
(15)xαy=xα(eβy)=(xαe)βy=xβy.Theorem 15.
LetS 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 Lemma14.Example 16.
Theorem15 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.
LetS be a Γ-semihypergroup and for every γ∈Γ, and semihypergroup (S,γ) has a scalar identity eγ∈S. Then, one has card(Γ)≤card(S).Proof.
By Lemma14, we have
(16)card(Γ)=card({eγ∣γ∈Γ})≤card(S).In general, the above theorem is not true for aΓ-semihypergroup with identity elements.Example 18.
LetS={0,1} and Γ={α,β,γ} be as follows:
(17)β01010101β010{0,1}{0,1}1{0,1}{0,1}γ01001110InΓ-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
(18)card(Γ)=3>card(S)=2.Definition 19.
LetS 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:
(19)f(xγy)=f(x)h(γ)f(y).
If we set h(γ)=γ′, then f(xγy)=f(x)γ′f(y).Lemma 20.
LetS be a Γ-semihypergroup, S′ be a Γ′-semihypergroup, and f:S→S′ be a homomorphism. Then one has the following. (1)
IfT 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

IfS 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 AR¯B 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 xγaR¯yγa 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 setAR¯¯B 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 xγaR¯¯yγa 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.LetS 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:
(20)R(a)γRR(b)={R(x)∣x∈aγb}.
Now, we set Γ/R={γ/R∣γ∈Γ}.Theorem 21.
LetS 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 xγyR¯xγy′ and xγy′R¯x′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,
(21)R(x′)γRR(y′)⊆R(x)γRR(y).
Also, by a similar way, we have
(22)R(x)γRR(y)⊆R(x′)γRR(y′).
Now, we see that for every α/R,β/R∈Γ/R and R(x),R(y),R(z)∈S/R,
(23)R(x)αR(R(y)βRR(z))=(R(x)αRR(y))βRR(z).
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
(24)(R(x)αRR(y))βRR(z)⊆R(x)αR(R(y)βRR(z)).
(
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, xγyR¯x′γy and so R is regular.Theorem 22.
LetS 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 thatπ 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, letR(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
(25)π(S)γRπ(x)=π(Sγx)=π(S)=SR=π(S)=π(x)γRπ(S)=π(xγS).
This implies that S/R is a Γ/R-hypergroup.Theorem 23.
LetS 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,
(26)xRy⟺f(x)=f(y),
is regular and the function ψ:f(S)→S/R,ψ(f(x))=R(x) is an isomorphism.Proof.
Letx,y∈S and xRy. Then, for every a∈S and γ∈Γ we have
(27)f(xγa)=f(x)γ′f(a)=f(y)γ′f(a)=f(yγa).
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:
(28)ψ(f(x)γ′f(y))=ψ(f(xγy))={R(z)∣z∈xγy}=R(x)γRR(y)=ψ(f(x)γRf(y)),(29)ψ-1(R(x)γRR(y))=ψ-1({R(z)∣z∈xγy})=⋃z∈xγyf(z)=f(xγy)=f(x)γ′f(y)=ψ-1(R(x))γ′ψ-1(R(y)).
Therefore, ψ is an isomorphism.Theorem 24.
LetS 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)
ifx1Ry1 and x2Ry2 then x1γx2R¯¯y1γy2  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.
LetS 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, xγuR¯¯xγu and uγxR¯¯uγx. 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.
Letγ∈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 xγuR¯¯yγu, thus, we have
(30)(xγu)αvR¯¯(yγu)αv⇒xγ(uαv)R¯¯yγ(uαv).
Since xγa⊆xγ(uαv) and yγa⊆yγ(uαv), then xγaR¯¯xγa. In the same way, we obtain aγxR¯¯aγy. Therefore, R is a strongly regular relation on a Γ-semihypergroup S.Theorem 26.
LetS be a Γ-semihypergroup. Then,(1)
ifH is a Γ′-semigroup and ϕ:S→H is a homomorphism, then the equivalence relation R associated to ϕ is strongly regular;(2)
ifS is a Γ-hypergroup and R a strongly regular relation, then S/R is a Γ/R-group.Proof.
(1): We havexRy⇔ϕ(x)=ϕ(y). Since H is a Γ′-semigroup, we obtain
(31)ϕ(xγa)∈ϕ(x)γ′ϕ(a),
for every γ∈Γ and a∈S, and xRy implies that
(32)|ϕ(xγa)|=1=|ϕ(yγa)|.
Therefore, xγaR¯¯yγa.
(2): It is obtained from Theorems22 and 24.Definition 27.
LetS 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 relationθ is the smallest equivalence relation on a Γ-semihypergroup S such that the quotient S/θ is a Γ/θ-semigroup.Definition 28.
LetS be a Γ-semihypergroup and ρ be a relation on S as follows:
x
ρ
y if and only if there exist n∈ℕ, x1,…xn∈S and γ1,…,γn-1∈Γ such that
(33){x,y}⊆x1γ1x2γ2⋯γn-2xn-1γn-1xn,
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:
(34)xβ∘y⟺∃x1,…,xn∈H,suchthat{x,y}⊆x1∘x2∘⋯∘xn.
Therefore, the relation ρ is a generalization of the relation β.Lemma 30.
LetS be a Γ-semihypergroup. Then, for every γ∈Γ one has βγ⊆ρ.Proof.
It is  straightforward.Lemma 31.
The relationρ is reflexive and symmetric on S and ρ* is an equivalence relation, moreover one has
(35)(⋃γ∈Γβγ)*⊆ρ*.Proof.
Clearly,ρ is reflexive and symmetric, so ρ* is an equivalence relation. Since for every γ∈Γ, βγ⊆ρ, then ⋃γ∈Γβγ⊆ρ and so (⋃γ∈Γβγ)*⊆ρ*.Example 32.
LetG be the Γ-hypergroup in Example 11. For every 1≤i≤n, we set βγi=βi. Then, we have the following:
(36)β1⊆β2⊆⋯⊆βk⊆ρ.
Also, it is not difficult to see that βk=ρ. Every (G,γi) is a hypergroup and so β=β* (see [27]). Therefore, we obtain ρ=βk=βk* and ρ is an equivalence relation on S.Theorem 33.
The relationρ* is a strongly regular equivalence relation on the Γ-semihypergroup S.Proof.
By Lemma31, ρ* is an equivalence relation. In order to prove that it is strongly regular, we show first that
(37)xρy⇒xγaρ¯¯yγa,aγxρ¯¯aγy,
for every a∈S, and γ∈Γ. If xρy, then there exist n∈ℕ, x1,…xn∈S and γ1,…,γn-1∈Γ such that
(38){x,y}⊆x1γ1x2γ2⋯γn-2xn-1γn-1xn.
For every a∈S and γ∈Γ and every u∈xγa and v∈yγa, we obtain:
(39)u∈xγa⊆x1γ1x2γ2⋯γn-2xn-1γn-1xnγa,v∈yγa⊆x1γ1x2γ2⋯γn-2xn-1γn-1xnγa.
Thus, uρv and so xγaρ¯¯yγa. In the same way, we can show that xρy⇒aγxρ¯¯aγy.
Now, ifxρ*y, then there exist m∈ℕ and x=w0,w1,…,wm-1,wm=y∈S such that x=w0ρw1ρ⋯ρwm-1ρwm=y, whence, we obtain
(40)xγa=w0γaρ¯¯w1γaρ¯¯⋯ρ¯¯wm-1γaρ¯¯wmγa=yγa.
Finally, for allv∈xγa=w0γa and for all w∈wmγa=yγa, taking z1∈w1γa, z2∈w2γa,…,zm-1∈wm-1γa, we have vρz1ρz2ρ⋯ρzm-1ρw and so vρ*w. Therefore,
(41)xρ*y⇒xγaρ*¯¯yγa.
Similarly, we can prove that xρ*y⇒aγxρ*¯¯aγy. Hence, ρ* is strongly regular.Corollary 34.
LetS be a Γ-semihypergroup. Then, the quotient S/ρ* is a Γ/ρ*-semigroup.Proof.
By Theorems24 and 33, the proof is completed.Theorem 35.
LetS 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.
LetR be a strongly regular relation. Now, let xρy so there exist n∈ℕ, x1,…xn∈S and γ1,…,γn-1∈Γ such that
(42){x,y}⊆x1γ1x2γ2⋯γn-2xn-1γn-1xn.
Since R is a strongly regular relation, then S/R is a Γ/R-semigroup and so
(43)R(x)=R(x1)γ¨1R(x2)γ¨2⋯γ¨n-2R(xn-1)γ¨n-1R(xn),R(y)=R(x1)γ¨1R(x2)γ¨2⋯γ¨n-2R(xn-1)γ¨n-1R(xn),
where γ¨i=γi/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,…,wm-1,wm=y∈S such that x=w0ρw1ρ⋯ρwm-1ρwm=y. Therefore, x=w0Rw1R⋯Rwm-1Rwm=y, and transitivity of R implies that xRy. Thus, ρ*⊆R.Theorem 36.
IfS is a Γ-hypergroup, then the quotient S/ρ* is a Γ/ρ*-group.Proof.
Letρ*(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.
IfS is a Γ-semihypergroup with an identity element e∈S, then card(Γ/ρ*)=1 and so S/ρ* is a semigroup.Proof.
Sinceρ*(e) is an identity element of Γ/ρ*-semigroup S/ρ*, then by Theorem 15, the proof is completed.Theorem 38.
IfS 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 everyγ∈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.



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*Source: 915250-2013-01-30.xml*