In 1981, Sen [1] introduced the concept and notion of the Γ-semigroup as a generalizationof plain semigroup and ternary semigroup. Many classical notions and results of the theoryof semigroups have been extended and generalized to Γ-semigroups. Green’s relations for Γ-semigroups (see [2–5]) play an important role in studying of the structure of Γ-semigroups aswell as in the case of plain semigroups. Croisot [6] gave a definition of (m, n)-regularity inplain semigroups and later, Kapp [7] generalized this definition in plain semigroups. We extendand generalize this definition introducing the (m, n)0-regularity in Γ-semigroups. In particular,in this paper, we investigate and characterize the (2, 2)0-regular class of Γ-semigroups usingGreen’s relations. We show that a Γ-semigroup M is 20-regular if and only if for all x ∈ M andγ ∈ Γ, xγx = 0 or xγx ∈ Hx. Furthermore, every regular D-class D union {0} of a 20-regularΓ-semigroup is itself a completely 0-simple Γ-semigroup. In generalizing Croisot’s theory ofdecomposition for Γ-semigroups, we introduce the absorbent Γ-semigroup and regular absorbentΓ-semigroup and prove that a regular Γ-semigroup with 0 is absorbent if and only if it is mutuallyannihilating collection of completely 0-simple sub-Γ-semigroups with a common zero. We prove(cf. Theorem 4.25) 7 equivalent conditions for such a decomposition for Γ-semigroups with zero.In this paper, continuing our earlier investigation [8] concerning quasi-ideals in Γ-semigroups,we try to approach the above problem by examining quasi-ideals using Green’s relations. Wecharacterize the absorbency in terms of a partial order on Green’s equivalence classes wherethis natural partial ordering is trivial. We show that for an absorbent Γ-semigroup with 0every H-class union 0 is a quasi-ideal (these are the only 0-minimal quasi-ideals). For regular
Received January 15, 2010, revised August 23, 2011, accepted December 26, 2011
610 Hila K. and Dine J.
or commutative Γ-semigroups we show (cf. Theorem 3.12 and Theorem 4.25) that the converseis true. An example (cf. Remark 3.14) answers to the question as to whether the converse isalways true.
In 1986, Sen and Saha [9] defined Γ-semigroup as a generalization of semigroup and ternarysemigroup as follows:
Definition 1.1 Let M and Γ be two non-empty sets. Denote by the letters of the Englishalphabet, a, b, c, . . . , the elements of M and with the letters of the Greek alphabet, α, β, γ, . . . ,
the elements of Γ. 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 = aα(bβc) for all a, b, c ∈ M and for all α, β ∈ Γ.
For non-empty subsets A and B of M and a non-empty subset Γ′ of Γ, let AΓ′B = {aγb :a ∈ A, b ∈ B and γ ∈ Γ′}. If A = {a}, then we also write aΓ′B instead of {a}Γ′B, and similarlyif B = {b} or Γ′ = {γ}.
A Γ-semigroup M is called commutative Γ-semigroup if for all a, b ∈ M and γ ∈ Γ, aγb =bγa. A non-empty subset K of a Γ-semigroup M is called a sub-Γ-semigroup of M if for alla, b ∈ K and γ ∈ Γ, aγb ∈ K.
Several examples of Γ-semigroups can be found in [2, 9–13].
Definition 1.2 Let M be a Γ-semigroup. A non-empty subset I of M is called a left (resp.right) ideal of M if MΓI ⊆ I (resp. IΓM ⊆ I). A non-empty subset I of M is called an idealof M if it is a left ideal as well as a right ideal of M .
An element a of a Γ-semigroup M is called idempotent if a = aγa, for some γ ∈ Γ. Anelement a of a Γ-semigroup M is called zero element of M if aγb = bγa = a, ∀ b ∈ M and∀ γ ∈ Γ and it is denoted by 0. A Γ-semigroup M is called left (resp. right) simple if it doesnot contain proper left (resp. right) ideals or equivalently, if for every left (resp. right) ideal A
of M , we have A = M .The element a of a Γ-semigroup M is called regular in M if a ∈ aΓMΓa, where aΓMΓa =
{(aαb)βa|a, b ∈ M, α, β ∈ Γ}. M is called regular if and only if every element of M is regular.
Definition 1.3 A Γ-semigroup M with zero element is called 0-simple (left 0-simple, right0-simple) if
1. MΓM �= {0}, and2. {0} is the only proper two sided (left, right)-ideal of M .
Definition 1.4 A two sided (left, right) ideal I of a Γ-semigroup M is called 0-minimal if1. I �= {0}, and2. {0} is the only two sided (left, right) ideal of M contained in I.
For each element a of a Γ-semigroup M , the left ideal MΓa∪{a} containing a is the smallestleft ideal of M containing a, for if A is any other left ideal containing a, then MΓa ∪ {a} ⊆ A
and this ideal is denoted by (a)l and called the principal left ideal generated by the element a.Similarly, for each a ∈ M , the smallest right ideal containing a is aΓM ∪ {a} which is denotedby (a)r and called the principal right ideal generated by the element a. The principal ideal ofM generated by the element a is denoted by (a) and (a) = {a}∪MΓ{a}∪{a}ΓM ∪MΓ{a}ΓM .
On Green’s Relations, 20-regularity and Quasi-ideals in Γ-semigroups 611
Let M be a Γ-semigroup and α be a fixed element of Γ. We define a ◦ b in M by a ◦ b =aαb, ∀ a, b ∈ M . In [9], the authors have shown that M is a semigroup and denoted thissemigroup by Mα. They have shown that if Mα is a group for some α ∈ Γ, then Mα is a groupfor all α ∈ Γ. A Γ-semigroup M is called a Γ-group if Mα is a group for some (hence for all)α ∈ Γ.
In [3], the authors defined the Green’s equivalences on Γ-semigroup as follows:Let M be a Γ-semigroup. Let a, b ∈ M ,
aLb ⇔ (a)l = (b)l, aRb ⇔ (a)r = (b)r, aJ b ⇔ (a) = (b),
aHb ⇔ aLb and aRb, aDb ⇔ aLc and cRb for some c ∈ M.
It is clear that a Γ-semigroup M is left (right) simple if and only if it consists of a singleL[R] class, and that M is simple if and only if it consists of a single J -class. We say thata Γ-semigroup M is bisimple if it consists of a single D-class. Since D ⊆ J , every bisimpleΓ-semigroup is also simple.
Definition 1.5 Let M a Γ-semigroup. If La and Lb are L-classes containing a and b of aΓ-semigroup M respectively, then La ≤ Lb if (a)l ⊆ (b)l. Then “ ≤ ” is a partial order in M/Lwhich is the set of L-classes of M . Similarly, Ra ≤ Rb and Ja ≤ Jb are defined in M/R andM/J .
Definition 1.6 A Γ-semigroup M is said to satisfy minL or minR condition if every non-empty set of L-classes or of R-classes possess a minimal member respectively.
Definition 1.7 A Γ-semigroup M is called completely 0-simple if M is 0-simple and it satisfiesthe minL and minR conditions.
2 Green’s Relations and 20-regularity
The following lemma which we may call the Green’s lemma for Γ-semigroups has been provedin [3].
Lemma 2.1 ([3, Lemma 3.6]) Let a, b two elements of a Γ-semigroup M such that aRb. Ifaαs = b and bβs′ = a where s, s′ ∈ M, α, β ∈ Γ, the mappings
σ : La → Lb
x �→ xαsand
σ′ : Lb → La
y �→ yβs′
are mutually inverse, R-preserving, one to one maps from La to Lb and from Lb to La respec-tively.
In [14], the authors proved the following Green’s theorem for Γ-semigroups.
Theorem 2.2 If the elements a, b and aγb (γ ∈ Γ) of a Γ-semigroup M all belong to the sameH-class H of M , then H is a subgroup of the semigroup Mγ . Moreover, for every two elementsh1, h2 ∈ H, the element h1γh2 belongs to H.
Also, they proved the following corollary and theorem.
Corollary 2.3 If an H-class H of a Γ-semigroup M contains an idempotent e = eαe, α ∈ Γ,then H is a subgroup of Sα.
Theorem 2.4 Let M be an arbitrary Γ-semigroup. If the idempotent eαe, α ∈ Γ, togetherwith a, b ∈ M all belong to some H-class H, then eαa = aαe = a and aαb ∈ H.
612 Hila K. and Dine J.
Lemma 2.5 Let M be a Γ-semigroup. The set product LΓR of any L-class L and any R-classR of M is always contained in a single D-class of M .
Proof The statement of the lemma is equivalent to asserting that if a, a′, b, b′ ∈ M such thataLa′ and bRb′, then aγbDa′γb′ for all γ ∈ Γ. Since L is a right congruence (see [3, Lemma 3.2]),aLa′ implies aγbLa′γb for all γ ∈ Γ. Since R is a left congruence (see [3, Lemma 3.2]), bRb′
implies a′γbRa′γb′ for all γ ∈ Γ. But aγbLa′γb and a′γbRa′γb′ imply aγbDa′γb′ for all γ ∈ Γ. �Now we state the following lemma from [15].
Lemma 2.6 ([15, Lemma 5.3]) Let M be a Γ-semigroup. If a, b ∈ M , then aγb ∈ Ra ∩Lb forsome γ ∈ Γ if and only if Rb ∩ La contains a γ-idempotent. In such a case,
Haγb = aγHb = HaγHb = Haγb = Ra ∩ Lb.
In [16], Seth proved the following lemma:
Lemma 2.7 ([16, Lemma 2.9]) Let a, b be any two elements of a completely 0-simple Γ-semigroup M and γ be any element of Γ. Then either aγb = 0 or aγb ∈ Ra ∩ Lb.
Definition 2.8 Let m and n be nonnegative integers with m + n > 1. A Γ-semigroupM will be in the class of (m, n)-Γ-semigroups, written M ∈ (m, n) if and only if for ev-ery x ∈ M , there exist u ∈ M and α, β, α1, . . . , αm−1, β1, . . . , βn−1 ∈ Γ such that x =xα1x · · ·αm−1xαuβxβ1x · · ·βn−1x. M is then called (m, n)-regular.
Since we can always adjoin a zero to a Γ-semigroup, M , we will consider M to have a zero, 0,in what follows.
Definition 2.9 Let m and n be nonnegative integers with m + n > 1. A Γ-semigroup M willbe in the class of (m, n)0-Γ-semigroups, written M ∈ (m, n)0, if and only if for every x ∈ M
one of the following holds :1. m > 0 and xα1x · · ·αm−1x = 0,2. n > 0 and xβ1x · · ·βn−1x = 0,3. x = xα1x · · ·αm−1xαuβxβ1x · · ·βn−1x for some u ∈ M and α, β, α1, . . . , αm−1, β1, . . . ,
βn−1 ∈ Γ where x0 is suppressed in the equation when necessary.We will say that M is (m, n)0-regular whenever M ∈ (m, n)0 and that M is n0-regular when
M ∈ (n, n)0.
It is clear that (m, n) ⊆ (m, n)0. Indeed, if M is a Γ-semigroup with no nilpotent elements(other than perhaps 0), we have M ∈ (m, n) if and only if M ∈ (m, n)0.
As an extension and generalization of Croisot’s (2, 2)-regular semigroup, we give the follow-ing definition.
Definition 2.10 A Γ-semigroup M with 0 is said to be 20-regular whenever for each x ∈ M
either xγx = 0 for some γ ∈ Γ or x ∈ xΓxΓMΓxΓx.
It is clear that the Γ-semigroups M such that M ∈ (1, 1) are regular Γ-semigroups. Foranalogy, we have
Definition 2.11 A left regular Γ-semigroup is a Γ-semigroup M such that M ∈ (0, 2), thatis, for all x ∈ M , there exist u ∈ M and α, β ∈ Γ, such that x = uαxβx.
Definition 2.12 A right regular Γ-semigroup is a Γ-semigroup M such that M ∈ (2, 0), thatis, for all x ∈ M , there exist u ∈ M and α, β ∈ Γ, such that x = xβxαu.
On Green’s Relations, 20-regularity and Quasi-ideals in Γ-semigroups 613
It is obvious that the class of regular Γ-semigroups does not coincide with the class of Γ-semigroups that are at the same time left regular and right regular. In fact, two of the threeconditions (1, 1), (0, 2), (2, 0) imply the third and there is no other connections between thoseconditions.
As an application of the result proved in Theorem 1 in [17], we have the following lemmas:
Lemma 2.13 Let M be a Γ-semigroup. The following are equivalent :1. M is left regular.2. aL(aγa), ∀ a ∈ M, ∀ γ ∈ Γ.
Lemma 2.14 Let M be a Γ-semigroup. The following are equivalent :1. M is right regular.2. aR(aγa), ∀ a ∈ M, ∀ γ ∈ Γ.
Theorem 2.15 A Γ-semigroup M is a union of Γ-groups if and only if it is at the same timea left and right regular Γ-semigroup.
Proof It can be easily obtained by using Theorem 2.2, Lemmas 2.13 and 2.14. �
Corollary 2.16 Let M be a Γ-semigroup. The following statements are equivalent :1. M ∈ (2, 2).2. M ∈ (2, 0) ∩ (0, 2).3. M is a union of Γ-groups.
Proof It follows easily from Theorem 2.15. �
Proposition 2.17 (n, n)0 = (0, n)0 ∩ (n, 0)0 for n ≥ 2.
Proof See Theorem 2.15. �
Definition 2.18 (1) A Γ-semigroup M is said to be left-0-regular if M ∈ (0, 2)0.(2) A Γ-semigroup M is said to be right-0-regular if M ∈ (2, 0)0.(3) A Γ-semigroup M is said to be intra-0-regular if for each x ∈ M either xγx = 0 for
some γ ∈ Γ or x = uαxβxρv for some u, v ∈ M and α, β, ρ ∈ Γ.
Definition 2.19 (1) An ideal I of a Γ-semigroup (with 0) is said to be 0-semiprime if xγx ∈I\{0}, for all γ ∈ Γ, implies x ∈ I.
(2) A Γ-semigroup M is said to be (left, right) 0-semiprime whenever every (left, right)two-sided ideal of M is 0-semiprime.
It can be easily proved the following proposition.
Proposition 2.20 Let M be a Γ-semigroup with 0. The following statements are equivalent :1. M is (left, right) intra-0-regular.2. M is (left, right) 0-semiprime.3. If x ∈ M and xγx �= 0, ∀ γ ∈ Γ, then [xLxγx, xRxγx]xJ xγx,∀x ∈ Γ.4. If x ∈ M and xγx �= 0, ∀ γ ∈ Γ, then [x ∈ MΓxγx, x ∈ xγxΓM ]x ∈ MΓxγxΓM, ∀ γ ∈ Γ.
Proposition 2.21 Let M be a Γ-semigroup with 0. All left, right and two-sided ideals of M
are 0-semiprime if and only if M is 20-regular.
Proof This follows directly from Proposition 2.17 with n = 2 and Proposition 2.20. �
614 Hila K. and Dine J.
Theorem 2.22 Let M be a Γ-semigroup with 0. M is 20-regular if and only if for all x ∈ M
and γ ∈ Γ, xγx = 0 or xγx ∈ Hx.
Proof Assume M is 20-regular, that is, M ∈ (2, 2). Let x ∈ M and γ ∈ Γ. Let us supposethat xγx �= 0. By Corollary 2.16 and Theorem 2.15, it follows that M is a left and right regularΓ-semigroup. By Lemmas 2.13 and 2.14, it follows that xH(xγx), that is, xγx ∈ Hx.
Conversely, suppose that for all x ∈ M and γ ∈ Γ, xγx = 0 or xγx ∈ Hx. In the former case,the first part of Definition 2.9 is satisfied; in the latter case, by Theorem 2.2, Hx is subgroup ofMγ and so, the equation x = xγxγuγxγx is the solvable for u in Hx, that is, x ∈ xΓxΓMΓxΓx.Thus in either case M is 20-regular. �
Let D be a D-class. Then either every element of D is regular or no element of D isregular [4]. We call the D-class regular if all its elements are regular.
Corollary 2.23 Let M be a 20-regular Γ-semigroup. Then all the irregular elements of M liein D-classes, D, such that DΓD = {0}.Proof Let D be an irregular D-class of M , which means that no element of D is regular. Leta, b ∈ D and let x ∈ Rb ∩ La. By Theorem 2.22, since M is 20-regular, xγx = 0 for someγ ∈ Γ or else Hx would be a Γ-group by Corollary 2.3, and D then would not be irregular. ByLemma 2.5, LaΓRb ⊆ D′ a D-class. Since xγx = 0 ∈ LaΓRb for some γ ∈ Γ, D′ must be thezero D-class {0} and aαb = 0, for all α ∈ Γ. It thus follows that DΓD = {0}. �
Theorem 2.24 Let M be a 20-regular Γ-semigroup and suppose D is a nonzero regular D-classunion {0}. Then D is itself a completely 0-simple Γ-semigroup.
Proof D is a Γ-semigroup. Indeed, we will show that if a, b ∈ D\{0}, either aγb = 0 oraγb ∈ Ra ∩ Lb for some γ ∈ Γ. Let a, b ∈ D\{0}. Then LaΓRb ⊆ D′ where D′ is a D-class byLemma 2.5. Let c ∈ Rb ∩ La which is non-empty so that we have cαc ∈ LaΓRb ⊆ D′ for someα ∈ Γ. If cαc = 0, then D′ = {0} since D′ is a D-class and thus aγb = 0 ∈ D. But if cαc �= 0,then cαc ∈ Hc by Theorem 2.22, since M is 20-regular. By Corollary 2.3, Hc = Rb ∩ La is aΓ-group. Hence by Lemma 2.6 we have aγb ∈ Ra ∩ Lb ⊆ D. If either a or b is 0, then surelyaγb ∈ D. In any case aγb ∈ (Ra ∩ Lb) ⊆ D or aγb = 0 ∈ D.
Since D\{0} is regular, it contains nonzero idempotents by [2, Lemma 3] and hence DΓD �=0. We will now show that each nonzero idempotent in D is primitive (cf. [16, p. 224]). Lete, f ∈ M be respective δ-idempotent and ρ-idempotent of M , δ, ρ ∈ Γ. We must show thatif 0 �= e ≤ f , then e = f . In fact, we will show for any two idempotent e, f ∈ D, eveneδf = fρe �= 0 implies e = f . As above, if eδf �= 0, then we must have eδf ∈ Re ∩Lf . Likewisefρe �= 0 implies fρe ∈ Rf ∩ Le. Hence eδf = fρe ∈ Re ∩ Le ∩ Rf ∩ Lf = He ∩ Hf . Sinceeδf = fρe is an idempotent and a Γ-group can contain at most one idempotent, it follows byCorollary 2.3 that e = eδf = fρe = f .
It remains to show that D, as a Γ-semigroup is 0-simple. Indeed, we will show that ifM1Γa = M1Γb for a, b ∈ D\{0}, then D1Γa = D1Γb. Suppose aLb in M . Then either a = b
and the result follows immediately or there is an x in M such that xγa = b for some γ ∈ Γ.Since a and b are regular, we can find [2, Lemma 3], R-equivalent idempotents e and f for a andb respectively. Similarly, let g be an L-equivalent μ-idempotent for a and b. From xγa = b, wededuce that fρxγeδa = b. By [16, Lemma 2.2] and [4, Theorem 3.5], we can find a (μ, δ)-inversea′ for a in Le ∩ Rg. Hence 0 �= fρxγe = (fρxγe)δe = (fρxγeδa)μa′ = bμa′ and so bμa′ �= 0.
On Green’s Relations, 20-regularity and Quasi-ideals in Γ-semigroups 615
Thus by the first part of the proof fρxγe = bμa′ ∈ Rb ∩La′ ⊆ D\{0}. So the equation zγa = b
is solvable in D. Likewise yγ1b = a, γ1 ∈ Γ is also solvable in D. This clearly is sufficient toshow D1Γa = D1Γb. Dually, cΓM1 = dΓM1 implies cΓD1 = dΓD1 for c, d ∈ D\{0}. It followsthat D is 0-bisimple and hence 0-simple. Whence we have shown that D is completely 0-simple,and the proof is complete. �
Theorem 2.25 Let M be a regular Γ-semigroup with 0. Then the following are equivalent :1. M is 20-regular.2. M is the 0-disjoint union of sub-Γ-semigroups which are themselves completely 0-simple
Γ-semigroups.3. M is left 0-regular and right 0-regular.4. All left and right ideals of M are 0-semiprime.
Proof (1) ⇒ (2) If M is 20-regular, then by Theorem 2.24 each nonzero D-class union 0 is acompletely 0-simple Γ-semigroup. Since M is regular by our overall assumption and D is anequivalence relation, we have a 0-disjoint union of the D-classes of M of the type desired in (2).
Conversely, if we assume (2) holds, then for any a ∈ M , a and aγa, γ ∈ Γ both belongto a sub-Γ-semigroup which is completely 0-simple. In such sub-Γ-semigroup aγa ∈ Ha ∪ {0}by Lemma 2.7 (Ha here denotes an H-class with respect to Green’s relation defined on thesub-Γ-semigroup). If aγa �= 0, then Ha is a Γ-group by Theorem 2.2 and in this case Ha iseasily seen to be contained in an H-class of M , hence aγa ∈ HM
a where HMa is the H-class of a
in M . If aγa = 0, then aγa ∈ HMa ∪ {0}. In either case we can conclude by Theorem 2.22 that
M is 20-regular.(1) ⇔ (3) This is just Proposition 2.17 with n = 2.(1) ⇔ (4) This is just Proposition 2.21. �
3 Green’s Relations and Quasi-ideals
Definition 3.1 A Γ-semigroup M with 0 is said to be absorbent whenever aγb ∈ (Ra ∩ Lb)0
for every a, b ∈ M and γ ∈ Γ.
Remark 3.2 It is clear that an absorbent Γ-semigroup is 20-regular, it suffices to set b = a
in the above definition.
Let M be a Γ-semigroup with 0. Referred to Definition 1.9, we will say that the set ofR-[L]-classes of M is trivially partially ordered whenever Rx ≤ Ry[Lu ≤ Lv] implies that eitherx = 0 or xRy[u = 0 or uLv]. We will say that M has trivial class order whenever both the setof R-classes and L-classes are trivially partially ordered.
Theorem 3.3 A Γ-semigroup M with 0 is absorbent if and only if M has trivial class order.
Proof Let M be an absorbent Γ-semigroup with 0. Suppose Rx ≤ Ry. If x = 0, thenRx = {0}; if x = y, then Rx = Ry, in either case we are done. Suppose then that x �= y andx �= 0 �= y. From Definition 1.9, since rx ≤ Ry, we have xΓM1 ⊆ yΓM1 and this implies thatx = yγs for some s ∈ M and γ ∈ Γ. Then by absorbency x ∈ Ry ∩ Ls and hence xRy andRx = Ry. Dually, one shows that Lu ≤ Lv implies either u = 0 or Lu = Lv. Whence M hastrivial class order.
Conversely, suppose that M has trivial class order. Let a, b ∈ M . Clearly, Raγb ≤ Ra andLaγb ≤ Lb for all γ ∈ Γ. Thus either aγb = 0 or Raγb = Ra and Laγb = Lb. Whence, in either
616 Hila K. and Dine J.
case aγb ∈ (Ra ∩ Lb)0 and it follows that M is absorbent. �
Definition 3.4 Let M be a Γ-semigroup and Q a non-empty subset of M . Then Q is calledquasi-ideal of M if
QΓM ∩ MΓQ ⊆ Q.
Example 3.5 Let M be a semigroup and Γ be any non-empty set. Define a mapping M ×Γ × M → M by aγb = ab, ∀ a, b ∈ M and γ ∈ Γ. Then M is a Γ-semigroup. Let Q be aquasi-ideal of M . Thus MQ ∩ QM ⊆ Q. We have MΓQ ∩ QΓM = MQ ∩ QM ⊆ Q. Hence, Q
is a quasi-ideal of M .
This example implies that the class of quasi-ideals in Γ-semigroups is a generalization ofquasi-ideals in semigroups.
Theorem 3.6 If M is an absorbent Γ-semigroup, then each H-class union {0} is a quasi-ideal.
Proof Let H be an H-class union {0}. Suppose x ∈ HΓM ∩MΓH. If x �= 0, then x = hαs =s′βh′ for some h, h′ ∈ H, s, s′ ∈ M and α, β ∈ Γ. It follows that Rx ≤ Rh and Lx ≤ Lh′ . Bythe above theorem, M has trivial class order, so that either x = 0 or x �= 0 and Rx = Rh andLx = Lh′ . Since hHh′ in the latter case, we have x ∈ Rx ∩ Lx = Rh ∩ Lh′ = H\{0}. In eithercase x ∈ H and we can conclude HΓM ∩MΓH ⊆ H, so that H is a quasi-ideal by definition. �
Proposition 3.7 If M is a Γ-semigroup with 0 such that each H-class union {0} is a quasi-ideal, then M is 20-regular.
Proof Let x ∈ M and let H = H0x. Then HΓH ⊆ MΓH∩HΓM ⊆ H since H is a quasi-ideal.
Thus either xγx = 0 for some γ ∈ Γ or xγx ∈ H for some γ ∈ Γ. It follows from Theorem 2.22that M is 20-regular. �
Definition 3.8 ([8, p. 148]) A quasi-ideal Q of a Γ-semigroup M with 0 is said to be 0-minimalif there is no quasi-ideal Q1 of M such that {0} ⊂ Q1 ⊂ Q.
Proposition 3.9 Let M be a Γ-semigroup with 0. If a non-zero H-class union {0} is aquasi-ideal, then it is a 0-minimal quasi-ideal.
Proof Let H be a non-zero H-class union {0} which is a quasi-ideal of M . Suppose that Q isa quasi-ideal of M and 0 ⊂ Q ⊆ H. Now let q ∈ Q\{0} and let h ∈ H\{0}. Since q ∈ H\{0},an H-class, if q �= h, we can find an r, s ∈ M and α, β ∈ Γ such that rαq = qβs = h, i.e., qHh.Now h ∈ MΓq ∩ qΓM ⊆ MΓQ ∩ QΓM ⊆ Q which shows that H\{0} ⊆ Q. Whence Q = H
and H is, by definition, a 0-minimal quasi-ideal. �
Proposition 3.10 Let M be a Γ-semigroup with 0. Every quasi-ideal Q of M is the union ofits H-classes, i.e., Q =
⋃q∈Q Hq.
Proof Let Q be a quasi-ideal and let q ∈ Q. Let h ∈ Hq. If h �= q, there exist r, s ∈ M andα, β ∈ Γ such that rαq = qβs = h. Thus h ∈ MΓQ ∩ QΓM ⊆ Q. Hence Hq ⊆ Q and it followsthat Q =
⋃q∈Q Hq. �
Proposition 3.11 Let M be a Γ-semigroup with 0. If each H-class union {0} of M is anideal, then M is absorbent.
Proof Let a, b ∈ M . Then since H0a , H0
b are ideals by hypothesis, aγb ∈ H0aΓH0
b ⊆ (Ha ∩Hb)0 ⊆ (Ra ∩ Lb)0, γ ∈ Γ and M is thus absorbent. �
On Green’s Relations, 20-regularity and Quasi-ideals in Γ-semigroups 617
Lemma 3.12 For commutative Γ-semigroup, quasi-ideals and ideals coincide.
Proof Let M be a commutative Γ-semigroup. Then MΓQ = QΓM for any subset Q of M .Thus if Q is a quasi-ideal, we have QΓM ∩ MΓQ ⊆ Q. Hence since MΓQ = QΓM , it followsthat QΓM, MΓQ ⊆ Q and Q is an ideal. Clearly, any ideal is also a quasi-ideal. �
Theorem 3.13 A commutative Γ-semigroup M with 0 is absorbent if and only if each H-classunion {0} of M is a quasi-ideal.
Proof One implication is just Theorem 3.6.Conversely, suppose each H-class union {0} of M is a quasi-ideal. Then since M is a
commutative, each quasi-ideal is an ideal by Lemma 3.12. Applying Proposition 3.11, wecomplete the proof. �
Remark 3.14 From the above, we saw that the absorbency implies every H-class union 0 isa quasi-ideal (cf. Theorem 3.6). Also, for regular or commutative Γ-semigroups it is proved(cf. Theorem 3.13 and Theorem 4.25) that the converse is true. A natural question is posed:whether this is always true. The following examples answer to this question:
a) Let us consider the Γ-semigroup M = {0, a, b, aγb, bγa} with Γ = {γ}, where aγa =bγb = aγbγa = bγaγb = 0. M is not absorbent but each H-class union {0} is a quasi-ideal.
b) Let M = {0, x, y, z} and Γ = {α, β} with the multiplication defined by the followingtable:
α 0 x y z
0 0 0 0 0
x 0 x y x
y 0 x y x
z 0 x y x
β 0 x y z
0 0 0 0 0
x 0 x y x
y 0 x y y
z 0 x y z
M is a 20-regular Γ-semigroup, but it is not absorbent.
Theorem 3.15 ([8, Lemma 2.3 (4), Proposition 2.8, Corollary 2.9]) The intersection of a leftideal and a right ideal of a Γ-semigroup M is a quasi-ideal. Conversely, any quasi-ideal Q of M
can be obtained as the intersection of a left ideal and a right ideal (Q = (Q∪MΓQ)∩(Q∪QΓM)).
Theorem 3.16 Let M be a Γ-semigroup with 0. A quasi-ideal Q is 0-minimal if and only ifQ is an H-class union {0}.Proof Let Q be a quasi-ideal of M which is an H-class union {0}. Then by Proposition 3.9,it is 0-minimal.
Conversely, suppose Q is a 0-minimal quasi-ideal of M . Let a, b ∈ Q\{0}. Since M1Γa ∩aΓM1 and M1Γb ∩ bΓM1 are non-zero quasi-ideals by Theorem 3.14 which are contained inQ, we must have M1Γa ∩ aΓM1 = Q = M1Γb ∩ bΓM1. This clearly implies aHb. WhenceQ = H0
a . �
Corollary 3.17 A 0-minimal quasi-ideal is either a Γ-group with adjoined 0 or a null sub-Γ-semigroup.
Proof From the above theorem, a 0-minimal quasi-ideal Q is an H-class union {0}. Thus sinceQΓQ ⊆ Q for a quasi-ideal, it follows by Theorem 2.2 that either QΓQ = {0} or QΓQ = Q and
618 Hila K. and Dine J.
in that case Q is a Γ-group with adjoined 0. �
Theorem 3.18 Let L be a 0-minimal left ideal of a Γ-semigroup M . The eΓL = eΓMΓe = H0e
where 0 �= e = eαe for some α ∈ Γ.
Proof Let L be a 0-minimal left ideal and eαe = e �= 0 for some α ∈ Γ. Then we have {0} ⊂MΓe ⊆ L. Thus MΓe = L and eΓMΓe = eΓL ⊆ L. By the duality of [16, Lemma 4.3], L\{0} isan L-class. Hence if 0 �= h ∈ eΓL, we have hLe. Thus there is an r′ such that r′γh = e for someγ ∈ Γ. Let r = eαr′γe. Since h = hαe = eαh, we have rαh = eαr′γeαh = eαr′γh = eαe = e.Now, however, r ∈ eΓL\{0} and so rLe. But rαh = e and eαr = r imply rRe. Hence eHr.
Now (hαr)αh = h(rαh) = hαe = h and hα(r) = hαr imply hαrRh. Since both h andhαr ∈ L\{0}, hLhαr. Hence hαrHh. But (hαr)α(hαr) = hα(rαh)αr = (hαe)αr = hαr sothat Hh is a Γ-group by Theorem 2.2. But now h = eαh ∈ Re ∩ Lh = Re ∩ Le = He sinceLe ∩ Rh = Lh ∩ Rh = Hh is a Γ-group (cf. Lemma 2.6). Whence eΓL = H0
e . �
Corollary 3.19 ([8, Lemma 4.4]) Let L be a 0-minimal left ideal of a Γ-semigroup M and0 �= e = eαe for some α ∈ Γ. Then eΓL is a 0-minimal quasi-ideal.
Corollary 3.20 Let L be a 0-minimal left ideal of a Γ-semigroup M and 0 �= e = eαe forsome α ∈ Γ. Then eΓL = eΓMΓe = H0
e = MΓh ∩ hΓM for any h ∈ eΓMΓe\{0}.Proposition 3.21 If eαe = e �= 0 for some α ∈ Γ and eΓMΓe is a 0-minimal quasi-ideal ofa Γ-semigroup M , then eΓMΓe = H0
e = MΓh ∩ hΓM for any h ∈ eΓMΓe\{0}.Proof Let h ∈ eΓMΓe\{0}. Thus h = eαh = hαe and hence MΓh ∩ hΓM ⊆ MΓ(eΓMΓe) ∩(eΓMΓe)ΓM ⊆ eΓMΓe is a non-trivial quasi-ideal contained in eΓMΓe. We must have there-fore MΓh ∩ hΓM = eΓMΓe. Since e ∈ eΓMΓe, there are r, s ∈ M, γ1, γ2 ∈ Γ such thatrγ1h = hγ2s = e. These equations and h = eαh = hαe imply that hHe so that eΓMΓe = H0
e . �
Corollary 3.22 If for some α ∈ Γ, 0 �= eαe = e ∈ Q a 0-minimal quasi-ideal, then Q =eΓMΓe = H0
e .
4 A Mutually Annihilating Sum of Completely 0-simple Γ-semigroups
Definition 4.1 Let M be a Γ-semigroup with 0. A collection of sub-Γ-semigroups {Mi}i∈Aof a Γ-semigroup M will be called mutually annihilating if 0 ∈ Mi for every i ∈ A and ifMiΓMj = {0} = MjΓMi for i �= j.
It is easy to check that the Mi are ideals in the above sum. It is then easy to verify thatM is also the 0-direct union of its ideals, Mi, according the following definition:
Definition 4.2 A Γ-semigroup M with 0 is said to be the 0-direct union of its ideals Mi (i ∈ A)if M =
⋃i∈A Mi and Mi ∩ (
⋃j �=i Mj) = {0}.
Proposition 4.3 Let M be an absorbent Γ-semigroups with 0. A regular nonzero D-classunion {0} of M is itself a completely 0-simple Γ-semigroup.
Proof This follows immediately from Theorem 2.24 by Remark 3.2. �
Lemma 4.4 Let M be an absorbent Γ-semigroup. Then the collection of D-classes union {0}of M is mutually annihilating.
Proof It readily follows from Remark 3.2, Corollary 2.23 and Theorem 2.24 that each D-classunion {0} is a Γ-semigroup. If aDb, then Ra ∩ Lb = ∅. It then follows from the definition of
On Green’s Relations, 20-regularity and Quasi-ideals in Γ-semigroups 619
absorbency that aγb = 0, γ ∈ Γ and thus (Da ∪ {0})Γ(Db ∪ {0}) = {0}. �
Theorem 4.5 A Γ-semigroup M with 0 is absorbent if and only if it is 20-regular and thecollection of its D-classes union {0} is mutually annihilating.
Proof One implication follows directly from Remark 3.2 and Lemma 4.4.Conversely, suppose M is 20-regular and that the product of any two distinct, D-classes is
{0}. Let a, b ∈ M . If aDb, then aγb = 0, γ ∈ Γ. On the other hand, if aDb and Da are irregular,then we have aγb = 0 by Corollary 2.23. But if Da is regular, then Da ∪ {0} is completely0-simple (Theorem 2.24) and we have aγb ∈ (Ra ∩ Lb)0 by Lemma 2.7 and M is absorbent.This completes the proof. �
Corollary 4.6 A regular Γ-semigroup with 0 is absorbent if and only if it is mutually anni-hilating collection of completely 0-simple sub-Γ-semigroup with a common zero.
Corollary 4.7 If M is an absorbent Γ-semigroup, then J = D on M . Indeed each D-classunion {0} is an ideal.
Proof Since M is absorbent, the last statement is immediate since the D-classes are mutuallyannihilating by Lemma 4.4. Suppose now that b �= 0 and aJ b. Then we can find u, v ∈M1, α, β ∈ Γ such that a = uαbβv. Now we have seen above that Db ∪{0} is an ideal, and fromaJ b, b �= 0, it follows that a �= 0. Thus from a = uαbβv, we can conclude a ∈ Db. Whence aDb
and the proof is complete since J0 = D0 = {0}. �
Lemma 4.8 Let L be a 0-minimal left ideal of a Γ-semigroup M with 0, and let c ∈ M . ThenLΓc is either 0 or a minimal 0-minimal left ideal of M .
Proof It is the same as the proof of [8, Lemma 4.6]. �
Lemma 4.9 If I is a 0-minimal ideal of a Γ-semigroup M with 0 such that IΓI �= {0}, andif L is a non-zero left ideal of M contained in I, then LΓL �= {0}.Proof Since LΓM is an ideal of M contained in I, we must have either LΓM = I or LΓM ={0}. If LΓM = {0}, then L is an ideal of M , whence L = I, and so IΓI = LΓI ⊆ LΓM = {0},contrary to hypothesis. Hence LΓM = I, and from I = IΓI = LΓMΓLΓM ⊆ LΓLΓI, weconclude that LΓL �= {0}. �
Now we will prove an improved version of the [15, Theorem 3.1].
Theorem 4.10 Let M be a Γ-semigroup with 0. Let I be a 0-minimal ideal of M containingat least one 0-minimal left ideal of M . Then I is the union of all the 0-minimal left ideals ofM contained in I.
Proof Let A be the union of all the 0-minimal left ideals of M contained in I. We will showthat A = I. It is clear that A is a left ideal of M . We proceed to show that A is also a rightideal. Let a ∈ A and c ∈ M . By definition of A, a ∈ L for some 0-minimal left ideal L of M
contained in I. By Lemma 4.9, LΓc = {0} or LΓc is a 0-minimal left ideal of M . Moreover,LΓc ⊆ IΓc ⊆ I, and hence LΓc ⊆ A. Consequently aαc ∈ A for some α ∈ Γ. Now A �= {0}since it contains at least one 0-minimal left ideal of M . Hence A is a non-zero two-sided idealof M contained in I, whence A = I by the 0-minimality of I. �
Lemma 4.11 Let M be a 0-simple Γ-semigroup containing a 0-minimal left ideal and a 0-minimal right ideal. Then to each 0-minimal left ideal L of M corresponds at least one 0-
620 Hila K. and Dine J.
minimal right ideal R of M such that LΓR �= {0}.Proof It is clear that LΓM is an ideal of M , and hence LΓM = M or LΓM = {0}. IfLΓM = {0}, then LΓL = {0}, contrary to Lemma 4.9. Hence LΓM = M . In particular,LΓc �= {0} for some c ∈ M . By the dual of Theorem 4.10, M is the union of 0-minimal rightideals. Hence c ∈ R for some 0-minimal right ideal R of M , and evidently LΓR �= {0}. �
The following theorem is an immediate corollary of [16, Lemma 4.2] and [16, Theorem 4.4].
Theorem 4.12 A 0-simple Γ-semigroup is completely 0-simple if and only if it contains atleast one 0-minimal left ideal and at least one 0-minimal right ideal.
Corollary 4.13 A completely 0-simple Γ-semigroup is the union of its 0-minimal left (right)ideals.
We cite some results without proof.
Lemma 4.14 ([16, Lemma 2.5]) A completely 0-simple Γ-semigroup M is regular.
Lemma 4.15 ([8, Lemma 2.3 (4), Lemma 2.4]) Let e be an α-idempotent element, L a leftideal, R a right ideal of a Γ-semigroup M with 0. Then R ∩ L, eΓL and RΓe are quasi-idealsof M .
Lemma 4.16 ([8, Lemma 4.4]) Let L be a (0)-minimal left ideal of M and 0 �= e = eαe ∈ L,α ∈ Γ. Then, (eΓL)\{0} is a sub-Γ-group of M , moreover a 0-minimal quasi-ideal of M .
Theorem 4.17 ([8, Theorem 4.1]) Let M be a Γ-semigroup. Then, for each 0-minimal leftideal L and 0-minimal right ideal R, the intersection Q = L∩R is either a 0-minimal quasi-idealof M or Q = {0}.Lemma 4.18 Every non-zero left (right) ideal of a regular Γ-semigroup M with 0 contains atleast one non-zero idempotent element.
Proof Let L be a non-zero left ideal of M and a ∈ L, a �= 0. By regularity, there is an x ∈ M
and α, β ∈ Γ such that a = aαxβa. It is clear that xβa �= 0 is an α-idempotent element andxβa ∈ MΓa ⊆ MΓI ⊆ I.
Analogously for the right ideals of M . �Let M be a Γ-semigroup with 0 and A ⊆ M . We shall say that A is nilpotent if for some
integer k ≥ 1 the relation (AΓ)k−1A = {0} holds. The union of all nilpotent left ideals of M iscalled the radical of M .
Lemma 4.19 If M is a Γ-semigroup with radical 0, then 0 is the unique nilpotent right idealof M .
Proof If R �= {0} were a nilpotent right ideal of M , then the ideal R ∪ MΓR �= {0} would bea nilpotent ideal of M . �
Corollary 4.20 A regular Γ-semigroup M with 0 has zero radical.
Corollary 4.21 Every 0-minimal left (right) ideal L(R) of a regular Γ-semigroup M with 0is of the form L = MΓe (R = fΓM) with e = eαe (f = fβf), for some α, β ∈ Γ.
Theorem 4.22 Let M be a Γ-semigroup with radical 0. Then MΓe (eαe = e, α ∈ Γ) is a0-minimal left ideal if and only if eΓM is a 0-minimal right ideal of M .
Let MΓe be a 0-minimal left ideal of M where e = eαe, α ∈ Γ. Then by Lemma 4.16, eΓMΓe
On Green’s Relations, 20-regularity and Quasi-ideals in Γ-semigroups 621
is a 0-minimal quasi-ideal of M . Let R denote a right ideal of M with {0} ⊂ R ⊆ eΓM . HenceeΓR = R. By Lemma 4.15, eΓRΓe (⊆ eΓMΓe) is a quasi-ideal of M and so eΓRΓe = eΓMΓe
holds. Hence e ∈ eΓRΓe ⊆ eΓR = R which implies eΓM ⊆ R. Therefore, R = eΓM .
Theorem 4.23 Let M be a Γ-semigroup with radical 0. Then we can write every 0-minimalquasi-ideal Q of M in the form Q = L ∩ R, where L is a 0-minimal left ideal and R is a0-minimal right ideal of M .
Proof It is a straightforward proof. �
Definition 4.24 The quasi-ideals Qij (i, j ∈ B) of a Γ-semigroup M with 0 are said to forma complete system whenever the following three conditions hold :
1) Qij = {0} or Qij is a 0-minimal quasi-ideal of M.
2) If Qij �= {0}, then it is of the form eiΓMΓej, where ei and ej are α-idempotent andβ-idempotents respectively.
3) If Qij �= {0}, then QjiΓQij �= {0}.Theorem 4.25 Let M be a non-trivial Γ-semigroup with 0. The following statements areequivalent :
I. M is regular and the union of its 0-minimal left ideals.II. M is the union of 0-minimal left ideals of the form MΓei, where eiαei = ei, i ∈ B for
some α ∈ Γ (i.e., ei is an α-idempotent).III. M is the 0-direct union of two-sided ideals which are completely 0-simple sub-Γ-
semigroups of M .IV. M is the union of quasi-ideals which form a complete system.V. M is regular and the union of its 0-minimal quasi-ideals.VI. M is regular and absorbent.VII. M is regular and every H-class union {0} is a quasi-ideal.
Proof I ⇒ II By Corollary 4.21, every 0-minimal left ideal Li of M is of the form Li = MΓei,where ei is an α-idempotent in Li.
II ⇒ III First, we show that the radical of M is 0. Let sγei ( �= 0) (sγei ∈ MΓei) for someγ ∈ Γ be an arbitrary element of the ideal I ( �= {0}) of M . This implies I ∩MΓei �= {0}. Withrespect to the 0-minimality of the left ideal MΓei, it must hold ei ∈ MΓei ⊆ I. Thus I cannotbe nilpotent and the radical of M is indeed 0.
As the left ideals MΓei (i ∈ B) are 0-minimal,
either (MΓei)Γ(MΓej) = {0} or (MΓei)Γ(MΓej) = MΓej (4.1)
holds. It is easy to see that the relations ≡ defined by
MΓei ≡ MΓej ⇔ (MΓei)Γ(MΓej) = MΓej (4.2)
is an equivalence relation in the set of the 0-minimal left ideals MΓei(i ∈ B). Let Si denotethe union of all the left ideals belonging to the equivalence class Ki. Thus
M =⋃
i∈A
Si, (4.3)
where A denotes the index set of the difference classes.
622 Hila K. and Dine J.
First, we show that Si =⋃
MΓel∈KiMΓel is a 0-simple two-sided ideal of M . By (4.1)
and (4.2), it follows that
SiΓSj =
⎧⎨
⎩
Si, if i = j,
0, if i �= j.(4.4)
Hence, by (4.3), it follows that
MΓSi = SiΓM = Si. (4.5)
Let V be an ideal of Si. With respect to (4.3) and (4.4), V is an ideal of M and thereforeV ΓV �= {0} holds. Hence
{0} �= V ΓV ⊆ V ΓSi =⋃
MΓel∈Ki
V ΓMΓel. (4.6)
This implies V ΓMΓet �= {0} for some MΓet ∈ Ki, whence V ΓMΓet = MΓet ⊆ V follows.By (4.2), we obtain
V ⊇ V ΓSi ⊇ MΓetΓSi =⋃
MΓel∈Ki
(MΓet)Γ(MΓel) =⋃
MΓel∈Ki
MΓel = Si
establishing the 0-simplicity of Si.As Si is a 0-simple ideal of M ,
Si ∩( ⋃
i�=j∈A
Sj
)
= {0} or Si
holds. The second case implies Si ⊆⋃
i�=j∈A Sj . By (4.4), we have
SiΓSi ⊆ SiΓ( ⋃
i�=j∈A
Sj
)
=⋃
i�=j∈A
SiΓSj = {0},
which is a contradiction. Thus for every Si (i ∈ A),
Si ∩( ⋃
i�=j∈A
Sj
)
= {0}
must hold and therefore (4.3) is a 0-direct union.If MΓel is a 0-minimal left ideal contained in Si, then elΓM is a 0-minimal right ideal in
Si. By Theorem 4.12, Si is a completely 0-simple sub-Γ-semigroup of M .III ⇒ IV Let M be a 0-direct union of its ideal Si (i ∈ A), where Si are completely 0-
simple sub-Γ-semigroup of M . By Lemma 4.14, Si (i ∈ A) are regular Γ-semigroups, thereforeM is itself regular. Corollary 4.13 implies that every Si (i ∈ A) is the union of its 0-minimal leftideals and the union of its 0-minimal right ideals. Since M is the 0-direct union of the 0-simpleideals Si (i ∈ A), all the left (right) ideals of Si (i ∈ A) are left (right) ideals of M . Thus M isthe union of its 0-minimal left ideals and the union of its 0-minimal right ideals.
The regularity of M implies that the 0-minimal left ideals of M are of the form MΓet
with α-idempotent elements et �= 0 (t ∈ B). From Corollary 4.20 and Theorem 4.22, we getthat etΓM (t ∈ B) are the 0-minimal right ideals of M . So we can write M =
⋃t∈B MΓet =
⋃t∈B etΓM . Hence since MΓM = M ,
M = MΓM =( ⋃
t∈B
etΓM
)
Γ( ⋃
t′∈B
MΓet′
)
=⋃
t,t′∈B
etΓMΓet′ . (4.7)
On Green’s Relations, 20-regularity and Quasi-ideals in Γ-semigroups 623
By Lemma 4.15, etΓMΓet′ (t, t′ ∈ B) are quasi-ideals of M satisfying {0} ⊆ etΓMΓet′ ⊆etΓM ∩ MΓet′ . This and Theorem 4.17 imply that etΓMΓet (t, t′ ∈ B) are either {0} or0-minimal quasi-ideals of M .
We have to verify only the condition 3). Let etΓMΓet′ �= {0}. MΓetΓMΓet′ is a left idealnon-zero contained in the 0-minimal left ideal MΓet′ . Hence MΓetΓMΓet′ = MΓet′ , whence(etΓMΓet)Γ(etΓMΓet′) = et′ΓMΓet′ �= {0}.
IV ⇒ V We have only to show the regularity of M . By supposition,
M =⋃
t,t′∈B
Qtt′ =⋃
t,t′∈B
etΓMΓet′ .
Let a = etγ1sγ2et′ ( �= 0) be an arbitrary element of M for γ1, γ2 ∈ Γ. By 3), the hypothesisetΓMΓet′ �= {0} implies et′ΓMΓet �= {0}. By Lemma 4.15, etγ1sγ2et′βet′ΓMΓet is a quasi-idealof M . The 0-minimality of the quasi-ideal etΓMΓet implies that either
The first possibility implies etΓMΓetγ1sγ2et′βet′ΓMΓet = {0}. Since the quasi-idealsetΓMΓetγ1sγ2et′ ( �= 0) is contained in the 0-minimal quasi-ideal etΓMΓet′ , we get
etΓMΓetγ1sγ2et′ = etΓMΓet′ .
Thus etΓMΓet′βet′ΓMΓet = {0} holds, which contradicts the condition 3).Thus we necessarily have etγ1sγ2et′βet′ΓMΓet = etΓMΓet. This implies the existence of
an element et′γ3zγ4et ∈ et′ΓMΓet with etγ1sγ2et′βet′γ3zγ4et = et. Hence
etγ1sγ2et′βet′γ3zγ4etαetγ1sγ2et′ = etγ1sγ2et′ ,
or otherwise aβ(et′γ3zγ4et)αa = a, which says that a is a regular element of M . This provesour assertion.
V ⇒ I Corollary 4.20 and Theorem 4.23 imply that we can write every 0-minimal quasi-ideal Qi (i ∈ A) of M in the form Qi = Li ∩ Ri, where Li = MΓei (eiαei = ei, α ∈ Γ) is a0-minimal left ideal and Ri = fiΓM (fiβfi = fi, β ∈ Γ) is a 0-minimal right ideal of M . ThusM =
⋃i∈A Qi =
⋃i∈A[Li ∩ Ri] ⊆
⋃i∈A Li ⊆ M . Hence M =
⋃i∈A Li =
⋃i∈A MΓei.
III ⇒ VI It follows immediately from Corollary 4.6 and by the equivalence of Definitions 4.1and 4.2.
VI ⇒ VII It follows by Theorem 3.6.VII ⇒ V It follows by Proposition 3.9 since any Γ-semigroup is the union of its H-classes.Thus the proof is complete. �
Acknowledgements The authors are highly grateful to referees for their valuable commentsand suggestions.
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