# On Naturally Ordered Abundant Semigroups with an Adequate Monoid Transversal

Authors: Wei Chen
Journal: Discrete Dynamics in Nature and Society
 (2019)

---

## Abstract

In this paper, we study a class of naturally ordered abundant semigroups with an adequate monoid transversal, namely, naturally ordered concordant semigroups with an adequate monoid transversal. After giving some properties of such semigroups, we obtain a structure theorem for naturally ordered concordant semigroups with an adequate monoid transversal.


---

## Body

## 1. Introduction

Suppose thatS is a regular semigroup and S∘ is an inverse subsemigroup of S. Then S∘ is called an inverse transversal of S, if S∘ contains a unique inverse x∘ of each element x of S. The structure theorems for regular semigroups with an inverse transversal have been given by many authors (see [1–4]). An analogue of an inverse transversal which is termed an adequate transversal was introduced for abundant semigroups by El-Qallali (see [5]). An ordered semigroup (S,≤) isnaturally ordered if, for all e,f∈E(S), eωf implies e≤f, where ω is the natural partial order on E(S). Recently, Blyth and Almeida Santos have investigated naturally ordered regular semigroups with an inverse monoid transversal and determined the structure of such regular semigroups. Since abundant semigroups generalize regular semigroups, it is natural that, in the first instance, we should approach their structure theory by looking for generalization of results from the theory of regular semigroups. The following papers contain some work along these lines: [5–20]. As we shall discuss below, this paper is one of a sequence in which we concentrate on the structure of a class of naturally ordered abundant semigroups with an adequate monoid transversal.We proceed as follows: Section2 presents some necessary notation and known results. In Section 3, we obtain some characterizations for naturally ordered concordant semigroups with an adequate transversal. In Section 4, we give several order properties of naturally ordered concordant semigroups with an adequate transversal. In Section 5, we give some characterizations of the regularity of Green ∗-relations. In Section 6, we establish a structure theorem for naturally ordered concordant semigroups with an adequate monoid transversal, which generalizes the main result of [21].


## 2. Preliminaries

In what follows, we shall use the notion and notation of   [6, 21]. Other undefined terms can be found in [11, 22, 23]. Here we provide some known results repeatedly used without mention in the following. First we recall some of the basic facts about the relations L∗ and R∗.Lemma 1 (see [11]).
Leta,b be elements of a semigroup S. Then the following statements are equivalent:
(1)aL∗b(aR∗b);
(2) for allx,y∈S1, ax=ay(xa=ya)⇔bx=by(xb=yb).As an easy but useful consequence, we have the following.Corollary 2 (see [11]).
Leta,e2=e∈S. Then the following statements are equivalent:
(1)aL∗e(aR∗e);
(2)ae=a(ea=a) and for all x,y∈S1, ax=ay(xa=ya)⇒ex=ey(xe=ye).It is well known thatL∗ is a right congruence while R∗ is a left congruence. In general, L⊆L∗ and R⊆R∗. But when a,b are regular elements, aLb [resp., aRb] if and only if aL∗b [resp., aR∗b]. For convenience, a+ [resp. a∗] is denoted the typical idempotent related a by the relation R∗ [resp. L∗]. A semigroup S is calledabundant if each L∗-class and each R∗-class contains at least an idempotent. An abundant semigroup is calledquasi-adequate if the set of idempotents forms a subsemigroup. Moreover, a quasi-adequate semigroup is calledadequate if its band of idempotents is a semilattice. In this case, each L∗-class and each R∗-class contains exactly one idempotent.In this paper,≤n denotes the natural partial order on an abundant semigroup S, defined by the rule that a≤nb if and only if for some idempotents e and f, a=eb=bf. And ω(a) stands for the set {b∈S∣b≤na}. Following [9], S isidempotent-connected(for short, IC) provided, for every element a of S, and for some [for all] a+,a∗, there exists a bijection θ:<a+>→<a∗> such that xa=a(xθ) for all x of <a+>, where <a+> (<a∗>) is the subsemigroup of S generated by the set {y∈E(S)∣y=ya+=a+y} (resp., {y∈E(S)∣y=ya∗=a∗y}).Lemma 3 (see [18]).
Leta,b be elements of a semigroup S. Then the following statements are equivalent:
(1)S is IC;
(2) for each element a ofS two conditions hold:
(i) for some[for all]a∗ and for all e∈ω(a∗) there exists an idempotent f∈ω(a+) such that ae=fa.
(ii) for some[for all]a+ and for all h∈ω(a+) there exists an idempotent g∈ω(a∗) such that ha=ag.An abundant semigroup is calledconcordant if it is IC and satisfies the regularity condition.IfS is an abundant semigroup and U is an abundant subsemigroup of S; then we say that U is a ∗-subsemigroup of S if L∗(U)=L∗(S)∩(U×U) and R∗(U)=R∗(S)∩(U×U).Lemma 4 (see [19]).
LetS be an abundant semigroup. If e is an idempotent of S, then eSe is a ∗-subsemigroup of S.LetS∘ be an adequate ∗-subsemigroup of S and E∘ be the idempotent semilattice of S∘. As in [5], S∘ is called anadequate transversal for S if, for each element x in S, there exists a unique element x¯∈S∘ and idempotents e,f in E such that x=ex¯f, where eLx¯+, fRx¯∗ for x¯+,x¯∗ in E∘. It is straightforward to show that such e and f are uniquely determined by x (see [5]). Hence we normally denote e by ex, f by fx. We define (1)I=ex∣x∈S,Λ=fx∣x∈S,L=exx¯∣x∈S,R=x¯fx∣x∈S. The adequate transversal S∘ is said to be aquasi-ideal if S∘SS∘⊆S∘ or, equivalently, ΛI⊆S∘ andleft simplistic if S∘IS∘⊆S∘ or equivalently, S∘I⊆S∘. Let S∘ be a ∗-adequate subsemigroup of S and a∈S∘. Throughout this paper, we denote by a+[resp., a∗] the unique idempotent of La∗[resp. Ra∗] in S∘. For our purpose, we list the following results.Lemma 5 (see [6]).
LetS∘ be an adequate transversal of an abundant semigroup S and x,y∈S. Then
(1)exR∗x and fxL∗x;
(2)ex¯Lex and fx¯Rfx;
(3)ex¯=ex¯=x¯+=fex, fx¯=fx¯=x¯∗=efx;
(4) ifx∈S∘ then ex=x+∈E∘, x¯=x, fx=x∗∈E∘;
(5) ifx∈E∘ then ex=x¯=fx=x;
(6) ifx∈I, y∈Λ then ex=x,x¯=fx=ex¯ and ey=y¯=fy¯,fy=y.Lemma 6 (see [7]).
LetS∘ be an adequate transversal of an abundant semigroup S and x,y∈S. Then
(1)xR∗y⇔ex=ey;
(2)xL∗y⇔fx=fy.Lemma 7 (see [20]).
LetS be an abundant semigroup with a quasi-ideal adequate transversal S∘. For any x,y∈S, then
(1)xy¯=x¯fxeyy¯;
(2)exy=ex(x¯fxey)+;
(3)fxy=(fxeyy¯)∗fy.LetS be an abundant semigroup with an adequate transversal S∘. If x∈Reg(S), then, by Lemma 5(1) and [23, Theorem 2.3.4], there exists a unique x∘∈V(x) with xx∘=ex and x∘x=fx.Theorem 8 (see [6]).
LetS be an abundant semigroup with an adequate transversal S∘. If x∈Reg(S) then x∘∈S∘,x¯=x∘∘, and x∘=x∘∘∘.Lemma 9 (see [6]).
LetS be an abundant semigroup with a quasi-ideal adequate transversal S∘. If Reg(S) is a subsemigroup of S then I is a left regular subband of S and Λ is a right regular subband of S.Lemma 10 (see [6]).
LetS be an abundant semigroup with an adequate transversal S∘. Then I∩Λ=E∘,L∩R=S∘,I=L∩E(S),Λ=R∩E(S).Theorem 11 (see [6]).
LetS be an abundant semigroup with an adequate transversal S∘. If I is a band, then the following statements are equivalent:
(1)S∘ is left simplistic;
(2)E∘ is a right ideal of I;
(3)I is a normal band.Theorem 12 (see [6]).
LetS be an abundant semigroup with an adequate transversal S∘. Then S∘ is both left and right simplistic if and only if S∘ is a quasi-ideal of S.


## 3. Characterization

The object of this section is to give characterizations of naturally ordered concordant semigroups with an adequate transversal.Lemma 13 (see [17]).
LetS be an ordered abundant semigroup. Then S is naturally ordered if and only if, for all e∈E(S), eSe is R∗-unipotent, L∗-unipotent, and naturally ordered.As a consequence of Lemma13, we have the following.Corollary 14.
LetS be an ordered abundant semigroup which satisfies the regularity condition. Then S is naturally ordered if and only if S is locally adequate and, for all e∈E(S), eSe is naturally ordered.Lemma 15 (see [16]).
LetS be an IC abundant semigroup. Then the following statements are equivalent:
(1)S is locally adequate;
(2)≤n is compatible with multiplication.Combining with Corollary14 and Lemma 15, we have the following.Theorem 16.
A concordant semigroup can be naturally ordered if and only if it is locally adequate.Theorem 17.
LetS be a concordant semigroup with a quasi-ideal adequate transversal S∘. Then S is locally adequate.Proof.
Suppose thatS∘ is a quasi-ideal adequate transversal of a concordant semigroup S. Let T=Reg(S) and let U=T∩S∘. It is clear that U is a regular subsemigroup of T. It is easy to see that U={x∘:x∈T}={x∈T:x=x¯=x∘∘}. Since S∘ is adequate, then we can easily deduce that U is inverse and that U is an inverse transversal of the regular semigroup T. It is reasonably clear that E(T)=E(S), E(U)=E(S∘). Since S is a concordant semigroup, UTU⊆T. Because S∘ is a quasi-ideal, UTU⊆S∘SS∘⊆S∘. Thus TUT⊆T∩S∘=U. This implies that U is a quasi-ideal inverse transversal of T. By [21, Theorem 3], T is locally inverse. Let e∈E(S)=E(T). Then eTe is an inverse subsemigroup of T and, by [18, Lemma 1.5], eSe is concordant. Assume that x∈Reg(eSe). Then x∈T. Thus x=exe∈eTe. It follows that Reg(eSe)⊆eTe⊆eSe. Since S is concordant, eTe⊆Reg(eSe) and so Reg(eSe)=eTe. Therefore, eSe is adequate. This shows that S is locally adequate.Furthermore, we have the following result, which generalizes [21, Theorem 3].Theorem 18.
LetS be a concordant semigroup with an adequate transversal S∘. Then the following statements are equivalent:
(1)S can be naturally ordered;
(2)S is locally adequate;
(3)S∘ is a quasi-ideal of S.Proof.
We need to only prove that (2) implies (3). Suppose thatS is locally adequate. Then, by Lemma 15, ≤n is compatible with multiplication. Hence, by Lemma 9 and [23, Exercise 4.7.18], I is a normal band whence by Theorem 11, S∘ is left simplistic. Similarly, S is right simplistic. Hence, by Theorem 12, S∘ is a quasi-ideal.


## 4. Order Properties

Let(S;≤) be a naturally ordered concordant semigroup with an adequate transversal S∘. By Theorem 18 above the adequate transversal S∘ is a quasi-ideal. We now consider relationships between the imposed order ≤ and the natural order ≤n on S. The proof of the following lemma is essentially the same as that of [21, note].Lemma 19.
Let(S;≤) be a naturally ordered concordant semigroup with an adequate transversal S∘. Then the orders ≤ and ≤n coincide on E∘.In the following results, we shall see that the order≤ also extends the natural order ≤n on the adequate transversal S∘; and, as a consequence, the assignment x↦x¯ is always isotone.Lemma 20.
Let(S;≤) be a naturally ordered concordant semigroup with an adequate transversal S∘. If x,y∈S are such that x≤ny then ex≤ey,fx≤fy.Proof.
Suppose thatx,y∈S are such that x≤ny; then there exist e,f∈E(S) such that x=ey=yf. Observe that x=yf gives ex=eyf=ey(y¯fyef)+ by Lemma 7(2), whence we have eyex=ex; and, by Lemma 9, exey=exeyex=ex. Hence we have ex≤ney and so ex≤ey. Similarly fx≤fy.The next results generalize [21, Theorem 5 and its Corollary].Theorem 21.
Let(S;≤) be a naturally ordered concordant semigroup with an adequate transversal S∘. If x,y∈S are such that x¯≤ny¯, then x¯≤y¯.Proof.
Suppose thatx,y∈S are such that x¯≤ny¯. Then, by Lemma 20, in E∘ we have x¯+=ex¯≤ney¯=y¯+. Consequently, x¯+=x¯+y¯+=y¯+x¯+ and x¯+≤y¯+. Since x¯≤ny¯, there exists f′∈E such that x¯=f′y¯=x¯+f′y¯+y. Let f=x¯+f′y¯+. Since f′x¯=x¯, f′x¯+=x¯+ and so x¯+f′∈E(S). Because x¯+f′y¯+x¯+f′y¯+=x¯+f′x¯+f′y¯+=x¯+f′y¯+∈S∘, f=x¯+f′y¯+∈E(S∘). Since E(S∘) is a semilattice, f=x¯+f=fx¯+, which implies that f≤nx¯+. Hence, f≤x¯+≤y¯+ by Lemma 19. It follows that x¯=fy¯≤y¯+y¯=y¯.Corollary 22.
Let(S;≤) be a naturally ordered concordant semigroup with an adequate transversal S∘. If x,y∈S are such that x≤y, then x¯≤y¯.Proof.
Ifx,y∈S are such that x≤y, then, using the fact that S∘ is a quasi-ideal, we have (2)x¯=ex¯xfx¯≤ex¯yfx¯=ex¯eyy¯fyfx¯.By Theorems 11 and 12, ex¯ey,fyfx¯∈E∘. Writing g=ex¯ey and b=y¯fyfx¯, we have gb=b+gb+b. Since S is IC, there exists h∈E(S) such that gb=b+gb+b=bh. Hence ex¯eyy¯fyfx¯≤ny¯fyfx¯. Similarly y¯fyfx¯≤ny¯. Since ex¯eyy¯fyfx¯,y¯fyfx¯∈S∘ by Lemma 5(3) and S∘ is a quasi-ideal, by Lemma 5(4) and Theorem 21, x¯≤y¯.By the proof of [21, Theorem 8], we have the following result:Lemma 23.
Let(S;≤) be a naturally ordered concordant semigroup with an adequate transversal S∘. If e,f∈E(S) are such that e≤nf then e=fe∘e=ee∘f=fe∘f.Theorem 24.
Let(S;≤) be a naturally ordered concordant semigroup with an adequate transversal S∘. The following conditions are equivalent:
(1) Ifx,y∈Reg(S) are such that x≤y, then y∘≤x∘;
(2) Ife,f∈E(S) are such that e≤f, then f∘≤e∘;
(3)S is completely J∗-simple;
(4) The adequate transversalS∘ is a cancellative monoid.Proof.
(1)⇒(2) It is trivial.
(2)⇒(3) Let e,f∈E(S) be such that e≤nf. Then e≤f. Hence by the hypothesis and Lemma 23, f=ff∘f≤fe∘f=e. Thus f=e. Therefore, ≤n is equality on E(S) and by [11, Corollary 5.2] S is completely J∗-simple.
(3)⇒(4) Suppose that S is completely J∗-simple. Then by [11, Corollary 5.2], we may assume S is a Rees matrix semigroup μ(T;I,Λ;P) without zero over a cancellative monoid T where each entry in P is a unit of T. Since S∘ is an adequate semigroup, S∘ contains an idempotent. Let (i,pλi-1,λ) and (j,pμj-1,μ) be idempotents of S∘. Then they are inverses of (i,pμi,μ). Since S∘ is an adequate transversal of S, by [7, Corollary 2.4], (i,pλi-1,λ)=(j,pμj-1,μ), so that S∘ has only one idempotent. Thus S∘ is a cancellative monoid.
(4)⇒(1) If x,y∈Reg(S) are such that x≤y then, by the Corollary 22, we have x∘∘=x¯≤y¯=y∘∘. Because S∘ is a cancellative monoid, Reg(S∘) is a group. Thus y∘≤x∘.Theorem 25.
Let(S;≤) be a naturally ordered concordant semigroup with an adequate transversal S∘. The following conditions are equivalent:
(1) The adequate transversalS∘ is a monoid;
(2)(S∘;≤) has a biggest idempotent;
(3)(∃ξ∈E∘)S∘=ξSξ;
(4)(∃ξ∈E∘)(∀x∈S)x¯=ξxξ;
(5)(∃ξ∈E∘)L=Sξ,R=ξS;
(6)L[resp. R] is an idempotent-generated principal left [resp. right] ideal.Proof.
(1)⇒(2) If (1) holds then for every e∈E∘ we have e≤n1S∘. Thus e≤1S∘.
(2)⇒(1) Suppose that (2) holds and let ξ=max⁡E∘. If x¯∈S∘, then by Lemma 5 (3) x¯+=ex¯≤ξ. Since ≤ coincides with ≤n on E∘, x¯+≤nξ. Thus ξx¯=ξx¯+x¯=x¯+x¯=x¯. Similarly, x¯ξ=x¯. Consequently, ξ=1S∘.
(1)⇒(3) If (1) holds then we have (3)1S∘S1S∘⊆S∘SS∘⊆S∘=1S∘S∘1S∘⊆1S∘S1S∘ whence S∘=ξSξ, where ξ=1S∘.
(3)⇒(4) Let T=Reg(S); let U=T∩S∘. It is clear that U is a regular subsemigroup of T. It is easy to see that U={x∘:x∈T}={x∈T:x=x¯=x∘∘}. Since S∘ is adequate then we can easily deduce that U is inverse and that U is an inverse transversal of the regular semigroup T. It is reasonably clear that E(T)=E(S), E(U)=E(S∘). Since S∘=ξSξ, U=ξTξ. By [6, Theorem 18(4)], for all x∈T,x∘∘=ξxξ. Hence we have, for all x∈S, (4)ξxξ=ξexx¯fxξ=ξexξx¯ξfxξ=ex∘∘x¯fx∘∘=ex¯x¯fx¯=ex¯x¯fx¯=x¯.
(4)⇒(1) It is trivial.
(4)⇒(5) If (4) holds, then for every x∈S we have exx¯=exξxξ∈Sξ. Hence L⊆Sξ. Similarly R⊆ξS. By Lemma 7, we have (5)exξxξ¯=exx¯fxξ+x¯fxξ=exx¯fxξ=xξ.
ThusSξ⊆L. Similarly ξS⊆R and we have (5).
(5)⇒(3) If (5) holds, then, by Lemma 10, S∘=L∩R=Sξ∩ξS=ξSξ.
(5)⇒(6) This is clear.
(6)⇒(5) The proof is essentially the same as that of [21, Theorem 10].


## 5. The Regularity ofL∗ and R∗

Lemma20 and [21, Definition 3.2] motivate the next definition.Definition 26.
Let(S;≤) be a naturally ordered concordant semigroup with an adequate transversal S∘. L∗[resp. R∗] is said to be regular if (6)∀x,y∈Sx≤y⇒fx≤fyresp.ex≤ey.Definition 27 (see [21]).
An equivalence relationΘ on an ordered set A is said to satisfy the link property if (7)aΘx≤b⇒∃y∈Aa≤yΘb; and the dual link property if (8)a≤xΘb⇒∃y∈AaΘy≤b.Theorem 28.
Let(S;≤) be a naturally ordered concordant semigroup with an adequate transversal S∘. If L∗[resp. R∗] is regular then L∗[resp. R∗] satisfies the dual link property.Proof.
Ifa≤xL∗b, then since L∗ is regular, we have fa≤fx=fb by Lemma 6 and so bfa≤b. Since fafxfa≤nfa, whence fafxfa≤fa. Since also fa≤fx, it follows that fa=fafafa≤fafxfa and so fafxfa=fa. If y,z∈S such that bfay=bfaz, then we have fxfay=fxfaz. Hence fafxfay=fafxfaz and so fay=faz. Thus ay=az. If c,d∈S such that ac=ad, then we have fac=fad and so bfac=bfad. It follows that aL∗bfa≤b. Similarly, when regular, R∗ satisfies the dual link property.Definition 29.
Let(S;≤) be a naturally ordered concordant semigroup with an adequate transversal S∘. L∗ is said to be lower Λ-stable if(9)∀x∈Sx≤l∈Λ⇒fx≤l;and upper Λ-stable if (10)∀x∈Sx≥l∈Λ⇒fx≥l. Dually, we define R∗ to be lower and upper I-stable.Theorem 30.
Let(S;≤) be a naturally ordered concordant semigroup with an adequate transversal S∘. The following conditions are equivalent:
(1)L∗ is regular;
(2)L∗ satisfies the dual link property and is lower Λ-stable.Proof.
(1)⇒(2) If L∗ is regular then, by Theorem 28, it satisfies the dual link property. Moreover, if x≤l∈Λ, then fx≤fl=l by Lemma 5(6) and so L∗ is lower Λ-stable.
(2)⇒(1) Let x,y∈S be such that x≤y. Because x≤yL∗fy and L∗ satisfies the dual link property, there exists z∈S such that xL∗z≤fy. Then fx=fz. Since L∗ is lower Λ-stable, fz≤fy. Thus L∗ is regular.Theorem 31.
Let(S;≤) be a naturally ordered concordant semigroup with an adequate transversal S∘. If S is upper directed then the following conditions are equivalent:
(1)L∗ is regular;
(2)L∗ satisfies the link property and is upper Λ-stable.Proof.
(1)⇒(2) Suppose that aL∗x≤b and let k∈S be such that a,b≤k. Then fa=fx≤fb whence a=afa≤kfb. Also, fb≤fk which, by [21, Theorem 1] and Lemma 9, gives fb=fbfkfb=fkfb whence, by Lemmas 5 and 7, fkfb=(fkfb¯)∗fb=fkfb¯fb=fkfb=fb. Hence a≤kfbL∗b. It is easy to see that L∗ is upper Λ-stable.
(2)⇒(1) Let x,y∈S be such that x≤y. Since fxL∗x≤y and L∗ satisfies the link property, there exists z∈S such that fx≤zL∗y. Then fz=fy. Because L∗ is upper Λ-stable, fx≤fz. Thus L∗ is regular.


## 6. Structure Theorem

The main objective in this section is to give a structure theorem for a naturally ordered concordant semigroup with an adequate monoid transversal.LetS be an abundant semigroup with a set of idempotents E. An idempotent u of S is calledmedial if, for all e∈E,e=eue.We have the following result, which generalizes [21, Theorem 17].Theorem 32.
LetL and R be concordant semigroups with a common medial idempotent ξ and a common adequate submonoid T=ξLξ=ξRξ. Define a mapping R×L→Reg(T) by (a,x)↦a∘x with the following properties:
(1)(∀a,b∈R)(∀y,z∈L)a(a∘y)b∘z=[(a∘y)bξ]∗(b∘z);
(2)(∀a,b∈R)(∀y,z∈L)a∘y(b∘z)z=(a∘y)[ξy(b∘z)]+;
(3)(∀a∈R)(∀x∈L)(∀t∈T)a∘t=(a∘ξ)(ξ∘t),t∘x=(t∘ξ)(ξ∘x);
(4)(∀a∈R)(∀x∈L)a(a∘ξ)=ξaξ,(ξ∘x)x=ξxξ;
(5)(∀a∈R,f∈E(La∗))(∀x∈L,e∈E(Rx∗))f∘x=a∘x=a∘e
On the setL|×|ξR={(x,a)∈L×R∣ξx=aξ} define a multiplication by (11)x,ay,b=xa∘yy,aa∘yb. Then L|×|ξR is a locally adequate concordant semigroup that has an adequate monoid transversal.
Moreover, every such semigroup is obtained in this way. More precisely, ifS is a locally adequate concordant semigroup with an adequate monoid transversal S∘, that is a monoid with identity, ξ then S∘=ξSξ, ξ is a medial idempotent of both Sξ and ξS, the mapping ξS×Sξ→Reg(S∘) given by (a,x)↦faex satisfies properties (1) to (5) above, and there is a semigroup isomorphism S≅Sξ|×|ξξS.Proof.
It is easy to see that the multiplication onL|×|ξR is well-defined.
We proceed in the following stages.
(i)L|×|ξR is a semigroup.
For associativity of the multiplication, let(x,a),(y,b),(z,c)∈L|×|ξR. It is easy to see that the first component of [(x,a)(y,b)](z,c) is x(a∘y)y[a(a∘y)b∘z]z which by (1) is x(a∘y)y[a(a∘y)bξ]∗(b∘z)z. Using the fact that bξ=ξy and ξ is an identity of T, we see that this reduces to x(a∘y)y(b∘z)z. On the other hand, the first component of (x,a)[(y,b)(z,c)] is x[a∘y(b∘z)z]y(b∘z)z which by (2) is x(a∘y)[ξy(b∘z)]+y(b∘z)z. It also reduces to x(a∘y)y(b∘z)z. Thus the first components of the products are the same and, similarly, so are the second components. Thus, with the above multiplication, L|×|ξR is a semigroup.
(ii)L|×|ξR is an abundant semigroup.
Suppose that(x,a)∈L|×|ξR. Let e∈E(L),f∈E(R) be such that eR∗x,fR∗a. It is easy to see e=eξ and f=ξf. We have for some g∈E(Lx∗), ξx=ξxgξg=ξxξg=(ξ∘x)xg=(ξ∘x)x=ξxξ. Similarly, aξ=ξaξ. Since ξeR∗ξx and ξeξe=ξe, it follows that ξe=(ξxξ)+. If b,c∈R are such that baξ=caξ then we have for some h∈E(La∗), bahξh=cahξh. Hence ba=ca and so aR∗aξ. Thus fξR∗fR∗aR∗aξ and so fξ=ξfξ=(ξaξ)+=(ξxξ)+=ξe whence (e,fξ)∈L|×|ξR. We have (12)e,fξe,fξ=efξ∘ee,fξfξ∘efξ=efξ∘ξξ∘ee,fξfξ∘ξξ∘efξby3=eξefξ∘ξξeξ,ξfξξ∘eξeby4=efξfξ∘ξξeξ,ξfξξeξby4=eξfξeξ,ξfξfξby4=eξeξe,fξ=e,fξwhence (e,fξ)∈E(L|×|ξR). We also have (13)e,fξx,a=efξ∘xx,fξfξ∘xa=efξ∘ξξ∘xx,fξfξ∘ξξ∘xahξhby3=eξefξ∘ξξxξ,ξfξξ∘xξxhby4=efξfξ∘ξξxξ,ξfξξxξhby4=eξfξxξ,ξfξaξhby4=eξeξx,fξa=x,a.If (y,b),(z,c)∈L|×|ξR are such that (y,b)(x,a)=(z,c)(x,a), then y(b∘x)x=z(c∘x)x and b(b∘x)a=c(c∘x)a. Hence by (5) and eR∗x,fξR∗a, y(b∘e)e=z(c∘e)e and b(b∘e)fξ=c(c∘e)fξ. Thus we have (14)y,be,fξ=yb∘ee,bb∘efξ=zc∘ee,cc∘efξ=z,ce,fξwhence (x,a)R∗(e,fξ)=(e,ξe).
Dually,(hξ,h)=(ξg,h)∈E(L|×|ξR) and (x,a)L∗(hξ,h), where g∈E(Lx∗),h∈E(La∗). This shows that L|×|ξR is abundant.
(iii)L|×|ξR is IC.
Let(e,ξe)∈E(R(x,a)∗) and (e′,f′)∈ω((e,ξe)) where e∈E(Rx∗),ξe∈E(Ra∗). Then (e′,f′)(e′,f′)=(e′,f′) and (e′,f′)(e,ξe)=(e′,f′)=(e,ξe)(e′,f′). It follows that e′(f′∘e)e=e′, e(ξe∘e′)e′=e′, and f′(f′∘e)ξe=f′, whence e′e=e′, ee′=e′, and f′=f′ξ=ξe′. Hence, we have, for some α∈E(Re′∗), (15)e′f′∘e′e′=e′⇒e′ξe′∘e′e′=e′⇒αξαe′ξe′∘ξξ∘e′e′=e′by3⇒αξe′ξξe′ξ=e′by4⇒αξe′e′=e′⇒e′e′=e′and so e′∈ω(e). Suppose that k∈ω(e), it is easy to see that (k,ξk)∈ω(e,ξe). Thus (16)e′,f′∈ωe,ξe⇔e′∈ωe,f′=ξe′. Dually, let (hξ,h)∈E(L(x,a)∗) where h∈E(La∗),hξ∈E(Lx∗). Then we have (17)g′,h′∈ωhξ,h⇔h′∈ωh,g′=h′ξ. Let (k,ξk)∈ω((e,ξe)). We have (18)k,ξkx,a=kξk∘xx,ξkξk∘xa=kξk∘ξξ∘xx,ξkξk∘ξξ∘xahξhby3=kξkξk∘ξξxξ,ξkξξ∘xξxhby4=kξkξxξ,ξkξaξhby4=kξkξx,ξka=kx,ξka.Since R is IC and ξk∈ω(ξe), by Lemma 3, there exists l∈ω(h) such that ξka=al. Hence we have (19)x,alξ,l=xa∘lξlξ,aa∘lξl=xa∘ξξ∘lξlξ,aa∘ξξ∘lξlξlby3=eξexa∘ξξlξ,ξaξlξlby4=eξaξlξ,alξlby4=eξalξ,al=eξkaξ,ξka=eξekξx,ξka=kx,ξka.Thus (k,ξk)(x,a)=(x,a)(lξ,l). Similarly, for every (l′ξ,l′)∈ω(hξ,h), there exists (k′,ξk′)∈ω(e,ξe) such that (x,a)(l′ξ,l′)=(k′,ξk′)(x,a). This shows that L|×|ξR is IC.
(iv)L|×|ξR satisfies the regularity condition.
If(x,a)∈Reg(L|×|ξR), then (y,b) exists such that (x,a)(y,b)(x,a)=(x,a). Hence we have (20)xa∘yy,aa∘ybx,a=x,a⇒xa∘yyaa∘yb∘xx,aa∘ybaa∘yb∘xa=x,a⇒xa∘yya∘ybξ∗b∘xx,aa∘yba∘ybξ∗b∘xa=x,aby1⇒xa∘yyb∘xx,aa∘ybb∘xa=x,a⇒xa∘yyb∘xx=x,aa∘ybb∘xa=awhence x∈Reg(L) and a∈Reg(R). Since T is an adequate semigroup, Reg(T) is an inverse semigroup. If (x,a)∈L|×|ξR, x∈Reg(L), and a∈Reg(R), then we can define β=(ξxξ)-1=(ξaξ)-1∈T. Then the first component of the product(x,a)(β,β)(x,a) is as in (i), (21)xa∘βββ∘xx=eξexa∘βββ∘ξξ∘xxby3=eaξa∘ξξ∘βξβξξxξby4=eξaξβξxξby4=eξxξ=xwhere e∈E(Rx∗). In a similar way we can see that the second component of the product is a. Thus we see that (x,a)(β,β)(x,a)=(x,a). Therefore, (22)x,a∈RegL×LξR⇔x∈RegL,a∈RegR.If (x,a),(y,b)∈Reg(L|×|ξR), then x,y∈Reg(L) and a,b∈Reg(R). Since L and R are both concordant, x(a∘y)y∈Reg(L) and a(a∘y)b∈Reg(R). Hence (x,a)(y,b)=(x(a∘y)y,a(a∘y)b)∈Reg(L|×|ξR). This proves that S satisfies the regularity condition. Thus S is a concordant semigroup.
(v)T^=diag(T×T) is a ∗-adequate submonoid of L|×|ξR.
It is easy to see thatT^ is a subsemigroup of L|×|ξR. Since T=ξLξ=ξRξ, by Lemma 4 and T is a ∗-adequate submonoid of both L and R. If (γ,γ)∈T^, then, by (ii), (γ+,γ+)R∗(γ,γ)L∗(γ∗,γ∗). Hence T^ is abundant. The proof of the following results is essentially the same as that of [21, Theorem 17]: E(T^) is a semilattice and (23)γ,γ∈ET^⇔γ∈ET.Therefore, T^ is an adequate semigroup. Let (x,a)∈E(L|×|ξR) and (γ,γ)∈E(T^) be such that (x,a)≤n(γ,γ). Then (x,a)(γ,γ)=(x,a)=(γ,γ)(x,a). Hence we have (24)x,aγ,γ=x,a⇒xa∘γγ,aa∘γγ=x,a⇒xa∘ξξ∘γγ,aa∘ξξ∘γγ=x,a⇒xξγξ,ξaξγξ=x,a⇒xγ=x,aγ=a.Similarly, we have γx=x,γa=a. Hence (25)γx=x⇒γξx=x⇒γaξ=x⇒aξ=x⇒ξaξ=x Similarly, a=ξxξ. Thus a=x=ξxξ and so (x,a)∈E(T^). Therefore, by [9, Lemma 1.6], T^ is a ∗-adequate subsemigroup of L|×|ξR. It is easy to see that (ξ,ξ) is an identity element for T^.
(vi)T^ is a quasi-ideal adequate transversal of L|×|ξR.
Suppose that(x,a)∈L|×|ξR and (e,ξe)R∗(x,a)L∗(fξ,f) where e∈E(Rx∗),f∈E(La∗). By the proof of (ii), eLξe=(ξxξ)+ and fRfξ=(ξxξ)∗. Hence (e,ξe)L(ξe,ξe), (fξ,f)R(fξ,fξ), and we have (26)e,ξeξxξ,ξxξfξ,f=eξe∘ξxξξxξ,ξeξe∘ξxξξxξfξ,f=eξeξe∘ξξ∘ξxξξxξ,ξeξe∘ξξ∘ξxξξxξfξ,f=eξeξxξ,ξeξxξfξ,f=x,ξxξfξ,f=xξxξ∘fξfξ,ξxξξxξ∘fξf=eξxξξxξ∘ξξ∘fξfξ,ξxξξxξ∘ξξ∘fξfξf=eξxξfξ,ξxξfξf=eξxξ,aξf=x,a.Suppose that there exist (β,β)∈T^,(e1,f1),(e2,f2)∈E(L|×|ξR) such that (x,a)=(e1,f1)(β,β)(e2,f2), (e1,f1)L(β,β)+=(β+,β+), and (e2,f2)R(β,β)∗=(β∗,β∗). Then we have for α∈E(Re1)(27)e1,f1β+,β+=e1,f1⇒e1f1∘β+β+,f1f1∘β+β+=e1,f1⇒αξe1f1∘ξξ∘β+β+,f1f1∘ξξ∘β+β+=e1,f1⇒αf1f1∘ξξβ+ξ,ξf1ξβ+ξ⇒e1ξβ+ξ,ξf1ξβ+ξ=e1,f1⇒e1β+=e1,f1β+=f1. Similarly, β+e1=β+, β+f1=β+ and so e1e1=e1β+e1=e1β+=e1, f1f1=f1β+f1=f1β+=f1∈E(T). It follows that ξe1Le1Lβ+ and f1=β+. Hence f1=ξe1=β+. In a similar way we have e2=f2ξ=β∗ and f2∈E(Rβ∗). Thus (28)e1,f1β,βe2,f2=e1,ξe1β,βf2ξ,f2=e1ξe1∘ββ,ξe1ξe1∘ββf2ξ,f2=e1ξe1ξe1∘ξξ∘ββ,ξe1ξe1∘ξξ∘ββf2ξ,f2=e1ξe1ξβξ,ξe1ξβξf2ξ,f2=e1β,ξe1βf2ξ,f2=e1βξe1β∘f2ξf2ξ,ξe1βξe1β∘f2ξf2=e1βξe1β∘ξξ∘f2ξf2ξ,ξe1βξe1β∘ξξ∘f2ξf2=e1ξe1βξe1β∘ξξf2ξ,ξe1βξξ∘f2ξf2ξf2=e1ξe1βf2ξ,ξe1βf2ξf2=e1βf2ξ,ξe1βf2.It follows that e1βf2ξ=x and so β=ξe1βf2ξ=ξxξ. This shows that T^ is an adequate transversal. That T^ is a quasi-ideal holds by essentially the argument of [21, Theorem 17]
In summary, from the above and Theorem18 we have that L|×|ξR is a locally adequate concordant semigroup with an adequate monoid transversal.
To show that every such semigroup is obtained in this way, letS be a locally adequate concordant semigroup with an adequate monoid transversal S∘. Let the identity element of S∘ be ξ. By Theorem 25(3) we have S∘=ξSξ. Moreover, by Theorem 25(5), we see that L=Sξ is a semigroup with right identity ξ, and R=ξS is a semigroup with left identity ξ. Thus ξ is a medial idempotent of both Sξ and ξS. Moreover, ξLξ=ξSξ=ξRξ. Since E(L)=I(S) and E(R)=Λ(S), for every x∈L, xR∗ex∈E(L) and xL∗fx∈E(R). Hence x=xξL∗fxξ∈E(L) and so fx=fx(fxξ)=fxξ∈T⊆L. Hence L is abundant. Let e∈E(L), g∈E(S) be such that g≤ne. Then g=ege=egeξ∈L. Hence E(L) is an order-ideal of E(S) and so L is a ∗-subsemigroup of S. Let x∈L. Since S is IC, there is a bijection θ:<ex>→<fx> satisfying for all y∈<ex>,yx=x(yθ). We have <ex>⊆exSex⊆L, and <fx>⊆fxSfx⊆L. Hence L is IC and so L is a concordant semigroup. Similarly, R is a concordant semigroup. We are therefore in the initial conditions of the first part. Consider therefore the mapping ξS×Sξ→Reg(S∘)=Reg(ξSξ) given by (29)a,x↦a∘x=faex.To see that this satisfies property (1) above, observe that (30)aa∘yb∘z=afaeyb∘z=aeyb∘z=faeybez=faeeybeyb¯∗feybez=faeeyey¯eb+ey¯ebb¯∗ey¯ebb¯∗fbez=faeyey¯ebb¯∗ey¯ebb¯∗fbez=faeybξ∗ξeybξ∗fbez.Since (31)faeybξξeybξ∗=faeyξeybξξeybξ∗=faeyξeybξ=faeybξ,(faeybξ)∗(ξeybξ)∗L∗faeybξ(ξeybξ)∗=faeybξ. So (faeybξ)∗(ξeybξ)∗=(faeybξ)∗. Hence (32)aa∘yb∘z=faeybξ∗fbez=a∘ybξ∗b∘z.It is readily verified that (2)-(5) also hold.
Consider now the mappingϑ:S→Sξ|×|ξξS given by ϑ(x)=(xξ,ξx). For all x,y∈S, we have (33)ϑxϑy=xξ,ξxyξ,ξy=xξξx∘yξyξ,ξxξx∘yξξy=xξfξxeyξyξ,ξxfξxeyξξy=xξξexx¯∗fxyξ,ξxeyy¯fyξ+ξy=xξξexξxξ∗fxyξ,ξxeyξyξfyξ+ξy=exξxξξexxξ∗fxyξ,ξxeyξyξ+ξyξfy=exξxξξxξ∗fxyξ,ξxeyξyξfy=exξxξfxyξ,ξxeyyfy=xyξ,ξxy=ϑxyand so ϑ is a morphism.
Ifϑ(x)=ϑ(y), then xξ=yξ and ξx=ξy. Hence x¯=ξxξ=ξyξ=y¯. Thus exξ=ex(x¯fxξ)+=ex(ξxξfxξ)+=ex(ξxfxξ)+=ex(ξxξ)+=exx¯+=ex. Similarly, fξx=fx. Consequently, x=exx¯fx=exξx¯fξx=eyξy¯fξy=eyy¯fy=y. Thus ϑ is injective.
To see thatϑ is also subjective, let (xξ,ξy)∈Sξ|×|ξξS. Then ξxξ=ξyξ, whence x¯=y¯. Now let t=exy. Then tξ=exξyξ=exξxξ=exxξ=xξ and ξt=ξexy=ξexξy=ex¯y=ex¯y=ey¯y=ey¯y=ξeyξy=ξeyy=ξy. Thus ϑ(t)=(tξ,ξt)=(xξ,ξy).
It follows from the above thatS≅Sξ|×|ξξS.We can use the above theorem to establish a structure theorem for naturally ordered concordant semigroups with an adequate monoid transversal, which generalizes [21, Theorem 18].Theorem 33.
LetS be a naturally ordered concordant semigroup with an adequate transversal S∘ that is a monoid with identity element ξ. Let Sξ|×|ξξS consist of the subset of the Cartesian ordered set Sξ×ξS given by Sξ|×|ξξS={(xξ,ξx)∣x∈S} together with the multiplication (xξ,ξx)(yξ,ξy)=(xyξ,ξxy). Then, Sξ|×|ξξS is an ordered concordant semigroup. Moreover, if either L∗ or R∗ is regular on S then there is an ordered semigroup isomorphism: (34)S≅Sξ×ξξS.Proof.
SinceS is locally adequate by Theorem 18, it follows from Theorems 25(5) and 32 that there is an algebraic isomorphism ϑ:S→Sξ|×|ξξS given by ϑ(x)=(xξ,ξx). Suppose now that, for example, R∗ is regular on S. If ϑ(x)≤ϑ(y), then xξ≤yξ and ξx≤ξy. Hence, x=exx=exξx=exξξx≤eyξξy=y. Thus, x≤y⇔ϑ(x)=ϑ(y). Therefore, the isomorphism ϑ is an order isomorphism. Similarly, if L∗ is regular, then the isomorphism ϑ is also an order isomorphism.



---

*Source: 9867312-2019-07-04.xml*