The product of quasi-ideal refined generalised quasi-adequate transversals

Xiangjun Kong; Pei Wang; Yonghong Wu

doi:10.1515/math-2019-0004

Artikel Open Access

The product of quasi-ideal refined generalised quasi-adequate transversals

Xiangjun Kong , Pei Wang und Yonghong Wu

Veröffentlicht/Copyright: 28. Februar 2019

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Aus der Zeitschrift Open Mathematics Band 17 Heft 1

Abstract

As the real common generalisations of both orthodox transversals and adequate transversals in the abundant case, the concept of refined generalised quasi-adequate transversal, for short, RGQA transversal was introduced by Kong and Wang. In this paper, an interesting characterization for a generalised quasi-adequate transversal to be refined is acquired. It is shown that the product of every two quasi-ideal RGQA transversals of the abundant semigroup S satisfying the regularity condition is a quasi-ideal RGQA transversal of S and that all quasi-ideal RGQA transversals of S compose a rectangular band. The related results concerning adequate transversals are generalised and enriched.

Keywords: abundant semigroup; product; RGQA transversal; quasi-ideal

MSC 2010: 20M10

1 Introduction

Suppose that S is a regular semigroup with a subsemigroup S^o. We denote the intersection of V(a) and S^o by V_S^o(a) and that I = {aa^o : a ∈ S, a^o ∈ V_S^o(a)} and Λ = {a^oa : a ∈ S, a^o ∈ V_S^o(a)}. An inverse transversal of the semigroup S is a subsemigroup S^o that contains exactly one inverse of every element of S, that is, S is regular and S^o inverse with |V_S^o(a)| = 1. This important concept was introduced by Blyth and McFadden [1]. Thereafter, this class of regular semigroups excited many semigroup researchers’ attention and a good many important results were obtained (see [1,2,3,4] and their references). Tang [4] showed that for S a regular semigroup with an inverse transversal S^o, then I and Λ are both bands with I left regular and Λ right regular. These two bands play an important role in the study of regular semigroups with inverse transversals. Other important subsets of S are R = {x ∈ S: x^ox = x^ox^oo} and L = {x ∈ S: xx^o = x^oox^o}. They are subsemigroups with R and L left and right inverse respectively. The concept of an orthodox transversal was introduced by Chen [5] as an interesting generalisation of inverse transversals, and a structure for a regular semigroup with a quasi-ideal orthodox transversal was established. Chen and Guo [6] further considered the general case of an orthodox transversal and acquired many properties focused on the sets I and Λ. In [7, 8], Kong and Zhao introduced two interesting sets R and L and established the structure for a regular semigroup with quasi-ideal orthodox transversals. In 2014, Kong [9] introduced the concept of a generalised orthodox transversal and Kong and Meng [10] acquired the characterization for a generalised orthodox transversal to be an orthodox transversal and obtain a concrete characterization of the maximum idempotent-separating congruence on a regular semigroup with an orthodox transversal. If the concept of transversals could be introduced in the range of E-inversive semigroups, then the congruences [11, 12] on them will be characterized more neatly. More recently, Kong [13] investigated the weakly simplistic orthodox transversal and obtained the sufficient and necessary condition for the orthodox transversal S^o to be weakly simplistic.

The concept of an adequate transversal, was introduced in the range of abundant semigroups by El-Qallali [14] as the generalisation of the concept of an inverse transversal. Chen, Guo and Shum [15, 16] obtained some important results about a quasi-ideal adequate transversal. Kong [17] explored some properties about adequate transversals. Kong and Wang [18] considered the product of quasi-ideal adequate transversals and proposed the open problem of the isomorphism of adequate transversals. The concept of a quasi-adequate transversal was introduced by Ni [19] and followed by Luo, Kong and Wang [20, 21], their work mainly focused on the properties and the structure of multiplicative quasi-adequate transversals. Unfortunately, quasi-adequate transversals are neither the generalisation of an orthodox transversal nor adequate transversals. Inspired by the characterization of orthodox transversals [10], the concept of a RGQA transversal was introduced by Kong and Wang [22]. It was demonstrated that RGQA transversals are the real common generalisations of both orthodox transversals and adequate transversals in the abundant case.

In the present paper, we continue along the lines of [3, 18, 22] by exploring the relationship between the quasi-ideal RGQA transversals of the abundant semigroup. The main result of this paper is that the product of two quasi-ideal RGQA transversals of an abundant semigroup S satisfying the regularity condition is a quasi-ideal RGQA transversal of S and that all quasi-ideal RGQA transversals of S compose a rectangular band. The related results concerning adequate transversals are generalised and enriched.

2 Preliminaries

The Miller-Clifford theorem is crucial in the study of semigroups.

Lemma 2.1

[23] (1) Suppose that e and f are idempotents of a semigroup S. For every a in R_e ∩ L_f, there exists a unique b in R_f ∩ L_e such that ab = e and ba = f;

(2) Suppose that a, b ars elements of a semigroup S. Then ab ∈ R_a ∩ L_b if and only if there exists an idempotent in L_a ∩ R_b.

Definition 2.2

[5] Let S^o be an orthodox subsemigroup of the regular semigroup S. S^o is called be an orthodox transversal of S, if the following two conditions are satisfied:

(∀ a ∈ S) V_S^o(a) ≠ ∅;
For any a, b ∈ S, if {a, b} ∩ S^o ≠ ∅, then V_S^o(a) V_S^o(b) ⊆ V_S^o(ba).

Lemma 2.3

[10] Let S^o be an orthodox subsemigroup of the regular semigroup S with V_S^o(a) ≠ ∅ for every a ∈ S. Then the sufficient and necessary condition for S^o to be an orthodox transversal of S is

(∀a,b∈S) [VSo(a)∩VSo(b)≠∅⇒VSo(a)=VSo(b)].

A subsemigroup T of a semigroup S is called a quasi-ideal of S, if it satisfies TST ⊆ T.

Let S and S^o be semigroups. In this paper, the set of idempotents of S and S^o are denoted by E and E^o respectively to avoid confusion. If the product of any two elements of E is regular, then S is called satisfying the regularity condition.

On any semigroup S the relation 𝓛^* is defined by a𝓛^*b if and only if {∀x, y ∈ S¹, ax = ay ⇔ bx = by}. The relation 𝓡^* is dually defined. Certainly, 𝓛^* and 𝓡^* are right and left congruences respectively with 𝓛 ⊆ 𝓛^* and 𝓡 ⊆ 𝓡^*. If a, b are regular in S, then a𝓛^*b (a𝓡^*b) if and only if a𝓛b (a𝓡b). A semigroup is said to be abundant [24] if every 𝓛^*- class and each 𝓡^*- class contains an idempotent. An abundant semigroup S is said to be quasi-adequate [25] (adequate semigroup) if its idempotents constituate a band (semilattice). A band is said to be a rectangular band if it satisfies abc = ac. Suppose that S is an abundant semigroup with U an abundant subsemigroup, U is a ^*-subsemigroup of S iff 𝓛^*(U) = 𝓛^*(S) ∩ (U × U) and 𝓡^*(U) = 𝓡^*(S) ∩ (U × U).

Let S^o be a ^*-adequate subsemigroup of the abundant semigroup S. S^o is said to be an adequate transversal of S, if for every x ∈ S there exist two idempotents e, f in S and a unique element x in S^o such that x = exf, with e𝓛x⁺ and f𝓡x^*. (see [14] for detail).

Suppose that S is an abundant semigroup having the set of idempotents E and S^o a quasi-adequate ^*-subsemigroup of S having the set of idempotents E^o. S^o is said to be a generalised quasi-adequate transversal of S if C_S^o(x) = {x^o ∈ S^o | x = ex^of, e𝓛x^o+, f𝓡x^o* for some x^o+, x^o* ∈ E^o} ≠ ∅. Let

Ix={e∈E|(∃xo∈CSo(x))x=exof,eLxo+,fRxo∗for somexo+,xo∗∈Eo},Λx={f∈E|(∃xo∈CSo(x))x=exof,eLxo+,fRxo∗for somexo+,xo∗∈Eo},I=⋃x∈SIx,Λ=⋃x∈SΛx.

By Ni^[19], the generalised quasi-adequate transversal S^o is said to be a quasi-adequate transversal of S if it satisfies (∀e ∈ E) (∀g ∈ E^o), C_S^o(e) C_S^o(g) ⊆ C_S^o(ge) and C_S^o(g)C_S^o(e) ⊆ C_S^o(eg).

Definition 2.4

[22] Suppose that S^o is a generalised quasi-adequate transversal of the abundant semigroup S. If for all a, b ∈ RegS, V_S^o(a) ∩ V_S^o(b) ≠ ∅ implies that V_S^o(a) = V_S^o(b), then S^o is said to be a a RGQA transversal of S.

Lemma 2.5

[22] Suppose that S is an abundant semigroup having a generalised quasi-adequate transversal S^o. Then S^o is refined if and only if IE^o, E^oΛ ⊆ E and for all e ∈ I, λ ∈ Λ, f ∈ E^o, if fe, λf are regular, then they are idempotent.

Lemma 2.6

[22] Let S^o be a refined generalised quasi-adequate transversal of the abundant semigroup S. Then

S^o is an orthodox transversal of S if and only if S is a regular semigroup.
S^o is an adequate transversal of S if and only if S^o is an adequate semigroup.

Therefore, by Lemma 2.6 we can say that RGQA transversals are the real common generalisation of an orthodox transversal and adequate transversals in the abundant case.

3 A characterization for a generalised quasi-adequate transversal to be refined

Lemma 2.5 gives the important equivalent conditions for a generalised quasi-adequate transversal to be refined. Now we supplement Lemma 2.5 with another characterization of RGQA transversals within the class of abundant semigroups. It is analogous to the definition of an orthodox transversal in an interesting manner.

Theorem 3.1

Suppose that S is an abundant semigroup having a generalised quasi-adequate transversal S^o. Then S^o is refined if and only if for any regular elements a ∈ S, b ∈ S^o, if ba is regular, then V_S^o(a) V_S^o(b) ⊆ V_S^o(ba); and if ab is regular, then V_S^o(b) V_S^o(a) ⊆ V_S^o(ab).

Proof

(Necessity) For any regular elements a ∈ S, b ∈ S^o, we may take a^o ∈ V_S^o(a), b^o ∈ V_S^o(b), if the generalised quasi-adequate transversal S^o is refined, then by Lemma 2.5, aa^ob^ob ∈ IE^o ⊆ E. If ba is regular, take (ba)^o ∈ V_S^o(ba), then

(bobaao)(a(ba)ob)(bobaao)=bo(baaoa)(ba)o(bboba)ao=bo(ba)(ba)o(ba)ao=bo(ba)ao=bobaao.

Thus b^obaa^o is regular and so b^obaa^o ∈ E^oI ⊆ E. Therefore

aobo⋅ba⋅aobo=ao(aaobob)(aaobob)bo=ao⋅aaobob⋅bo=aobo

ba⋅aobo⋅ba=b(bobaao)(bobaao)a=b⋅bobaao⋅a=ba

and so V_S^o(a) V_S^o(b) ⊆ V_S^o(ba). Similarly, if ab is regular, then V_S^o(b) V_S^o(a) ⊆ V_S^o(ab).

(Sufficiency) For any regular elements s₁, s₂ ∈ S^o, if V(s₁) ∩ V(s₂) ≠ ∅, take s ∈ V(s₁) ∩ V(s₂) and s1o ∈ V_S^o(s₁). From s₂s𝓛s₁s𝓡s₁ s1o , by Lemma 2.1, s₂𝓡s₂ss₁ s1o 𝓛 s1o and (s2ss1s1o)2=s2(ss1s1os2ss1)s1o=s2ss1s1o since by the assumption s1o s₂ ∈ V_S^o(ss₁). Similarly, s₂𝓛 s1o s₁ss₂𝓡 s1o with s1o s₁ss₂ ∈ E. Thus

s1os2s1o=s1os2ss1ss2s1o=s1o(s2ss1s1o)s1ss2s1o=(s1os1ss2)s1o=s1o

and

s2s1os2=s2(ss2s1os1)s1o(s1s1os2s)s2=s2ss1s1os1ss2=s2ss1ss2=s2ss2=s2.

Hence s1o ∈ V_S^o(s₂), that is, V_S^o(s₁) ∩ V_S^o(s₂) ≠ ∅. Therefore V_S^o(s₁) = V_S^o(s₂) since the regular elements of S^o form an orthodox subsemigroup of S.

For any e ∈ S, if V_S^o(e) ∩ E^o ≠ ∅, take f ∈ V_S^o(e) ∩ E^o. Then for any e^o ∈ V_S^o(e), we have e ∈ V(f) ∩ V(e^o) and so by the above result, V_S^o(f) = V_S^o(e^o). Consequently, e^o is an inverse of f in S^o and e^o ∈ E^o since S^o is quasi-adequate. That is, if V_S^o(e) ∩ E^o ≠ ∅, then V_S^o(e) ⊆ E^o.

Let e, f ∈ I with e𝓛f. Let h ∈ E^o be such that h𝓛e𝓛f, then h ∈ V_S^o(e) ∩ V_S^o(f). For any g ∈ V_S^o(e), by the result above we have g ∈ E^o. It is clear that ghg ∈ V_S^o(gfg) and ghg ∈ V_S^o(geg) = V_S^o(g). Then gfg and g have a common inverse ghg and so ghg ⋅ gfg ⋅ ghg = ghg. Thus gfg = g. Since ge𝓛e𝓛f, by Lemma 2.1, fg𝓡f and so fgf = f. Thus g ∈ V_S^o(f) and so V_S^o(e) ⊆ V_S^o(f). Similarly, the reverse inclusion is hold and hence V_S^o(e) = V_S^o(f). Dually, if e, f ∈ Λ with e𝓡f, then V_S^o(e) = V_S^o(f).

It is a routine matter to show that for any regular element a ∈ S with a^o ∈ V_S^o(a), then V_S^o(a) = V_S^o(a^oa)a^oV_S^o(aa^o).

If regular elements a, b ∈ S with V_S^o(a) ∩ V_S^o(b) ≠ ∅, we can take x^o ∈ V_S^o(a) ∩ V_S^o(b). It follows that

VSo(a)=VSo(xoa)xoVSo(axo) and VSo(b)=VSo(xob)xoVSo(bxo).

Since ax^o, bx^o ∈ I with ax^o𝓛bx^o, we have V_S^o(ax^o) = V_S^o(bx^o). Dually, V_S^o(x^oa) = V_S^o(x^ob). Therefore V_S^o(a) = V_S^o(b) and so S^o is refined.□

4 The main result

In 1986, Saito [3] had acquired the result that the product of two quasi-ideal inverse transversals of the regular semigroup S is again a quasi-ideal inverse transversal of S. This important result was generalised to adequate transversals by Kong and Wang [18]. In this section we obtain the main result that the product of any two quasi-ideal RGQA transversals of the abundant semigroup S satisfying the regularity condition is a quasi-ideal RGQA transversal of S. And that, if S has quasi-ideal RGQA transversals, then all quasi-ideal RGQA transversals of S compose a rectangular band.

Let A and B be subsets of a semigroup S and AB for {ab : a ∈ A, b ∈ B}. It is obvious that (∀A, B, C ⊆ S) (AB)C = A(BC), and we denote it by ABC.

Lemma 4.1

Suppose that S^o is a quasi-ideal RGQA transversal of the abundant semigroup S and A a subset of S. Then

ASS^o = AS^o and S^oSA = S^oA;
AS^o and S^oA are both subsemigroups and quasi-ideals of S;
For any regular element x ∈ S, if |V(x) ⋂ A| ≥ 1, then |V(x) ⋂ AS^o| ≥ 1 and

|V(x) ⋂ S^oA| ≥ 1.

Proof

Suppose that a ∈ A, x ∈ S and s ∈ S^o. Then a = e_aaf_a with f_a𝓡a^* and so axs = aa^*f_axs ∈ AS^oSS^o ⊆ AS^o. It is obvious that as = af_as ∈ ASS^o and thus ASS^o = AS^o. Similarly, S^oSA = S^oA.
Obviously, AS^o ⋅ AS^o ⊆ A ⋅ S^oSS^o ⊆ AS^o, thus AS^o is closed and a subsemigroup of S. Similarly, AS^o ⋅ S ⋅ AS^o ⊆ A ⋅ S^oSS^o ⊆ AS^o and so AS^o is a quasi-ideal of S. There is a dual result for S^oA
For any regular element x ∈ S, take x′ ∈ V(x) ⋂ A, then for any x^o ∈ V_S^o(x), x′xx^o ∈ V(x) ⋂ ASS^o = V(x) ⋂ AS^o, that is |V(x) ⋂ AS^o| ≥ 1. Similarly, |V(x) ⋂ S^oA| ≥ 1.□

Lemma 4.2

Let S^o, S^◻ be quasi-ideal RGQA transversals of an abundant semigroup S. For every regular element a ∈ S, V_S^◻S^o(a) = V_S^◻(a) ⋅ a ⋅ V_S^o(a).

Proof

Take a^◻ ∈ V_S^◻(a), a^o ∈ V_S^o(a), then a^◻ aa^o ∈ S^◻ SS^o = S^◻S^o and a^◻ aa^o ∈ V(a), thus V_S^◻(a) ⋅ a ⋅ V_S^o(a) ⊆ V_S^◻S^o(a). For any x^◻y^o ∈ V_S^◻S^o(a), we have

a=ax◻yoa, x◻yo=x◻yo⋅a⋅x◻yo.

Thus

x◻yo=x◻yo⋅aa◻aaoa⋅x◻yo=x◻yoaa◻⋅a⋅aoax◻yo.

and

x◻yoaa◻∈S◻SS◻⊆S◻, aoax◻yo∈SoSSo⊆So,

Meanwhile

a⋅x◻yoaa◻⋅a=a⋅x◻yo⋅a=a,x◻yoaa◻⋅a⋅x◻yoaa◻=x◻yoax◻yoaa◻=x◻yoaa◻.

Hence x^◻y^oaa^◻ ∈ V_S^◻(a), Similarly, a^oax^◻y^o ∈ V_S^o(a). Therefore V_S^◻S^o(a) ⊆ V_S^◻(a) ⋅ a ⋅ V_S^o(a) and hence V_S^◻S^o(a) = V_S^◻(a) ⋅ a ⋅ V_S^o(a).□

Lemma 4.3

Suppose that S^o is a quasi-ideal RGQA transversal of the abundant semigroup S. For any x, y ∈ S, there exist x ∈ C_S^o(x), y ∈ C_S^o(y) such that x = e_xxf_x, e_x𝓛x⁺, f_x𝓡x^* for some x⁺, x^* ∈ E^o and y = e_yyf_y, e_y𝓛y⁺, f_y𝓡y^* for some y⁺, y^* ∈ E^o. Then

xf_xe_yy ∈ C_S^o(xy);
e_x(xf_xe_y)⁺ ∈ I_xy;
(f_xe_yy)^*f_y ∈ Λ_xy.

Proof

Certainly

xy=exx¯fxeyy¯fy=ex(x¯fxey)+(x¯fxeyy¯)(fxeyy¯)∗fy,

where e_x(xf_xe_y)⁺ ∈ IE^o ⊆ E, (f_xe_yy)^*f_y ∈ E^oΛ ⊆ E and xf_xe_yy ∈ S^o since S^o is a quasi-ideal. Since 𝓡^*, 𝓡 are left congruences and 𝓛^*, 𝓛 are right congruences, it is easy to see

ex(x¯fxey)+ L x¯+(x¯fxey)+ R∗ x¯+(x¯fxey)=x¯fxeyy¯+ R∗ x¯fxeyyo,(fxeyy¯)∗fy R (fxeyy¯)∗y¯∗ L∗ (fxeyy¯)y¯∗=x¯∗fxeyy¯ L∗ x¯fxeyy¯.

Thus the needed results are proved.□

Theorem 4.1

Let S^o be a quasi-ideal RGQA transversal of the abundant semigroup S satisfying the regularity condition. Then both I and Λ are bands.

Proof

Since S^o is a RGQA transversal of S and S satisfying the regularity condition, then by Lemma 2.5 IE^o ⊆ E and E^o I ⊆ E. Let e, f ∈ I with e𝓛 e^* ∈ E^o. Obviously, e^*f = e^*ff^* ∈ S^oSS^o ⋂ E^oI ⊆ S^o ⋂ I = E^o since S^o is a quasi-ideal. Thus ef = ee^*f ∈ IE^o ⊆ E and ef 𝓛 e^*f ∈ E^o and so ef ∈ I. Therefore I is a band and there is a dual result for Λ.□

In the following S^o and S^◻ denote a pair of RGQA transversals of the abundant semigroup S and E_S^o and E_S^◻ denote the idempotents of them respectively to avoid confusion, and similar as E_S, E_S^◻S^o, I_o, I_◻, Λ_o and Λ_◻ speaking for themselves. For simplicity, in S^◻, a typical idempotent which 𝓛^*-related and 𝓡^*-related to a ∈ S^◻ are denoted by a^* and a⁺ respectively. For every x ∈ S, we denote x = e_xxf_x in S^o and x = i_xx̃λ_x in S^◻ as the decompositions and of x respectively. Then x̃ ∈ S^◻ has the same sense as in the definition of generalised quasi-adequate transversals, that is i_x, λ_x ∈ E_S and x̃^*, x̃⁺ ∈ E_S^◻ with x̃^* 𝓛^* x̃𝓡^*x̃⁺ and i_x𝓛x̃⁺, λ_x𝓡x̃^*, and also i_x𝓡^*x𝓛^*λ_x.

Let S^◻ and S^o be RGQA transversals of the abundant semigroup S. Write

I(S◻,So)={aao: a∈Reg(S◻),ao∈VSo(a)},Λ(So,S◻)={a◻a: a∈Reg(So),a◻∈VS◻(a)}.

Theorem 4.2

Let S^◻ and S^o be a pair of quasi-ideal RGQA transversals of the abundant semigroup S satisfying the regularity condition. Then I(S^◻, S^o) = Λ(S^o, S^◻) = I_o ⋂ Λ_◻.

Proof

For any aa^o ∈ I(S^◻, S^o), where a ∈ Reg(S^◻), a^o ∈ V_S^o(a), certainly, a ∈ V_S^◻(a^o) and so aa^o = a^o◻a^o ∈ Λ(S^o, S^◻). Thus I(S^◻, S^o) ⊆ Λ(S^o, S^◻) and dually Λ(S^o, S^◻) ⊆ I(S^◻, S^o). Therefore,

I(S◻,So)=Λ(So,S◻)=Σ, say

It is clear from the above definitions that Σ ⊆ I_o ⋂ Λ_◻.

For the reverse part, let x ∈ I_o ⋂ Λ_◻. Since x ∈ Λ_◻, we have x = x^◻x for some x^◻ ∈ V_S^◻(x) with x^◻ ∈ E(S^◻) and so x^◻ = xx^◻. Similarly, x ∈ I_o implied that x = xx^o for some x^o ∈ V_S^o(x) with x^o ∈ E(S^o) and so x^o = x^ox. Let x^◻o ∈ V_S^o(x^◻). From x^o 𝓛 x 𝓡 x^◻ 𝓡 x^◻ x^◻o, by Lemma 2.1, we deduce that x^o 𝓡 x^ox^◻x^◻o 𝓛 x^◻x^◻o 𝓛 x^◻o with x^ox^◻x^◻o ∈ E^oI_o ⊆ E^o since S^o is a quasi-ideal and S satisfies the regularity condition. Thus x^o 𝓛 x^◻ox^o 𝓡 x^◻o. Certainly, x^o 𝓡 x^o x^◻ 𝓛 x^◻ and so by Lemma 2.1, x^◻o 𝓡 x^◻ox^ox^◻ 𝓛 x^o x^◻ and x^◻ox^ox^◻ 𝓗^* x^◻ox^◻ ∈ I_o ⋂ Λ_◻. Consequently, x^◻x^◻ox^o 𝓗x and so x^◻x^◻ox^o = x since x ∈ E and x^◻x^◻o ⋅ x^o ∈ I_oE^o ⊆ E. Also (x^◻ox^ox^◻)² = x^◻ox^o(x^◻x^◻ox^o)x^◻ = x^◻ox^oxx^◻ = x^◻ox^ox^◻ and x^◻ox^ox^◻ ∈ E. Therefore

x◻⋅x◻oxo⋅x◻=xx◻=x◻ and x◻oxo⋅x◻⋅x◻oxo=x◻oxox=x◻oxo

and so x^◻ox^o ∈ V_S^o(x^◻). Hence x = x^◻ ⋅ x^◻ox^o ∈ I(S^◻, S^o) = Σ.□

Theorem 4.3

Suppose that S^◻ and S^o are quasi-ideal RGQA transversals of the abundant semigroup S satisfying the regularity condition. Then S^◻ S^o is a quasi-ideal RGQA transversal of S.

Proof

It is easy to see that S^◻S^o is a quasi-ideal and a subsemigroup of S.

For any x ∈ S^◻S^o, there exist s^◻ ∈ S^◻, t^o ∈ S^o such that x = s^◻t^o. We deduce from S^o is a quasi-ideal of S that e_s^◻(s^◻f_s^◻e_t^o)⁺ ∈ I_s^◻t^o = I_x and we denote it by e_x. It is obvious that i_s^◻ ∈ E_S^◻ since (s^◻) ∈ S^◻ and so from e_s^◻𝓡^*i_s^◻ ∈ E_S^◻ we deduce that e_s^◻ ∈ I_o ⋂ Λ_◻. It follows from Theorem 4.2 that there exists a ∈ Reg(S^◻) such that e_s^◻ = aa^o and so

ex=es◻(s◻¯fs◻eto)+=aao(s◻¯fs◻eto)+∈S◻So.

Dually, λ_x ∈ S^◻S^o. Thus e_x, λ_x ∈ E_S^◻S^o, and it follows from e_x𝓡^*x𝓛^*λ_x that S^◻S^o is abundant. Through simple calculation we way show that e_x𝓡^*(S)x𝓛^*(S)λ_x, hence S^◻S^o is a abundant ^*-subsemigroup of S.

Suppose that e is an idempotent of S^◻S^o. Then e can decompose into as for some a ∈ S^◻, s ∈ S^o. Since (sas)(asa)(sas) = sas, (asa)(sas)(asa) = asa and sas ∈ S^o, we have sas ∈ V_S^o(asa), and so e = asasas = asa(asa)^o. By asa ∈ S^◻, every idempotent of S^◻S^o is of the form bb^o for some b in Reg(S^◻). Suppose that e and f are idempotents of S^◻S^o. Then e = bb^o and f = cc^o for some b, c ∈ Reg(S^◻) with b^o ∈ V_S^o(b) and c^o ∈ V_S^o(c). For any l ∈ E_S^o, by the regularity condition, lcc^o is regular and so lcc^o ∈ E_S since S^o is a RGQA transversal of S. Thus lcc^o ∈ E_S ⋂ S^o = E_S^o since S^o is also a quasi-ideal of S. Therefore ef = bb^occ^o = bb^o(b^o*cc^o) ∈ I_oE_S^o ⊆ E_S and S^◻S^o is a quasi-adequate semigroup.

For every x ∈ S, there exist a, b ∈ Reg(S) with e_x = aa^o, λ_x = b^◻b, where a^o ∈ V_S^o(a), b^◻ ∈ V_S^◻(b). Then

x=exxλx=aaoxb◻b=aao(ao◻aoxb◻b◻o)b◻b,

where a^o◻ ∈ V_S^◻(a^o), b^◻o ∈ V_S^o(b^◻), and so

ex=aaoLao◻ao∈ES◻So, λx=b◻bRb◻b◻o∈ES◻So.

Since a^o◻a^oxb^◻b^◻oλ_x = a^o◻a^oxλ_x = a^o◻a^ox, hence a^o◻a^oxb^◻b^◻o𝓡^*a^o◻a^ox. Since x𝓡^*e_x with 𝓡^* a left congruence, it follows that

ao◻aoxR∗ao◻aoex=ao◻ao∈ES◻So.

Similarly,

ao◻aoxb◻b◻oL∗xb◻b◻oL∗b◻b◻o∈ES◻So.

Therefore, x = e_x(a^o◻a^oxb^◻ b^◻o)λ_x, where e_x, λ_x ∈ E_S, e_x𝓛(a^o◻a^oxb^◻ b^◻o)⁺ = a^o◻a^o ∈ E_S^◻S^o and λ_x𝓡(a^o◻a^oxb^◻b^◻o)^* = b^◻ b^◻o ∈ E_S^◻S^o. Consequently, S^◻S^o is a generalised quasi-adequate transversal of S.

If a ∈ S, b ∈ S^◻S^o are regular, take a′ ∈ V_S^◻S^o(a), b′ ∈ V_S^◻S^o(b), it follows from Lemma 4.2 that there exist a^◻ ∈ V_S^◻(a), a^o ∈ V_S^o(a), b^◻ ∈ V_S^◻(b), b^o ∈ V_S^o(b), such that a′ = a^◻aa^o, b′ = b^◻bb^o. Since b ∈ S^◻S^o, b^o ∈ V_S^o(b), we have b ∈ V_S^◻S^o(b^o). By Lemma 4.2, there exist (b^o)^◻ ∈ V_S^◻(b^o), (b^o)^o ∈ V_S^o(b^o), such that b = (b^o)^◻b^o(b^o)^o. Thus

a′abb′=a◻aaoabb◻bbo=a◻abbo=a◻a(bo)◻bo(bo)obo=a◻a(bo)◻bo∈Λ◻Λ◻⊆Λ◻,

and a′abb′ is idempotent. Meanwhile

bb′a′a=bb◻bboa◻aaoa=bboa◻a=(bo)◻bo(bo)oboa◻a=(bo)◻boa◻a∈Λ◻Λ◻⊆Λ◻,

and bb′a′a ∈ E(S). Therefore

ab(b′a′)ab=a(a′abb′)(a′abb′)b=(aa′a)(bb′b)=ab,b′a′(ab)b′a′=b′(bb′a′a)(bb′a′a)a′=(b′bb′)(aa′a)=b′a′,

that is,

VS◻So(b)VS◻So(a)⊆VS◻So(ab).

Similarly,

VS◻So(a)VS◻So(b)⊆VS◻So(ba).

Therefore by Theorem 3.1 S^◻S^o is refined. Combining with S^◻S^o a quasi-ideal implied that S^◻S^o is a quasi-ideal RGQA transversal of S.□

Theorem 4.4

Suppose that S is an abundant semigroup satisfying the regularity condition. If S has quasi-ideal RGQA transversals, then all quasi-ideal RGQA transversals of S compose a rectangular band.

Proof

Suppose that S^o is a quasi-ideal RGQA transversal of S, then S^oS^o = S^o. In fact, for any s^o ∈ S^o, s^o = s^o(s^o)^* ∈ S^oS^o, hence S^o ⊆ S^oS^o and it is clear that the reverse inclusion valids. By Theorem 4.3, all quasi-ideal RGQA transversals of S compose a semigroup and so compose a band.

Let S^o, S^◻, S^⋄ be quasi-ideal RGQA transversals of S. For any s^o ∈ S^o, x ∈ S, t^◻ ∈ S^◻, we have

soxt◻=soxet◻(t◻¯)+t◻∈SoSSSoS◻⊆SoS◻,sot◻=so(so)∗t◻∈SoSoS◻⊆SoSS◻,

where t^◻𝓡^*e_t^◻ ∈ E_S and e_t^◻𝓛(t^◻)⁺ ∈ E_S^o. Thus S^oSS^◻ = S^oS^◻ and so

SoS⋄S◻⊆SoSS◻=SoS◻.

For any s^o ∈ S^o, t^◻ ∈ S^◻, we have

sot◻=sofso(fso)⋄fsot◻∈SoSS⋄SS◻=SoS⋄S◻,

where (f_s^o)^⋄ denotes an inverse of f_s^o in S^⋄. Hence S^oS^⋄S^◻ = S^oS^◻ and so all quasi-ideal RGQA transversals of S compose a rectangular band.□

5 An Example

In [18], Kong and Wang gave an example to show that S satisfy the regularity condition cannot be removed in Theorem 3.3 of [18]. In fact this example also demonstrates that S satisfying the regularity condition cannot be removed in Theorem 4.3. We only need to notice that Example 1 in [18]

S^oS^◻ = {i, j, o, w} is not a quasi-adequate subsemigroup of S as ji = w ∉ E_S^oS^◻ and therefore not a RGQA transversal of S.

Acknowledgement

The first author is a postdoctoral researcher of the Postdoctoral Station of Qufu Normal University and a Visiting Research Fellow of Curtin University. This research is supported by Shandong Province NSF (ZR2016AM02), the NSFC (11471186), Project Higher Educational Science and Technology Program of Shandong Province(J18KA248) and Scientific Foundation of Qufu Normal University (xkj201509).

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Erhalten: 2018-09-13

Angenommen: 2018-12-07

Online erschienen: 2019-02-28

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https://doi.org/10.1515/math-2019-0004

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abundant semigroup; product; RGQA transversal; quasi-ideal

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