On 𝓠-regular semigroups

Xinyang Feng

doi:10.1515/math-2018-0048

Article Open Access

On 𝓠-regular semigroups

Xinyang Feng

Published/Copyright: May 30, 2018

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$Open Mathematics$

From the journal Open Mathematics Volume 16 Issue 1

Abstract

In this paper, we give some characterizations of 𝓠-regular semigroups and show that the class of 𝓠-regular semigroups is closed under the direct product and homomorphic images. Furthermore, we characterize the 𝓠-subdirect products of this class of semigroups and study the E-unitary 𝓠-regular covers for 𝓠-regular semigroups, in particular for those whose maximum group homomorphic image is a given group. As an application of these results, we claim that the similar results on V-regular semigroups also hold.

Keywords: 𝓠-regular semigroup; 𝓠-subdirect product; Surjective 𝓠-subhomomorphism; E-unitary 𝓠-regular cover

MSC 2010: 20M10

1 Introduction and Preliminaries

Let S be a regular semigroup. Let V(a) be the set of all inverses of a for each a ∈ S. We also use E(S) to denote the set of all idempotents in S. It is well known that a regular semigroup S is orthodox if and only if for all a, b ∈ S,

V(b)V(a)⊆V(ab).

The concept of 𝓟-regular semigroups was first introduced by Yamada and Sen [1]. In [2], Zhang and He characterized the structure of 𝓟-regular semigroups. In fact, the class of 𝓟-regular semigroups is a generalization of orthodox semigroups and regular *-semigroups.

On the other hand, by Onstad [3], a regular semigroup is said to be a V-regular semigroup if for all a, b ∈ S,

V(ab)⊆V(b)V(a).

According to the definition of orthodox semigroups, V-regular semigroup is a dual form of orthodox semigorup. Evidently, a regular semigroup S is both orthodox and V-regular if and only if for all a, b ∈ S,

V(ab)=V(b)V(a).

The class of inverse semigroups forms the most important class of regular semigroups which satisfy the above condition. As a generalization of inverse semigroups and orthodox semigroups, Gu and Tang [4] investigated Vⁿ-semigroups and showed that the class of Vⁿ-semigroups is closed under direct products and homomorphic images.

In V-regular semigroups case, Nambooripad and Pastijn [5] gave a characterization of this class of semigroups and Zheng and Ren [6] described congruences on V-regular semigroups in terms of certain congruence pairs. Then, Li [7] generalized the concept of V-regular semigroups and investigated 𝓠-regular semigroups which is a dual form of 𝓟-regular semigroups.

A regular semigroup S is called 𝓠-regular if for any a ∈ S there exists a non-empty set V_Q ⊆ V(a) satisfying the following conditions:

aa⁺ ∈ V_Q(aa⁺) and a⁺a ∈ V_Q(a⁺a) for any a ∈ S and a⁺ ∈ V_Q(a),
V_Q(ab) ⊆ V_Q(b)V_Q(a) for all a, b ∈ S, where a⁺ satisfying the above conditions is called a 𝓠-inverse of a and V_Q(a) denotes the set of all 𝓠-inverses of a.

In [7], Li also gave an example of 𝓠-regular semigroup and showed that this class of semigroups properly contains the class of V-regular semigroups. At the same time, an equivalent characterization of 𝓠-regular semigroups was obtained.

A regular semigroup S is 𝓠-regular if and only if there exists a set Q ⊆ E(S) satisfying the following conditions:

Q ∩ L_a ≠ ∅ and Q ∩ R_a ≠ ∅ for any a ∈ S;
ω^l|_Q = ω|_Q ∘ 𝓛|_Q and ω^r|_Q = ω|_Q ∘ 𝓡|_Q;
For any a, b ∈ S, if (ab)⁺ ∈ V(ab) such that (ab)⁺(ab), (ab)(ab)⁺ ∈ Q, then there exists e₁ ∈ Q ∩ L_a and f₂ ∈ Q ∩ R_a such that b(ab)⁺a = f₂e₁.

In this paper, a 𝓠-regular semigroup S will be denoted by S(Q). A subset Q of E(S) satisfying (1)-(3) is called a charcateristic set (for simple a C-set) of S.

Some results on subdirect products of inverse semigroups were characterized by McAlister and Reilly [8]. In [9], Nambooripad and Veeramony discussed the subdirect products of regular semigroups. Mitsch [10] studied the subdirect products of E-inversive semigroups. In [11], Zheng characterized the subdirect products of 𝓟-regular semigroups. Throughout this paper, we investigate some properties of 𝓠-regular semigroups at first and characterize the 𝓠-subdirect products of this class of semigroups. Furthermore, we introduce the concept of the E-unitary 𝓠-regular covers for 𝓠-regular semigroups, in particular for those whose maximum group homomorphic image is a given group. Finally, we deduce the similar results on V-regular semigroups also hold up.

For notations and definitions given in this paper, the reader is referred to Howie [12].

2 On 𝓠-regular semigroups

As a dual form of 𝓟-regular semigroups, Li [7] introduced the concept of 𝓠-regular semigroups. In this section, we give some characterizations of 𝓠-regular semigroups. In particular, we show that the class of 𝓠-regular semigroups is closed under the homomorphic images.

Proposition 2.1

LetS(Q) be a 𝓠-regular semigroup. For anyp ∈ Q,

VQ(p)=(Q∩Lp)(Q∩Rp).

Proof

Since p ∈ Q ⊆ E(S), Q ∩ L_p ≠ ∅, Q ∩ R_p ≠ ∅. For any h ∈ V_Q(p), we have h = hph = (hp)(ph) ∈ (Q ∩ L_p)(Q ∩ R_p), and so V_Q(p) ⊆ (Q ∩ L_p)(Q ∩ R_p).

Conversely, let f ∈ Q ∩ L_p, g ∈ Q ∩ R_p. Then

p(fg)p=(pf)(gp)=p⋅p=p,(fg)p(fg)=f(gp)fg=fpfg=fg.

So fg ∈ V(p). Additionally,

(fg)p=f(gp)=fp=f∈Q,p(fg)=(pf)g=pg=g∈Q.

Thus fg ∈ V_Q(p), namely, (Q ∩ L_p)(Q ∩ R_p) ⊆ V_Q(p). Therefore, V_Q(p) = (Q ∩ L_p)(Q ∩ R_p). □

Proposition 2.2

([13]). LetS(Q) be a 𝓠-regular semigroup. Ifa ∈ S(Q), a⁺ ∈ V_Q(a), then

VQ(a)=VQ(a+a)a+VQ(aa+).

Theorem 2.3

LetS(Q) be a 𝓠-regular semigroup. Ife ∈ Q ∩ L_a, f ∈ Q ∩ R_a, for anya ∈ S, then there exists a uniquea⁺ ∈ V_Q(a) such thata⁺a = e, aa⁺ = f.

Proof

For any a ∈ S, let e ∈ Q ∩ L_a and f ∈ Q ∩ R_a. And so there is an inverse a′ of a in R_e ∩ L_f such that a′a = e ∈ Q, aa′ = f ∈ Q. Thus a′ ∈ V_Q(a). If there exists a a⁺ ∈ V_Q(a) such that a⁺a = e, aa⁺ = f, then

a′=a′aa′=a′f=a′aa+=ea+=a+aa+=a+.

□

Corollary 2.4

LetS(Q) be a 𝓠-regular semigroup. For anya, b ∈ S,

(a, b) ∈ 𝓛 if and only if there existsa⁺ ∈ V_Q(a), b⁺ ∈ V_Q(b) such thata⁺a = b⁺b;
(a, b) ∈ 𝓡 if and only if there existsa⁺ ∈ V_Q(a), b⁺ ∈ V_Q(b) such thataa⁺ = bb⁺;
(a, b) ∈ 𝓗 if and only if there existsa⁺ ∈ V_Q(a), b⁺ ∈ V_Q(b) such thataa⁺ = bb⁺, a⁺a = b⁺b.

Proof

(1) For any a⁺ ∈ V_Q(a), we have a⁺a ∈ Q ∩ L_a. If (a, b) ∈ 𝓛 and there exists a e ∈ Q ∩ R_b, then e 𝓡 b 𝓛 a 𝓛 a⁺a. By Theorem 2.3, there exists a unique b⁺ ∈ V_Q(b) such that b⁺b = a⁺a.

On the other hand, if there exist a⁺ ∈ V_Q(a), b⁺ ∈ V_Q(b) such that b⁺b = a⁺a, then a 𝓛 a⁺a = b⁺b 𝓛 b, that is a 𝓛 b.

By using similar arguments as the above, we can proof (2) and (3). □

Corollary 2.5

LetS(Q) be a 𝓠-regular semigroup. Ife, f ∈ Q, then (e, f) ∈ 𝓓 if and only if there exista ∈ S, a⁺ ∈ V_Q(a) such thata⁺a = f, aa⁺ = e.

Proof

If (e, f) ∈ 𝓓, there exists a ∈ R_e ∩ L_f. By Theorem 2.3, there exists a⁺ ∈ V_Q(a) such that a⁺a = f, aa⁺ = e.

Conversely, if there exist a ∈ S, a⁺ ∈ V_Q(a) such that a⁺a = f, aa⁺ = e, then e 𝓡 a, a 𝓛 f. Hence e 𝓓 f. □

Proposition 2.6

LetS(Q) be 𝓠-regular semigroup. Ifq 𝓡 p (q 𝓛 p) for any p, q ∈ Q, then q ∈ V_Q(p).

Proof

Let q ∈ R_p. Since p, q ∈ Q ⊆ E(S), so pq = q, qp = p. Hence pqp = qp = p, qpq = pq = q, namely q ∈ V(p). For another, pq = q ∈ Q, qp = p ∈ Q. Hence q ∈ V_Q(p). In the case of p 𝓛 q, the proof is similar. □

Corollary 2.7

LetS(Q) be a 𝓠-regular semigroup. Then the following statements are equivalent:

for anyq, p ∈ Q, ifV_Q(q) ∩ V_Q(p) ≠ ∅, thenV_Q(q) = V_Q(p);
for anye, f ∈ E(S), ifV_Q(e) ∩ V_Q(f) ≠ ∅, thenV_Q(e) = V_Q(f);
for anya, b ∈ S(Q), ifV_Q(a) ∩ V_Q(b) ≠ ∅, thenV_Q(a) = V_Q(b).

Proof

We only need to proof (1) ⇒ (3). Let c ∈ V_Q(a) ∩ V_Q(b). By Proposition 2.2,

VQ(a)=VQ(ca)cVQ(ac),VQ(b)=VQ(cb)cVQ(bc).

Since ca, ac, cb, bc ∈ Q and ca 𝓡 c 𝓡 cb, ac 𝓛 c 𝓛 bc, by Proposition 2.6,

ca∈VQ(ca)∩VQ(cb),ac∈VQ(ac)∩VQ(bc).

By (1),

VQ(ca)=VQ(cb),VQ(ac)=VQ(bc).

Thus, V_Q(a) = V_Q(b). □

Proposition 2.8

LetS(Q) be a 𝓠-regular semigroup andρan idempotent-separating congruence onS. For anya, b ∈ S, ifaρb, thena⁺ρb⁺for somea⁺ ∈ V_Q(a), b⁺ ∈ V_Q(b).

Proof

Since ρ is an idempotent-separating congruence on S, aρb implies that a 𝓗 b. By Corollary 2.4, there exist a⁺ ∈ V_Q(a), b⁺ ∈ V_Q(b) such that aa⁺ = bb⁺, a⁺a = b⁺b. Thus

a+=a+aa+=a+bb+,b+=b+bb+=a+ab+.

Since ρ is a congruence, a⁺ab⁺ρa⁺bb⁺, that is b⁺ρa⁺. □

Theorem 2.9

LetS(Q) be a 𝓠-regular semigroup andTa regular semigroup. Ifψ : S(Q) → Tis a semigroup homomorphism andQ̄ = {qψ : q ∈ Q}, thenT(Q̄) is 𝓠-regular.

Proof

For any a ∈ S, there exists a a⁺ ∈ V_Q(a) such that aa⁺, a⁺a ∈ Q. And so

(aψ)(a+ψ)(aψ)=(aa+a)ψ=aψ,(a+ψ)(aψ)(a+ψ)=(a+aa+)ψ=a+ψ.

That is a⁺ψ ∈ V(aψ). Since (aψ)(a⁺ψ) = (aa⁺)ψ ∈ Q̄, a⁺ψ ∈ V_Q̄(aψ) ≠ ∅. It is easy to see that

(aψ)(a+ψ)∈VQ((aψ)(a+ψ))and(a+ψ)(aψ)∈VQ((a+ψ)(aψ)).

On the other hand, for any (ab)⁺ ∈ V_Q(ab), since S is 𝓠-regular, there exist a⁺ ∈ V_Q(a), b⁺ ∈ V_Q(b) such that

(ab)+ψ=(b+a+)ψ=(b+ψ)(a+ψ).

Thus V_Q̄((ab)ψ) ⊆ V_Q̄(bψ)V_Q̄(aψ) and T(Q̄) is 𝓠-regular. □

3 𝓠-subdirect products of 𝓠-regular semigroups

Some results on subdirect products of regular semigroups and E-inversive semigroups were characterized by Nambooripad [9] and Mitsch [10]. In this section, we show that the class of 𝓠-regular semigroups is closed under the direct product and characterize the 𝓠-subdirect products of 𝓠-regular semigroups.

For arbitrary semigroup S, Petrich [14] introduced the concept of filters of S, that is, a subsemigroup F of S such that ab ∈ F implies a ∈ F and b ∈ F.

Example 3.1

LetS = {a, b, c, d, e} be the semigroup with operation defined by

⋅	a	b	c	d	e
a	a	b	c	c	c
b	a	b	c	c	c
c	a	b	c	c	c
d	a	b	c	e	d
e	a	b	c	d	e

ThenF = {d, e} is a filter ofS.

In what follows, we introduce the concept of 𝓠-inverse filters of a 𝓠-regular semigroup.

Definition 3.2

A regular subsemigroupTof 𝓠-regular semigroupS(Q) is called a 𝓠-inverse filter, if

T(Q ∩ E_T) is a 𝓠-regular semigroup;
For anyt₁, t₂ ∈ Tandt1+∈VQS(t1),t2+∈VQS(t2),ift1+t2+∈T,thent1+,t2+ ∈ T.

Definition 3.3

LetS₁(Q₁) andS₂(Q₂) be two 𝓠-regular semigroups. A homomorphismfofS₁(Q₁) intoS₂(Q₂) is called a 𝓠-homomorphism ifQ₁f = Q₂ ∩ S₁(Q₁)f. A 𝓠-homomorphismf : S₁(Q₁) → S₂(Q₂) is called a 𝓠-isomorphism iffis bijective, in such the case, S₁(Q₁) is called 𝓠-isomorphic toS₂(Q₂), and denoted byS₁(Q₁) ≅QS₂(Q₂).

Proposition 3.4

LetS₁(Q₁) andS₂(Q₂) be two 𝓠-regular semigroups. IfS(Q) = S₁(Q₁) × S₂(Q₂), whereQ = {(p₁, p₂): p₁ ∈ Q₁, p₂ ∈ Q₂}, thenS(Q) is a 𝓠-regular semigroup.

Proof

Obviously, S(Q) is a semigroup. If (s, t) ∈ S, where s ∈ S₁, t ∈ S₂, since S₁, S₂ are all 𝓠-regular, there exists s′ ∈ V_S₁(s), t′ ∈ V_S₂(t) such that (s′, t′) ∈ S. It is easy to see that (s′, t′) ∈ V(s, t), and so S is regular. Let

VQ1(s)×VQ2(t)={(s+,t+):s+∈VQ1(s),t+∈VQ2(t)}.

Clearly, it is non-empty and contained in V(s, t). For any (s⁺, t⁺) ∈ V_Q₁(s) × V_Q₂(t), we have

(s,t)(s+,t+)∈VQ1(ss+)×VQ2(tt+),(s+,t+)(s,t)∈VQ1(s+s)×VQ2(t+t).

For another, let (s, t), (x, y) ∈ S. For any ((sx)⁺, (ty)⁺) ∈ V_Q₁(sx) × V_Q₂(ty), since S₁, S₂ are all 𝓠-regular, there exists s⁺ ∈ V_Q₁(s), x⁺ ∈ V_Q₁(x), t⁺ ∈ V_Q₂(t), y⁺ ∈ V_Q₂(y) such that

((sx)+,(ty)+)=(x+s+,y+t+)=(x+,y+)(s+,t+)∈(VQ1(x)×VQ2(y))(VQ1(s)×VQ2(t)).

Hence, S(Q) is a 𝓠-regular semigroup. □

Definition 3.5

LetS₁(Q₁), S₂(Q₂) be 𝓠-regular semigroups andS(Q) be the direct product of them, whereQ = {(p₁, p₂): p₁ ∈ Q₁, p₂ ∈ Q₂}. IfT(Q′) is a 𝓠-inverse filter ofS(Q) and the projections

f1:T(Q′)→S1(Q1),(s1,s2)↦s1,f2:T(Q′)→S2(Q2),(s1,s2)↦s2

are all surjective 𝓠-homomorphisms, thenT(Q′) is called a 𝓠-subdirect product ofS₁(Q₁) andS₂(Q₂).

Definition 3.6

LetS(Q₁) andT(Q₂) be two 𝓠-regular semigroups. A mappingφ : S → 2^T(the power set of T) is called a surjective 𝓠-subhomomorphism ofS(Q₁) ontoT(Q₂), if it satisfies the following

for anys ∈ S, sφ ≠ ∅;
for anys₁, s₂ ∈ S, (s₁φ)(s₂φ) ⊆ (s₁s₂)φ;
⋃_s∈Ssφ = T;
for anyt ∈ sφ (s ∈ S, t ∈ T), there exists⁺ ∈ V_Q₁(s), t⁺ ∈ V_Q₂(t) such thatt⁺ ∈ s⁺φ;
for anyp₁ ∈ Q₁, there existsp₂ ∈ Q₂such thatp₂ ∈ p₁φ; and for anyp₂ ∈ Q₂, there existsp₁ ∈ Q₁such thatp₂ ∈ p₁φ;
for anyt₁ ∈ s₁φ, t₂ ∈ s₂φ, ift1+t2+∈(s1+s2+)φ,thent1+∈s1+φ,t2+∈s2+φ,whereti+∈VQ2(ti),si+ ∈ V_Q₁(s_i), i = 1, 2.

Theorem 3.7

LetS(Q₁) andT(Q₂) be 𝓠-regular semigroups, φa surjective 𝓠-subhomomorphism ofS(Q₁) ontoT(Q₂). If

π=π(S,T,φ)={(s,t)∈S×T:t∈sφ},Q={(p1,p2)∈Q1×Q2:p2∈p1φ},

thenπ(Q) is a 𝓠-subdirect product ofS(Q₁) andT(Q₂).

Conversely, every 𝓠-subdirect product ofS(Q₁) andT(Q₂) can be constructed in this way.

Proof

If (s₁, t₁), (s₂, t₂) ∈ π, then t₁ ∈ s₁φ, t₂ ∈ s₂φ. By (2), t₁t₂ ∈ (s₁s₂)φ, and so (s₁s₂, t₁t₂) ∈ π. Hence π is a subsemigroup of S × T. If (s, t) ∈ π, then t ∈ sφ. By (4), there exist s⁺ ∈ V_Q₁(s) and t⁺ ∈ V_Q₂(t) such that t⁺ ∈ s⁺φ, and so (s⁺, t⁺) ∈ π. It is easy to see that (s⁺, t⁺) ∈ V(s, t). Thus π is a regular semigroup. Let

VQ(s,t)={(s+,t+)∈VQ1(s)×VQ2(t):t+∈s+φ}.

If (s, t) ∈ π, then t ∈ sφ. By (4), V_Q(s, t) ≠ ∅. For any (s, t) ∈ π, (s⁺, t⁺) ∈ V_Q(s, t),

(s,t)(s+,t+)=(ss+,tt+),

where s⁺ ∈ V_Q₁(s), t⁺ ∈ V_Q₂(t), t⁺ ∈ s⁺φ. Since S(Q₁), T(Q₂) are 𝓠-regular, it follows that ss⁺ ∈ V_Q₁(ss⁺), tt⁺ ∈ V_Q₂(tt⁺). And tt⁺ ∈ (ss⁺)φ, hence

(s,t)(s+,t+)∈VQ(ss+,tt+)=VQ((s,t)(s+,t+)).

Similarly, (s⁺, t⁺)(s, t) ∈ V_Q((s⁺, t⁺)(s, t)).

On the other hand, let (s₁, t₁), (s₂, t₂) ∈ π. For any

((s1s2)+,(t1t2)+)∈VQ(s1s2,t1t2),

since S(Q₁) and T(Q₂) are all 𝓠-regular semigroups, there exist s1+ ∈ V_Q₁(s₁), s2+ ∈ V_Q₁(s₂), t1+ ∈ V_Q₂(t₁), t2+ ∈ V_Q₂(t₂) such that

((s1s2)+,(t1t2)+)=(s2+s1+,t2+t1+)=(s2+,t2+)(s1+,t1+).

Since t2+t1+∈(s2+s1+)φ, by (6), t2+∈s2+φ,t1+∈s1+φ. And so

(s2+,t2+)∈VQ(s2,t2),(s1+,t1+)∈VQ(s1,t1).

Hence,

VQ(s1s2,t1t2)⊆VQ(s2,t2)VQ(s1,t1).

that is, π(Q) is a 𝓠-regular semigroup.

For any

(s1+,t1+)∈VQ1(s1)×VQ2(t1),(s2+,t2+)∈VQ1(s2)×VQ2(t2).

If (s1+,t1+)(s2+,t2+) ∈ π, namely (s1+s2+,t1+t2+) ∈ π, then t1+t2+∈(s1+s2+)φ. By (6), t1+∈s1+φ,t2+∈s2+φ, and so (s1+,t1+),(s2+,t2+) ∈ π. Hence π(Q) is a 𝓠-inverse filter of S(Q₁) × T(Q₂).

By (1) and (5), the projection f₁: π → S, (s, t) ↦ s is a surjective 𝓠-homomorphism. By (3) and (5), the projection f₂: π → T, (s, t) ↦ t is a surjection 𝓠-homomorphism. Therefore, π(Q) is a 𝓠-subdirect product of S(Q₁) and T(Q₂).

Conversely, let H(Q) be a 𝓠-subdirect product of S(Q₁) and T(Q₂). Let

φ:S→2T,s↦{t∈T:(s,t)∈H}.

Since H(Q) is a 𝓠-subdirect product of S(Q₁) and T(Q₂), there exists t ∈ T such that (s, t) ∈ H for any s ∈ S. Thus t ∈ sφ. Hence (1) holds. Since H is a subsemigroup of S(Q₁) × T(Q₂), (2) holds. Since the projection f₂: H → T is surjective, (3) holds. If t ∈ sφ (s ∈ S, t ∈ T), then (s, t) ∈ H. Since H(Q) is 𝓠-regular, there exists

(s+,t+)∈(VQ1(s)×VQ2(t))∩H.

Hence s⁺ ∈ V_Q₁(s), t⁺ ∈ V_Q₂(t), and t⁺ ∈ s⁺φ, so (4) holds. For any p₁ ∈ Q₁, there exists (p₁, p₂) ∈ Q = H ∩ (Q₁ × Q₂), and so p₂ ∈ Q₂ and p₂ ∈ p₁φ. Additionally, since f₂ is a surjective 𝓠-homomorphism, there exists

(p1,p2)∈Q=H∩(Q1×Q2)

for any p₂ ∈ Q₂. Hence p₁ ∈ Q₁ and p₂ ∈ p₁φ. Thus (5) holds. Since H(Q) is a 𝓠-inverse filter of S(Q₁) × T(Q₂), (6) holds. By the proof above, φ is a surjective 𝓠-subhomomorphism of S(Q₁) onto T(Q₂). Obviously, H = π(S, T, φ). □

4 E-unitary covers for 𝓠-regular semigroups

McAlister and Reilly [8] have given E-unitary covers of inverse semigroups by using surjective subhomomorphisms of inverse semigroups. Mitsch [10] has given some sufficient conditions on E-unitary, E-inverse covers of E-inversive semigroups.

In this section, we introduce the concept of the E-unitary 𝓠-regular covers of 𝓠-regular semigroups, in particular for those whose maximum group homomorphic image is a given group.

Let S be a regular semigroup. A subset H of S is said to be

full if E(S) ⊆ H;
self-conjugate if aHa′ ⊆ H and a′Ha ⊆ H for all a ∈ S and all a′ ∈ V(a).

Let U be the minimum full and self-conjugate subsemigroup of S.

Lemma 4.1

([15]). LetSbe a regular semiroup. The minimum group congruenceσonSis given by

σ={(a,b)∈S×S:(∃x,y∈U)xa=by}.

In [12], a subset A of a semigroup S is called right unitary, if

(∀a∈A)(∀s∈S)sa∈A⇒s∈A.

A is called left unitary, if

(∀a∈A)(∀s∈S)as∈A⇒s∈A.

A right and left unitary subset is called unitary.

A regular semigroup S is called E-unitary, if the set E(S) of idempotents of S is unitary.

Definition 4.2

LetT(Q₁) andS(Q) be two 𝓠-regular semigroups. T(Q₁) is called anE-unitary 𝓠-regular cover ofS(Q), ifT(Q₁) isE-unitary, and there exists an idempotent separating 𝓠-homomorphism ofT(Q₁) ontoS(Q).

A group G is a 𝓠-regular semigroup whose C-set is {1}, where 1 is the identity element of G.

Definition 4.3

LetS(Q) be a 𝓠-regular semigroup andGbe a group. A surjective 𝓠-subhomomorphismφofS(Q) ontoGis called unitary, if

(∀s∈S)1∈sφ⇒s∈E(S),

where 1 is the identity element ofG.

Theorem 4.4

LetS(Q) be a 𝓠-regular semigroup andGbe a group. If there exists a unitary surjetive 𝓠-subhomomorphismφofS(Q) ontoG, thenπ (S, G, φ) is anE-unitary 𝓠-regular cover ofS(Q).

Proof

By Theorem 3.7, T(Q₁) = π(S, G, φ) is 𝓠-regular whose C-set is

Q1={(p,1)∈S×G:p∈Q,1∈pφ}.

For any (s, g) ∈ T, (e, 1) ∈ E_T, if

(s,g)(e,1)∈ET⊆{(t,1)∈S×G:t∈ES},

then (se, g) ∈ E_T, and so g = 1, and (s, 1) ∈ T(Q₁). By the definition of T(Q₁), 1 ∈ sφ. Since φ is unitary, s ∈ E_S. Hence (s, g) ∈ E_T, so T(Q₁) is E-unitary.

Define f : T(Q₁) → S(Q), (s, g) ↦ s. Then f is a surjective 𝓠-homomorphism. Obviously, f is idempotent separating. Hence π(S, G, φ) is an E-unitary 𝓠-regular cover of S(Q). □

Definition 4.5

A 𝓠-regular semigroupT(Q₁) is called anE-unitary 𝓠-regular cover ofS(Q) throughG, if

T(Q₁) is anE-unitary 𝓠-regular cover ofS(Q);
T/σ≅QG, whereσis the minimum group congruence onT(Q₁).

Lemma 4.6

LetS(Q₁), T(Q₂) be two 𝓠-regular semigroups andf : S(Q₁) → T(Q₂) be a 𝓠-homomorphism. ThenS(Q₁)/kerf≅QS(Q₁)f.

Proof

It is easy to see that the mapping ϕ: S(Q₁)/kerf → S(Q₁)f is an isomorphism. Since f is a 𝓠-homomorphism, we have

(Q1kerf)ϕ=Q1f=Q2∩S(Q1)f=Q2∩(S(Q1)kerf)ϕ.

Thus ϕ is also a 𝓠-homomorphism and S(Q₁)/kerf≅QS(Q₁)f. □

Theorem 4.7

If there exists a unitary surjective 𝓠-subhomomorphismφofS(Q) ontoG, thenT(Q₁) = π(S, G, φ) is anE-unitary 𝓠-regular cover ofS(Q) throughG.

Proof

By Theorem 4.4, T(Q₁) = π(S, G, φ) is an E-unitary 𝓠-regular cover of S(Q). We prove that T/σ≅QG as follow.

It follows from Theorem 3.7 that T(Q₁) is a 𝓠-subdirect product of S(Q) and G. Hence G is the 𝓠-homomorphic image of T(Q₁) under the projection f₂ : T(Q₁) → G, (a, g) ↦ g. Since G is a group, kerf₂ is a group congruence on T(Q₁), and so σ ⊆ kerf₂. Conversely, if (a, g) ∈ (b, g)kerf₂, then (a, g)f₂ = (b, h)f₂, so that g = h. Since T(Q₁) is 𝓠-regular, for any (b, g) = (b, h) ∈ T(Q₁) there exists (x, y) ∈ V_Q₁(b, g) such that

(b,g)(x,y)=(bx,gy)∈ET⊆{(e,1)∈T:e∈ES}.

Hence bx ∈ E_S and gy = 1, that is g⁻¹ = y. So (x, g⁻¹) ∈ T(Q₁) and (bx, 1) ∈ Q₁. Now (xa, 1) = (x, y)(a, g) ∈ T. Hence, by the definition of T, 1 ∈ (xa)φ. Notice that φ is unitary, we have xa ∈ E_S. Thus (xa, 1) ∈ E_T, and so

(bx,1)(a,g)=(bxa,g)=(b,h)(xa,1).

Since (bx, 1), (xa, 1) ∈ E_T ⊆ U, by Lemma 4.1, (a, g)σ (b, h). Thus kerf₂ ⊆ σ. Hence σ = kerf₂. Therefore, it follows from lemma 4.6 that T/σ = T/kerf₂≅QG. □

Remark 4.8

As we know, the class of 𝓠-regular semigroups properly contains the class ofV-regular semigroups. In fact, if we restrict theC-setQof a 𝓠-regular semigroupSto the set of idempotents, then the subdirect products ofV-regular semigroups can be constructed in a similar way and we can also use this contruction to study theE-unitary cover forV-regular semigroups.

Acknowledgement

Research Supported by Fundamental Research Funds for the Central Universities.

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Received: 2016-11-01

Accepted: 2018-03-28

Published Online: 2018-05-30

This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 License.

Articles in the same Issue

https://doi.org/10.1515/math-2018-0048

Keywords for this article

𝓠-regular semigroup; 𝓠-subdirect product; Surjective 𝓠-subhomomorphism; E-unitary 𝓠-regular cover

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