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# Open Mathematics

### formerly Central European Journal of Mathematics

Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo

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Volume 16, Issue 1

# On 𝓠-regular semigroups

Xinyang Feng
Published Online: 2018-05-30 | DOI: https://doi.org/10.1515/math-2018-0048

## 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.

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 aS. 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, bS,

$V(b)V(a)⊆V(ab).$

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, bS,

$V(ab)⊆V(b)V(a).$

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, bS,

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

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 aS there exists a non-empty set VQV(a) satisfying the following conditions:

1. aa+VQ(aa+) and a+aVQ(a+a) for any aS and a+VQ(a),

2. VQ(ab) ⊆ VQ(b)VQ(a) for all a, bS, where a+ satisfying the above conditions is called a 𝓠-inverse of a and VQ(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 QE(S) satisfying the following conditions:

1. QLa ≠ ∅ and QRa ≠ ∅ for any aS;

2. ωl|Q = ω|Q ∘ 𝓛|Q and ωr|Q = ω|Q ∘ 𝓡|Q;

3. For any a, bS, if (ab)+V(ab) such that (ab)+(ab), (ab)(ab)+Q, then there exists e1QLa and f2QRa 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

Let S(Q) be a 𝓠-regular semigroup. For any pQ,

$VQ(p)=(Q∩Lp)(Q∩Rp).$

#### Proof

Since pQE(S), QLp ≠ ∅, QRp ≠ ∅. For any hVQ(p), we have h = hph = (hp)(ph) ∈ (QLp)(QRp), and so VQ(p) ⊆ (QLp)(QRp).

Conversely, let fQLp, gQRp. Then

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

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

Thus fgVQ(p), namely, (QLp)(QRp) ⊆ VQ(p). Therefore, VQ(p) = (QLp)(QRp). □

#### Proposition 2.2

([13]). Let S(Q) be a 𝓠-regular semigroup. If aS(Q), a+VQ(a), then

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

#### Theorem 2.3

Let S(Q) be a 𝓠-regular semigroup. If eQLa, fQRa, for any aS, then there exists a unique a+VQ(a) such that a+a = e, aa+ = f.

#### Proof

For any aS, let eQLa and fQRa. And so there is an inverse a′ of a in ReLf such that aa = eQ, aa′ = fQ. Thus a′ ∈ VQ(a). If there exists a a+VQ(a) such that a+a = e, aa+ = f, then

$a′=a′aa′=a′f=a′aa+=ea+=a+aa+=a+.$ □

#### Corollary 2.4

Let S(Q) be a 𝓠-regular semigroup. For any a, bS,

1. (a, b) ∈ 𝓛 if and only if there exists a+VQ(a), b+VQ(b) such that a+a = b+b;

2. (a, b) ∈ 𝓡 if and only if there exists a+VQ(a), b+VQ(b) such that aa+ = bb+;

3. (a, b) ∈ 𝓗 if and only if there exists a+VQ(a), b+VQ(b) such that aa+ = bb+, a+a = b+b.

#### Proof

(1) For any a+VQ(a), we have a+aQLa. If (a, b) ∈ 𝓛 and there exists a eQRb, then e 𝓡 b 𝓛 a 𝓛 a+a. By Theorem 2.3, there exists a unique b+VQ(b) such that b+b = a+a.

On the other hand, if there exist a+VQ(a), b+VQ(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

Let S(Q) be a 𝓠-regular semigroup. If e, fQ, then (e, f) ∈ 𝓓 if and only if there exist aS, a+VQ(a) such that a+a = f, aa+ = e.

#### Proof

If (e, f) ∈ 𝓓, there exists aReLf. By Theorem 2.3, there exists a+VQ(a) such that a+a = f, aa+ = e.

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

#### Proposition 2.6

Let S(Q) be 𝓠-regular semigroup. If q 𝓡 p (q 𝓛 p) for any p, qQ, then qVQ(p).

#### Proof

Let qRp. Since p, qQE(S), so pq = q, qp = p. Hence pqp = qp = p, qpq = pq = q, namely qV(p). For another, pq = qQ, qp = pQ. Hence qVQ(p). In the case of p 𝓛 q, the proof is similar. □

#### Corollary 2.7

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

1. for any q, pQ, if VQ(q) ∩ VQ(p) ≠ ∅, then VQ(q) = VQ(p);

2. for any e, fE(S), if VQ(e) ∩ VQ(f) ≠ ∅, then VQ(e) = VQ(f);

3. for any a, bS(Q), if VQ(a) ∩ VQ(b) ≠ ∅, then VQ(a) = VQ(b).

#### Proof

We only need to proof (1) ⇒ (3). Let cVQ(a) ∩ VQ(b). By Proposition 2.2,

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

Since ca, ac, cb, bcQ 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, VQ(a) = VQ(b). □

#### Proposition 2.8

Let S(Q) be a 𝓠-regular semigroup and ρ an idempotent-separating congruence on S. For any a, bS, if a ρ b, then a+ ρ b+ for some a+VQ(a), b+VQ(b).

#### Proof

Since ρ is an idempotent-separating congruence on S, a ρ b implies that a 𝓗 b. By Corollary 2.4, there exist a+VQ(a), b+VQ(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+ ρ a+bb+, that is b+ ρ a+. □

#### Theorem 2.9

Let S(Q) be a 𝓠-regular semigroup and T a regular semigroup. If ψ : S(Q) → T is a semigroup homomorphism and = { : qQ}, then T() is 𝓠-regular.

#### Proof

For any aS, there exists a a+VQ(a) such that aa+, a+aQ. And so

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

That is a+ψV(). Since (aψ)(a+ψ) = (aa+)ψ, a+ψV() ≠ ∅. 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)+VQ(ab), since S is 𝓠-regular, there exist a+VQ(a), b+VQ(b) such that

$(ab)+ψ=(b+a+)ψ=(b+ψ)(a+ψ).$

Thus V((ab)ψ) ⊆ V()V() and T() 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 abF implies aF and bF.

#### Example 3.1

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

Then F = {d, e} is a filter of S.

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

#### Definition 3.2

A regular subsemigroup T of 𝓠-regular semigroup S(Q) is called a 𝓠-inverse filter, if

1. T(QET) is a 𝓠-regular semigroup;

2. For any t1, t2T and $\begin{array}{}{t}_{1}^{+}\in {V}_{Q}^{S}\left({t}_{1}\right),{t}_{2}^{+}\in {V}_{Q}^{S}\left({t}_{2}\right),\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}if\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{t}_{1}^{+}{t}_{2}^{+}\in T,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}then\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{t}_{1}^{+},\phantom{\rule{thinmathspace}{0ex}}{t}_{2}^{+}\end{array}$T.

#### Definition 3.3

Let S1(Q1) and S2(Q2) be two 𝓠-regular semigroups. A homomorphism f of S1(Q1) into S2(Q2) is called a 𝓠-homomorphism if Q1f = Q2S1(Q1)f. A 𝓠-homomorphism f : S1(Q1) → S2(Q2) is called a 𝓠-isomorphism if f is bijective, in such the case, S1(Q1) is called 𝓠-isomorphic to S2(Q2), and denoted by S1(Q1) $\begin{array}{}\stackrel{\mathcal{Q}}{\cong }\end{array}$ S2(Q2).

#### Proposition 3.4

Let S1(Q1) and S2(Q2) be two 𝓠-regular semigroups. If S(Q) = S1(Q1) × S2(Q2), where Q = {(p1, p2): p1Q1, p2Q2}, then S(Q) is a 𝓠-regular semigroup.

#### Proof

Obviously, S(Q) is a semigroup. If (s, t) ∈ S, where sS1, tS2, since S1, S2 are all 𝓠-regular, there exists s′ ∈ VS1(s), t′ ∈ VS2(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+) ∈ VQ1(s) × VQ2(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)+) ∈ VQ1(sx) × VQ2(ty), since S1, S2 are all 𝓠-regular, there exists s+VQ1(s), x+VQ1(x), t+VQ2(t), y+VQ2(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

Let S1(Q1), S2(Q2) be 𝓠-regular semigroups and S(Q) be the direct product of them, where Q = {(p1, p2): p1Q1, p2Q2}. If T(Q′) is a 𝓠-inverse filter of S(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, then T(Q′) is called a 𝓠-subdirect product of S1(Q1) and S2(Q2).

#### Definition 3.6

Let S(Q1) and T(Q2) be two 𝓠-regular semigroups. A mapping φ : S → 2T (the power set of T) is called a surjective 𝓠-subhomomorphism of S(Q1) onto T(Q2), if it satisfies the following

1. for any sS, sφ ≠ ∅;

2. for any s1, s2S, (s1φ)(s2φ) ⊆ (s1s2)φ;

3. sS sφ = T;

4. for any tsφ (sS, tT), there exist s+VQ1(s), t+VQ2(t) such that t+s+φ;

5. for any p1Q1, there exists p2Q2 such that p2p1φ; and for any p2Q2, there exists p1Q1 such that p2p1φ;

6. for any t1s1φ, t2s2φ, if $\begin{array}{}{t}_{1}^{+}{t}_{2}^{+}\in \left({s}_{1}^{+}{s}_{2}^{+}\right)\phi ,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}then\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{t}_{1}^{+}\in {s}_{1}^{+}\phi ,{t}_{2}^{+}\in {s}_{2}^{+}\phi ,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}where\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{t}_{i}^{+}\in {V}_{{Q}_{2}}\left({t}_{i}\right),{s}_{i}^{+}\end{array}$VQ1(si), i = 1, 2.

#### Theorem 3.7

Let S(Q1) and T(Q2) be 𝓠-regular semigroups, φ a surjective 𝓠-subhomomorphism of S(Q1) onto T(Q2). If

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

then π(Q) is a 𝓠-subdirect product of S(Q1) and T(Q2).

Conversely, every 𝓠-subdirect product of S(Q1) and T(Q2) can be constructed in this way.

#### Proof

If (s1, t1), (s2, t2) ∈ π, then t1s1φ, t2s2φ. By (2), t1t2 ∈ (s1s2)φ, and so (s1s2, t1t2) ∈ π. Hence π is a subsemigroup of S × T. If (s, t) ∈ π, then tsφ. By (4), there exist s+VQ1(s) and t+VQ2(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 tsφ. By (4), VQ(s, t) ≠ ∅. For any (s, t) ∈ π, (s+, t+) ∈ VQ(s, t),

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

where s+VQ1(s), t+VQ2(t), t+s+φ. Since S(Q1), T(Q2) are 𝓠-regular, it follows that ss+VQ1(ss+), tt+VQ2(tt+). And tt+ ∈ (ss+)φ, hence

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

Similarly, (s+, t+)(s, t) ∈ VQ((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 $\begin{array}{}{s}_{1}^{+}\end{array}$VQ1(s1), $\begin{array}{}{s}_{2}^{+}\end{array}$VQ1(s2), $\begin{array}{}{t}_{1}^{+}\end{array}$VQ2(t1), $\begin{array}{}{t}_{2}^{+}\end{array}$VQ2(t2) such that

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

Since $\begin{array}{}{t}_{2}^{+}{t}_{1}^{+}\in \left({s}_{2}^{+}{s}_{1}^{+}\right)\phi ,\end{array}$ by (6), $\begin{array}{}{t}_{2}^{+}\in {s}_{2}^{+}\phi ,\phantom{\rule{thinmathspace}{0ex}}{t}_{1}^{+}\in {s}_{1}^{+}\phi .\end{array}$ 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 $\begin{array}{}\left({s}_{1}^{+},{t}_{1}^{+}\right)\left({s}_{2}^{+},{t}_{2}^{+}\right)\end{array}$π, namely $\begin{array}{}\left({s}_{1}^{+}{s}_{2}^{+},{t}_{1}^{+}{t}_{2}^{+}\right)\end{array}$π, then $\begin{array}{}{t}_{1}^{+}{t}_{2}^{+}\in \left({s}_{1}^{+}{s}_{2}^{+}\right)\phi .\end{array}$ By (6), $\begin{array}{}{t}_{1}^{+}\in {s}_{1}^{+}\phi ,\phantom{\rule{thinmathspace}{0ex}}{t}_{2}^{+}\in {s}_{2}^{+}\phi ,\end{array}$ and so $\begin{array}{}\left({s}_{1}^{+},{t}_{1}^{+}\right),\phantom{\rule{thinmathspace}{0ex}}\left({s}_{2}^{+},{t}_{2}^{+}\right)\end{array}$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 tT such that (s, t) ∈ H for any sS. Thus tsφ. Hence (1) holds. Since H is a subsemigroup of S(Q1) × T(Q2), (2) holds. Since the projection f2: HT is surjective, (3) holds. If tsφ (sS, tT), then (s, t) ∈ H. Since H(Q) is 𝓠-regular, there exists

$(s+,t+)∈(VQ1(s)×VQ2(t))∩H.$

Hence s+VQ1(s), t+VQ2(t), and t+s+φ, so (4) holds. For any p1Q1, there exists (p1, p2) ∈ Q = H ∩ (Q1 × Q2), and so p2Q2 and p2p1φ. Additionally, since f2 is a surjective 𝓠-homomorphism, there exists

$(p1,p2)∈Q=H∩(Q1×Q2)$

for any p2Q2. Hence p1Q1 and p2p1φ. 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

1. full if E(S) ⊆ H;

2. self-conjugate if aHa′ ⊆ H and aHaH for all aS and all a′ ∈ V(a).

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

#### Lemma 4.1

([15]). Let S be a regular semiroup. The minimum group congruence σ on S is 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

Let T(Q1) and S(Q) be two 𝓠-regular semigroups. T(Q1) is called an E-unitary 𝓠-regular cover of S(Q), if T(Q1) is E-unitary, and there exists an idempotent separating 𝓠-homomorphism of T(Q1) onto S(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

Let S(Q) be a 𝓠-regular semigroup and G be a group. A surjective 𝓠-subhomomorphism φ of S(Q) onto G is called unitary, if

$(∀s∈S)1∈sφ⇒s∈E(S),$

where 1 is the identity element of G.

#### Theorem 4.4

Let S(Q) be a 𝓠-regular semigroup and G be a group. If there exists a unitary surjetive 𝓠-subhomomorphism φ of S(Q) onto G, then π (S, G, φ) is an E-unitary 𝓠-regular cover of S(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) ∈ ET, if

$(s,g)(e,1)∈ET⊆{(t,1)∈S×G:t∈ES},$

then (se, g) ∈ ET, and so g = 1, and (s, 1) ∈ T(Q1). By the definition of T(Q1), 1 ∈ sφ. Since φ is unitary, sES. Hence (s, g) ∈ ET, 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 semigroup T(Q1) is called an E-unitary 𝓠-regular cover of S(Q) through G, if

1. T(Q1) is an E-unitary 𝓠-regular cover of S(Q);

2. T/σ $\begin{array}{}\stackrel{\mathcal{Q}}{\cong }\end{array}$ G, where σ is the minimum group congruence on T(Q1).

#### Lemma 4.6

Let S(Q1), T(Q2) be two 𝓠-regular semigroups and f : S(Q1) → T(Q2) be a 𝓠-homomorphism. Then S(Q1)/kerf $\begin{array}{}\stackrel{\mathcal{Q}}{\cong }\end{array}$ S(Q1)f.

#### Proof

It is easy to see that the mapping ϕ: S(Q1)/kerfS(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 $\begin{array}{}\stackrel{\mathcal{Q}}{\cong }\end{array}$ S(Q1)f. □

#### Theorem 4.7

If there exists a unitary surjective 𝓠-subhomomorphism φ of S(Q) onto G, then T(Q1) = π(S, G, φ) is an E-unitary 𝓠-regular cover of S(Q) through G.

#### 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 $\begin{array}{}\stackrel{\mathcal{Q}}{\cong }\end{array}$ G 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, ker f2 is a group congruence on T(Q1), and so σkerf2. 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) ∈ VQ1(b, g) such that

$(b,g)(x,y)=(bx,gy)∈ET⊆{(e,1)∈T:e∈ES}.$

Hence bxES 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 xaES. Thus (xa, 1) ∈ ET, and so

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

Since (bx, 1), (xa, 1) ∈ ETU, by Lemma 4.1, (a, g)σ (b, h). Thus ker f2σ. Hence σ = kerf2. Therefore, it follows from lemma 4.6 that T/σ = T/kerf2 $\begin{array}{}\stackrel{\mathcal{Q}}{\cong }\end{array}$ G. □

#### Remark 4.8

As we know, the class of 𝓠-regular semigroups properly contains the class of V-regular semigroups. In fact, if we restrict the C-set Q of a 𝓠-regular semigroup S to the set of idempotents, then the subdirect products of V-regular semigroups can be constructed in a similar way and we can also use this contruction to study the E-unitary cover for V-regular semigroups.

## Acknowledgement

Research Supported by Fundamental Research Funds for the Central Universities.

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Accepted: 2018-03-28

Published Online: 2018-05-30

Citation Information: Open Mathematics, Volume 16, Issue 1, Pages 522–530, ISSN (Online) 2391-5455,

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