EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 11, No. 3, 2018, 589-597 ISSN 1307-5543 – www.ejpam.com Published by New York Business Global The strong semilattice of π-groups Jiangang Zhang1,∗, Yuhui Yang2, Ran Shen3 1 Department of Mathematics , Shanghai Normal University, Shanghai, 200234, China 2 Department of Mathematics, Lvliang University, Shanxi Lvliang, 033000, China 3 College of Science, Donghua University, Shanghai, 201620, China Abstract. A semigroup is called a GV-inverse semigroup if and only if it is isomorphic to a semi- lattice of π-groups. In this paper, we give the sufficient and necessary conditions for a GV-inverse semigroup to be a strong semilattice of π-groups. Some conclusions about Clifford semigroups are generalized. Key Words and Phrases: GV-inverse semigroup, π-group, strong semilattice, homomorphism 1. Introduction First of all, we give the basic definition for this paper. Let Y be a semilattice. For each α ∈ Y , let Sα be a semigroup and assume that Sα ∩ Sβ = ∅ if α 6= β. For each pair α, β ∈ Y such that α ≥ β, there exists a homomorphism φα,β : Sα → Sβ such that: (C1) φα,α = 1Sα for any α ∈ Y . (C2) For any α, β, γ ∈ Y with α ≥ β ≥ γ, φα,βφβ,γ = φα,γ . Define a multiplication on S = ∪α∈Y Sα, in terms of the multiplications in the components Sα and the homomorphisms φα,β, for each x in Sα and y in Sβ, xy = xφα,αβyφβ,αβ . Then S with the multiplication defined above is a strong semilattice Y of semigroup Sα, to be denoted by S[Y ;Sα, φα,β]. The homomorphisms φα,β are called the structure homomorphisms of S. And if Sα ∈ H for all α ∈ Y and some class of semigroups H, then S is a strong semilattice of type H. As is well known, a semigroup is a Clifford semigroup if and only if it is isomorphic to a strong semilattice of groups. A semigroup S is a π-group if it is a nil-extension of a group, which means that there exists a subgroup G of S and G is an ideal, and for any a ∈ S, there exists a number n ∈ N such that an ∈ G, where N is the natural number set. A semigroup is called a GV-inverse semigroup if and only if it is isomorphic to a semilattice of π-groups. It is natural to ask how about the strong semilattice of π-groups. In this paper, we give the sufficient and necessary conditions for a GV-inverse semigroup to be a strong semilattice of π-groups. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v11i3.3274 Email addresses: jgzhang@shnu.edu.cn (J. Zhang), ranshen@dhu.edu.cn (R. Shen) http://www.ejpam.com 589 c© 2018 EJPAM All rights reserved. J. Zhang, Y. Yang, R. Shen / Eur. J. Pure Appl. Math, 11 (3) (2018), 589-597 590 2. Preliminaries The class of π-regular semigroups is one of the important classes of non-regular semi- groups. Recall that a semigroup S is said to be a π-regular semigroup if for any a ∈ S, there exists a positive integer m such that am ∈ amSam. Denote r(a) = min{m ∈ N : am ∈ amSam} and call it the least regular index of a. A π-regular semigroup is called a GV- semigroup if every regular element is completely regular. Furthermore, a GV-semigroup S is called a GV-inverse semigroup if every regular element of S possesses a unique inverse. GV-semigroups and GV-inverse semigroups are the generalizations of completely regular semigroups and Clifford semigroups in the range of π-regular semigroups respectively. Throughout this paper, we denote the set of all regular elements of a π-regular semi- group S by RegS. We will write maps on the right of the objects on which they act. Let S be a π-regular semigroup. Generalized Green’s equivalences were defined by: aL∗b⇔ S1ar(a) = S1br(b), aR∗b⇔ ar(a)S1 = br(b)S1, aJ ∗b⇔ S1ar(a)S1 = S1br(b)S1 H∗ = L∗ ∩R∗, D∗ = L∗ ∨R∗. If S is a regular semigroup, then K = K∗ on S for any K ∈ {H,L,R,D,J }. The class of π-regular semigroups and some of its subclasses have been studied in in [1], [2], [4], [5], [6]. For notations and terminologies not mentioned here, the reader is referred to [1] and [3]. 3. Strong Semilattice of π-groups In this section, we characterize the strong semilattice of π-groups. At first, we give some characterizations of GV-inverse semigroups. Lemma 1. ([1]) Let S be a semigroup and x be an element of S such that xn lies in a subgroup G of S for some positive integer n. If e is the identity of G, then (i) ex = xe ∈ G. (ii) xm ∈ G for every m ≥ n and m ∈ N . Lemma 2. ([1]) Let S be a semigroup. Then the following conditions are equivalent: (i) S is a GV-inverse semigroup. (ii) S is π-regular, and a = axa implies that ax = xa. (iii) S is a semilattice of π-groups. For convenience, we always denote a GV-inverse semigroup by S = ∪α∈Y Sα in this section, where Y is a semilattice, Sα is a π-group for each α ∈ Y by Lemma 2. Further, we write Sα = Gα ∪ Qα, where Gα is the group kernel of Sα, and the identity of Gα is denoted by eα for any α ∈ Y . And Qα = Sα\Gα is the set of non-regular elements of Sα and it is a partial semigroup by the definition of π-group. Certainly, if Sα is just a group, then Qα is an empty set. According to the results in [1], we know that H∗ = L∗ = R∗ = D∗ = J ∗ on a GV-inverse semigroup. On the other hand, we have the following results. J. Zhang, Y. Yang, R. Shen / Eur. J. Pure Appl. Math, 11 (3) (2018), 589-597 591 Lemma 3. [5] Let S be a GV-inverse semigroup. Then for any K ∈ {H,L,R,D,J }, K ⊆ K∗ on S. Lemma 4. Let S be a GV-inverse semigroup and RegS be an ideal of S. For any α ∈ Y , if a ∈ Gα, then Ha = La = Ra = Ja = Gα; if a ∈ Qα, then Ja = {a}. Proof. For any α ∈ Y , let a, b ∈ Qα. Suppose that aRb. Then there exist s ∈ Sβ and t ∈ Sγ such that a = bs, b = at for some β, γ ∈ Y with β, γ ≥ α. Further, a = bs = ats = a(ts)2 = · · · = a(ts)m, b = at = bst = b(st)2 = · · · = b(st)m where m ≥ max{r(st), r(ts)}. By Lemma 1, (ts)m, (st)m ∈ Gβγ ⊆ RegS. Since RegS is an ideal of S, we get that s = t = 1 and a = b, so Ra = {a}. Similarly, Ka = {a} for any K ∈ {H,L,R,D,J }, a ∈ Qα. (1) On the other hand, by Lemma 2, it is easy to see that if a ∈ Gα, Ha = La = Ra = Ja = Gα. (2) Let S be a GV-inverse semigroup. Define a mapping ψ from S into RegS as follows: for any a ∈ S, ψ : S → RegS; a 7→ aeα, if a ∈ Sα where eα is the unique idempotent of π-group Sα. Then it is obvious that ψ|Gα = 1Gα . For any a, b ∈ S, define the following relation: aψ̃b if and only if aψ = bψ. Then it is clear that ψ̃ is an equivalence on S and ψ̃ |RegS= ε, where ε is the equality relation. Since aψ̃aeα for any a ∈ Sα, ψ̃ |S= ε if and only if S is a Clifford semigroup. In general, ψ̃ |S\RegS is not necessary the equality relation. Next, we give an example. Example 1. Let S = {e, a, b} with the following Cayley table a b e a b e e b e e e e e e e It is clear that S is a π-group and RegS = {e} and aψ = ae = e = be = bψ, but a 6= b. Lemma 5. Let S be a GV-inverse semigroup. For any a, b ∈ Sα, if aψ = bψ, then there exists M ∈ N such that am = bm for any m ∈ N with m ≥ M . In particular, (aψ)r(a) = ar(a). And if aψ = ar(a), then ar(a) 2−r(a) = eα. Proof. Let a, b ∈ Sα. If aψ = bψ, then aeα = beα, and (aeα)n = (beα)n for any n ∈ N . By Lemma 1, aneα = bneα. Take M = max{r(a), r(b)}, then for any m ∈ N with m ≥M , am, bm ∈ RegS and am = bm. In particular, (aψ)r(a) = ar(a)eα = ar(a). If aψ = ar(a), then aψ = (aψ)r(a) and (aψ)r(a)−1 = eα, and hence ar(a) 2−r(a) = eα. J. Zhang, Y. Yang, R. Shen / Eur. J. Pure Appl. Math, 11 (3) (2018), 589-597 592 Lemma 6. Let S be a GV-inverse semigroup and RegS be an ideal of S. Then for any α, β ∈ Y and a ∈ Sα, eβ ∈ Sβ, aeβ = eβa, which means that E(S) ⊆ C(S), where C(S) is the center of S. Proof. By Lemma 2, RegS is a Clifford subsemigroup of S, then aeβ, eβa ∈ Gαβ. And hence aeβ = eαβ(aeβ)eαβ = (eαβa)eαβ = eαβ(eβa)eαβ = eβa. Lemma 7. Let S be a GV-inverse semigroup and RegS be an ideal of S. Then ψ is a homomorphism and ψ̃ is the least Clifford congruence on S. Proof. Let a, b ∈ S and a ∈ Sα, b ∈ Sβ. Then aψ = aeα and bψ = beβ. By Lemma 6, we can get that aψbψ = aeαbeβ = aeαeβb = aeαβb = abeαβ = (ab)ψ. So ψ is a homomorphism and it is easy to understand that ψ̃ is a congruence on S. For any a ∈ S, it is easy to see that aψ̃aψ(∈ RegS) and so ψ̃ is a regular congruence. Further, since aψ̃b if and only if a = b for any a, b ∈ RegS, ψ̃ is the least Clifford congruence on S. Definition 1. Let S be a partial semigroup and T be a semigroup. A mapping f : S → T is called a partial semigroup homomorphism if (ab)f = (af)(bf) for any a, b ∈ S and ab ∈ S. Now we give the main result of this paper. Theorem 1. Let S = ∪α∈Y Sα be a GV-inverse semigroup. If the following conditions are satisfied: (i) RegS = ∪α∈YGα is a Clifford subsemigroup of S, denoted by G[Y ;Gα, θα,β], and it is an ideal of S, which means that S is a nil-extension of a Clifford semigroup. (ii) For any α, β ∈ Y with α ≥ β, if Qα 6= ∅, there is a partial semigroup homomor- phism ϕα,β : Qα → Sβ such that (1) ϕα,α = 1Qα for any α ∈ Y . (2) For any α, β, γ ∈ Y with γ ≤ αβ and a ∈ Qα, b ∈ Qβ, if ab /∈ Qαβ, then (aϕα,γ)(bϕβ,γ) /∈ Qγ. (3) For any α, β, γ ∈ Y with α ≥ β ≥ γ and a ∈ Qα, if aϕα,β ∈ Qβ, then aϕα,βϕβ,γ = aϕα,γ. If aϕα,β /∈ Qβ, then aϕα,γ /∈ Qγ. (4) For any α, β ∈ Y and a ∈ Qα, b ∈ Qβ, if ab ∈ Qαβ, then ab = (aϕα,αβ)(bϕβ,αβ). (iii) For any α, β ∈ Y with α ≥ β, ϕα,βψ = ψθα,β, where ψ is the homomorphism in Lemma 7. Define a mapping φα,β : Sα → Sβ for any α, β ∈ Y with α ≥ β and a ∈ Sα, aφα,β = { aθα,β, a ∈ Gα, aϕα,β, a ∈ Qα. Then S is a strong semilattice of π-groups, denoted by S[Y ;Sα, φα,β]. Conversely, every strong semilattice of π-groups can be so obtained. J. Zhang, Y. Yang, R. Shen / Eur. J. Pure Appl. Math, 11 (3) (2018), 589-597 593 Proof. Let S be a GV-inverse semigroup and the given conditions are satisfied. In order to prove that S is a strong semilattice of π-groups, we firstly show that the mapping φα,β defined is a homomorphism from Sα to Sβ. For any α, β ∈ Y with α ≥ β, suppose that a, b ∈ Sα. Case 1: if a ∈ Qα, b ∈ Qα and ab ∈ Qα, since ϕα,β is a partial semigroup homomor- phism, (ab)φα,β = (ab)ϕα,β = (aϕα,β)(bϕα,β) = (aφα,β)(bφα,β). Case 2: if a ∈ Qα, b ∈ Qα and ab ∈ Gα, since Gα is the group kernel of Sα, (ab)φα,β = (abeα)φα,β = ((aeα)(beα))θα,β = (aeα)θα,β(beα)θα,β = (aψθα,β)(bψθα,β) = (aϕα,βψ)(bϕα,βψ)(by Condition (iii)) = (aϕα,βeβ)(bϕα,βeβ) = aϕα,βbϕα,βeβ = aϕα,βbϕα,β = aφα,βbφα,β (by Condition (ii)(2)). Case 3: if a ∈ Qα, b ∈ Gα, then ab ∈ Gα since Gα is the group kernel of Sα, (ab)φα,β = (ab)θα,β = (a(eαb))θα,β = ((aeα)b)θα,β = (aeα)θα,βbθα,β = (aψθα,β)bθα,β = (aϕα,βeβ)(bθα,β) (by Condition (iii)) = (aϕα,β)(eβbθα,β) = aϕα,βbθα,β = aφα,βbφα,β. Case 4: if a ∈ Gα, b ∈ Qα, then ab ∈ Gα, similar to the above case, (ab)φα,β = (ab)θα,β = ((aeα)b)θα,β = (a(eαb))θα,β = aθα,β(eαb)θα,β = aθα,β(beα)θα,β (by Lemma 1) = aθα,βbϕα,βeβ = aθα,βbϕα,β (by Condition (iii)) = aφα,βbφα,β. Case 5: if a ∈ Gα, b ∈ Gα, then ab ∈ Gα, by Condition (i), (ab)φα,β = (ab)θα,β = aθα,βbθα,β = aφα,βbφα,β. And so the mapping φα,β defined is a homomorphism from Sα to Sβ. By Condition (i), (ii)(1) and the definition of φα,β, it is clear that φα,α = 1Sα . For any α, β, γ ∈ Y with α ≥ β ≥ γ, let a ∈ Qα. If aϕα,β ∈ Qβ, by Condition (ii)(3), aφα,βφβ,γ = aϕα,βφβ,γ = aϕα,βϕβ,γ = aϕα,γ = aφα,γ . If a ∈ Qα and aϕα,β /∈ Qβ, then aϕα,γ /∈ Qγ by Condition (ii)(3), aφα,βφβ,γ = aϕα,βθβ,γ = (aϕα,βeβ)θβ,γ = (aψθα,β)θβ,γ = (aeα)θα,βθβ,γ (by Condition (iii)) = (aeα)θα,γ = (aψ)θα,γ = aϕα,γψ (by Condition (i), (iii)) = aϕα,γeγ = aϕα,γ = aφα,γ . J. Zhang, Y. Yang, R. Shen / Eur. J. Pure Appl. Math, 11 (3) (2018), 589-597 594 Let a ∈ Gα. aφα,βφβ,γ = aθα,βφβ,γ = aθα,βθβ,γ = aθα,γ = aφα,γ . Now we consider the multiplication on S. For any α, β ∈ Y , suppose that a ∈ Sα and b ∈ Sβ. Case i: if a ∈ Qα, b ∈ Qβ, and ab ∈ Qαβ, then by Condition (ii)(4), ab = aϕα,αβbϕβ,αβ = aφα,αβbφβ,αβ . Case ii: if a ∈ Qα, b ∈ Qβ, and ab ∈ Gαβ, then ab = (ab)eαβ = a(beαβ) = aeαβ(beαβ) (since RegS is an ideal of S) = aeαeαβbeβeαβ = (aeα)eαβ(beβ)eαβ = (aeα)θα,αβ(beβ)θβ,αβ (since RegS is a Clifford subsemigroup) = (aψθα,αβ)(bψθβ,αβ) = (aϕα,αβeαβ)(bϕβ,αβeαβ) (by Condition (iii)) = aϕα,αβ(eαβbϕβ,αβeαβ) = aϕα,αβbϕβ,αβeαβ = aϕα,αβbϕβ,αβ (by Condition (ii)(2)) = aφα,αβbφβ,αβ . Case iii: if a ∈ Qα, b ∈ Gβ, then ab ∈ Gαβ by Condition (i), and ab = (ab)eαβ = aeαβ(beαβ) = (aeαeαβ)beαβ = (aeα)θα,αβbθβ,αβ = (aψθα,αβ)bθβ,αβ = (aϕα,αβψ)bθβ,αβ = aϕα,αβeαβbθβ,αβ (by Condition (iii)) = aϕα,αβbθβ,αβ = aφα,αβbφβ,αβ . Case iv: if a ∈ Gα, b ∈ Qβ, ab ∈ Gαβ, then similar to the above case ab = (ab)eαβ = aeαβ(beαβ) = aeαβbeβeαβ = aθα,αβ(beβ)θβ,αβ = aθα,αβ(bψθβ,αβ) = aθα,αβ(bϕβ,αβψ) = aθα,αβbϕβ,αβeαβ (by Condition (iii)) = aθα,αβbϕβ,αβ = aφα,αβbφβ,αβ . Case v: if a ∈ Gα, b ∈ Gβ, then ab ∈ Gαβ, since RegS is a Clifford subsemigroup of S, ab = aθα,αβbθβ,αβ = aφα,αβbφβ,αβ . By now we have proved that GV-inverse semigroup S is a strong semilattice of π-groups if it satisfies the conditions. Conversely, suppose that GV-inverse semigroup S is a strong semilattice of π-groups, denoted by S[Y ;Sα, φα,β]. For any α, β ∈ Y and a ∈ Sα, b ∈ Gβ ⊆ RegS, ab = aφα,αβbφβ,αβ . Because the structure homomorphisms φα,β preserve the Green’s relations on S, bφβ,αβ ∈ Gαβ by Lemma 4. On the other hand, Gαβ is an ideal of Sαβ, and so ab ∈ Gαβ ⊆ RegS. Similarly, we can prove that ab ∈ RegS for any a ∈ RegS, b ∈ S. And so RegS is an ideal of S. Since RegS = ∪α∈YGα, it is clear that RegS is a Clifford subsemigroup of S. Denote φα,β|Gα by θα,β, then RegS = G[Y ;Gα, θα,β]. J. Zhang, Y. Yang, R. Shen / Eur. J. Pure Appl. Math, 11 (3) (2018), 589-597 595 By Lemma 2, Qα is a partial semigroup for any α ∈ Y . Denote φα,β|Qα by ϕα,β, then it is easy to understand that ϕα,β is a partial semigroup homomorphism from Qα to Sβ such that Condition (ii) and (iii) are satisfied. In fact, for any α, β, γ ∈ Y with γ ≤ αβ and a ∈ Qα, b ∈ Qβ, since (aϕα,γ)(bϕβ,γ) = (aφα,γ)(bφβ,γ) = (aφα,αβφαβ,γ)(bφβ,αβφαβ,γ) = (aφα,αβbφβ,αβ)φαβ,γ = (ab)φαβ,γ . If ab /∈ Qαβ, then (aϕα,γ)(bϕβ,γ) = (ab)φαβ,γ /∈ Qγ . So Condition (ii)(2) holds. And it is easy to see that Condition (ii)(1),(3),(4) hold by the definition of strong semilattice. Further, for any α, β ∈ Y with α ≥ β and a ∈ Qα, since aeα is regular, aψθα,β = (aeα)θα,β = (aeα)φα,β = (aφα,β)(eαφα,β) = aϕα,βeβ = aϕα,βψ. So each strong semilattice of π-groups satisfies the conditions. The proof is completed. At last, we consider the homomorphisms between two strong semilattices of π-groups. Let S[Y ;Sα, φα,β] and T [M ;Tα, ωα,β] be two strong semilattices of π-groups. Denote the set of non-regular elements of S and T by QS and QT respectively. Every homomorphism f from S to T induces a homomorphism from Y to M , we denoted this semilattice homo- morphism by fL, and it is obvious that eαf = eαfL for any α ∈ Y . On the other hand, since f |Sα∈ Hom(Sα, TαfL), we can get a family of π-group homomorphisms denoted by {fα : α ∈ Y }. Theorem 2. Let S and T be be two strong semilattices of π-groups as defined above. Given a semilattice homomorphism fL : Y →M and a family of π-group homomorphisms {fα : α ∈ Y }, where fα ∈ Hom(Sα, TαfL). Define f : S → T by sf = sfα for any α ∈ Y, s ∈ Sα. And the follwing conditions are satisfied. (1) f |QS is a partial semigroup homomorphism from QS to T . And for any a, b ∈ QS, if ab /∈ QS, then (af)(bf) /∈ QT . (2) φα,βfβψ = fαωαfL,βfLψ for any α, β ∈ Y with α ≥ β. Then f is a homomorphism. Conversely, every homomorphism between two strong semilattices of π-groups satisfies the conditions. Proof. It is obvious that f is a map from S to T . We only need to prove that f is a homomorphism. Let a ∈ Sα and b ∈ Sβ. Case 1: if ab ∈ Qαβ, then a ∈ Qα, b ∈ Qβ, then by Condition (1), afbf = (ab)f . Case 2: if ab /∈ Qαβ, then (ab)f /∈ QT , (ab)f = (ab)fαβ = (aφα,αβbφβ,αβ)fαβ = (aφα,αβfαβ)(bφβ,αβfαβ)e(αβ)fL = (aφα,αβfαβe(αβ)fL)(bφβ,αβfαβe(αβ)fL) = (aφα,αβfαβψ)(bφβ,αβfαβψ). On the other hand, afbf = (afα)(bfβ) = (afαωαfL,(αβ)fL)(bfβωβfL,(αβ)fL) = (afαωαfL,(αβ)fLe(αβ)fL)(bfβωβfL,(αβ)fLe(αβ)fL) (by Condition (1)) = (afαωαfL,(αβ)fLψ)(bfβωβfL,(αβ)fLψ). J. Zhang, Y. Yang, R. Shen / Eur. J. Pure Appl. Math, 11 (3) (2018), 589-597 596 By Condition (2), we get afbf = (ab)f . Conversely, let f be a homomorphism from S to T , fL and {fα : α ∈ Y } be the corresponding semilattice homomorphism and the family of π-group homomorphisms as considered above this theorem. That Condition (1) holds is clear. We only need to prove the equation holds in Condition (2). At first, we give the following fact. Let R[Z;Rα, χα,β] be a strong semilattices of π-groups. For any α, β ∈ Z with α ≥ β and for any a ∈ Rα, a(ψχα,β) = (aeα)χα,β = aχα,βeαχα,β = aχα,βeβ = a(χα,βψ). And so ψχα,β = χα,βψ. (3) Now we return to the proof. For any α, β ∈ Y with α ≥ β and for any a ∈ Sα, (aeβ)f = (aeαeβ)f = a(ψφα,βfβ), (af)(eβf) = (afα)(eβfL) = (afα)(eαfL)(eβfL) = a(fαψωαfL,βfL). And hence, ψφα,βfβ = fαψωαfL,βfL . (4) Specially, for any α ∈ Y and a ∈ Sα, a(ψfα) = (aeα)fα = afαeαfL = a(fαψ). Which means that ψfα = fαψ. 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