The strong semilattice of π-groups

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 Zhang 1,, Yuhui Yang 2, Ran Shen 3 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 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. 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 S α be a semigroup and assume that S α S β = if α β. For each pair α, β Y such that α β, there exists a homomorphism φ α,β : S α S β such that: (C1) φ α,α = 1 Sα for any α Y. (C2) For any α, β, γ Y with α β γ, φ α,β φ β,γ = φ α,γ. Define a multiplication on S = α Y S α, in terms of the multiplications in the components S α and the homomorphisms φ α,β, for each x in S α and y in S β, xy = xφ α,αβ yφ β,αβ. Then S with the multiplication defined above is a strong semilattice Y of semigroup S α, to be denoted by S[Y ; S α, φ α,β ]. The homomorphisms φ α,β are called the structure homomorphisms of S. And if S α 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 a n 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 semigroups. 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 a m a m Sa m. Denote r(a) = min{m N : a m a m Sa m } and call it the least regular index of a. A π-regular semigroup is called a GVsemigroup 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 semigroup 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 S 1 a r(a) = S 1 b r(b), ar b a r(a) S 1 = b r(b) S 1, aj b S 1 a r(a) S 1 = S 1 b r(b) S 1 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 x n 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) x m 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 S α in this section, where Y is a semilattice, S α is a π-group for each α Y by Lemma 2. Further, we write S α = G α Q α, where G α is the group kernel of S α, and the identity of G α is denoted by e α for any α Y. And Q α = S α \G α is the set of non-regular elements of S α and it is a partial semigroup by the definition of π-group. Certainly, if S α is just a group, then Q α 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 G α, then H a = L a = R a = J a = G α ; if a Q α, then J a = {a}. Proof. For any α Y, let a, b Q α. Suppose that arb. Then there exist s S β and t S γ 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 G βγ RegS. Since RegS is an ideal of S, we get that s = t = 1 and a = b, so R a = {a}. Similarly, K a = {a} for any K {H, L, R, D, J }, a Q α. (1) On the other hand, by Lemma 2, it is easy to see that if a G α, H a = L a = R a = J a = G α. (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 ae α, if a S α where e α is the unique idempotent of π-group S α. Then it is obvious that ψ Gα = 1 Gα. For any a, b S, define the following relation: a ψb if and only if aψ = bψ. Then it is clear that ψ is an equivalence on S and ψ RegS = ε, where ε is the equality relation. Since a ψae α for any a S α, ψ S = ε if and only if S is a Clifford semigroup. In general, ψ 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 b. Lemma 5. Let S be a GV-inverse semigroup. For any a, b S α, if aψ = bψ, then there exists M N such that a m = b m for any m N with m M. In particular, (aψ) r(a) = a r(a). And if aψ = a r(a), then a r(a)2 r(a) = e α. Proof. Let a, b S α. If aψ = bψ, then ae α = be α, and (ae α ) n = (be α ) n for any n N. By Lemma 1, a n e α = b n e α. Take M = max{r(a), r(b)}, then for any m N with m M, a m, b m RegS and a m = b m. In particular, (aψ) r(a) = a r(a) e α = a r(a). If aψ = a r(a), then aψ = (aψ) r(a) and (aψ) r(a) 1 = e α, and hence a r(a)2 r(a) = e α.

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 S α, e β S β, ae β = e β a, 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 ae β, e β a G αβ. And hence ae β = e αβ (ae β )e αβ = (e αβ a)e αβ = e αβ (e β a)e αβ = e β a. Lemma 7. Let S be a GV-inverse semigroup and RegS be an ideal of S. Then ψ is a homomorphism and ψ is the least Clifford congruence on S. Proof. Let a, b S and a S α, b S β. Then aψ = ae α and bψ = be β. By Lemma 6, we can get that aψbψ = ae α be β = ae α e β b = ae αβ b = abe αβ = (ab)ψ. So ψ is a homomorphism and it is easy to understand that ψ is a congruence on S. For any a S, it is easy to see that a ψaψ( RegS) and so ψ is a regular congruence. Further, since a ψb if and only if a = b for any a, b RegS, ψ 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 S α be a GV-inverse semigroup. If the following conditions are satisfied: (i) RegS = α Y G α is a Clifford subsemigroup of S, denoted by G[Y ; G α, θ α,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 Q α, there is a partial semigroup homomorphism ϕ α,β : Q α S β such that (1) ϕ α,α = 1 Qα for any α Y. (2) For any α, β, γ Y with γ αβ and a Q α, b Q β, if ab / Q αβ, then (aϕ α,γ )(bϕ β,γ ) / Q γ. (3) For any α, β, γ Y with α β γ and a Q α, if aϕ α,β Q β, then aϕ α,β ϕ β,γ = aϕ α,γ. If aϕ α,β / Q β, then aϕ α,γ / Q γ. (4) For any α, β Y and a Q α, b Q β, if ab Q αβ, then ab = (aϕ α,αβ )(bϕ β,αβ ). (iii) For any α, β Y with α β, ϕ α,β ψ = ψθ α,β, where ψ is the homomorphism in Lemma 7. Define a mapping φ α,β : S α S β for any α, β Y with α β and a S α, { aθα,β, a G aφ α,β = α, aϕ α,β, a Q α. Then S is a strong semilattice of π-groups, denoted by S[Y ; S α, φ α,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 φ α,β defined is a homomorphism from S α to S β. For any α, β Y with α β, suppose that a, b S α. Case 1: if a Q α, b Q α and ab Q α, since ϕ α,β is a partial semigroup homomorphism, (ab)φ α,β = (ab)ϕ α,β = (aϕ α,β )(bϕ α,β ) = (aφ α,β )(bφ α,β ). Case 2: if a Q α, b Q α and ab G α, since G α is the group kernel of S α, (ab)φ α,β = (abe α )φ α,β = ((ae α )(be α ))θ α,β = (ae α )θ α,β (be α )θ α,β = (aψθ α,β )(bψθ α,β ) = (aϕ α,β ψ)(bϕ α,β ψ)(by Condition (iii)) = (aϕ α,β e β )(bϕ α,β e β ) = aϕ α,β bϕ α,β e β = aϕ α,β bϕ α,β = aφ α,β bφ α,β (by Condition (ii)(2)). Case 3: if a Q α, b G α, then ab G α since G α is the group kernel of S α, (ab)φ α,β = (ab)θ α,β = (a(e α b))θ α,β = ((ae α )b)θ α,β = (ae α )θ α,β bθ α,β = (aψθ α,β )bθ α,β = (aϕ α,β e β )(bθ α,β ) (by Condition (iii)) = (aϕ α,β )(e β bθ α,β ) = aϕ α,β bθ α,β = aφ α,β bφ α,β. Case 4: if a G α, b Q α, then ab G α, similar to the above case, (ab)φ α,β = (ab)θ α,β = ((ae α )b)θ α,β = (a(e α b))θ α,β = aθ α,β (e α b)θ α,β = aθ α,β (be α )θ α,β (by Lemma 1) = aθ α,β bϕ α,β e β = aθ α,β bϕ α,β (by Condition (iii)) = aφ α,β bφ α,β. Case 5: if a G α, b G α, then ab G α, by Condition (i), (ab)φ α,β = (ab)θ α,β = aθ α,β bθ α,β = aφ α,β bφ α,β. And so the mapping φ α,β defined is a homomorphism from S α to S β. By Condition (i), (ii)(1) and the definition of φ α,β, it is clear that φ α,α = 1 Sα. For any α, β, γ Y with α β γ, let a Q α. If aϕ α,β Q β, by Condition (ii)(3), aφ α,β φ β,γ = aϕ α,β φ β,γ = aϕ α,β ϕ β,γ = aϕ α,γ = aφ α,γ. If a Q α and aϕ α,β / Q β, then aϕ α,γ / Q γ by Condition (ii)(3), aφ α,β φ β,γ = aϕ α,β θ β,γ = (aϕ α,β e β )θ β,γ = (aψθ α,β )θ β,γ = (ae α )θ α,β θ β,γ (by Condition (iii)) = (ae α )θ α,γ = (aψ)θ α,γ = aϕ α,γ ψ (by Condition (i), (iii)) = aϕ α,γ e γ = aϕ α,γ = aφ α,γ.

J. Zhang, Y. Yang, R. Shen / Eur. J. Pure Appl. Math, 11 (3) (2018), 589-597 594 Let a G α. aφ α,β φ β,γ = aθ α,β φ β,γ = aθ α,β θ β,γ = aθ α,γ = aφ α,γ. Now we consider the multiplication on S. For any α, β Y, suppose that a S α and b S β. Case i: if a Q α, b Q β, and ab Q αβ, then by Condition (ii)(4), ab = aϕ α,αβ bϕ β,αβ = aφ α,αβ bφ β,αβ. Case ii: if a Q α, b Q β, and ab G αβ, then ab = (ab)e αβ = a(be αβ ) = ae αβ (be αβ ) (since RegS is an ideal of S) = ae α e αβ be β e αβ = (ae α )e αβ (be β )e αβ = (ae α )θ α,αβ (be β )θ β,αβ (since RegS is a Clifford subsemigroup) = (aψθ α,αβ )(bψθ β,αβ ) = (aϕ α,αβ e αβ )(bϕ β,αβ e αβ ) (by Condition (iii)) = aϕ α,αβ (e αβ bϕ β,αβ e αβ ) = aϕ α,αβ bϕ β,αβ e αβ = aϕ α,αβ bϕ β,αβ (by Condition (ii)(2)) = aφ α,αβ bφ β,αβ. Case iii: if a Q α, b G β, then ab G αβ by Condition (i), and ab = (ab)e αβ = ae αβ (be αβ ) = (ae α e αβ )be αβ = (ae α )θ α,αβ bθ β,αβ = (aψθ α,αβ )bθ β,αβ = (aϕ α,αβ ψ)bθ β,αβ = aϕ α,αβ e αβ bθ β,αβ (by Condition (iii)) = aϕ α,αβ bθ β,αβ = aφ α,αβ bφ β,αβ. Case iv: if a G α, b Q β, ab G αβ, then similar to the above case ab = (ab)e αβ = ae αβ (be αβ ) = ae αβ be β e αβ = aθ α,αβ (be β )θ β,αβ = aθ α,αβ (bψθ β,αβ ) = aθ α,αβ (bϕ β,αβ ψ) = aθ α,αβ bϕ β,αβ e αβ (by Condition (iii)) = aθ α,αβ bϕ β,αβ = aφ α,αβ bφ β,αβ. Case v: if a G α, b G β, then ab G αβ, since RegS is a Clifford subsemigroup of S, ab = aθ α,αβ bθ β,αβ = aφ α,αβ bφ β,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 ; S α, φ α,β ]. For any α, β Y and a S α, b G β RegS, ab = aφ α,αβ bφ β,αβ. Because the structure homomorphisms φ α,β preserve the Green s relations on S, bφ β,αβ G αβ by Lemma 4. On the other hand, G αβ is an ideal of S αβ, and so ab G αβ 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 Y G α, it is clear that RegS is a Clifford subsemigroup of S. Denote φ α,β Gα by θ α,β, then RegS = G[Y ; G α, θ α,β ].

J. Zhang, Y. Yang, R. Shen / Eur. J. Pure Appl. Math, 11 (3) (2018), 589-597 595 By Lemma 2, Q α is a partial semigroup for any α Y. Denote φ α,β Qα by ϕ α,β, then it is easy to understand that ϕ α,β is a partial semigroup homomorphism from Q α to S β such that Condition (ii) and (iii) are satisfied. In fact, for any α, β, γ Y with γ αβ and a Q α, b Q β, since (aϕ α,γ )(bϕ β,γ ) = (aφ α,γ )(bφ β,γ ) = (aφ α,αβ φ αβ,γ )(bφ β,αβ φ αβ,γ ) = (aφ α,αβ bφ β,αβ )φ αβ,γ = (ab)φ αβ,γ. If ab / Q αβ, then (aϕ α,γ )(bϕ β,γ ) = (ab)φ αβ,γ / Q γ. 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 Q α, since ae α is regular, aψθ α,β = (ae α )θ α,β = (ae α )φ α,β = (aφ α,β )(e α φ α,β ) = aϕ α,β e β = aϕ α,β ψ. 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 ; S α, φ α,β ] and T [M; T α, ω α,β ] be two strong semilattices of π-groups. Denote the set of non-regular elements of S and T by Q S and Q T respectively. Every homomorphism f from S to T induces a homomorphism from Y to M, we denoted this semilattice homomorphism by f L, and it is obvious that e α f = e αfl for any α Y. On the other hand, since f Sα Hom(S α, T αfl ), we can get a family of π-group homomorphisms denoted by {f α : α Y }. Theorem 2. Let S and T be be two strong semilattices of π-groups as defined above. Given a semilattice homomorphism f L : Y M and a family of π-group homomorphisms {f α : α Y }, where f α Hom(S α, T αfl ). Define f : S T by sf = sf α for any α Y, s S α. And the follwing conditions are satisfied. (1) f Q S is a partial semigroup homomorphism from Q S to T. And for any a, b Q S, if ab / Q S, then (af)(bf) / Q T. (2) φ α,β f β ψ = f α ω αfl,βf L ψ 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 S α and b S β. Case 1: if ab Q αβ, then a Q α, b Q β, then by Condition (1), afbf = (ab)f. Case 2: if ab / Q αβ, then (ab)f / Q T, (ab)f = (ab)f αβ = (aφ α,αβ bφ β,αβ )f αβ = (aφ α,αβ f αβ )(bφ β,αβ f αβ )e (αβ)fl = (aφ α,αβ f αβ e (αβ)fl )(bφ β,αβ f αβ e (αβ)fl ) = (aφ α,αβ f αβ ψ)(bφ β,αβ f αβ ψ). On the other hand, afbf = (af α )(bf β ) = (af α ω αfl,(αβ)f L )(bf β ω βfl,(αβ)f L ) = (af α ω αfl,(αβ)f L e (αβ)fl )(bf β ω βfl,(αβ)f L e (αβ)fl ) (by Condition (1)) = (af α ω αfl,(αβ)f L ψ)(bf β ω βfl,(αβ)f L ψ).

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, f L and {f α : 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; R α, χ α,β ] be a strong semilattices of π-groups. For any α, β Z with α β and for any a R α, And so a(ψχ α,β ) = (ae α )χ α,β = aχ α,β e α χ α,β = aχ α,β e β = a(χ α,β ψ). ψχ α,β = χ α,β ψ. (3) Now we return to the proof. For any α, β Y with α β and for any a S α, And hence, (ae β )f = (ae α e β )f = a(ψφ α,β f β ), (af)(e β f) = (af α )(e βfl ) = (af α )(e αfl )(e βfl ) = a(f α ψω αfl,βf L ). Specially, for any α Y and a S α, Which means that ψφ α,β f β = f α ψω αfl,βf L. (4) a(ψf α ) = (ae α )f α = af α e αfl = a(f α ψ). By the equations (3), (4) and (5), it is easy to understand that The proof is completed. ψf α = f α ψ. (5) φ α,β f β ψ = ψφ α,β f β = f α ψω αfl,βf L = f α ω αfl,βf L ψ. (6) Corollary 1. Let G 1 [Y 1 ; G α, φ α,β ] and G 2 [Y 2 ; G α, ω α,β ] be two Clifford semigroups. Given a semilattice homomorphism f L : Y 1 Y 2 and a family of group homomorphisms {f α : α Y 1 }, where f α Hom(G α, G αfl ). Define f : G 1 G 2 by sf = sf α for any s G α. And for any α, β Y with α β, the follwing equation is satisfied. φ α,β f β = f α ω αfl,βf L. Then f is a homomorphism. Conversely, every homomorphism between this two Clifford semigroups satisfies the conditions.

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