Period Math Hung (2016) 72:12–22 DOI 10.1007/s10998-015-0107-y

Global determinism of semigroups having regular globals Aiping Gan1,2 · Xianzhong Zhao1 · Miaomiao Ren1

Published online: 16 December 2015 © Akadémiai Kiadó, Budapest, Hungary 2015

Abstract The aim of this paper is to study the global determinism of the class A of all semigroups having regular globals. It is known from PeliKán (Periodica Math Hungarica 4:103–106, 1973) and Pondˇelíˇcek (On semigroups having regular globals, 1976) that A can be divided into two subclasses: the class A2 of all semigroups having idempotent globals and the class A3 of all semigroups having regular but non-idempotent globals. We prove that A2 is globally determined and that A3 satisfies the strong isomorphism property. This shows that A is globally determined. Keywords

Idempotent semigroup · Regular power semigroup · Global

Mathematics Subject Classification

06A12 · 20M17

1 Introduction and preliminaries The power semigroup, or global, of a semigroup S is the semigroup P(S) of all non-empty subsets of S equipped with the multiplication AB = {ab : a ∈ A, b ∈ B} for all A, B ∈ P(S). There are a series of papers in the literature considering power semigroups and related varieties of semigroups, see e.g., [4–8] and dozens of references contained in these papers.

B

Xianzhong Zhao [email protected] Aiping Gan [email protected] Miaomiao Ren [email protected]

1

Department of Mathematics, Northwest University, Xi’an, Shaanxi 710127, People’s Republic of China

2

College of Mathematics and Information Science, Jiangxi Normal University, Nanchang, Jiangxi 330022, People’s Republic of China

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Recall that a class C of semigroups is said to be globally determined if any two members of C having isomorphic globals must themselves be isomorphic; and that a class K of semigroups is said to satisfy the strong isomorphism property [14] if for each pair S, S  ∈ K and each

isomorphism θ from P(S) onto P(S  ), we have θ (S) = S  , where S = {{s} : s ∈ S} ⊆ P(S), S  = {{s  } : s  ∈ S  } ⊆ P(S  ), and hence S ∼ = S  (if a singleton {x} is identified with x, then  S and S in this definition can be replaced by S and S  , respectively). Obviously, if a class of semigroups satisfies the strong isomorphism property, then it is globally determined, but the converse is not necessarily true. Tamura [9] asked in 1967 whether the class of all semigroups is globally determined. The question was negatively answered by Mogiljanskaja [11] in 1973. Gan and Zhao [12] proved that the class of all Clifford semigroups satisfies the strong isomorphism property. This is the case for the following classes: all semilattices [14]; all completely regular periodic monoids with irreducible identity (in particular, all idempotent semigroups with identity) [16]; etc. It is also known that some classes of semigroups, such as finite semigroups [6]; left zero semigroups and rectangular groups [13]; completely simple semigroups and completely 0simple semigroups [15], are globally determined, but do not satisfy the strong isomorphism property. Notice that the Generalized Continuum Hypothesis is assumed in [13,15]. In this paper we study the global determinism of the class A of all semigroups having regular globals. Rédei [3] introduced and studied so called breakable semigroups. Recall that a semigroup is breakable if all non-empty subsets of it are subsemigroups. It is easy to see that a semigroup S is breakable if and only if ab ∈ {a, b} for any a, b ∈ S. Breakable semigroups were also considered in [10] as ‘semigroups of primes’. The following is shown: Result 1.1 ([3, Theorem 50, p. 81] and [10, Theorem 5.4]) The following conditions are equivalent for any semigroup S: (i) S is breakable. (ii) P(S) is an idempotent semigroup, i.e., X = X 2 for all X ∈ P(S). (iii) S = γ ∈ Sγ , where  is a chain, the semigroups Sγ (γ ∈ ) are pairwise disjoint left or right zero semigroups, and ab = ba = a for any a ∈ Sα , b ∈ Sβ with α < β in . We denote the class of all semigroups satisfying one of the conditions in Result 1.1 by A2 . That is to say, A2 is the class of all semigroups having idempotent globals. Pelikán [7] introduced and studied a semigroup S with the following property (An ) :

(∀ a1 , a2 , . . . , an ∈ S) a1 a2 · · · an ∈ {a1 , a2 , . . . , an }.

Result 1.1 tells us that A2 is exactly the class of all semigroups with property (A2 ). Also, it is easy to see that every semigroup in A2 satisfies (An ) for any integer n ≥ 2. Theorems 1 and 3 in [7] show that a semigroup satisfies (A2 ) ((A3 ), resp.) if it satisfies (A2k ) ((A2k+1 ), resp.) for some positive integer k. The following result will be needed later. Result 1.2 ([7, Theorem 2 and Remark] and [8, Theorem 1]) The following conditions are equivalent for any semigroup S: (i) (ii) (iii) (iv)

S satisfies property (A3 ). X = X 3 for all X ∈ P(S), and consequently, P(S) is completely regular. P(S) is regular.  S is breakable or has the following decomposition: S = γ ∈ Sγ , where  is a chain with a maximum element ε, the semigroups Sγ (γ  = ε) are pairwise disjoint left or right zero subsemigroups of S, Sε is isomorphic to the cyclic group of order two, and ab = ba = a for any a ∈ Sα , b ∈ Sβ with α < β in .

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We denote the class of all semigroups satisfying one of the conditions in Result 1.2 by A and write A3 = A \ A2 . That is to say, A is the class of all semigroups having regular globals and A3 is the class of all semigroups having regular but non-idempotent globals.  For a semigroup S in A, by Results 1.1 and 1.2, we may assume that S = γ ∈ Sγ , where  is a chain, the semigroups Sγ (γ ∈ ) are pairwise disjoint subsemigroups of S. For any A ∈ P(S), let Aα denote the set A ∩ Sα and I d(A) the subset {α ∈  : Aα  = ∅} of . It is easy to see that (∀A, B ∈ P(S)) I d(AB) = I d(A)I d(B).

(1.1)

For convenience, we frequently identify an element a of the semigroup S with the singleton member {a} of P(S), so that for any x ∈ S and any isomorphism ψ from P(S) onto P(S  ), we simply write ψ({x}) as ψ(x). We need the following notation:  |A| denotes the cardinality of a set A.  P(S) = {A ∈ P(S) : |I d(A)| = 1}.  E P(S) = {A ∈ P(S) : A2 = A}, that is to say, E P(S) denotes the set of all idempotents of P(S).   E P(S) = {A ∈ E P(S) : (∀B ∈ P(S))B 2 = A ⇒ B = A}.  It is obvious that P(S) = E P(S) = E P(S) if S is a semigroup in A2 . In addition to this introduction and preliminaries, this paper comprises two sections. In Sect. 2, suppose that S, S  are semigroups in A2 , and that ψ is an isomorphism from P(S) onto P(S  ). By virtue of the technique used in [14], we show that ψ| P(S) is an isomorphism from P(S) onto P(S  ). Based on this result, we prove that A2 is globally determined. In  Sect. 3 we first characterize the subsets E P(S) and E P(S) of P(S) for a semigroup S in A3 . Next, we show that A3 satisfies the strong isomorphism property, and conclude that A is globally determined. For other notations and terminology not given in this paper, the reader is referred to the books [1,2].

2 On semigroups having idempotent globals In this section we study the global determinism of the class A2 of all semigroups having idempotent globals. Assume that S, S  are semigroups in A2 and that ψ is an isomorphism from P(S) onto P(S  ). By Result 1.1, we may write   Sγ and S  = Sγ  , S= γ ∈

γ  ∈ 

are chains, the semigroups Sγ (γ ∈ ) (Sγ  (γ  ∈   ), resp.) are pairwise where  and disjoint left or right zero subsemigroups of S (S  , resp.). Notice that P(S) is an idempotent semigroup. We have the natural partial order ≤ on P(S) and write A < B whenever A ≤ B and A  = B. By identifying an element a of the semigroup S with the singleton {a}, we can find that the restriction ≤ | S of ≤ to S is exactly the natural partial order on S. Recall that for any a, b ∈ S, we say that b covers a in the poset (S, ≤) if a < b and there exists no c ∈ S such that a < c < b. Similarly, for any A, B ∈ P(S), we write A  B (A ≺ B, resp.) if A < B and there exists no C ∈ P(S) (C ∈ P(S), resp.) such that A < C < B. It is obvious that A  B implies A ≺ B and A  B implies ψ(A)  ψ(B) for any A, B ∈ P(S). 

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Remark 2.1 Let A ∈ P(S). If α, β ∈ I d(A) such that α < β, then, by Result 1.1 (iii), Aα < Aβ . Proposition 2.2 Let A, B ∈ P(S) with A ≤ B. (i) If I d(A) = I d(B) = {α}, then A = B. (ii) Bα ⊆ Aα for any α ∈ I d(A) ∩ I d(B). Proof Suppose that A, B ∈ P(S) with A ≤ B. Then A = AB = B A. (i) It is obvious. (ii) By Result 1.1, we have B = B 2 , and so A = AB = AB 2 = B AB. Assume that α ∈ I d(A) ∩ I d(B) and bα ∈ Bα . Choose an element aα ∈ Aα . Since Sα is either a left zero or a right zero semigroup, we have bα = bα aα bα ∈ (B AB)α = Aα . This shows that Bα ⊆ Aα , as required.   The following lemmas are analogous to Lemmas 1–3 in [14]. Lemma 2.3 Let A ∈ P(S) and α ∈ I d(A). If α is not a maximal element of I d(A), then (∀aα ∈ Aα ) A  A \ {aα }. Proof Suppose that A ∈ P(S), α ∈ I d(A) and aα ∈ Aα . If α is not a maximal element of I d(A), i.e., there exists β ∈ I d(A) such that α < β, then, by Result 1.1(iii), we have aα = aα aβ ∈ A(A \ {aα }) and aα = aβ aα ∈ (A \ {aα })A for any aβ ∈ Aβ . This implies that A = A(A \ {aα }) = (A \ {aα })A, i.e., A < A \ {aα }. If A ≤ C ≤ A \ {aα } for some C ∈ P(S), then A = AC = C A and C = (A \ {aα })C = C(A \ {aα }). It follows that C = (A \ {aα })C ⊆ AC = A. Also, for every u ∈ A \ {aα }, since A = AC, there exist a ∈ A, c ∈ C such that u = ac. Thus u = uc ∈ (A \ {aα })C = C. This shows that A \ {aα } ⊆ C. We now have A \ {aα } ⊆ C ⊆ A. Hence C is equal to either A or A \ {aα }. This completes the proof.   Lemma 2.4 Let A ∈ P(S). If I d(A) has a maximum element ω, then (∀B ∈ P(S)) Aω ≺ B ⇒ A  A ∪ B. Proof Assume that A ∈ P(S) and that I d(A) has a maximum element ω. If Aω ≺ B for some B ∈ P(S), then it is easy to see that A < A ∪ B and α < β for any α ∈ I d(A), where {β} = I d(B). Suppose that A < C ≤ A ∪ B for some C ∈ P(S). Then A = AC = C A,

C = C(A ∪ B) = (A ∪ B)C.

(2.1)

This implies that C = C(A ∪ B) = C A ∪ C B = A ∪ C B,

(2.2)

and so A ⊆ C. Thus I d(A) ⊆ I d(C) and Aα ⊆ Cα for any α ∈ I d(A). Also, since A < C, by Proposition 2.2 we have Cα ⊆ Aα for any α ∈ I d(A) ∩ I d(C) = I d(A). Hence Aα = Cα for any α ∈ I d(A). Since A  = C, we have I d(C \ A)  = ∅.

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To show that C = A ∪ B, we must show that Cγ = B for any γ ∈ I d(C \ A). In fact, for any γ ∈ I d(C \ A), we have that ω < γ , for otherwise, by (1.1) and (2.1), γ = ωγ ∈ I d(A)I d(C) = I d(AC) = I d(A), a contradiction. Hence, by Remark 2.1, Aω = Cω < Cγ . For any cγ ∈ Cγ , since Cγ ∩ A = ∅, we have cγ ∈ / A and so cγ ∈ C B by (2.2). It follows by (1.1) that γ ∈ I d(C B) = I d(C)I d(B) = I d(C)β. Thus γ ≤ β. This, together with the fact that α < β for any α ∈ I d(A), gives ξ ≤ β for any ξ ∈ I d(A) ∪ I d(C \ A) = I d(C). Hence cγ ∈ Cγ B and so Cγ ⊆ Cγ B. In addition, by (2.2) we have Cγ B ⊆ (C B)γ ⊆ Cγ . This implies that Cγ B = Cγ . Dually, we can show that BCγ = Cγ . Thus BCγ = Cγ B = Cγ , i.e., Cγ ≤ B. We now have Aω < Cγ ≤ B for any γ ∈ I d(C \ A). Since Aω ≺ B we conclude that Cγ = B. This implies that C = A ∪ B. Hence A  A ∪ B as required.   Lemma 2.5 Let A ∈ P(S). Then (i) (∀B ∈ P(S)) A ≺ B ⇒ A  A ∪ B. (ii) (∀C ∈ P(S))[A  C ⇒ (∃B ∈ P(S))(C = A ∪ B and A ≺ B)]. Proof Suppose that A ∈ P(S). Then I d(A) = {α} and A = Aα ⊆ Sα for some α ∈ Y . (i) It is a direct consequence of Lemma 2.4. (ii) Assume that A  C for some C ∈ P(S). Then A  = C and A = AC = C A. It follows by (1.1) that α ≤ γ for any γ ∈ I d(C). Also, it is easy to verify that A ≤ A ∪ C ≤ C. This implies that either A = A ∪C or C = A ∪C, since A  C. If A = A ∪C, then C ⊆ A ⊆ Sα . Since A ≤ C, we have by Proposition 2.2 (i) that A = C, contradicting the fact that A  = C. Thus C = A ∪ C and so α ∈ I d(C) and A = Aα ⊆ Cα . In addition, since A ≤ C, we can find by Proposition 2.2 (ii) that Cα ⊆ Aα = A. Hence A = Cα and so |I d(C)| ≥ 2, since A  = C. In the following we shall show that |I d(C)| = 2. Suppose now that |I d(C)| ≥ 3. Then there exist β, δ ∈ I d(C)  such that α < β < δ, since α ≤ γ for any γ ∈ I d(C) and I d(C) is a chain. Let H = γ ∈I d(C),γ ≤β Cγ . By Result 1.1, it is easy to verify that A < H < C, contradicting the fact that A  C. Thus |I d(C)| = 2. Since α ∈ I d(C), we may assume that I d(C) = {α, β} with α < β. Thus C = Cα ∪Cβ = A ∪ Cβ . Let B = Cβ . Then, by Remark 2.1, A < B. If A < K < B for some K ∈ P(S), then it is easy to see that A < A ∪ K < A ∪ B = C, contradicting the fact that A  C. This shows that A ≺ B.   Recall that P(S) = {A ∈ P(S) : |I d(A)| = 1}. Next we shall show that the restriction ψ| P(S) of ψ to P(S) is a bijection from P(S) onto P(S  ).

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Let A, B, C, D ∈ P(S) and B  = C. We use the notation of a topknot, which is introduced in [14], to describe the configuration of arrows as shown on the diagram below.

B

•

•D

A•

•

C Diagram 1 Proposition 2.6 ψ| P(S) is a bijection from P(S) onto P(S  ). Proof Assume that A ∈ P(S) and I d(A) = {α}. To show that ψ(A) ∈ P(S  ), we must show that |I d(ψ(A))| = 1. In fact, we need only consider the following two cases. Case 1: |I d(ψ(A))| ≥ 3. In this case, there exist a ∈ ψ(A)ω , b ∈ ψ(A)δ and c ∈ ψ(A)γ such that δ < ω, γ < ω in   and b  = c. By Lemma 2.3, we get a topknot of ψ(A) (see Diagram 2). ψ(A) \ {b}

A∪B

•

ψ(A) •

•

• ψ(A) \ {b, c} •

A•

•W

•

ψ(A) \ {c}

A∪C

Diagram 2

Diagram 3

Since ψ(A)  ψ(A) \ {b} and ψ(A)  ψ(A) \ {c}, we have A  ψ −1 (ψ(A) \ {b}) and A  ψ −1 (ψ(A) \ {c}). It follows by Lemma 2.5 that ψ −1 (ψ(A) \ {b}) = A ∪ B and ψ −1 (ψ(A) \ {c}) = A ∪ C for some B, C ∈ P(S) with A ≺ B, A ≺ C. Applying ψ −1 to Diagram 2, and letting W = ψ −1 (ψ(A) \ {b, c}), we get a topknot of ψ −1 (ψ(A)) = A (see Diagram 3). In the following we shall show that A ∪ B ∪ C = W . Indeed, by Result 1.1 (iii), we have ψ(A) = (ψ(A) \ {b})(ψ(A) \ {c}) = (ψ(A) \ {c})(ψ(A) \ {b}). Applying

ψ −1

to both sides of the above equation, we immediately have A = (A ∪ B)(A ∪ C) = (A ∪ C)(A ∪ B).

(2.3)

This implies that A ∪ B = (A ∪ B)(A ∪ B ∪ C) = (A ∪ B ∪ C)(A ∪ B), and A ∪ C = (A ∪ C)(A ∪ B ∪ C) = (A ∪ B ∪ C)(A ∪ C). That is to say, A ∪ B ≤ A ∪ B ∪ C and A ∪ C ≤ A ∪ B ∪ C.

(2.4)

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Also, since A ∪ B  W and A ∪ C  W , we have A ∪ B = (A ∪ B)W = W (A ∪ B) and A ∪ C = (A ∪ C)W = W (A ∪ C). It follows that A ∪ B ∪ C = (A ∪ B ∪ C)W = W (A ∪ B ∪ C), i.e., A ∪ B ∪ C ≤ W.

(2.5)

By (2.4) and (2.5), we conclude that A ∪ B ∪ C = W , since A ∪ B  W and A ∪ C  W . Notice that B, C ∈ P(S). We may assume that I d(B) = {β} and I d(C) = {γ }. Then α < β and α < γ , since A ≺ B and A ≺ C. By (1.1) and (2.3), we obtain that I d(A) = I d((A ∪ B)(A ∪ C)) = I d(A ∪ B)I d(A ∪ C), that is to say, {α} = {α, β}{α, γ }. Thus α = βγ . By Result 1.1, β and γ are comparable. This implies that α = βγ = β or α = βγ = γ , contradicting the fact that α < β and α < γ . Case 2: |I d(ψ(A))| = 2. Without loss of generality, we assume that I d(ψ(A)) = {δ, ω} with δ < ω. If |ψ(A)δ | ≥ 2, then there exist a ∈ ψ(A)ω and b, c ∈ ψ(A)δ such that b  = c. This is similar to Case 1. If |ψ(A)δ | = 1, say ψ(A)δ = {u}, then, by Lemma 2.3, ψ(A)  (ψ(A) \ {u}) = ψ(A)ω and so A  ψ −1 (ψ(A)ω ). It follows by Lemma 2.5 that ψ −1 (ψ(A)ω ) = A ∪ U and A ≺ U for some U ∈ P(S) with α < μ, where {μ} = I d(U ). Thus, for any a ∈ A, again by Lemma 2.3, A ∪ U  (A ∪ U ) \ {a} and so ψ(A ∪ U ) = ψ(A)ω  ψ((A ∪ U ) \ {a}). This implies by Lemma 2.5 that ψ((A ∪ U ) \ {a}) = ψ(A)ω ∪ V and ψ(A)ω ≺ V for some V ∈ P(S  ) and ω < ν, where {ν} = I d(V ). Thus, by Lemma 2.4, ψ(A)  ψ(A) ∪ V . It follows by Lemma 2.3 that ψ(A) ∪ V  ψ(A)ω ∪ V and so ψ −1 (ψ(A) ∪ V )  ψ −1 (ψ(A)ω ∪ V ) = (A ∪ U ) \ {a}. Let ψ −1 (ψ(A) ∪ V ) = D. By the above arguments, we have the following topknots (see Diagrams 4 and 5): ψ(A)ω

A∪U

•

ψ(A)•

•

• ψ(A)ω ∪ V

•

ψ(A) ∪ V Diagram 4

• (A ∪ U ) \ {a}

A•

•

D Diagram 5

In the following we shall show that A ∪ U ∪ D = A ∪ U . In fact, since ψ(A)  ψ(A)ω  ψ(A)ω ∪ V , we have ψ(A)(ψ(A)ω ) = (ψ(A)ω )ψ(A) = ψ(A) and ψ(A)ω (ψ(A)ω ∪ V ) = (ψ(A)ω ∪ V )ψ(A)ω = ψ(A)ω . It follows that ψ(A) = ψ(A)ω (ψ(A) ∪ V ) = (ψ(A) ∪ V )ψ(A)ω . Applying

ψ −1

to both sides of the above equation, we immediately have A = (A ∪ U )D = D(A ∪ U ).

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(2.6)

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This implies that A ∪ U = (A ∪ U ∪ D)(A ∪ U ) = (A ∪ U )(A ∪ U ∪ D), i.e., A ∪ U ≤ A ∪ U ∪ D.

(2.7)

Also, since A ∪ U  (A ∪ U ) \ {a} and D  (A ∪ U ) \ {a}, we have A ∪ U = (A ∪ U )((A ∪ U ) \ {a}) = ((A ∪ U ) \ {a})(A ∪ U ) and D = D((A ∪ U ) \ {a}) = ((A ∪ U ) \ {a})D. It follows that A ∪ U ∪ D ≤ (A ∪ U ) \ {a}.

(2.8)

Since A ∪ U  (A ∪ U ) \ {a}, we have by (2.7) and (2.8) that A ∪ U ∪ D = A ∪ U or (A ∪U )\{a}. Notice that A and U are disjoint. We have (A ∪U )\{a}  A ∪U ⊆ A ∪U ∪ D, and so A ∪ U ∪ D = A ∪ U . We now claim that I d(D) = {α} = I d(A). In fact, for any ξ ∈ I d(D), since I d(U ) = {μ}, it follows by Result 1.1 that μ and ξ are comparable. By (1.1) and (2.6), we have I d(A) = I d((A ∪ U )D) = I d(A ∪ U )I d(D), and so {α} = {α, μ}{ξ }. This implies that μξ = α. Thus μ = α or ξ = α, since μ and ξ are comparable. Notice that α < μ. We have that ξ = α and so I d(D) = {α} = I d(A). Since A  D, by Proposition 2.2 (i) we obtain A = D, a contradiction. Hence |I d(ψ(A))| = 1 and so ψ(A) ∈ P(S  ). Similarly, we can show that ψ −1 (B) ∈ P(S) for any B ∈ P(S  ). This completes the proof.   Proposition 2.7 There exists a semilattice isomorphism θ from  onto   such that ψ| P(Sα ) is an isomorphism from P(Sα ) onto P(Sθ (α) ) for every α ∈ . Proof Take any α ∈  and b ∈ Sα . Then {b} ∈ P(S). It follows by Proposition 2.6 that ψ(b) ∈ P(S  ). Thus I d(ψ(b)) = {α  } for some α  ∈   . In the following we shall show that ψ(A) ∈ P(Sα  ) for any A ∈ P(Sα ). In fact, for any A ∈ P(Sα ), since Sα is either a left or right zero semigroup, we have that A = A2 = Ab A and b = b Ab. Thus ψ(A) = ψ(A)ψ(b)ψ(A) and ψ(b) = ψ(b)ψ(A)ψ(b). It follows by (1.1) that I d(ψ(A)) = I d(ψ(A))I d(ψ(b)) = I d(ψ(b)) = {α  }, and so ψ(A) ∈ P(Sα  ). This shows that ψ| P(Sα ) is a mapping from P(Sα ) to P(Sα  ). Applying the above arguments to ψ −1 , we can also prove that ψ −1 | P(S   ) is a mapping from P(Sα  ) to α P(Sα ). Thus ψ| P(Sα ) is an isomorphism from P(Sα ) onto P(Sα  ). Define a mapping θ from  to   by θ (α) = I d(ψ(aα )), where aα is an element in Sα and we identify the singleton I d(ψ(aα )) with the element it contains. It is easy to see that θ is bijective and θ −1 (β  ) = I d(ψ −1 (bβ  )), where bβ  is an element in Sβ  and we identify the singleton I d(ψ −1 (bβ  )) with the element it contains. Also, for any α, β ∈  and for any aα ∈ Sα , bβ ∈ Sβ , we have by the definition of θ that θ (αβ) = I d(ψ(aα bβ )), since aα bβ ∈ Sα Sβ ⊆ Sαβ . Thus, by (1.1), θ (αβ) = I d(ψ(aα bβ )) = I d(ψ(aα ))I d(ψ(bβ )) = θ (α)θ (β).

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We now have shown that θ is a semilattice isomorphism from  onto   and ψ| P(Sα ) is an isomorphism from P(Sα ) onto P(Sθ (α) ). This completes the proof.   Theorem 2.8 The class A2 of all semigroups having idempotent globals is globally determined.   Proof Assume that S = γ ∈ Sγ , S  = γ  ∈  Sγ  are semigroups in A2 and that ψ is an isomorphism from P(S) onto P(S  ). By Proposition 2.7, there exists a semilattice isomorphism θ from  onto   such that ψ| P(Sα ) is an isomorphism from P(Sα ) onto P(Sθ (α) ) for every α ∈ . Since the class of all rectangular bands is globally determined (see [13]), Sα is isomorphic to Sθ (α) . Let ηα be an isomorphism from Sα onto Sθ (α) and let η = α∈ ηα , that is to say, η(a) = ηα (a) if a ∈ Sα . In the remainder we shall show that η is an isomorphism from S onto S  . In fact, it is easy to see that η is bijective. Suppose now that a ∈ Sα and b ∈ Sβ . Then η(a) = ηα (a) ∈ Sθ (α) , η(b) = ηβ (b) ∈ Sθ (β) . Notice that α and β are comparable. We consider the following cases: • α = β. Then η(ab) = ηα (ab) = ηα (a)ηα (b) = η(a)η(b). • α < β. Then θ (α) < θ (β), since θ is a semilattice isomorphism from  onto   . Thus, by Result 1.1, we have that ab = a and ηα (a)ηβ (b) = ηα (a). It follows that η(ab) = η(a) = ηα (a) = ηα (a)ηβ (b) = η(a)η(b). • α > β. This case is similar to the preceding one. Therefore, η is an isomorphism from S onto S  . This completes the proof.

 

3 On semigroups having regular but non-idempotent globals In this section we study the global determinism of the class A3 of all semigroups having regular but non-idempotent globals. Assume that T, T  are semigroups in A3 and that ϕ is an isomorphism from P(T ) onto P(T  ). By Result 1.2, we may write   T = Tγ and T  = Tγ  , γ ∈Y

γ  ∈Y 

where Y (Y  , resp.) is a chain with maximum element ε (ε  , resp.), Tε , Tε are cyclic groups of order two, and the semigroups Tγ (γ  = ε) (Tγ  (γ   = ε  ), resp.) are pairwise disjoint left or right zero subsemigroups of T (T  , resp.). Let Tε = {1, a} and Tε = {1 , a  }, where 1 (1 , resp.) denotes the identity of the group Tε (Tε , resp.). It is easy to see that 1 (1 , resp.) is also the identity of the semigroup T (T  , resp.). To show that A3 satisfies the strong isomorphism property, i.e., to show that ϕ|T is an  isomorphism from T onto T  , we first characterize the subsets E P(T ) and E P(T ) of P(T ). Recall that E P(T ) denotes the set of all idempotents of P(T ) and  E P(T ) = {A ∈ E P(T ) : (∀B ∈ P(T ))B 2 = A ⇒ B = A}.  P(T ), resp.) onto Obviously, ϕ| E P(T ) (ϕ| EP(T ) , resp.) is an isomorphism from E P(T ) ( E    P(T ), resp.). E P(T ) ( E Lemma 3.1 Let A ∈ P(T ). Then (i) A ∈ E P(T ) ⇔ A ∈ P(T \ {a}) or {1, a} ⊆ A.

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(ii) A ∈  E P(T ) ⇔ A ∈ P(T \ {1, a}) or {1, a} ⊆ A. Proof Suppose that A ∈ P(T ). (i) By Result 1.2, it is easy to verify that (i) holds.  (ii) Assume that A ∈ E P(T ). Then A ∈ E P(T ) and so A ∈ P(T \ {a}) or {1, a} ⊆ A by (i). We need only consider the case A ∈ P(T \ {a})). If A ∈ / P(T \ {1, a}), then a ∈ / A and 1 ∈ A. Let B = {a} ∪ (A \ {1}). It is easy to verify that A = B 2 but A  = B, contradicting  the fact that A ∈ E P(T ). Thus A ∈ P(T \ {1, a}). Conversely, if A ∈ P(T \ {1, a}) or {1, a} ⊆ A, then A ∈ E P(T ) by (i). Assume that  C ∈ P(T ) and C 2 = A. To show that A ∈ E P(T ), we must show that C = A. In fact, if C ∈ E P(T ), then C = C 2 = A. If C ∈ / E P(T ), then it follows by (i) that a ∈ C and 1 ∈ / C. This implies that A = C 2 = {1} ∪ (C \ {a}), contradicting the fact that A ∈ P(T \ {1, a}) or {1, a} ⊆ A. This completes the proof.   Following Gould and Iskra [16], we denote the set of all idempotents of a semigroup S by E(S), and for each e ∈ E(S), we denote the maximal subgroup H-class of S containing e by He (S). The following lemma will be useful to us, which implies that the class of all groups satisfies the strong isomorphism property. Lemma 3.2 [16, Lemma 2.1] Let S be a semigroup and e ∈ E(S). Then He (P(S)) = He (S). Theorem 3.3 The class A3 of all semigroups having regular but non-idempotent globals satisfies the strong isomorphism property.   Proof Assume that T = γ ∈Y Tγ , T  = γ  ∈Y  Tγ  are semigroups in A3 and that ϕ is an isomorphism from P(T ) onto P(T  ). Since 1 (1 , resp.) is the identity of the semigroup T (T  , resp.), we have that 1 (1 , resp.) is the identity of the power semigroup P(T ) (P(T  ), resp.). Thus ϕ(1) = 1 . It follows by Lemma 3.2 that ϕ(a) = a  . Let A ∈ P(T ). In the following we shall show that a ∈ A implies a  ∈ ϕ(A). In fact, we need only consider the following cases: • a ∈ A and 1 ∈ / A. It follows by Lemma 3.1 (i) that A2  = A and so (ϕ(A))2  = ϕ(A). Again by Lemma 3.1 (i) we have a  ∈ ϕ(A).  • A = {1, a}. It follows by Lemma 3.1 (ii) that A ∈ E P(T ). Since ϕ| EP(T ) is an    isomorphism from E P(T ) onto E P(T  ), ϕ(A) ∈ E P(T  ). Again by Lemma 3.1 (ii), we have that ϕ(A) ∈ P(T  \{1 , a  }) or {1 , a  } ⊆ ϕ(A). This implies that {1 , a  }ϕ(A) = ϕ(A). Thus (ϕ −1 ({1 , a  })){1, a} = {1, a}, and so ϕ −1 ({1 , a  }) ⊆ {1, a}. Also, since ϕ(1) = 1 and ϕ(a) = a  , we obtain that ϕ −1 ({1 , a  }) = {1, a}. Hence ϕ({1, a}) = {1 , a  }. • {1, a} ⊆ A and A  = {1, a}. It is easy to verify that A = {1, a}(A \ {1}). By the preceding cases, we have that a  ∈ ϕ(A \ {1}) and ϕ({1, a}) = {1 , a  }. Thus a  = 1 a  ∈ ϕ({1, a})ϕ(A \ {1}) = ϕ(A). We have shown that a ∈ A implies a  ∈ ϕ(A). Similarly, we can show that a  ∈ ϕ(A) implies a ∈ A. Therefore, a ∈ A if and only if a  ∈ ϕ(A). This implies that ϕ| P(T \{a}) is an isomorphism from P(T \ {a}) onto P(T  \ {a  }). Notice that both T \ {a} and T \ {a  } are idempotent semigroups with identity, and the class of all idempotent semigroups with identity

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satisfies the strong isomorphism property (see [16]). We have that ϕ|T \{a} is an isomorphism from T \ {a} onto T  \ {a  }, which, together with the fact that ϕ(a) = a  , gives ϕ|T is an isomorphism from T onto T  . This completes the proof.   By Theorems 2.8 and 3.3, we immediately have Theorem 3.4 The class A of all semigroups having regular globals is globally determined. Acknowledgments The authors are particularly grateful to the referee for his valuable comments and suggestions which lead to a substantial improvement of this paper. This work is supported by National Natural Science Foundation of China (11261021, 11571278) and Grant of Natural Science Foundation of Shannxi Province (2011JQ1017) and Natural Science Foundation of Jiangxi Province (20142BAB201002).

References 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16.

J.M. Howie, Fundamentals of Semigroup Theory (Clarendon Press, Oxford, 1995) M. Petrich, N.R. Reilly, Completely Regular Semigroups (Wiley-Interscience, New York, 1999) L. Rédei, Algebra I (Pergamon Press, Oxford, 1967) J. Almeida, On power varieties of semigroups. J. Algebra 120, 1–17 (1989) J.E. Pin, Power semigroups and related varieties of finite semigroups, in Semigroups and Their Applications, ed. by Simon M. Goberstein, Peter M. Higgins (Reidel, Dordrecht, 1987), pp. 139–152 T. Tamura, On the recent results in the study of power semigroups, in Semigroups and Their Applications, ed. by Simon M. Goberstein, Peter M. Higgins (Reidel, Dordrecht, 1987), pp. 191–200 J. Pelikán, On semigroups, in which products are equal to one of the factors. Periodica Math. Hungarica 4, 103–106 (1973) B. Pondˇelíˇcek, On semigroups having regular globals. Colloquia Math. Soc. János Bolyai. 20. Algebraic Theory of Semigroup, Szeged (Hungary), 453–461 (1976) T. Tamura, Unsolved problems on semigroups. Kokyuroku, Kyoto Univ. 31, 33–35 (1967) T. Tamura, J. Shafer, Power semigroups. Math. Jpn. 12, 25–32 (1967) E.M. Mogiljanskaja, Non-isomorphic semigroups with isomorphic semigroups of subsets. Semigroup Forum 6, 330–333 (1973) A.P. Gan, X.Z. Zhao, Global determinism of Clifford semigroups. J. Aust. Math. Soc. 97, 63–77 (2014) T. Tamura, Power semigroups of rectangular groups. Math. Jpn. 4, 671–678 (1984) Y. Kobayashi, Semilattices are globally determined. Semigroup Forum 29, 217–222 (1984) T. Tamura, Isomorphism Problem of Power Semigroups of completely 0-simple semigroups. J. Algebra 98, 319–361 (1986) M. Gould, J.A. Iskra, Globals of completely regular periodic semigroups. Semigroup Forum 29, 365–374 (1984)

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