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Algebra Colloquium 14 : 2 (2007) 215–228

Algebra Colloquium

Algebra Colloq. 2007.14:215-228. Downloaded from www.worldscientific.com by CHINESE UNIVERSITY OF HONG KONG on 09/09/18. Re-use and distribution is strictly not permitted, except for Open Access articles.

c 2007 AMSS CAS ° & SUZHOU UNIV

On Superabundant Semigroups Whose Set of Idempotents Forms a Subsemigroup∗ X.M. Ren Department of Mathematics Xi’an University of Architecture and Technology, Xi’an 710055, China E-mail: [email protected]

K.P. Shum Faculty of Science, The Chinese University of Hong Kong Shatin, N.T., Hong Kong (SAR), China E-mail: [email protected] Received 20 April 2005 Revised 4 July 2006 Communicated by A.C. Kim Abstract. In this paper, we study the structure of a superabundant semigroup S whose set of idempotents E(S) forms a subsemigroup. We call such a semigroup a cyber-group because it is a generalization of orthogroups in the class of completely regular semigroups studied by Petrich and Reilly. We show that a cyber-group can be expressed as a semilattice of rectangular monoids. Thus, our result generalizes the well-known result obtained by Petrich in 1987 for orthogroups. Some properties of cyber-groups are given and some special superabundant semigroups are discussed. 2000 Mathematics Subject Classification: 20M15 Keywords: cyber-groups, orthogroups, superabundant semigroups, rectangular monoids, left C-a semigroups

1 Introduction Recall that an abundant semigroup S is a semigroup in which every L∗ -class and every R∗ -class of S contain an idempotent. An abundant semigroup S is called superabundant if every H∗ -class of S contains an idempotent. It is well known that L∗ = L, R∗ = R and H∗ = H hold in a regular semigroup S. Clearly, a superabundant semigroup is a natural generalization of a completely regular semigroup in the class of abundant semigroups. ∗

The first author is supported by the National Natural Science Foundation of China (grant no. 10671151); and the second author is partially supported by a UGC (HK) grant.


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Abundant semigroups and superabundant semigroups were first studied by Fountain in [3]. Later on, the class of abundant semigroups and some of its special subclasses have been extensively investigated by many authors (for example, see [2, 4, 12]). In particular, we notice that Guo, Guo and Shum [4] have investigated left C-a semigroups S which are superabundant and in which the relation L∗ is a semilattice congruence on S. They have shown that the set of all idempotents of a left C-a semigroup forms a left regular band B, that is, aba = ab holds for all a, b ∈ B. Among regular semigroups, the class of completely regular semigroups is the most important and most interesting class of regular semigroups. In the literature, a completely regular semigroup S is called an orthogroup if all its idempotents form a subsemigroup of S (see [9]). As an analogue of orthogroups in the class of completely regular semigroups, we now call a superabundant semigroup whose set of idempotents E(S) forms a subsemigroup of S a “cyber-group” in the class of superabundant semigroups. Of course, a cyber-group is obviously not necessarily a completely regular semigroup. Although the structure theory of superabundant semigroups has recently been fully developed in [11], the structure of cyber-groups has not yet been known. However, the structure of left cyber-groups which is a kind of abundant semigroups whose set of idempotents is a left regular band has been recently investigated by Guo and Shum in [5]. In this paper, we observe that orthogroups are special cyber-groups and give a structure theorem for cyber-groups. Our result on cyber-groups gives a substantial generalization of the structure theorem of orthogroups obtained by Petrich [8] in 1987 (see also [15] and [16]). As a special case of cyber-groups, we also re-obtain a structure theorem for left C-a semigroups (dually, right C-a semigroups) in the literature (see [4, 13, 14]). Some non-trivial examples of cyber-groups are provided in this paper. For notations and terminologies not mentioned in this paper, the reader is referred to [1, 3, 5, 6, 7, 9, 10]. 2 Preliminaries We first recall some basic results of abundant semigroups given by Fountain in [3]. Lemma 2.1. [3] Let S be a semigroup, a, b ∈ S, and S 1 the semigroup S adjoined with the identity 1. Then the following statements hold : (i) a L∗ b if and only if ax = ay ⇔ bx = by for all x, y ∈ S 1 . (ii) For any idempotent e ∈ S, e L∗ a if and only if ae = a and ax = ay ⇒ ex = ey for all x, y ∈ S 1 . Trivially, we see that L ⊆ L∗ holds on a semigroup S, and for any regular elements a, b ∈ S, (a, b) ∈ L∗ if and only if (a, b) ∈ L. The dual results for R∗ also hold in a semigroup S. Lemma 2.2. [3] A semigroup S is a superabundant semigroup if and only if S is a semilattice Y of completely J ∗ -simple semigroups Sα (α ∈ Y ) such that for any α ∈ Y and a ∈ Sα , we have L∗a (S) = L∗a (Sα ) and Ra∗ (S) = Ra∗ (Sα ).


Superabundant Semigroups

217

Suppose that µ(T ; I, Λ; P ) is a Rees matrix semigroup and P is a Λ × I matrix over the cancellative monoid T . According to [11], the sandwich matrix P is said to be normalized at 1 if there exists an element 1 ∈ I ∩ Λ such that p1i = pλ1 = e for all i ∈ I and λ ∈ Λ, where e is the identity of the cancellative monoid T . In this case, the Rees matrix semigroup µ(T ; I, Λ; P ) is said to be normalized. Lemma 2.3. [11] Let T be a cancellative monoid with identity e, and I, Λ nonempty sets. Let P = (pλi ) be a Λ × I matrix, where each entry in P is a unit of T . Then the normalized Rees matrix semigroup µ(T ; I, Λ; P ) is a completely J ∗ -simple semigroup. Conversely, every completely J ∗ -simple semigroup is isomorphic to a normalized Rees matrix semigroup µ(T ; I, Λ; P ) over a cancellative monoid T . We now formulate the following definition. Definition 2.4. Let S be a superabundant semigroup. If the set of all idempotents of S forms a subsemigroup of S, then we call S a cyber-group. If T is a cancellative monoid, I and Λ are a left zero band and a right zero band, respectively, then the product of T , I and Λ is called a rectangular monoid. We now give the following description for rectangular monoids. Lemma 2.5. Let S be a completely J ∗ -simple semigroup. Then the set of idempotents of S forms a subsemigroup of S if and only if S is a rectangular monoid. Proof. We only need to show the necessity part because the sufficiency part is almost immediate. Let S be a completely J ∗ -simple semigroup. Then by Lemma 2.3, S can be expressed as a normalized Rees matrix semigroup µ(T ; I, Λ; P ) over a cancellative monoid T in which every entry in P is a unit. Now let E(S) be the set of all idempotents of S. Suppose that a = (1, p−1 λ1 , λ) −1 and b = (i, p−1 , 1) ∈ E(S). Since P is normalized at 1 ∈ Λ × I, we have p 1i 1i = −1 pλ1 = e, where e is the identity element of the monoid T . Because E(S) forms a subsemigroup of S, we have ab = (1, e, λ)(i, e, 1) = (1, epλi e, 1) ∈ E(S) and hence pλi = p11 = e for all i ∈ I and λ ∈ Λ. To finish our proof, it suffices to show that the mapping φ : µ(T ; I, Λ; P ) → I × T × Λ given by (i, x, λ)φ = (i, x, λ) is a semigroup isomorphism. The proof is straightforward and hence S ' I ×T ×Λ. 2 By using the above lemmas, we are now able to give the following theorem for cyber-groups. Theorem 2.6. A semigroup S is a cyber-group if and only if S is a semilattice Y of rectangular monoids Sα = Iα × Tα × Λα (α ∈ Y ) such that for any α ∈ Y and a = (i, x, λ) ∈ Sα , we have a H∗ (S) a0 , where a0 = (i, eα , λ) and eα is the identity of Tα . Proof. Suppose that S is a cyber-group. Then S is of course a superabundant semigroup. By Lemmas 2.2 and 2.5, S is a semilattice Y of rectangular monoids


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Sα = Iα × Tα × Λα . Let a = (i, x, λ) ∈ Sα = Iα × Tα × Λα and a0 = (i, eα , λ), where eα is the identity of the cancellative monoid Tα . Then we can easily verify a H∗ (Sα ) a0 and so a H∗ (S) a0 by Lemma 2.2. Conversely, let α ∈ Y and a ∈ Sα . Then there exists an idempotent a0 of S such that a H∗ (S) a0 . This implies that S is a superabundant semigroup. We still need to prove that E(S) forms a subsemigroup of S. Suppose that S is a semilattice Y of Sα (α ∈ Y ), where each Sα is a rectangular monoid. Let e, f ∈ E(S). Then e ∈ E(Sα ) and f ∈ E(Sβ ) for some α, β ∈ Y and (ef )0 ∈ E(Sαβ ). Since ef · f = ef and ef L∗ (S) (ef )0 , it follows by Lemma 2.1 that (ef )0 f = (ef )0 . Similarly, we have e(ef )0 = (ef )0 . Let g = f (ef )0 and h = (ef )0 e. Then h, g ∈ Sαβ , g 2 = f (ef )0 f (ef )0 = f (ef )0 = g and h2 = (ef )0 e(ef )0 e = h. Consequently, we have g, h ∈ E(Sαβ ). But since hg = (ef )0 ef (ef )0 = ef , we deduce ef ∈ E(Sαβ ) ⊆ E(S) because E(Sαβ ) is a rectangular band. Thus, our proof is completed. 2 We now give a non-trivial example of cyber-groups, which shows that the class of cyber-groups contains the class of orthogroups as one of its proper subclasses. In fact, to construct such an example is not easy because we have to provide sufficiently many idempotents for a semigroup S so that it will be a superabundant semigroup. In general, a conceptual example will be difficult to illustrate the distinction between cyber-groups and orthogroups unless all the information of idempotents in the semigroup is given and described. Example 2.7. We start with the following elements:     a11

1 0 = 0 0



a21

0 1 = 0 0

1 0 0 0

0 0 0 0

0 0 , 0 0

0 1 0 0

0 0 0 0

0 0 , 0 0

a12

1 0 = 0 0





a22

0 1 = 0 0

1 0 0 0

1 0 0 0

0 0 , 0 0

0 1 0 0

0 1 0 0

0 0 , 0 0



a13



1 0 = 0 0



a23

0 1 = 0 0



1 0 0 0

1 0 0 0

1 0 , 0 0

0 1 0 0

0 1 0 0

0 1 . 0 0



Then we form the sets Sij = {3n aij | n ∈ N}, where i = 1, 2 and j = 1, 2, 3. It can be easily verified that the set Sα = S11 ∪ S12 ∪ S13 ∪ S21 ∪ S22 ∪ S23 is indeed a semigroup under the usual matrix multiplication in which the elements a11 , a12 , a13 , a21 , a22 and a23 are idempotents. Clearly, the sets S11 , S12 , S13 , S21 , S22 and S23 are subsemigroups of the semigroup Sα , and Sij is generated by aij . Also, we can easily see that the set of all idempotents of Sα forms a rectangular band. We now construct a semigroup Sβ = {e11 , e12 , e13 , e21 , e22 , e23 , a, b, c, d, e, f, g, h, s, t, u, v} (see Table 1). It is easy to check that Sβ is a completely regular semigroup and E(Sβ ) = {e11 , e12 , e13 , e21 , e22 , e23 } forms a rectangular band in Sβ under the semigroup multiplication so that Sβ is a rectangular group.

219



Table 1. The Cayley table of Sβ ∗ e11 e12 e13 e21 e22 e23 a b c d e f g h s t u v

e11 e11 e11 e11 e21 e21 e21 a a a d d d g g g t t t

e12 e12 e12 e12 e22 e22 e22 b b b e e e h h h u u u

e13 e13 e13 e13 e23 e23 e23 c c c f f f s s s v v v

e21 e11 e11 e11 e21 e21 e21 a a a d d d g g g t t t

e22 e12 e12 e12 e22 e22 e22 b b b e e e h h h u u u

e23 e13 e13 e13 e23 e23 e23 c c c f f f s s s v v v

a a a a d d d g g g t t t e11 e11 e11 e21 e21 e21

b b b b e e e h h h u u u e12 e12 e12 e22 e22 e22

c c c c f f f s s s v v v e13 e13 e13 e23 e23 e23

d a a a d d d g g g t t t e11 e11 e11 e21 e21 e21

e b b b e e e h h h u u u e12 e12 e12 e22 e22 e22

f c c c f f f s s s v v v e13 e13 e13 e23 e23 e23

g g g g t t t e11 e11 e11 e21 e21 e21 a a a d d d

h h h h u u u e12 e12 e12 e22 e22 e22 b b b e e e

s s s s v v v e13 e13 e13 e23 e23 e23 c c c f f f

t g g g t t t e11 e11 e11 e21 e21 e21 a a a d d d

u h h h u u u e12 e12 e12 e22 e22 e22 b b b e e e

v s s s v v v e13 e13 e13 e23 e23 e23 c c c f f f

We define a multiplication ∗ on S = Sα ∪ Sβ by extending the matrix multiplications on Sα and Sβ as follows: For any a ∈ Sα and b ∈ Sβ , we define a∗b = b∗a = b. Then S equipped with the above multiplication ∗ becomes a semigroup. By Lemma 2.1 and its dual, we can verify that the L∗ -classes of S are the sets {3n a11 , 3n a21 }, {3n a12 , 3n a22 }, {3n a13 , 3n a23 }, {e11 , e21 , a, d, g, t}, {e12 , e22 , b, e, h, u}, {e13 , e23 , c, f, s, v}, where n ∈ N. Also, the R∗ -classes of S are the sets {3n a11 , 3n a12 , 3n a13 }, {3n a21 , 3n a22 , 3n a23 }, {e11 , e12 , e13 , a, b, c, g, h, s}, {e21 , e22 , e23 , d, e, f, t, u, v}, where n ∈ N. Consequently, the H∗ -classes of S are the sets {3n a11 }, {3n a12 }, {3n a13 }, {3n a21 }, {3n a22 }, {3n a23 }, {e11 , a, g}, {e12 , b, h}, {e13 , c, s}, {e21 , d, t}, {e22 , e, u}, {e23 , f, v}, where n ∈ N. This shows that the semigroup S is indeed abundant and every H∗ class of S contains an idempotent, and in particular, S is superabundant. Since the set of all idempotents of S forms a subsemigroup, S is a cyber-group. Because every element of S \ {Sβ ∪ {a11 , a12 , a13 , a21 , a22 , a23 }} is non-regular, S itself is of course not an orthogroup. This example illustrates that the class of orthogroups is indeed a proper subclass of the class of cyber-groups. 3 Structure of Cyber-groups In this section, we describe the structure of cyber-groups.


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Theorem 3.1. Let Y be a semilattice. For each α in Y , let Sα = Iα ×S Tα × Λα be a rectangular monoid such that Sα ∩ Sβ = ∅ if α 6= β. Let T = α∈Y Tα be a strong semilattice of cancellative monoids Tα (α ∈ Y ) and form the disjoint · S set union S = α∈Y Sα . Suppose that 1α is a fixed element in Iα ∩ Λα . For any α, β ∈ Y with α > β, we define the mappings ϕα,β : Sα × Iβ → Iβ ,

(a, i) 7→ ha, ii

φα,β : Λβ × Sα → Λβ ,

(λ, a) 7→ [λ, a]

and satisfying the following requirements (I)–(III): (I) Let a = (i, x, λ) ∈ Sα and b = (j, y, µ) ∈ Sβ (α, β ∈ Y ). If k ∈ Iα and ν ∈Λ Sα , then ha, ki = i and [ν, a] = λ. Define a multiplication ◦ on the set S = α∈Y Sα by a ◦ b = (ha, hb, 1αβ ii, xy, [[1αβ , a], b]).

(1)

(II) If α, β, γ ∈ Y with αβ > γ, k ∈ Iγ and ν ∈ Λγ , then ha, hb, kii = ha ◦ b, ki and [[ν, a], b] = [ν, a ◦ b]. (III) For any a = (i, x, λ) ∈ Sα , denote a0 = (i, eα , λ), where eα is the identity of the cancellative monoid Tα . Then for any u, v ∈ S 1 , we assume that (i) a ◦ u = a ◦ v implies a0 ◦ u = a0 ◦ v; (ii) u ◦ a = v ◦ a implies u ◦ a0 = v ◦ a0 . Thus, the set S under the above multiplication ◦ forms a cyber-group, denoted by S = S(Y, Sα , ϕα,β , φα,β ), and the multiplication ◦ restricted to each rectangular monoid Sα coincides with the given multiplication on Sα . Conversely, every cyber-group is isomorphic to some S(Y, Sα , ϕα,β , φα,β ) which can be constructed in the above manner. Proof. We first prove the necessity part of Theorem 3.1. Suppose that a = (i, x, λ) ∈ Sα , b = (j, y, µ) ∈ Sβ and c = (k, z, ν) ∈ Sγ for some α, β, γ ∈ Y with αβγ = δ. Then by the above formula (1), we have a ◦ b = (ha, hb, 1αβ ii, xy, [[1αβ , a]b]), where xy is the semigroup product of x and y in T , and hence by the requirement (II) and formula (1), we have (a ◦ b) ◦ c = (ha ◦ b, hc, 1δ ii, (xy)z, [[1δ , a ◦ b], c]) = (ha, hb, hc, 1δ iii, xyz, [[[1δ , a], b], c]). Similarly, we have a ◦ (b ◦ c) = (ha, hb ◦ c, 1δ ii, x(yz), [[1δ , a], b ◦ c]) = (ha, hb, hc, 1δ iii, xyz, [[[1δ , a], b], c]).

221



Thus, (a ◦ b) ◦ c = a ◦ (b ◦ c) and hence S forms a semigroup under the above multiplication. Now by our requirement (I), we see that for any a = (i, x, λ), b = (j, y, µ) ∈ Sα , we have a ◦ b = (ha, hb, 1α ii, xy, [[1α , a], b]) = (i, xy, µ) = (i, x, λ)(j, y, ν) = ab. This implies that the multiplication ◦ on S restricted to each Sα (α ∈ Y ) is exactly the same as the given multiplication on Sα . Indeed, we have already shown that S is a semilattice of rectangular monoids Sα (α ∈ Y ). In order to show that S is a cyber-group, by Theorem 2.6, we only need to prove that for any α ∈ Y and a = (i, x, λ) ∈ Sα , we have a H∗ (S) a0 , where a0 = (i, eα , λ) and eα is the identity of Tα . Suppose that a = (i, x, λ) ∈ Sα = Iα × Tα × Λ for some α ∈ Y . Then we can verify a H∗ (Sα ) a0 , where a0 = (i, eα , λ) ∈ E(S) and eα is the identity of the cancellative monoid Tα . By the requirement (III) and Lemma 2.1 together with its dual, we have a L∗ (S) a0 and a R∗ (S) a0 , and thereby, a H∗ (S) a0 . Thus, the proof of the necessity part is completed. To prove the sufficiency part of Theorem 3.1, we need several lemmas. Lemma 3.2. Suppose that S is a cyber-group. Then the following statements hold : (i) There exists a semilattice Y and aSfamily of rectangular monoids Sα = Iα × Tα × Λα (α ∈ Y ) such that S = α∈Y Sα is a semilattice Y of rectangular monoids Sα . S (ii) If we express the set T by T = α∈Y Tα , where each Tα is a cancellative monoid with the identity eα , then T becomes a strong semilattice [Y ; Tα , θα,β ] of Tα (α ∈ Y ) such that for any α, β ∈ Y with α > β, a = (i, g, λ) ∈ Sα = Iα × Tα × Λα and j ∈ Iβ , we have (i, g, λ)(j, eβ , 1β ) = (−, gθα,β , 1β ) ∈ Sβ .

(2)

Proof. Suppose that S is a cyber-group. Then by Theorem 2.6, S is a semilattice Y of rectangular monoids Sα (α ∈ Y ), each Sα is a product Iα × Tα × Λα , where Iα is a left zero band, Tα is a cancellative monoid and Λα is a right zero band. This establishes the first part of Lemma 3.2. For any α, β ∈ Y with α > β and a = (i, g, λ) ∈ Sα = Iα × Tα × Λα , it is easy to see that (i, g, λ)(1β , eβ , 1β ) ∈ Sβ , where eβ is the identity element of the cancellative monoid Tβ . S Now we consider the set T = α∈Y Tα and define a mapping θα,β : Tα → Tβ ,

g 7→ gθα,β

such that (i, g, λ)(1β , eβ , 1β ) = (−, gθα,β , −).

(3)

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X.M. Ren, K.P. Shum


Then we can easily see that (i, g, λ)(1β , eβ , 1β ) = (−, gθα,β , 1β ).

(4)

Since Sβ is a rectangular monoid and (1β , eβ , 1β )(i, g, λ) ∈ Sβ , there exists a unique element x in Tβ such that (1β , eβ , 1β )(i, g, λ) = (1β , x, −).

(5)

Now by using (4) and (5), we immediately have (1β , eβ , 1β )[(i, g, λ)(1β , eβ , 1β )] = (1β , eβ , 1β )(−, gθα,β , 1β ) = (1β , gθα,β , 1β ) and [(1β , eβ , 1β )(i, g, λ)](1β , eβ , 1β ) = (1β , x, −)(1β , eβ , 1β ) = (1β , x, 1β ). Hence, we have x = gθα,β and (1β , eβ , 1β )(i, g, λ) = (1β , gθα,β , −).

(6)

Next, we can show that for any j ∈ Iβ , we have a(j, eβ , 1β ) = (−, gθα,β , 1β ), where a = (i, g, λ) ∈ Sα . Since a(j, eβ , 1β ) ∈ Sβ , there exist a unique element u ∈ Tβ and an element k ∈ Iβ such that a(j, eβ , 1β ) = (k, u, 1β ). By using this equality and (6), we deduce (1β , eβ , 1β )[a(j, eβ , 1β )] = (1β , eβ , 1β )(k, u, 1β ) = (1β , u, 1β ) and [(1β , eβ , 1β )a](j, eβ , 1β ) = (1β , gθα,β , −)(j, eβ , 1β ) = (1β , gθα,β , 1β ). Thus, we have u = gθα,β , and hence (2) is proved. We still need to prove that the mapping θα,β is indeed a semigroup homomorphism. Since Sα is a rectangular monoid, it is clear that θα,α is the identity mapping on Tα . For any α, β ∈ Y with α > β and a = (i, g, λ), b = (i0 , h, λ0 ) ∈ Sα , by (2), we can immediately deduce = = = = = and

a[b(1β , eβ , 1β )] (i, g, λ)[(i0 , h, λ0 )(1β , eβ , 1β )] (i, g, λ)(j, hθα,β , 1β ) (for some j ∈ Iβ ) (i, g, λ)(j, eβ , 1β )(j, hθα,β , 1β ) (−, gθα,β , 1β )(j, hθα,β , 1β ) (−, gθα,β hθα,β , 1β ) (ab)(1β , eβ , 1β ) = [(i, g, λ)(i0 , h, λ0 )](1β , eβ , 1β ) = (i, gh, λ0 )(1β , eβ , 1β ) = (−, (gh)θα,β , 1β ).

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Hence, we obtain (gh)θα,β = gθα,β hθα,β and this proves that θα,β is a homomorphism from Tα into Tβ . For any α, β, γ ∈ Y with α > β > γ, by (2), we obtain (i, g, λ)[(1β , eβ , 1β )(1γ , eγ , 1γ )] = (i, g, λ)(k, eγ , 1γ ) (for some k ∈ Iγ ) = (−, gθα,γ , 1γ ) and

[(i, g, λ)(1β , eβ , 1β )](1γ , eγ , 1γ ) = (−, gθα,β , 1β )(1γ , eγ , 1γ ) = (−, (gθα,β )θβ,γ , 1γ ) = (−, gθα,β θβ,γ , 1γ ).

This shows gθα,β θβ,γ = gθα,γ . Thus, S the mappings θα,β (α > β) are indeed the structure homomorphisms of T = α∈Y Tα , and consequently, T forms a strong semilattice of cancellative monoids Tα (α ∈ Y ). 2 S Lemma 3.3. Let Y be a semilattice and S = α∈Y Sα a semilattice Y of rectangular monoids Sα = Iα × Tα × Λα (α ∈ Y ). Then for any α, β ∈ Y with α > β, there exist the following mappings ϕα,β : Sα × Iβ → Iβ , (a, i) 7→ ha, ii, φα,β : Λβ × Sα → Λβ , (λ, a) 7→ [λ, a] such that for any a = (i, g, λ) ∈ Sα , j ∈ Iβ and µ ∈ Λβ , we have (1β , eβ , µ)a = (1β , gθα,β , [µ, a]), a(j, eβ , 1β ) = (ha, ji, gθα,β , 1β ),

(7) (8)

where eβ is the identity element of the cancellative monoid Tβ . S Proof. Suppose that S = α∈Y Sα is a semilattice decomposition of rectangular monoids Sα = Iα × Tα × Λα (α ∈ Y ). Let α, β ∈ Y with α > β. Then for any a = (i, g, λ) ∈ Sα , j ∈ Iβ and µ ∈ Λβ , it is immediate to see that a(j, eβ , 1β ), (1β , eβ , µ)a ∈ Sβ , where eβ is the identity element of the cancellative monoid Tβ . Now we consider the following mappings ϕα,β : Sα × Iβ → Iβ , (a, j) 7→ ha, ji, φα,β : Λβ × Sα → Λβ , (µ, a) 7→ [µ, a] satisfying a(j, eβ , 1β ) = (ha, ji, −, −), (1β , eβ , µ)a = (−, −, [µ, a]).

(9) (10)

By multiplying both sides of (10) with the idempotent (1β , eβ , 1β ), we immediately obtain (1β , eβ , µ)a = (1β , −, [µ, a]).


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Since (1β , eβ , µ)a ∈ Sβ , there exists a unique element u in the cancellative monoid Tβ such that (1β , eβ , µ)a = (1β , u, [µ, a]). On the other hand, by Lemma 3.2 and (4), we can deduce [(1β , eβ , µ)a](1β , eβ , 1β ) = (1β , u, [µ, a])(1β , eβ , 1β ) = (1β , u, 1β ) and (1β , eβ , µ)[a(1β , eβ , 1β )] = (1β , eβ , µ)(ha, 1β i, gθα,β , 1β ) = (1β , gθα,β , 1β ). Consequently, we obtain u = gθα,β and so (1β , eβ , µ)a = (1β , gθα,β , [µ, a]). This shows that (7) holds. By a similar argument, using (9) and Lemma 3.2, we can show a(j, eβ , 1β ) = (ha, ji, gθα,β , 1β ). Thus, our (8) is proved. 2 Lemma 3.4. Suppose that the notation is the same as in Lemmas 3.2 and 3.3. Then (i) The mappings ϕα,β and φα,β satisfy the requirements (I) and (II). (ii) For any a = (i, g, λ) ∈ Sα and b = (j, h, µ) ∈ Sβ , we have ab = (ha, hb, 1αβ ii, gh, [[1αβ , a], b]) = a ◦ b. Proof. Suppose that α, β ∈ Y satisfy αβ = γ. Then for any a = (i, g, λ) ∈ Sα and b = (j, h, µ) ∈ Sβ , by Lemmas 3.2 and 3.3, we can deduce = = = =

ab(1γ , eγ , 1γ ) a(hb, 1γ i, hθβ,γ , 1γ ) a(hb, 1γ i, eγ , 1γ )(hb, 1γ i, hθβ,γ , 1γ ) (ha, hb, 1γ ii, gθα,γ , 1γ )(hb, 1γ i, hθβ,γ , 1γ ) (ha, hb, 1γ ii, gθα,γ hθβ,γ , [[1γ , a], b])(1γ , eγ , 1γ ).

Now by using similar arguments as given above, we also have (1γ , eγ , 1γ )ab = (1γ , eγ , 1γ )(ha, hb, 1γ ii, gθα,γ hθβ,γ , [[1γ , a], b]). Because Sγ is a rectangular monoid, we can easily verify ab = (ha, hb, 1γ ii, gθα,γ hθβ,γ , [[1γ , a], b]). (11) S Thus, by Lemma 3.2, T = α∈Y Tα is a strong semilattice of cancellative monoids Tα and gθα,γ hθβ,γ = gh. Hence, we have ab = (ha, hb, 1γ ii, gh, [[1γ , a], b]) = a ◦ b.

(12)

This proves the second part of Lemma 3.4. In order to show the first part, we suppose that α, β, γ ∈ Y with αβ > γ. Then for any a ∈ Sα and b ∈ Sβ , by Lemmas 3.2, 3.3 and the equality (12), we have ab(k, eγ , 1γ ) = (hab, ki, −, −) = (ha ◦ b, ki, −, −)

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for any k ∈ Iγ . On the other hand, we also deduce a[b(k, eγ , 1γ )] = a(hb, ki, −, −) = a(hb, ki, eγ , 1γ )(hb, ki, −, −) = (ha, hb, kii, −, −). Hence, we conclude ha ◦ b, ki = ha, hb, kii.

(13)

Similarly, for any λ ∈ Λγ , we can prove [λ, a ◦ b] = [[λ, a], b].

(14)

Thus, we have shown that the requirement (II) is satisfied by the mappings ϕα,β and φα,β . Also, it is easy to verify that the requirement (I) is satisfied by the mappings ϕα,β and φα,β . 2 Proof of the sufficiency part of Theorem 3.1. Let S be a cyber-group. In order to show that S is isomorphic to some semigroup S(Y, Sα , ϕα,β , φα,β ), by using Lemmas 3.2–3.4, we only need to prove that the requirement (III) is satisfied. Since S itself is a superabundant semigroup, for any a = (i, g, λ) ∈ Sα (α ∈ Y ), there exists a unique idempotent a0 = (i, eα , λ) ∈ Sα such that a H∗ (S) a0 . This means a L∗ (S) a0 , and hence a R∗ (S) a0 . By Lemmas 2.1, 2.2 and their dual, we can easily see that the requirement (III) holds on S. By summarizing the above results, we now see that the cyber-group S can be expressed by some S(Y, Sα , ϕα,β , φα,β ), as required. 2 Remark. The requirement (III) in Theorem 3.1 is particularly imposed to ensure that the semigroup S constructed above is abundant, and so in general, our hypothesis cannot be weakened. In order to illustrate this situation, we give the following example. Example 3.5. Let

Ã a=

1 0 0

1 0 0

1 0 0

!

and Sα = {5n a | n ∈ N}. Then the set Sα under the usual matrix multiplication forms a cancellative monoid with a its identity element. We now form a semigroup S = {b, c, d, e, f, g, a, an | n > 1} under the following multiplication (see Table 2), where an = 5n a for any n > 1, which extends the semigroup Sα . From Table 2, we can easily check that the semigroup Sβ = {b, c, d, e, f, g} is not only a subsemigroup of S but also a rectangular group. Clearly, Sα Sβ ⊆ Sβ and Sβ Sα ⊆ Sβ . Thus, Sα and Sβ are both special rectangular monoids, and consequently, S can be regarded as a semilattice of rectangular monoids.

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Table 2. The Cayley table of Sβ ∗ a am e f b c d g

a a am e f b c d g

an an am+n e e b c d d

e e e e e b b d d

f f f f f c c g g

b b b b b d d e e

c c c c c g g f f

d d d d d e e b b

g g g g g f f c c

Let x be any element in Sα \{a}. Then a R∗ (Sα ) x. However, since ex = f x and e = ea 6= f a = f , the elements x and a are clearly not R∗ -related in S. This means that the semigroup S does not satisfy the requirement (III)(ii) in Theorem 3.1 for cyber-groups. In this case, we can easily check (e, x) ∈ / R∗ (S) and (f, x) ∈ / R∗ (S). ∗ ∗ Consequently, we can see that the R -class Rx (S) of S containing the element x does not contain any idempotent of S. Thus, S itself is not an abundant semigroup, and as a consequence, S is clearly not a cyber-group. 4 Some Special Cases In this section, we consider two special cases of Theorem 3.1. Case 1. Left C-a semigroups. A left C-a semigroup is a superabundant semigroup S in which the relation L∗ is a semilattice congruence on S. This class of semigroups was first studied by Guo, Guo and Shum in [4]. The authors in [12] also described this kind of semigroups by using another approach (where such a semigroup is called a super R∗ -unipotent semigroup). It can be easily checked that the set of all idempotents of a left C-a semigroup is a left regular band so that a left C-a semigroup is always a special cyber-group. Lemma 4.1. [12] The following conditions are equivalent on a semigroup S: (i) S is a left C-a semigroup. (ii) S is a superabundant semigroup with eS ⊆ Se for any idempotent e ∈ S. (iii) S is a semilattice of Sα (α ∈ Y ), where each Sα is a direct product of a left zero band Iα and a cancellative monoid Tα . Also, Ha∗ (S) = Ha∗ (Sα ) for α ∈ Y and a ∈ Sα . In Theorem 3.1, if we take |Λα | = 1 for every α ∈ Y and denote by S = S(Y ; Sα , ϕα,β ) the semigroup obtained above, then as an easy consequence of Theorem 3.1, we have the following corollary which is the structure theorem of left C-a semigroups. Corollary 4.2. The semigroup S = S(Y ; Sα , ϕα,β ) is a left C-a semigroup. Conversely, every left C-a semigroup is isomorphic to some semigroup S(Y ; Sα , ϕα,β ). Case 2. Orthogroups.

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An interesting case of cyber-groups is when S becomes a completely regular semigroup. In this case, a cyber-group of course coincides with an orthogroup. Consequently, the structure theorem of Petrich in [8] for orthogroups follows immediately as a trivial corollary of our Theorem 3.1. Corollary 4.3. [8] Let Y be a semilattice. For each α in Y , form the product set Sα = Iα × Gα × Λα , where Iα is a left zero band, Gα is a group and Λα is a right · S zero band. Form S = α∈Y Sα . For each α ∈ Y , fix an element in Iα ∩ Λα , denoted by the same symbol α. Define h , i : Sα × Iβ → Iβ

and

[ , ] : Λ β × Sα → Λ β

whenever α > β. If G is a semilattice Y of groups Gα in which the multiplication is defined by juxtaposition, and for all a = (i, g, λ) ∈ Sα and b = (j, h, µ) ∈ Sβ , the following conditions are satisfied : (a) If k ∈ Iα and νS∈ Λα , then ha, ki = i and [ν, a] = λ. Define a multiplication on the set S = α∈Y Sα by a ◦ b = (ha, hb, αβii, xy, [[αβ, a], b]). (b) If αβ > γ, k ∈ Iγ and ν ∈ Λγ , then ha, hb, kii = ha ◦ b, ki

and

[[ν, a], b] = [ν, a ◦ b].

· S Then S = α∈Y Sα forms an orthogroup such that S / D ' Y and the multiplication of S restricted to each Sα coincides with the given multiplication on Sα . Conversely, every orthogroup is isomorphic to a semigroup constructed above.

References [1] A.H. Clifford, M. Petrich, Some classes of completely regular semigroups, J. Algebra 46 (1977) 462–480. [2] J.B. Fountain, Adequate semigroups, Proc. Edinburgh Math. Soc. 22 (1979) 113–125. [3] J.B. Fountain, Abundant semigroups, Proc. Lond. Math. Soc. 44 (3) (1982) 103–129. [4] X.J. Guo, Y.Q. Guo, K.P. Shum, Semi-spined structure of left C-a semigroups, in: Semigroups (Kunming, 1995), Springer-Verlag, Singapore, 1998, pp. 157–166. [5] X.J. Guo, K.P. Shum, On left cyber groups, Int. Math. J. 5 (2004) 705–717. [6] J.M. Howie, An Introduction to Semigroup Theory, Academic Press, London, 1976. [7] M. Petrich, The structure of completely regular semigroups, Trans. Amer. Math. Soc. 189 (1974) 211–236. [8] M. Petrich, A structure theorem for completely regular semigroups, Proc. Amer. Math. Soc. 99 (4) (1987) 617–622. [9] M. Petrich, N.R. Reilly, Completely Regular Semigroups, John Wiley & Sons, New York, 1999. [10] X.M. Ren, K.P. Shum, On generalized orthogroups, Comm. Algebra 29 (6) (2001) 2341–2361. [11] X.M. Ren, K.P. Shum, The structure of superabundant semigroups, Science in China (Ser. A) 47 (5) (2004) 756–771. [12] X.M. Ren, K.P. Shum, Y.Q. Guo, On super R∗ -unipotent semigroups, Southeast Asian Bull. Math. 22 (1998) 199–208.


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[13] M.K. Sen, X.M. Ren, K.P. Shum, A new structure theorem of LC-semigroups and a method for construction, Int. Math. J. 3 (3) (2003) 283–295. [14] K.P. Shum, X.M. Ren, Y.Q. Guo, On C ∗ -quasiregular semigroups, Comm. Algebra 27 (19) (1999) 4251–4274. [15] O. Steinfeld, On a generalization of completely 0-simple semigroups, Acta Sci. Math. (Szeged) 28 (1967) 135–145. [16] R.J. Warne, On the structure of semigroups which are unions of groups, Trans. Amer. Math. Soc. 186 (1973) 385–401.

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