On Cryptic Semisuperabundant Semigroups

Southeast Asian Bulletin of Mathematics (2011) 35: 237–247

Southeast Asian Bulletin of Mathematics c SEAMS. 2011 ⃝

On Cryptic Semisuperabundant Semigroups∗ Xiang-Zhi Kong

and

Zhi-Ling Yuan

School of Science, Jiangnan University, Wuxi 214122 China Email: [email protected]

Kar-Ping Shum Institute of Mathematics, Yunnan University, Kunming 650091 China Email: [email protected]

Received 9 May 2009 Accepted 9 May 2010 Communicated by Yuqi Guo AMS Mathematics Subject Classification(2000): 20M10 Abstract. By utilizing the well known ∼-Green’s relations, we will investigate some properties of semisuperabundant semigroups and describe the semilattice decomposition of cryptic semisuperabundant semigroups. In particular, we show that a semisuperabundant semigroup is a regular cryptic semisupersabundant semigroup if and only if it is a tight semilattice of completely J ∼ -simple semigroups. Our result in some sense generalizes the results of A.H. Clifford and M. Petrich on normal cryptogroups and regular cryptogroups and also extends a known theorem of J.B. Fountain on superabundant semigroups among the class of semisuperabundant semigroups. Keywords: ∼-Green’s relation; Tight semilattice; Natural partial order; Semilattice decomposition.

1. Introduction It is well known that the Green’s relations play an important role in the study of regular semigroups [1–5] and [8]. It was stated by Clifford in [1] that a semigroup ∗ This research is supported by the NNSF (No. 10871161) of China and PIRT Jiangnan of Jiangnan University.

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S is a completely regular semigroup if and only if S is a semilattice of completely simple semigroups [1] where a completely regular semigroup is in fact a semigroup whose every H-class contains an idempotent. M. Petrich and N. Rielly in [9], by using this decomposition, showed that a completely regular semigroup is a normal cryptogroup(i.e. a completely regular semigroup with its Green’s relation H a normal band congruence) if and only if it is a strong semilattice of completely simple semigroups [3]. In particular J.B. Fountain [3] generalized the known Clifford theorem by proving that that an abundant semigroup is a superabundant semigroup( that is, an abundant semigroup with its every H∗ -class contains an idempotent) if and only if it is a semilattice of completely J ∗ -simple semigroups. A generalized version of the well known Clifford theorem has also been recently re-established by X.M. Ren, Y.Q. Guo and K.P. Shum in [11]. We recall the generalized Green’s relations on a semigroup S which are usually used to study the abundant semigroups. The following ∗-Green’s relations are due to F.J. Pastijn [7] and deeply studied by J.B. Fountain [3]. Let S be an arbitrary semigroup. Define the following relations on S. L∗ = {(a, b) ∈ S × S : (∀x, y ∈ S 1 )ax = ay ⇔ bx = by}, R∗ = {(a, b) ∈ S × S : (∀x, y ∈ S 1 )xa = ya ⇔ xb = yb}, H∗ = L∗ ∩ R∗ , D∗ = L∗ ∨ R∗ , J ∗ = {(a, b) ∈ S × S : J ∗ (a) = J ∗ (b)}, where J ∗ (a) is the smallest ideal containing element a saturated by L∗ and R∗ , that is, J ∗ (a) is a union of some L∗ -classes and also a union of some R∗ -classes. e on a We note here that M.V. Lawson [6] generalized the relation R∗ to R semigroup S as following: e ⇔ (∀e ∈ E(S))ea = a ↔ eb = b, aRb where E(S) is the idempotents set of S. e and for regular elements a, b of semigroup S, It is easy to see that R∗ ⊆ R e M.V., Lawson also notices that R e is an equivalence but aRb if and only if aRb. not a left congruence. e = Le ∩ R, e D e = Le ∨ R, e Je = {(a, b) ∈ S × S : Dually,we can define Le and let H e e e J(a) = J(b)}, where J(a) is the smallest ideal containing element a saturated e by Le and R. One can easily see that there is at most one idempotent contained in each e e a ∩ E(S), for some a ∈ S, then we write e as x0 , for any x ∈ H ea. H-class. If e ∈ H 0 0 e Clearly, for any x ∈ Ha with a ∈ S, we have x = xx = x x. If a semigroup S is regular,then every L-class of S contains at least one idempotent, and so does every R-class of S. If S is a completely regular semigroup, then every H-class of S contains an idempotent, in such a case, every H-class is a group. A semigroup is abundant [7] if its every L∗ - and R∗ -class contains an idempotent. One can see that L∗ = L on egular elements of a semigroup.

On Cryptic Semisuperabundant Semigroups

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Hence, regular semigroups are obviously abundant semigroups. A semigroup S is called superabundant [7] if each of its H∗ -classes contains an idempotent, in such a case, every H∗ -class is a cancellative monoid, which is the generalization of completely regular semigroups among the class of abundant semigroups. A e e semigroup is called semiabundant if each L-class and each R-class contain an idempotent. Certainly an abundant semigroup is a semiabundant semigroup, however, the converse is not true since a monoid is semiabundant but not abune dant. We now call a semigroup semisuperabundant if each H-class contains an e idempotent, in such a case, every H-class is a monoid which is a generalization of completely regular semigroups and superabundant semigroups in the class of semiabundant semigroups. Recall that a regular band(normal band) is a band that satisfies the identity axya = axaya(axya = ayxa) and a semigroup is called cryptic if its ∼-Green’s e is a congruence. A regular(normal) cryptic semisuperabundant semirelation H e is a group is a semisuperabundant semigroup with its ∼-Green’s relation H regular(normal) band congruence. For further notations and terminology, the reader is referred to [2, 3]. For the other concepts that we do not given in this paper, we use the traditional equivalent terms in the literature, see [5] and [8].

2. Preliminaries Since a completely simple semigroup is a J -simple completely regular semigroup with Green’s relation H as a congruence [3] and a completely J ∗ -simple semigroup is a J ∗ -simple superabundant semigroup with H∗ a congruence, as generalization, we call a semisuperabundant semigroup S is a completely Jee is a congruence. simple semigroup if it is Je-simple and the ∼-Green’s relation H For Green’s relations and their generaizations on semigroups,the reader is referred to [14]. We first prove the following lemma concerning congruences on a semisuperabundant semigroup. e is a congruence Lemma 2.1. Let S be a semisuperabundant semigroup.Then H if and only if for any a, b ∈ S, (ab)0 = (a0 b0 )0 . e 0 and bHb e 0 . Since H e is a Proof. (Necessity). For any a, b ∈ S, we have aHa 0 0 0 0 0 0 0 e b . But we also have abH(ab) e congruence,abHa , and so (ab) = (a b ) since e every H-class contains a unique idempotent. e is an equivalence, we only need to show that H e is (Sufficiency). Since H 0 0 0 0 0 0 0 e compatible. Let (a, b) ∈ H and c ∈ S. Then (ca) = (c a ) = (c b ) = (cb)0 e is left compatible. Similarly, H e is right compatible and thus H e is a and so H congruence. e Lemma 2.2. If e, f are D-related idempotents of a semisuperabundant semigroup

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S, then eDf . e , there are elements a1 , · · · , ak of S such that eLa e 1 Proof. Since eDf 0e 0 0 e e e e Ra2 · · · ak Lf . Since S is a semisuperabundant semigroup, eLa1 Ra2 · · · ak Rf . e and L = L∗ = L. e Thus, eDf since for regular elements R = R∗ = R Corollary 2.3. If S is a semisuperabundant semigroup, then e = Le ◦ R e=R e ◦ L. e D e Proof. If a, b ∈ S and aDb,then, by Lemma 1.2,we have a0 Db0 . Thus, there are 0 0 e Rb e and aRd e Lb e and elements c, d in S with a LcRb and a0 RdLb0 . Then, aLc the result follows. Lemma 2.4. Let e, f be idempotents in a semisuperabundant semigroup S . If eJ f , then eDf on S. Proof. Since SeS = Sf S, there exist elements x, y, s, t in S such that f = set, e = xf y. Let h = (f y)0 and k = (se)0 . Then hf y = f y = f f y and so h = h2 = f h and sek = se = see so that we have k = k 2 = ke. It follows that hf, ek are idempotents with hf Rh and ekLk. Hence, ehf Reh and ekf Lkf . Now eh = xf yh = xf y = e and kf = kset = set = f so that eRef Lf , that is, eDf . The following propositions of a cryptic semisuperabundant semigroup were given in [9]. Proposition 2.5. If a is an element of a cryptic semisuperabundant semigroup e = Sa0 S. S, then J(a) e a left Remark 2.6. If we replace the condition ”cryptic” in the proposition by R e congruence and L a right congruence in the semisuperabundant semigroup S, then by using a similar proof given in [11], we obtain the same result. Moreover, the rest of the results in this section also hold with the same replacement. e Proposition 2.7. On a cryptic semisuperabundant semigroup S, Je = D. Proposition 2.8. A completely Je-simple S is primitive for idempotents. Proof. Let e, f be idempotents in S with e 6 f . Since S is completely Je-simple, it follows from Proposition 1.5 that f ∈ SeS. Now by the first part of Exercise 3 in [1, §8.4], there exists an idempotent g of S such that f Dg and g 6 e. Let a ∈ S be such that f LaRg. Then, f La0 Rg and since g 6 f we have a0 = ga0 (gf )a0 = g(f a0 ) = gf = g.


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Now we have g 6 f and gLf so that f = f g = g. But g 6 e so that e = f and all idempotent of S are primitive. Lemma 2.9. In a completely Je-simple semigroup S, the regular elements of S generate a completely simple subsemigroup. e Proof. Let a, b be regular elements of S. Since S consists of a single D-class (by Proposition 1.6), it follows from Corollary 1.3 that there is an element c ∈ S with e Rb. e Hence, we have aLc e 0 Rb. e Thus c0 b = b and aLc0 since a is regular. Now aLc we see that abLb and the regularity of ab follows from that of b. The property of completely simple of the subsemigroup generated by regular elements follows Proposition 1.6, Lemma 1.2 and Corollary 1.3 easily. Theorem 2.10. Let S be a cryptic semisuperabundant semigroup. Then S is a semilattice Y of completely Je-simple semigroups Sα (α ∈ Y ) such that for α ∈ Y e a (S) = R e a (Sα ). and a ∈ Sα , Lea (S) = Lea (Sα ), R e 2 so that by Proposition 1.5, J(a) e e 2 ). Now for Proof. If a ∈ S, then aHa = J(a 2 a, b ∈ S, (ab) ∈ SbaS, and so we have 2 e e e J(ab) = J((ab) ) ⊆ J(ba)

e e e and by symmetry, we get J(ab) = J(ba). By Proposition 1.5, J(a) = Sa0 S, e = Sb0 S so that if c ∈ J(a) e ∩ J(b), e J(b) then,c = xa0 y = zb0 t for some x, y, z, t ∈ 2 0 0 0 0 e 0 txa0 ) and J(b e 0 txa0 ) = J(a e 0 b0 tx) S. Now c = zb txa y ∈ Sb txa S ⊆ J(b 2 0 0 2 e e by the preceding paragraph. Hence, c ∈ J(a b ) and since cHc , we also have e 0 b0 ). Since aHa e 0 , bHb e 0 and H e is a congruence so that abHa e 0 b0 . Hence, c ∈ J(a e e ∩ J(b) e ⊆ J(ab) e c ∈ H(ab). Thus, J(a) and since the opposite inclusion is clear, e e e we have J(a) ∩ J(b) = J(ab).

e We now know the set Y of all ideals J(a)(a ∈ S) forms a semilattice under e set intersection and that the map a 7→ J(a) is a homomorphism from S onto Y . e The inverse image of J(a) is just the Je-class Jea which is thus a subsemigroup of S. Hence S is a semilattice Y of the semigroups Jea . e J). e Now, we let a, b be elements of the Je-class and suppose that (a, b) ∈ L( e J), e that is, a0 b0 = a0 , b0 a0 = b0 Certainly, a0 , b0 ∈ Je so that we have (a0 , b0 ) ∈ L( e e and (a0 , b0 ) ∈ L(S). It hence follows that (a, b) ∈ L(S) and consequently, since e e e e e La (S) ⊆ J, we have La (S) = La (J). By using a similar argument, we can show ea (S) = R ea (J). e that R e a (J) e =H e a (S) and so Je is semisuperabunFrom the last paragraph, we have H e then by Proposition 1.6, (a, b) ∈ D(S) e dant. Furthermore, if a, b ∈ J, so that, e e e e e e by Corollary 1.3, there is an element c in La (S) ∩ Rb (S) = La (J) ∩ Rb (J).Thus, e -related in Je so that Je is Je-simple. a, b are D

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While we consider the above theorem in regular semigroups and abundant semigroups, our theorem then turns to be the well known Theorem of Clifford in [1] and the theorem of J.B. Fountain in [3], respectively. Lemma 2.11. Let S = (Y ; Sα ) be a cryptic semisuperabundant semigroup. Then the following conditions hold: (i) Let a ∈ Sα and α > β. Then there exists b ∈ Sβ with a > b; e and a > b, c. Then b = c; (ii) Let a, b, c ∈ S, bHc, (iii) Let a ∈ E(S) and b ∈ S be such that a > b. Then b ∈ E(S). Proof. (i) Let b ∈ Sβ . By Lemma 1.1, a(aba)0 , (aba)0 a and (aba)0 are in the e same H-class and so a(aba)0 = (aba)0 a(aba)0 = (aba)0 a. Let b = a(aba)0 . Then b ∈ Sβ and a > b. (ii) By the definition of “>”, there exist e, f, g, h ∈ E(S) such that b = e 0 , we have eb0 = b0 . Similarly, ea = af , c = ga = ah. From eb = b and bHb 0 0 0 0 0 c h = c . Thus, ec = ec c = eb c = b c = c. Similarly, we have bh = b so that b = bh = eah = ec = c, as required. (iii) We have b = ea = af for some e, f ∈ E(S) whence b2 = (ea)(af ) = 2 ea f = b. Following Proposition 1.7, we can easily prove the following lemma. Lemma 2.12. Let ϕ be a homomorphism from a completely Je-simple semigroup S into another completely Je-simple semigroup T . Then (aϕ)0 = a0 ϕ. Remark 2.13. If ϕ is a homomorphism between two completely Je-simple semie R e are preserved, so that D e is pregroups. Then the ∼-Green’s relations L, served.By Proposition 1.7 and Lemma 1.10, we can show that a completely Je-simple semigroup is primitive.

3. Tight Semilattice of Semigroups In study the structure of semigroups, we often decompose a semigroup into a union of subsemigroups by a band congruence. Semilattices of some kind of semigroups were extensively considered in [12] and [13]. The strong semilattice decomposition of a semigroup is the best one, such as a normal band is a strong semilattice of rectangular bands and a normal cryptogroup is a strong semilattice of completely simple semigroups [3]. In this section, we introduce the concept of tight semilattice decomposition of semigroups so that we can study the structuure of a regular cryptic semisuperabundant semigroups. This tight semilattice decoposition of semigroups is a generalization of the well known strong semilattice decomposition of semigroups in some sense. For the sake of


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easy comparison, we cite again here the definition of strong semilattice decomposition of semigroups (see [5]). Definition 3.1. Let S = (Y ; Sα ) be a semilattice Y decomposition of semigroup S into subsemigroups Sα (α ∈ Y ). Suppose that for any α ≥ β on Y , there is a homomorphism Φα,β from Sα into Sβ and the following conditions hold in the semigroup S. (C1) for any α ∈ Y , Φα,α = 1Sα is the identity homomorphism on Sα ; (C2) for any α ≥ β ≥ γ on Y , Φα,β Φβ,γ = Φα,γ ; (C3) for any α, β ∈ Y and a ∈ Sα , b ∈ Sβ , ab = (aΦα,αβ )(bΦβ,αβ ). Then, S is said to be a strong semilattice of subsemigroups Sα and we denote it by S = [Y ; Sα , Φα,β ]. Definition 3.2. Let S = (Y ; Sα ) be a semilattice Y decomposition of semigroup S into subsemigroups Sα (α ∈ Y ). Suppose that for any α ≥ β on Y . Then, there is a family of homomorphisms {ϕd(α,β) : d(α, β) ∈ D(α, β)} = Φα,β from Sα into Sβ and the following conditions hold in the semigroup S. (C1) for any α ∈ Y , the index set |D(α, α)| = 1 is a singleton and Φα,α = 1Sα is the identity homomorphism on Sα ; (C2) for α ≥ β ≥ γ on Y , Φα,β Φβ,γ ⊆ Φα,γ , where Φα,β Φβ,γ = {ϕd(α,β) ϕd(β,γ) : d(α, β) ∈ D(α, β), d(β, γ) ∈ D(β, γ)}; (C3) for any α ∈ Y and a ∈ Sα , there exists unique ϕd(β,αβ) such that for all b ∈ Sβ , ab = (aϕd(α,αβ) )(bϕd(β,αβ) ). Now, we call S a tight semilattice of subsemigroups Sα (α ∈ Y ) and denote it by S = T [Y ; Sα , Φα,β ]. If S = [Y ; Sα , Φα,β ] is a strong semilattice of subsemigroups Sα with a structure homomorphism Φα,β , then it is obvious that a strong semilattice of semigroups must be a tight semilattice of semigroups, but from our results, we will see that the converse statement is not not nnecessarily hold.

4. Regular Cryptic Semisuperabundant Semigroups Following Theorem 1.9, a cryptic semisuperabundant semigroup S is a semilattice of completely Je-simple semigroups Sα (α ∈ Y ). In this section, we introduce a band congruence ρα,β on a regular cryptic semisuperabundant semigroup S = (Y ; Sα ) and the structure homomorphisms set Φα,β . Finally, we will show that a cryptic semisuperabundant semigroup is a regular cryptic semisuperabundant semigroup if and only if it is a tight semilattice of completely Je-simple semigroups. As a corollary, we show that a cryptic semisuperabundant semigroup is a normal cryptic semisuperabundant semigroup if and only if it is a strong semilattice of completely Je-simple semigroups.

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We remark here that the above result generalizes the known result of M.Petrich on normal cryptogroups [3]. In the following lemma, we consider the regular cryptic semisuperabundant semigroups. Lemma 4.1. Let S = (Y ; Sα ) be a regular cryptic semisuperabundant semigroup, For any α, β ∈ Y , we define ρα,β on Sβ as following: (x, y ∈ Sβ )(x, y) ∈ ρα,β ⇔ (axa)0 = (aya)0 , for some a ∈ Sα . Then, the following conditions hold: (i) ρα,β is a band congruence on Sβ and for x, y ∈ Sβ , (x, y) ∈ ρα,β if and only if for any b ∈ Sα , (bxb)0 = (byb)0 . (ii) for α ≥ β ≥ γ on Y , ρα,γ ⊆ ρβ,γ and ρα,α is the universal relation ωSα on Sα . (iii) for α ≥ β on Y and a ∈ Sα , b ∈ Sβ , abρα,β bρα,β ba. Proof. We only prove that (i), (ii) and (iii) can be similarly proved. Let x, y ∈ Sβ with (x, y) ∈ ρα,β .Then, there exists a ∈ Sα such that (axa)0 = (aya)0 . For any element b ∈ Sα , we have b(axa)0 b = b(aya)0 b. Thus (b(axa)0 b)0 = (b(aya)0 b)0 . By the property of regular bands and Lemma 1.1 and 1.8, we easily have (bxb)0 = (byb)0 . Hence, the proof is completed. Now, we denote the ρα,β -congruence classes by {Sd(α,β) : d(α, β) ∈ D(α, β)}, By Lemma 3.1, we see immediately that D(α, α) is a singleton. Lemma 4.2. The regular cryptic semisuperabundant semigroup has the following properties: (i) For any α > β on Y and d(α, β) ∈ D(α, β). Let a ∈ Sα , then, there exists a unique element ad(α,β) ∈ Sd(α,β) such that a > ad(α,β) . (ii) For any α > β on Y and a ∈ Sα , x ∈ Sd(α,β) , if a0 > e for some idempotent e ∈ Sd(α,β) , then eax = ax, xae = xa, ea = ae and (ea)0 = e. Proof. (i) By Lemma 3.1 (iii) and Lemma 1.10 (i), for any c ∈ Sd(α,β) , the element ad(α,β) = a(aca)0 = (aca)0 a ∈ Sd(α,β) such that a > ad(α,β) . it canbe easily seen that a0d(α,β) = (aca)0 . If there is another b ∈ Sd(α,β) such that a > b, then there are idempotents g, h ∈ E(S) such that b = ga = ah and so ba0 = b = a0 b, thus b0 a0 = b0 = a0 b0 e 0 , which implies b0 ≤ a0 and hence b0 = a0 b0 a0 = (aba)0 = (aca)0 = since bHb 0 e d(α,β) . Thus by Lemma 1.10 (ii), ad(α,β) = b as required. ad(α,β) , that is, bHa (ii) Since (a0 (ax)0 a0 )a0 = a0 (ax)0 a0 = a0 (a0 (ax)0 a0 ) and a0 (ax)0 a0 e H(a0 (ax)0 a0 )0 , we have (a0 (ax)0 a0 )0 a0 = (a0 (ax)0 a0 )0 = a0 (a0 (ax)0 a0 )0 , that is, a0 > (a0 (ax)0 a0 )0 . Also, since a ∈ Sα and x ∈ Sd(α,β) , we have ax ∈ Sd(α,β)


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and so that e = (a0 (ax)0 a0 )0 by (i). Thereby, we have eax = (a0 (ax)0 a0 )0 ax = (a0 (ax)0 a0 )0 a0 (ax)0 a0 ax = ax. Similarly, we have xae = xa. Since x is arbitrarily chosen element in Sd(α,β) , we can particularly choose x = e. In this way, we obtain that ea = ae and consequently, by Lemma 1.1, we have (ea)0 = (ea0 )0 = e. Lemma 4.3. Let S = (Y ; Sα ) be a regular cryptic semisuperabundant semigroup. For any α ≥ β on Y and d(α, β) ∈ D(α, β), define a mapping ϕd(α,β) from Sα into Sd(α,β) with aϕd(α,β) = ad(α,β) , where ad(α,β) is defined in Lemma 3.2. Write Φα,β = {ϕd(α,β) : d(α, β) ∈ D(α, β)}. Then (i) (ii) (iii) (iv)

ϕd(α,β) is a homomorphism. for α ∈ Y , ϕD(α,α) is the identity homomorphism of Sα . for α > β > γ on Y , Φα,β Φβ,γ ⊆ Φα,γ . for any α ∈ Y and a ∈ Sα , there exists unique ϕd(β,αβ) such that for all b ∈ Sβ , ab = (aϕd(α,αβ) )(bϕd(β,αβ) ).

Proof. (i) Following Lemma 3.2, ϕd(α,β) is well defined. For a, b ∈ Sα and c ∈ Sd(α,β) , by Lemma 3.2 again, (aϕd(α,β) )(bϕd(α,β) ) = (aca)0 ab(bcb)0 (aca)0 ab = ab(bcb)0 ≤ ab and so we can deduce that (ab)ϕd(α,β) = (aϕd(α,β) )(bϕd(α,β) ). (ii) this part follows easily since Sα is primitive. (iii) We only need prove that for any d(α, β) ∈ D(α, β), d(β, γ) ∈ D(α, γ), Sd(α,β) ϕd(β,γ) ⊆ Sd(α,γ) for some d(α, γ) ∈ D(α, γ). Let a ∈ Sα , b1 , b2 ∈ Sd(α,β) and c ∈ Sγ , we have (ab1 a)0 = (ab2 a)0 and b1 ϕd(β,γ) = b1 (b1 cb1 )0 , b2 ϕd(β,γ) = b2 (b2 cb2 )0 and so (a(b1 ϕd(β,γ) )a)0 = (ab1 (b1 cb1 )0 a)0 = (ab2 (b2 cb2 )0 a)0 = (a(b2 ϕd(β,γ) )a)0 , which implies Sd(α,β) ϕd(β,γ) ⊆ Sd(α,β) for some d(α, γ) ∈ D(α, γ). (iv) By Lemma 3.1, we can easily see that Sα b and Sα bSα fall in the same class Sd(α,αβ) and for any a ∈ Sα , by Lemma 3.2, aϕd(α,αβ) = (aba)0 a. Dually, aSβ and Sβ aSβ fall in the same class Sd(β,αβ) and bϕd(β,αβ) = b(bab)0 . Also we easily have (aba)0 a ∈ Sd(β,αβ) , b(bab)0 ∈ Sd(α,αβ) .Thus, by Lemma 3.2 (ii), we deducce that (aϕd(α,αβ) )(bϕd(β,αβ) ) = (aba)0 ab(bab)0 = ab(bab)0 = (aba)0 ab ≤ ab and so (aϕd(α,αβ) )(bϕd(β,αβ) ) = ab since Sαβ is primitive. From the above proof and considering the property of rectangular band, we can easily see the uniqueness of ϕd(β,αβ) by using the element b to multiply any element of Sα . Finally, we establish a characterization theorem for a cryptic semisuperabundant semigroup.

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Theorem 4.4. A semisuperabundant semigroup S is a regular cryptic semisuperabundant semigroup if and only if it is a tight semilattice of completely Je-simple semigroups. Proof. We are have already proved the necessity part of Lemma 3.2. Now, we prove the sufficiency of Lemma 3.2. Let S = T [Y ; Sα , Φα,β ] be a tight semilattice of completely Je-simple semie is a congruence and S/H e is a regular groups Sα . We only need to show that H band. For a ∈ Sα , b ∈ Sβ , Then, by Definition 2.2 and Lemma 1.11, we have (ab)0 == [(aϕd(α,αβ) )(bϕd(β,αβ) )]0 = [(a0 ϕd(α,αβ) )(b0 ϕd(β,αβ) )]0 = (a0 b0 )0 . e is a regular band, by a result of Thus, by Lemma 1.1, S is cryptic. To show S/H [6], we only need show that the Green’s relations L and R are both congruences e We only show that L is a congruence in S/H e as R is a congruence in on S/H. e can be proved in a similar fashion. By Theorem 1.9, we can let S/H e = S/H e (Y ; Sα /H) and let e, f ∈ Sα ∩ E(S), g ∈ Sβ ∩ E(S) with (e, f ) ∈ L. Then, we have ef = e and f e = f . By Definition 2.2, we can find homomorphisms ϕd(β,αβ) and ϕd′ (β,αβ) ∈ Φβ,αβ , ϕd(α,αβ) ∈ Φα,αβ such that e = [g(ef )](gf )H e (gegf )H e = [(gϕd(β,αβ) )((ef )ϕd(α,αβ) )][(gϕd′ (β,αβ) )(f ϕd(α,αβ) )]H e = (gϕd(β,αβ) )(f ϕd(α,αβ) )H and e = g(ef )H e = (gϕd(β,αβ) )((ef )ϕd(α,αβ) )H e = (gϕd(β,αβ) )(f ϕd(α,αβ) )H. e (ge)H e = (ge)H. e Analogously, we can also prove that (gf ge)H e= Thereby, (gegf )H e This proves that L is left compatible on S/H. e Since L is always right (gf )H. e , as required. Dually, R is compatible, we see that L is a congruence on S/H e Thus, by [10, II.3.6 Proposition], S/H e is a regular also a congruence on S/H. band and hence S is a regular cryptic semisuperabundant semigroup. Our proof is completed. Note that in a normal cryptic semisuperabundant semigroup S = (Y ; Sα ) the band congruence ρα,β defined in Lemma 3.1 is the universal relation ωSβ of Sβ for all α, β ∈ Y . We have the following corolllary. Corollary 4.5. A semisuperabundant semigroup S is a normal cryptic semisuperabundant semigroup if and only if it is a strong semilattice of completely Je-simple semigroups. If the semigroup S is abundant and regular respectively, then we deduce the following corollary.


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Corollary 4.6. A superabundant semigroup S is a normal cryptic superabundant semigroup if and only if it is a strong semilattice of completely J ∗ -simple semigroups. Corollary 4.7. A completely regular semigroup S is a normal cryptogroup if and only if it is a strong semilattice of completely simple semigroups [3].

References [1] A.H. Clifford, G.B. Preston, The Algebraic Theory of Semigroups, New York, American Mathematical Society, Providence, R. I., 1967. [2] A.H. Clifford, Semigroups admitting relative inverses, Ann. Math. 42 (1941) 1037– 1049. [3] J.B. Fountain, Abundant semigroups, Proc. London Math. Soc. 43 (3) (1982) 103–129. [4] Y.Q. Guo, K.P. Shum, C.M. Gong, on (∗, ∼)-Green’s relations and ortho-lcmonoids, Comm. Alg. 39 (1) (2011) 5-31. [5] J.M. Howie, Fundamental of Semigroup Theory, Oxford, Clarendon Press, 1995. [6] M.V. Lawson, Rees matrix semigroups, Proc. Edinburgh Math. Soc 33 (1990) 23– 39. [7] F.J. Pastijn, A Representation of a semigroup by a semigroup of matrices over a group with zero, Semigroup Forum 10 (1975) 238–249. [8] M. Petrich, Lectures in Semigroups, London, Wiley & Sons Inc., 1976. [9] M. Petrich and N.R. Reilly, Completely Regular Semigroups, New York, John Wiley & Sons, 1999. [10] M. Petrich, The structure of completely regular semigroups, Trans. Amer. Math. Soc. 189 (1974) 211–236. [11] X.M. Ren, K.P. Shum, Y.Q. Guo, A generalized Clifford theorem of semigroups, Sci. China Ser. A 53 (4) (2010) 1097–1101. [12] X.M. Ren, Strong semilattices of completely simple semigroups, Southeast Asian Bull. Math. 20 (4) (1996) 53–56. [13] K.P. Shum, X. M. Ren, On semilattices of quasi-rectangular groups, Southeast Asian Bull. Math. 22 (3) (1998) 307–318. [14] K.P. Shum, D. Lan, Y.Q. Guo, Green’s relations and their generalizations on semigroups, Discuss. Math. Gen. Algebra Appl. 30 (1) (2010) 71–89.