see that Ïa is a left congruence, we let (x, y) â Ïa and c â S. Then, by the definition of Ïa, we have (axa)0 = (aya)0. Since S is a regular cyber group, by Lemma ...
Semilattice structure of generalized rpp semigroups K. P. Shum Department of Mathematics, Hong Kong University Pokfulam Road, Hong Kong, China (SAR) Abstract In this survey article, we discuss the generation of some rpp semigroups, in particular, by using kitted semilattice of semigroups, we can describe the structure for right quasi-normal cyber groups and normal cyber groups. We shall show that our results generalize some known theorems of Petrich- Reilly on normal cryptic groups from the class of regular semigroups to the class of generalized abundant semigroups and also entrich some recent results of Guo-Shum on cyber groups.
2000 Mathematics Subject Classification: 20M10 Keywords Regular cyber group; Knitted semilattice; Cryptic group; H† -abundant semigroup.
1
Introduction
It is well known that regular semigroups are the core of semigroups because it was proved by Clifford that a completely regular semigroup whose idempotents are in the center can be expressed as union of groups. Thus, people usually regard regular semigroups as generalized groups. After the celebrated theorem of Clifford, regular semigroups were further generalized to legal semigroups ,left Clifford semigroups, quasi-regular semigroups and other kinds by many authors, for example, see [?],?].[?] etc. Later on, Clifford also proved that a completely regular semigroup with idempotents in the center can be expressed a strong semilattice of groups.This theorem can be regarded as a fundamental theorem of semigroups because it provide a method of constructing large size semigroups 1
by using small size semigroups or groups. In fact, many semigroups can be constructed by using this method. Other methods of construction such as spined product of semigroups and wreath product of semigroups were also given,see ?. [?] etc. On the other hand, in studying ring theory,people consider rpp rings which are the rings whose principal right ideal are projective. They notice that the class of rpp rings includes the class of regular rings as its subclass. Thus, rpp rings are also generations of regular rings. This result gives a hint that the generation of regular rings can also be obtained without using the definition related to regularity. In 1972, Fountain considered rpp monoids [?. He first proved that a C-rpp monoid can be expressed as a strong semilattice of left cancellative monoids. Thus, in comparing his theorem and the well known theorem of Clifford, one can easily see that rpp semigroup is a good generation of Clifford semigroups. In study rpp semigroups, J. B. Fountain [1] adopted the generalized Green relations, namely, the Green ∗-relations proposed by F. Pastijn [8]. The Green∗-relations are defined by: 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∗ . Let L and R be the usual Green relations. Then, one can easily see that R ⊆ R∗ ⊆ R† on a semigroup S and that there exists at most one idempotent in each H† -class. If e ∈ Ha† ∩ E(S), for some a ∈ S, then we write e as x0 , for any x ∈ Ha† . Thus, for any x ∈ Ha† with a ∈ S, we always have x = xx0 = x0 x. We now call a semigroup with a unique idempotent a unipotent semigroup( in brevity, we call it an up-semigroup). Examples of up-semigroups are plenty, for example, the groups; the (left or right) cancellative monoids and the semigroup S 1 such that S does not contains idempotent are up-semigroups. It is evident that H† = L† = R† on all up-semigroups. We can easily observe that if a semigroup S is regular then every R-class of S contains at least one idempotent, and so does every L-class of S. If a semigroup S is completely regular, then it is clear that every H-class of S contains an idempotent. Consequently, we call a semigroup S lpp if every R∗ -class of S contains at least one idempotent [1]. An rpp semigroup is the dual of an lpp semigroup (see [1] and [2]). According to Fountain [1], we call a semigroup S abundant if S is both an rpp and lpp semigroup. Obviously, R∗ = R 2
on all regular elements of a semigroup and hence regular semigroups are indeed special abundant semigroups. A semigroup S is also called superabundant [1] if every H∗ -class of S contains an idempotent. New structure theorems of superabundant semigroups have been recently obtained by Ren-Shum in [10] We now call a semigroup S a rectangular up-semigroup if S is the direct product of a up-semigroup and a rectangular band. According to M. Petrich-Reilly [9], a completely regular semigroup S is called cryptic if H† is a congruence on S. By using the notion of X. J. Guo and K. P. Shum [3], we call an H† -abundant semigroup S a cyber group if its idempotents set E(S) forms a band and H† is a band congruence on S. Clearly, a cyber group is a generalized cryptic group from the class of completely regular semigroups to the class of quasiabundant semigroups. Similarly, we can define the regular (respectively, right quasi-normal, normal) cyber groups if its set of idempotents form a regular (respectively, right quasi-normal, normal) band. Recall that a band is said to be right quasi-normal if it satisfies the identity xya = xaya [4]. The structure of left cyber groups has been studied by X. J. Guo and K. P. Shum [3]. The structure of completely regular semigroups with generalized strong semilattice has been recently studied by X.Z. Kong and K. P. Shum in [6]. However, there still lacks of adequate information for the structure of cyber groups in the class of H† -abundant semigroups although H† -abundant semigroups are the most natural generalization of Clifford semigroups from the class of completely regular semigroups to the class of generalized abundant semigroups. In this paper, we shall modify the concepts of refined semilattice and the quasi-strong semilattice of semigroups to knitted semilattice of semigroups and by using the knitted semilattice of semigroups, we shall give some structure theorems for regular ( respectively, right quasi-normal, normal) cyber groups. Thus, the results of M. Petrich and N. Reilly on regular cryptic groups [9] are generalized and strengthened. Also, the result given by X. J. Guo and K. P. Shum on left cyber groups [2] are enriched. Throughout this paper, X is a nonempty set. We always denote the universal relation on X by ωX and the identity relation on X by εX . We also denote the set of idempotents of a semigroup S by E(S). For notations and terminologies not given in this paper, the reader is referred to J. M. Howie [4].
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2
Knitted semilattices
We first introduce the concept of knitted semilattice of semigroups. Let S be a semigroup with a semilattice decomposition, denoted it by S = (Y ; Sα ), where Y is a semilattice and every Sα is a subsemigroup of S mounted on the vertices of Y . Let α > β on Y . Then we denote the index set between α and β by D(α, β). Definition 2.1 Let S = (Y ; Sα ) be a semigroup. Suppose that the following conditions hold in the semigroup S. (C1) For any α > β on Y , there exists a band congruence ρα,β on Sβ which partitions the semigroup Sβ into the disjoint union of the ρα,β -classes Sd(α,β) , where the index d(α, β) ∈ D(α, β). In this case, we call the semigroup Sβ a knot of S. (C2) For each d(α, β) ∈ D(α, β), there exists a homomorphism φd(α,β) : Sα −→ Sd(α,β) . In this case, the homomorphism φd(α,β) is called a string of S. Let Φα,β = {φd(α,β) |d(α, β) ∈ D(α, β)}. Then the set Φα,β is called the structure map of S. Now, we suppose that the band congruence ρα,β satisfies the following conditions: (i) (∀α ∈ Y ), ρα,α = ωSα is the universal relation, that is, D(α, α) is a singleton. Denote the element in D(α, α) by d(α, α). In this case, the string φd(α,α) : Sα −→ Sα is the identity automorphism of the semigroup Sα . (ii) (∀α, β, γ ∈ Y, α > β > γ), we have ρα,γ ⊆ ρβ,γ and 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)}. (iii) For any α, β ∈ Y and a ∈ Sα , aΦα,αβ = {aφd(α,αβ) : ∀d(α, αβ) ∈ D(α, αβ)} is contained in one of the ρβ,αβ -class. (iv) For α > β on Y and d(α, β) ∈ D(α, β), we have ax = (aφd(α,β) )x and xa = x(aφd(α,β) ), for any a ∈ Sα and x ∈ Sd(α,β) .
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Then, we write S = [Y ; Sα , ρα,β , Φα,β ] and call this a semilattice decomposition of the semigroup S = [Y ; Sα , ρα,β , Φα,β ] the knitted semilattice of S. Remark The usual strong semilattice of semigroups Sα is obviously a special knitted semilattice of semigroups Sα . For in this case, ρα,β is the universal relation ωSβ on Sβ for each α > β, and so each homomorphism set Φα,β is a singleton. However, it is noted that a knitted semilattice of semigroups is not necessarily a strong semilattice of semigroups. In fact, a knitted semilattice is a finer semilattice decomposition of semigroups and in particular, its decomposition depends entirely on the band congruences on the semigroups S. Definition 2.2 Let φ be a homomorphism which maps from a quasiabundant semigroup S to another quasiabundant semigroup T . Unlike regular semigroups, the homomorphic image of abundant semigroups are not necessarily abundant. Thus, if the Green †-relations L† and R† are preserved, then the homomorphism φ is said to be a “good” homomorphism. Similarly, the Congruence ρ on a quasiabundant semigroup S is said to be “ good” if the homomorphism φ : S −→ S/ρ is good. We can easily observe that if S is an H† -abundant semigroup then a congruence ρ on S is a good band congruence if and only if H† ⊆ ρ.
3
Properties of cyber groups
Throughout this section except Lemma 3.1 (i), S is a cyber group, that is, it is a cryptic H† -abundant semigroup whose set of idempotents E(S) forms a subsemigroup of S. We now give some properties of cyber groups. Lemma 3.1 Let S be an H† -abundant semigroup whose set of idempotents E(S) is a subsemigroup. Then the following statements hold (i) H† is a congruence on S if and only if for any a, b ∈ S, (ab)0 = a0 b0 . (ii) the relation J † defined by J † = {(x, y) ∈ S × S : x0 J (E(S))y 0 } is a semilattice congruence on S and every J † -class is a rectangular up-semigroup. Proof: (i) (=⇒) Let a, b ∈ S. Then aH† a0 and bH† b0 . Since H† is a congruence on S, abH† a0 b0 . However, abH† (ab)0 , and so (ab)0 = a0 b0 since every H † -class contains a 5
unique idempotent. This shows that E(S) is a band. (⇐=) Since H† is an equivalent relation on S, we only need to show that H† is a left congruence on S. Let (a, b) ∈ H† and c ∈ S. Then (ca)0 = c0 a0 = c0 b0 = (cb)0 and so H† is a left congruence. Dually, we can show that H† is right congruence, and hence H† is a congruence on S. (ii) It is easy to see that J † is an equivalence on S. Suppose that (a, b) ∈ J † . Then, a0 J (E(S))b0 . Now, we let c ∈ S, by (i), then (ca)0 = c0 a0 J (E(S))c0 b0 = (cb)0 since J (E(S)) is a congruence on E(S). Thus J † is a left congruence on S. Dually, J † is also a right congruence, and whence J † is a congruence on S. Since H† ⊆ J † and each H† -class of S is an up-semigroup, J (E(S)) is a semilattice congruence on E(S). Consequently, J † is a semilattice congruence on S. To show that every J † -class is a rectangular up-semigroup, we first denote the set of idempotents of aJ † by Ea for any a ∈ S. By the definitions of cyber group and J † , Ea = Ea0 is a J (E(S))-class of E(S), and consequently, a rectangular band. We can also easily see that Ha† is a up-semigroup. We now show that aJ † is isomorphic to Ea × Ha† . For this purpose, we define a mapping φ from aJ † to Ea × Ha† . Now,for any x ∈ aJ † , we let xφ = (x0 , a0 xa0 ). Then, a0 xa0 ∈ Ha† and hence φ is well defined. For any x, y ∈ aJ † , we have (xφ)(yφ) = (x0 , a0 xa0 )(y 0 , a0 ya0 ) = (x0 y 0 , a0 xa0 a0 ya0 ) = (x0 y 0 , a0 xx0 a0 y 0 ya0 ) = (x0 y 0 , a0 xya0 ) = (xy)φ, and φ is clearly a homomorphism. If (x0 , a0 xa0 ) = (y 0 , a0 ya0 ), then x0 = y 0 and a0 xa0 = a0 ya0 , and thereby x0 a0 xa0 x0 = y 0 a0 ya0 y 0 . This leads to x = y, and whence φ is injective. Furthermore, for any (e, b) ∈ Ea × Ha† , let x = ebe. Then xφ = ((ebe)0 , a0 (ebe)a0 ) = (e, b). Thus, φ is surjective and consequently, aJ † is isomorphic to a rectangular up-semigroup. 2 The proof of the following lemma is straightforward. Lemma 3.2 Let S = I × U × Λ be a rectangular up-semigroup, where I, U and Λ are left zero semigroup, up-semigroup and right zero semigroup ,respectively. Then for any (i, u, λ), (j, v, µ) ∈ S, (i, u, λ)R† (j, v, µ) if and only if i = j.
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¿From the above lemmas, we can easily see that the rectangular up-semigroups are primitive. We now consider the elements in a cyber group S = (Y ; Sα ). Lemma 3.3 Let S = (Y ; Sα ) be a cyber group. Then the following statements hold: (i) For a ∈ Sα and b ∈ Sβ , if a > b then there exist e, f ∈ E(Sβ ) with b = ea = af ; (ii) For a ∈ Sα , if α > β then there exists b ∈ Sβ with a > b; (iii) For a, b and c ∈ S, if bH† c and a > b, c then b = c; (iv) For a ∈ E(S) and b ∈ S,if a > b, then b ∈ E(S). Proof: (i) Let g, h ∈ E(S) be such that b = ga = ah. Then e = (ga)0 g, f = h(ah)0 ∈ E(Sβ ) and ea = (ga)0 ga = ga = b = ah = ah(ah)0 = af. (ii) Let b ∈ Sβ . Then by Lemma 3.1, a(aba)0 , (aba)0 a and (aba)0 are in the same H† -class and so a(aba)0 = (aba)0 a(aba)0 = (aba)0 a. Write b = a(aba)0 . Then we see immediately that b ∈ Sβ and a > b. (iii) By (i), there exist e, f, g, h ∈ E(Sβ ) such that b = ea = af , c = ga = ah. Thus ec = ega = ega0 a = ea = b since E(Sβ ) is a rectangular band. (iv) Since b = ea = af for some e, f ∈ E(S), b2 = (ea)(af ) = ea2 f = b. 2 Lemma 3.4 Let S be a regular cyber group (that is, a cyber group S such that E(S) is a regular band). For every a ∈ S, we define ρa on S by (b1 , b2 ) ∈ ρa if and only if (ab1 a)0 = (ab2 a)0 , (b1 , b2 ∈ S). Then the following properties hold: (i) ρa is a good band congruence on S; (ii) (∀a, a1 ∈ Sα ), ρa = ρa1 , that is, ρa depends only on the component semigroup Sα containing the element a rather than the element itself and we write ρa = ρα . 7
(iii) (∀α, β ∈ Y with α > β), ρα ⊆ ρβ and ρβ |Sα = ωSα , where ωSα is the universal relation on Sα . Proof: (i) It is easy to see that ρa is an equivalent relation on S, for all a ∈ S. To see that ρa is a left congruence, we let (x, y) ∈ ρa and c ∈ S. Then, by the definition of ρa , we have (axa)0 = (aya)0 . Since S is a regular cyber group, by Lemma 3.1(i) and the regularity of the band E(S), we obtain (acxa)0 = a0 c0 x0 a0 = a0 c0 a0 x0 a0 = a0 c0 a0 y 0 a0 = (acya)0 . Hence, (cx, cy) ∈ ρa . Dually, we can prove that ρa is a right congruence. Obviously, H† ⊆ ρa , and hence ρa is a good band congruence. (ii) Let (x, y) ∈ ρa . Then, by the definition of ρa , we have (axa)0 = (aya)0 and hence a01 (axa)0 a01 = a01 (aya)0 a01 . Since E(S) is a regular band and by Lemma 3.1, we obtain a01 (axa)0 a01 = a01 a0 x0 a0 a01 = a01 a0 a01 x0 a01 a0 a01 = a01 x0 a01 = (a1 xa1 )0 . Similarly, we have a01 (aya)0 a01 = (a1 ya1 )0 . Thus, (a1 xa1 )0 = (a1 ya1 )0 , and consequently, ρa ⊆ ρa1 . Dually, ρa1 ⊆ ρa . This proves that ρa = ρa1 . (iii) Let a ∈ Sα , b ∈ Sβ and α > β. To show that ρα ⊆ ρβ , we let (x, y) ∈ ρα = ρa , by (ii). Then, by the definition of ρa , we have (axa)0 = (aya)0 and hence b0 (axa)0 b0 = b0 (aya)0 b0 . Since α > β in Y and a ∈ Sα , b ∈ Sβ , (bab)0 = b0 . By Lemma 3.1(i) and the regularity of the band, we can show that (bxb)0 = (byb)0 , that is, (x, y) ∈ ρb = ρβ . Thus, ρα ⊆ ρβ , as required. Furthermore, it is trivial to see that ρβ |Sα = ωSα , the universal relation on Sα . 2 We now use the congruence ρα given in Lemma 3.4 to describe the structural homomorphisms for the semigroup S = (Y ; Sα ), where every Sα is a rectangular up-semigroup. We first let ρα,β = ρα |Sβ for α > β on Y , which is a good band congruence on Sβ . Then, by Lemma 3.4 (iii), ρα,α is the universal relation ωSα on Sα and ρα,γ ⊆ ρβ,γ for α > β > γ on Y . We denote all the ρα,β -classes by {Sd(α,β) : d(α, β) ∈ D(α, β)}, where D(α, β) is a non-empty index set for [α, β]. In particular, D(α, α) is a singleton and we write d(α, α) = D(α, α). Now,we have the following lemma. Lemma 3.5 Let S = (Y ; Sα ) be a regular cyber group. Then, for all α, β ∈ Y with α > β, the following statements hold for all d(α, β) ∈ D(α, β). 8
(i) for all a ∈ Sα , there exists a unique ad(α,β) ∈ Sd(α,β) satisfying a > ad(α,β) ; (ii) for all a ∈ Sα and 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; (iii) Let a ∈ Sα . Define φd(α,β) : Sα −→ Sd(α,β) by aφd(α,β) = ad(α,β) , where ad(α,β) ∈ Sd(α,β) and a > ad(α,β) . Then φd(α,β) is a homomorphism and ad(α,β) = a(aba)0 = (aba)0 a for any b ∈ Sd(α,β) . Proof: (i) We first claim that for any a ∈ Sα and b ∈ Sd(α,β) , ab ∈ Sd(α,β) , that is, (ab, b) ∈ ρα,β . In fact, since S = (Y, Sα ) is a regular cyber group, every Sα is a rectangular up-semigroup. Hence, (xax)0 = x0 , for all x ∈ Sα . This leads to (xabx)0 = (xaxbx)0 = (xbx)0 , by the regularity of the band E(S) and Lemma 3.1(i). Thereby, (ab, b) ∈ ρα,β . Similarly, we also have ba ∈ Sd(α,β) . By Invoking the above results, we have aba ∈ Sd(α,β) , for any b ∈ Sd(α,β) . Since H† is a band congruence on S, by Lemma 3.1(i) again, a(aba)0 , (aba)0 and (aba)0 a are in the same H† -class of S and hence, a(aba)0 = (aba)0 a(aba)0 = (aba)0 a. Let a(aba)0 = ad(α,β) . Then by the natural partial order imposed on the semigroup S, we have a > ad(α,β) . In order to show the uniqueness of ad(α,β) , we assume that there is another a∗d(α,β) ∈ Sd(α,β) satisfying a > a∗d(α,β) . Then, by the definition of “6”, we can write a∗d(α,β) = ea = af for some e, f ∈ E(S). Thus by Lemma 3.1(i), (a∗d(α,β) )0 = ea0 = a0 f , that is, a0 > (a∗d(α,β) )0 . Now by Lemmas 3.1(i) and 3.4, we have (a∗d(α,β) )0 = (a0 (a∗d(α,β) )0 a0 )0 = (aa∗d(α,β) a)0 = (aba)0 . This shows that (a∗d(α,β) , ad(α,β) ) ∈ H† and now by Lemma 3.3 (iii), a∗d(α,β) = ad(α,β) . Thus, the uniqueness of ad(α,β) is proved. (ii) It is easy to see that, by the definition of “6”, a0 > a0 (ax)0 a0 . Also, since a ∈ Sα and x ∈ Sd(α,β) , ax ∈ Sd(α,β) , by (i). Moreover, since Sd(α,β) is an ρα,β -congruence class, (ax)0 ∈ Sd(α,β) . Hence, by (i), a0 (ax)0 a0 ∈ Sd(α,β) and e = a0 (ax)0 a0 . Consequently, we deduce eax = a0 (ax)0 a0 ax = ax, and similarly, xae = xa. Since x can be arbitrarily chosen in Sd(α,β) , we can particularly choose x = e. In this case, we obtain ea = ae and by Lemma 3.1(i) again, we have (ea)0 = (ea0 )0 = e.
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(iii) By using the result in (i), we define φd(α,β) : Sα −→ Sd(α,β) by aφd(α,β) = ad(α,β) = a(aca)0 = (aca)0 a, for a ∈ Sα and c ∈ Sd(α,β) . Then, for any a, b ∈ Sα , by (ii),we have (aφd(α,β) )(bφd(α,β) ) = ad(α,β) bd(α,β) = (aca)0 ab(bcb)0 = (aca)0 (ab(bcb)0 ) = ab(bcb)0 . Similarly, we have (aφd(α,β) )(bφd(α,β) ) = (aca)0 ab. Hence, ab > (aφd(α,β) )(bφd(α,β) ). Thus, by the definition of φd(α,β) , we have (ab)φd(α,β) = (aφd(α,β) )(bφd(α,β) ). This shows that φd(α,β) is a homomorphism. 2 Lemma 3.6 Let S = (Y ; Sα ) be a regular cyber group and Φα,β = {φd(α,β) |d(α, β) ∈ D(α, β)} when α > β on Y , where D(α, β) is the non-empty index. Then the following statements hold: (i) φd(α,α) is the identity automorphism on Sα for each α ∈ Y ; (ii) (∀α, β, γ ∈ Y , α > β > γ), Φα,β Φβ,γ ⊆ Φα,γ ; (iii) (∀α, β ∈ Y )(a ∈ Sα ), aΦα,αβ is contained in one ρβ,αβ -class. Proof: (i) Let a ∈ Sα and aφd(α,α) = b. Then, by the definition of φd(α,α) , b = ea = af for some idempotents e, f ∈ E(Sα ). According to Lemma 3.2, we have (a, b) ∈ H† and therefore, a0 = b0 . Thus, we deduce b = a0 b = a0 ea = a0 ea0 a = a, and therefore, φd(α,α) is the identity automorphism of Sα . (ii) Let φd(α,β) : Sα −→ Sd(α,β) ⊆ Sβ and φd(β,γ) : Sβ −→ Sd(β,γ) ⊆ Sγ . Then we want to show that φd(α,β) φd(β,γ) = φd(α,γ) for some φd(α,γ) : Sα −→ Sd(α,γ) ⊆ Sγ . For this purpose, we let a ∈ Sα , b1 , b2 ∈ Sd(α,β) and c ∈ Sd(β,γ) . Then, by Lemma 3.5, we have b1 φd(β,γ) = b1 (b1 cb1 )0 and b2 φd(β,γ) = b2 (b2 cb2 )0 . Since b1 , b2 ∈ Sd(α,β) , by the definition of ρα,β , we have (b1 , b2 ) ∈ ρα,β . This leads to (ab1 a)0 = (ab2 a)0 . By the regularity of the band E(S), we deduce that (a(b1 φd(β,γ) )a)0 = (ab1 (b1 cb1 )0 a)0 = a0 b01 c0 b01 a0 = a0 b01 a0 c0 a0 b01 a0 = a0 b02 a0 c0 a0 b02 a0 = (ab2 (b2 cb2 )0 a)0 = (a(b2 φd(β,γ) )a)0 . 10
Thus, by the definition of ρα,γ , (b1 φd(β,γ) , b2 φd(β,γ) ) ∈ ρα,γ . In other words, there exists a ρα,γ -class Sd(α,γ) satisfying Sd(α,β) φd(β,γ) ⊆ Sd(α,γ) . Also, φd(α,β) φd(β,γ) is clearly a mapping which maps Sα into Sd(α,γ) . By the transitivity of “6”, we immediately obtain φd(α,β) φd(β,γ) = φd(α,γ) . This proves that Φα,β Φβ,γ ⊆ Φα,γ . (iii)We only need to show that for any φd(α,αβ) and φd0 (α,αβ) ∈ Φα,αβ , (aφd(α,αβ) , aφd0 (α,αβ) ) ∈ ρβ,αβ . For this purpose, we let x ∈ Sd(α,αβ) and x0 ∈ Sd0 (α,αβ) . Then, by Lemma 3.6, (iii), we have aφd(α,αβ) = a(axa)0 and aφd0 (α,αβ) = a(ax0 a)0 . Let b ∈ Sβ . Then, because Sαβ is a rectangular up-semigroup, and bab, aφd(α,αβ) , aφd0 (α,αβ) are elements in Sαβ , we obtain (bab, (bab)(aφd(α,αβ) )(bab)) ∈ H† and (bab, (bab)(aφd0 (α,αβ) )(bab)) ∈ H† . Since every H† -class of Sαβ is a up-semigroup, ((bab)(aφd(α,αβ) )(bab))0 = ((bab)(aφd0 (α,αβ) )(bab))0 . In other words, ((bab)(a(axa)0 )(bab))0 = ((bab)(a(ax0 a)0 )(bab))0 . Now, by using the regularity of the band E(S), we can further simplify the above equality to (b(a(axa)0 )b)0 = (b(a(ax0 a)0 )b)0 , that is, (b(aφd(α,αβ) )b)0 = (b(aφd0 (α,αβ) )b)0 . Hence, it follows that, by the definition of ρβ,αβ , (aφd(α,αβ) , aφd0 (α,αβ) ) ∈ ρβ,αβ . This completes the proof. 2 Lemma 3.7 Let S = (Y ; Sα ) be a regular cyber group. For any a ∈ Sα , b ∈ Sβ , let aΦα,αβ ⊆ Sd(β,αβ) and bΦβ,αβ ⊆ Sd(α,αβ) . Then ab = (aφd(α,αβ) )(bφd(β,αβ) ). Proof: Let c1 ∈ Sd(α,αβ) , c2 ∈ Sd(β,αβ) . Then (ac1 a)0 ∈ Sd(α,αβ) because Sd(α,αβ) is a ρα,αβ -equivalence class of Sαβ . Now, by Lemma 3.5, aφd(α,αβ) = (ac1 a)0 a and bφd(β,αβ) = b(bc2 b)0 for φd(α,αβ) ∈ Φα,αβ and φd(β,αβ) ∈ Φβ,αβ . Since we assume that aΦα,αβ ⊆ Sd(β,αβ) , aφd(α,αβ) = (ac1 a)0 a ∈ Sd(β,αβ) . Similarly, we have bφd0 (β,αβ) ∈ Sd(α,αβ) ∩ Sd(β,αβ) . Thus, by Lemma 3.5 (ii), we have (aφd(α,αβ) )(bφd(β,αβ) ) = (ac1 a)0 (ab(bc2 b)0 ) = ab(bc2 b)0 and also (aφd(α,αβ) )(bφd(β,αβ) ) = ((ac1 a)0 ab)(bc2 b)0 = (ac1 a)0 ab. However, by the definition of the natural partial order “6”, we have ab > (aφd(α,αβ) )(bφd(β,αβ) ). Thus ab = (aφd(α,αβ) )(bφd(β,αβ) ) because Sαβ is primitive. 2 ¿From Lemma 3.5 (i) and Lemma 3.7, we can easily see that the regular cyber group S = (Y ; Sα ) has the following property: for α > β on Y and d(α, β) ∈ D(α, β), ax = (aφd(α,β) )x and xa = x(aφd(α,β) ) for any a ∈ Sα and x ∈ Sd(α,β) . 11
4
Characterizations for cyber groups
We now use the knitted semilattice of rectangular up-semigroups to characterize the regular cyber groups. We consider the situation that the Green H† -relation is a right quasinormal or a normal band congruence. We first denote the band congruence ρα,β -classes by {Sd(α,β) |d(α, β) ∈ D(α, β)} and let Φα,β = {φd(α,β) : d(α, β) ∈ D(α, β)}, respectively in the knitted semilattice of rectangular up-semigroups S = [Y ; Sα , ρα,β , Φα,β ]. In establishing our main result, the following lemmas are crucial. Lemma 4.1 Let S = [Y ; Sα , ρα,β , Φα,β ] be a knitted semilattice of rectangular up-semigroups Sα (α ∈ Y ). Suppose that α > β on Y , a ∈ Sα and x ∈ Sd(α,β) , where d(α, β) ∈ D(α, β). Write a0 φd(α,β) = e. Then we have eax = ax, xae = xa, ea = aeand (ea)0 = e. Proof: Since S = [Y ; Sα , ρα,β , Φα,β ] is a knitted semilattice of rectangular up-semigroups, the structure mapping φd(α,β) is a homomorphism from Sα to Sd(α,β) ⊆ Sβ , we have eax = (ea)x = (e(aφd(α,β) ))x = ((a0 φd(α,β) )(aφd(α,β) ))x = (aφd(α,β) )x = ax.
Similarly, we have xae = xa. Now, ea = ae and (ea)0 = e, by the definition of knitted semilattice of semigroups. 2 Lemma 4.2 Let S = [Y ; Sα , ρα,β , Φα,β ] be a knitted semilattice of rectangular up-semigroups with ρα,β a good band congruence and E(S) forms a band. Suppose that α > β on Y with d(α, β) ∈ D(α, β) and a ∈ Sα , b ∈ Sd(α,β) . Then the following conditions are equivalent: (i) a > b; (ii) b = aφd(α,β) ; (iii) b = b0 a = ab0 . Proof: (i)=⇒ (ii) Assume that a > b. Then by using similar arguments as in the proof of Lemma 3.3 (1), we can let b = ea = af for some idempotents e, f ∈ Sβ . Suppose that e ∈ Sd0 (α,β) for some d0 (α, β) ∈ D(α, β). Then, we have b = ea = e(aφd0 (α,β) ) ∈ Sd0 (α,β) by 12
Definition 2.1. Hence, Sd0 (α,β) = Sd(α,β) so that e ∈ Sd(α,β) . Similarly, we can prove that f ∈ Sd(α,β) . Now, by b = ea = e(aφd(α,β) ), b = af = (aφd(α,β) )f , we have aφd(α,β) > b. Since Sβ is primitive,aφd(α,β) = b. (ii)=⇒ (iii) Since ρα,β is a good band congruence on Sβ , we immediately have H† |Sβ ⊆ ρα,β . Hence b0 ∈ Sd(α,β) , and thereby, we deduce that b0 a = b0 (aφd(α,β) )= b0 b = b. Dually, we have ab0 = b. (iii)=⇒ (i) The implication is obvious by the definition of “>”.
2
Lemma 4.3 Let S = [Y ; Sα , ρα,β , Φα,β ] be a knitted semilattice of rectangular up-semigroups with ρα,β a good band congruence and E(S) a band. Then for any a ∈ Sα , b ∈ Sβ ,we have ab = (aφd(α,αβ) )(bφd(β,αβ) ), where aΦα,αβ ⊆ Sd(β,αβ) , bΦβ,αβ ⊆ Sd(α,αβ) . Proof: Let aφd(α,αβ) = a1 and bφd(α,αβ) = b1 . Then by Lemma 4.2, we can easily see that a0 φd(α,αβ) = a01 , b0 φd(β,αβ) = b01 . Since ρα,αβ , ρβ,αβ are all good congruences, H† |Sαβ ⊆ ρα,αβ ∩ ρβ,αβ . Consequently, a01 , b01 ∈ Sd(α,αβ) ∩ Sd(β,αβ) . Now, by Lemma 4.1, we deduce that (aφd(α,αβ) )(bφd(β,αβ) ) = a01 abb01 = a01 ab = abb01 . This show that ab > (aφd(α,αβ) )(bφd(β,αβ) ) and so ab = (aφd(α,αβ) )(bφd(β,αβ) ) by the primitivity of the rectangular up-semigroup Sαβ . 2 We now formulate the following characterization theorem for regular cyber groups. Theorem 4.4 Let S be a semi-superabundant semigroup whose set of idempotents forms a subsemigroup. Then S is a regular cyber group if and only if S is a knitted semilattice of rectangular up-semigroups, that is, S = [Y ; Sα , ρα,β , Φα,β ] with ρα,β a good band congruence, for any α > β. Proof: By the definition of knitted semilattice of semigroups and the results obtained in Section 3, the necessity part of the theorem has been already proved. In proving the sufficiency part of the Theorem, we first show that the Green †-relation H† is a congruence on S. In fact, by Lemma 4.3, we can easily show that (ab)0 = a0 b0 , for all a, b ∈ S. Hence H† is a congruence on S,by Lemma 3.1. 13
To show that E(S) is a regular band, by using a result of M. Petrich in [12], we only need to show that the Green relations L and R are both congruences on E(S). We now show L is a congruence in E(S) as R is a congruence in E(S) is its dual. Since E(S) = (Y ; E(Sα )) , we can let e, f ∈ Sα ∩ E(S), g ∈ Sβ ∩ E(S) with (e, f ) ∈ L(E(S)). Then, we have ef = e and f e = f . By the definition of knitted semilattice of semigroups f S = [Y ; Sα , ρα,β , Φα,β ] and Lemma 4.3, we can find homomorphisms φef d(β,αβ) and φd(β,αβ) ∈ Φβ,αβ , φgd(α,αβ) ∈ Φα,αβ such that gegf = [g(ef )](gf ) g f g = [(gφef d(β,αβ) )((ef )φd(α,αβ) )][(gφd(β,αβ) )(f φd(α,αβ) )] g = (gφef d(β,αβ) )(f φd(α,αβ) )
and ge = g(ef ) g = (gφef d(β,αβ) )((ef )φd(α,αβ) ) g = (gφef d(β,αβ) )(f φd(α,αβ) ).
Thereby, gegf = ge. Analogously, we can also prove that gf ge = gf . This proves that L is a left congruence on E(S). Since L is always right congruence on S, L is a congruence on E(S), as required. Dually, R is also a congruence on E(S). Thus by [12] (see II.3.6 Proposition), E(S) is a regular band and hence S is a regular cyber group. Our proof is completed. 2 Recall that a right quasi-normal band is a band satisfying the identity yxa = yaxa. Also, a left quasi-normal band is a band satisfying the identity axy = axay. Thus, we can easily observe that both the right quasi-normal bands and the left quasi-normal bands are all special cases of the regular bands. Also, normal bands are special case of both right quasi-normal bands and left quasi-normal bands. Based on the above observation, we can establish the following theorem. Theorem 4.5 Let S be a semi-superabundant semigroup S whose set of idempotents E(S) forms a band. Then S is a right quasi-normal cyber group if and only if S is a knitted semilattice of rectangular up-semigroups, that is, S = [Y ; Sα , ρα,β , Φα,β ] with ρα,β a good band congruence and L† |Sβ ⊆ ρα,β for any α > β. The proof of this theorem is very similar to Theorem 4.4. We hence omit the proof.
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By using similar arguments as in Theorem 4.4, we can also obtain the following characterization theorem for normal cyber groups. Theorem 4.6 Let S be a semi-superabundant semigroup S whose set of idempotents E(S) forms a subsemigroup. Then S is a normal cyber group if and only if S can be expressed by a knitted semilattice of rectangular up-semigroups, that is, S = [Y ; Sα , ρα,β , Φα,β ] with ρα,β = ωSβ , for any α > β.
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