On normal cyber H+-groups

On normal cyber H+-groups Xiangzhi Kong Department of Mathematics Qufu Normal University Qufu, Shandong, 273165, China e-mail: [email protected]

K.P. Shum∗ Faculty of Science The Chinese University of Hong Kong Shatin, Hong Kong (SAR), China e-mail: [email protected]

Abstract We extend the set of usual Green relations to a new set of Green + − relations and the semilattice decomposition of a super H+ -abundant semigroups will be hence investigated. It was first shown by Petrich that a completely regular semigroup S is a normal crypto group if and only if S can be expressed as a strong semilattice of completely simple ∗

The research of the second author is partially supported by a UGC (HK) grant # 2160210/03-05

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semigroups.By generalizing this result to super H+ -abundant semigroups,we establish an analogous theorem that a super H+ -abundant semigroup S is a normal cyber H+ -group if and only if S is a strong semilattice of completely J + -simple semigroups.Our result also generalizes a corresponding result of super abundant semigroups recently obtained by Ren and Shum. Keywords Green + −relations; Cryptogroups; Super H+ −abundant semigroups; Normal bands; Normal cyber H+ -groups. Mathematics Subject Classification (2002): 20M10

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Introduction

It is well known that the Green relations play an important role in the theory of regular semigroups[1-7]. A classical theorem of Clifford says that a semigroup is a union of groups if and only if it is a semilattice of completely simple semigroups[1]. As inspired by this result, Petrich [6] shown that a completely regular semigroup S is a normal cryptogroup(that is, a completely regular semigroup whose Green relation H is a normal band congruence) if and only if S is a strong semilattice of completely simple semigroups.The classical theorem of Clifford was then generalized.Moreover,it was shown by Fountain [3] that an abundant semigroup S is a super abundant semigroup(that is, an abundant semigroup S whose every H∗ -class contains an unique idempotent of S) if and only if S is a semilattice of completely J ∗ -simple semigroups.Recently, Ren and Shum [9 ] has further studied the super abundant semigroup and gave a construction theorem for such semigroups based on the result of Fountain [3].Thus super abundant semigroup is indeed a proper generalization of a completely regular semigroup.

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In studying the abundant semigroups,Fountain [3] adopted the following Green∗ -relations which were first introduced by Pastijn [5]. 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∗ ; _ and D∗ = L∗ R∗ . In this paper, we give the definitions of the Green + -relations and study the semilattice decomposition of the super H+ -abundant semigroups. By using this decomposition, we are able to show that a super H+ -abundant semigroup is a normal cyber H+ −group if and only if it is a strong semilattice of completely J + -simple semigroups. This result hence generalizes and enriches the results of Petrich [7] and Ren-Shum [8]. We first extend the Green∗ −relations to the following Green+ −relations on a semigroup S: L+ = {(a, b) ∈ S × S : (∀e, f ∈ E(S 1 ))ae = af ⇔ be = bf }; R+ = {(a, b) ∈ S × S : (∀e, f ∈ E(S 1 ))ea = f a ⇔ eb = f b}; \ H+ = L+ R+ _ and D+ = L+ R+ . It can be easily seen that the relations L+ and R+ are also equivalent relations on a semigroup S.Same as the relations L∗ and R∗ on the semigroup S, the relation L+ is not necessarily a right congruence and the relation R+ is not necessarily a left congruence on S. We now denote the L+ -class + containing an element a of the semigroup S by L+ a (S) or by La if no ambiguity arises. It can be easily seen that L ⊆ L∗ ⊆ L+ . For the usual Green relation L and the Green like relations such as the L+ - relation and the L∗ relation,their dual relations are the corresponding R-relation;the R∗ -relation 3

and the R+ relation,respectively. We now consider the relations which are T related to the L+ -relation.In a semigroup S,if e ∈ Ha+ E(S) for some a ∈ S, then we write e as x0 , for any x ∈ Ha+ . Clearly, for any x ∈ Ha+ with a ∈ S, we have x = xx0 = x0 x. If S is a regular semigroup, then we know that every L-class of S contains at least one idempotent, and so does every R-class of S (see [3]). Moreover,if S is a completely regular semigroup, then every H-class of S contains an unique idempotent. A semigroup S is called abundant by Fountain [3] if every L∗ - and R∗ -class of S contains some idempotents. It is trivial to see that L∗ = L on the regular elements of a semigroup.Hence, all regular semigroups are special abundant semigroups. We call a semigroup S super abundant if each of its H∗ -classes contains an unique idempotent, and thereby a super abundant semigroup is a proper generalization of a completely regular semigroup.Following the terminologies of Fountain [3],we just call a semigroup S a semiabundant semigroup if each of its L+ -class and each of its R+ -class contains at least one idempotent. Naturally,we call a semigroup S a H+ -abundant semigroup if the relation H+ -is a congruence on S such that every H+ − class containing some idempotents of S. We can now easily see,by our definition, that the H+ -abundant semigroup is a special semiabundant semigroups. One can also easily see that the equalities R+ = R = R∗ and L+ = L∗ hold on regular semigroups.Similar to the concept of cryptogroups in the class of completely regular semigroups,we call a H+ − abundant semigroup S a cyber H+ −group if the set of all idempotents of S forms a subgroup of S. Recall that a band B is called a normal band if B satisfies the identity ef ge = egf e for all element e, f and g in B.According to Petrich and Reilly [8],we call a completely regular semigroup S a regular cryptogroup [6] if the Green relation H of S is a regular band congruence on S. Now, we call a semigroup S a super H+ -abundant semigroup if S is a H+ -abundant semigroup whose Green+ -relation H+ is a congruence on S and each H+ class of 4

S contains an unique idempotent of S. In addition,if the Green+ -relation on the semigroup S is a H+ normal band congruence on S ( that is, the set of all idempotents E(S) of S forms a normal band) then,following Guo and Shum in [4],we call a super H+ abundant semigroup S a normal cyber H+ −group.Clearly,the normal cyber H+ groups in the class of semiabundant semigroups are the analogues of the cryptogroups in the class of completely semigroups. The structure of normal crytogroups was investigated by Petrich in 1974[7],however,the structure of its analogues in the class of semiabundant semigroups,namely the normal cyber H+ −groups has not yet been known.In this paper, we will prove that the normal cyber H+ −group in the class of semiabundant semigroups can be expressed as a strong semilattice of completely J + -simple semigroups.Thus, we establish an analogous result of normal crytogroups in the class of semiabundant semigroups. For notations and terminologies not mentioned in this paper, the reader is referred to Clifford-Preston [1-2] and Howie[5].

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The Green+-relations

We have already defined the Green+ -relations,namely,the L+ , R+ and H+ , D+ relations on a semigroup S. To define the relation J + on S,we need to define a left+ (right + )-ideal L+ of a semigroup S which is a left(right) ideal + L(R) of S such that L+ a ⊆ L ( Ra ⊆ L), for all a ∈ L. A subset I of S is called an ideal+ if it is both a left ideal+ and a right ideal+ of S. We note that if S is a regular semigroup then every left(right, two-sided) ideal of S is a left(right, two-sided) ideal+ of S. We also observe that for an idempotent e in any semigroup S, the left(right) ideal Se(eS) of S is indeed a left(right) ideal+ of S. For if a ∈ Se, then a = ae, and hence for any element b in L+ a, we have b = be ∈ Se. Similar to Petrich in [7], we have the following lemmas. Lemma 2.1

(i) Let I1 ⊆ I2 ⊆ S with I2 an ideal+ of S and I1 an ideal+ 5

of I2 . If I12 = I1 , then I1 is an ideal+ of S. (ii) If I is a right ideal+ and J is a left ideal+ of an H+ -abundant semiT group S,then IJ = I J. Proof: We only prove (i) because (ii) can be easily proved. For any s ∈ S and a ∈ I1 , by I12 = I1 , we have a = bc for some b, c ∈ I1 ⊆ I2 . Thus sb ∈ I2 since I2 is an ideal of S.Similarly, sa = (sb)c ∈ I1 since I1 is also an ideal of I2 . Consequently, I1 is a left ideal of S. Dually, I1 is also a right ideal of S. This shows that I1 is an ideal of S. We now know that I1 is an ideal+ of S, by Exercise 4(a) of [1, §2.6]. 2 The proof of the following lemma is straightforward. Lemma 2.2 If {Iα : α ∈ Λ} is a set of ideals+ (lef t ideals+ , right ideals+ ) of a semigroup S, then (i) (ii)

T S

{Iα : α ∈ Λ} is an ideal+ (lef t ideal+ ,right ideal+ or empty). {Iα : α ∈ Λ} is an ideal+ (lef t ideal+ , right ideal+ ).

Let a be an element of a semigroup S. Then,in view of (i) and S is a ideal+ of itself, there is a smallest ideal+ , say,J + (a) containing a; a smallest left ideal+ , say, L+ (a) containing a; and a smallest right ideal+ , say, R+ (a) containing a. We now call J + (a), L+ (a), R+ (a) the principal ideal + , principal left ideal + , and principal right ideal + generated by a. It is clear that L+ (a) ⊆ J + (a) and R+ (a) ⊆ J + (a). Lemma 2.3 Let a be an element of a semigroup S. Then the following statements hold: 6

(i) b ∈ L+ (a) if and only if there are elements a0 , a1 , · · · , an ∈ S, x1 , · · · , xn ∈ S 1 such that a = a0 , b = an , and (ai , xi ai−1 ) ∈ L+ , for i = 1, · · · , n. (ii) b ∈ R+ (a) if and only if there are elements a0 , a1 , · · · , an ∈ S, x1 , · · · , xn ∈ S 1 such that a = a0 , b = an , and (ai , ai−1 xi ) ∈ R+ , for i = 1, · · · , n. (iii) b ∈ J + (a) if and only if there are elements a0 , a1 , · · · , an ∈ S, x1 , · · · , xn , y1 , · · · , yn ∈ S 1 such that a = a0 , b = an , and (ai , xi ai−1 yi ) ∈ D+ , for i = 1, · · · , n.

Proof: We only prove (i) since (ii) and (iii) can be similarly proved. Let I be the set of all b in S satisfying the given conditions. Now for any b ∈ I, we have a0 , a1 , · · · , an ∈ S, x1 , · · · , xn ∈ S 1 such that a = a0 , b = an , and (ai , xi ai−1 ) ∈ L+ for i = 1, · · · , n. If ai−1 ∈ L+ (a), then xi ai−1 ∈ L+ (a) since L+ (a) is a left ideal, and hence ai ∈ L+ (a) since L+ (a) is a lef t ideal+ . Because a0 = a ∈ L+ (a), we see that ai ∈ L+ (a) for all i = 0, 1, · · · , n. In particular,b ∈ L+ (a) and so I ⊆ L+ (a). Now if b ∈ I,then because that I is a left ideal of S,we have sb ∈ I, for all s ∈ S so that Ib+ ⊆ I. Hence I is a left ideal+ and by a ∈ I, we have I + (a) ⊆ I. 2

Corollary 2.4 Let a, b be elements of a semigroup S.Then we have (i) aL+ b if and only if L+ (a) = L+ (b). (ii) aR+ b if and only if R+ (a) = R+ (b).

Proof: It is clear from Lemma 2.3 that if aL+ b, then L+ (a) = L+ (b).Suppose that L+ (a)= L+ (b). Then b ∈ L+ (a) and so by Lemma 2.3, there are a0 , a1 , · · · , an ∈ S, x1 , · · · , xn ∈ S 1 such that a = a0 , b = an , and (ai , xi ai−1 ) ∈ 7

L+ for i = 1, · · · , n. Let e, f ∈ E(S 1 ) be such that ae = af . If ai−1 e = ai−1 f , then xi ai−1 e = xi ai−1 f and since ai L+ xi ai−1 we get ai e = ai f . Since a = a0 , it follows that be = bf . A similar argument shows that if e, f ∈ E(S 1 ) and be = bf , then ae = af . Hence aL+ b. Also,(ii) follows similarly. 2 We have already pointed out that for every idempotent e in the semigroup S,the left ideal Se is a left ideal+ of S. Thus, if a is a regular element in S, then aLe for some idempotent e and so Sa = Se = L+ (e) = L+ (a).

The following corollary is obvious.

Corollary 2.5 A semigroup S is semiabundant if and only if for every element a in S there exist idempotents e, f such that L+ (a) = Se, R+ (a) = f S.

Corollary 2.6 Let S be a semiabundant semigroup and a an element of S. Then for an idempotent e ∈ S, a L+ e if and only if a ∈ Se and Se is contained in every idempotent-generated left ideal containing a.

Proof: Suppose that a L+ e. Then L+ (a) = Se so that a ∈ Se.If a ∈ Sf for some idempotent f , then af = a and so ef = e. This shows that Se ⊆ Sf . Conversely, suppose that a ∈ Se and for any idempotent f ∈ E(S), a ∈ Sf implies that Se ⊆ Sf . Then we have L+ (a) ⊆ Se since Se is a left ideal+ . Because S is a semiabundant semigroup, L+ (a) = Sf for some idempotent f ∈ E(S). Hence, Se = Sf and so a L+ e. 2 We now define the relation J + by using an analogous characterization of L+ and R+ given in Corollary 2.6. That is, we can define the relation J + on a semigroup S by aJ + b if and only if (a) = J + (b). It is clear that L+ ⊆ J + and R+ ⊆ J + so that D+ ⊆ J + . We have already noted that in a regular 8

semigroup, every ideal of S is an ideal+ of S and in this case, we always have J+ = J. The followings are some properties of the H+ −abundant semigroups. Lemma 2.7 (i) If e, f are D+ -related idempotents of a super H+ -abundant semigroup S, then eDf . (ii) If S is an super H+ -abundant semigroup, then D+ = L+ ◦ R+ = R+ ◦ L+ . (iii) Let e, f be idempotents in an H+ -abundant semigroup S.If eJ f , then eDf . (iv) For a J + -simple cyber H+ -abundant semigroup S,we have J + = D+ . Proof: (i) Since e D+ f , there are a1 , · · · , ak of S such that eL+ a1 R+ a2 · · · ak L+ f . Since S is a H+ -abundant semigroup, we have eL+ a01 R+ a02 · · · a0k L+ f . Thus eDf . (ii) If a, b ∈ S and aD+ b,then by Lemma 2.7, we have a0 Db0 . Thus, there exist elements c, d in S with a0 LcRb0 and a0 RdLb0 . Consequently,we have aL+ cR+ b and aR+ dL+ b and hence the result. (iii) Since SeS = Sf S, there are elements x, y, s, t in S such that f = set, e = xf y. Let h = (f y)0 and k = (se)0 . Then by hf y = f y = f f y,we have h = h2 = f h, and sek = se = see and so we have k = k 2 = ke. It follows that hf, ek are idempotents of S 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 . 9

(iv) It is clear that a0 ∈ J + (a) so that Sa0 S ⊆ J + (a). We now proceed to show that the ideal Sa0 S is actually an ideal+ of S. If a = aa0 ∈ Sa0 S, then the result follows. On the other hand,if b = xa0 y ∈ Sa0 S( for some x, y ∈ S) and k = (a0 y)0 . Then a0 a0 y = ka0 y so that a0 (a0 y)0 = k 2 = k. Since H+ is a congruence on S, xa0 yH+ xk. Now, let h = (xk)0 = (xa0 y)0 . Then xkh = xkk so that h = h2 = hk = ha0 k ∈ Sa0 S. Thus, + 0 0 if c ∈ L+ b and d ∈ Rb , then c = ch, d = hd ∈ Sa S and hence Sa S is an ideal+ of S, as required. (v) Let a, b ∈ S with aJ + b. Then, by Lemma 2.10, we have Sa0 S = Sb0 S. By Lemma 2.10 again, we have a0 Da0 and hence the result. 2

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Normal cyber H+-groups

If a super H+ -abundant semigroup S is J + -simple,then S is said to be a completely J + -simple semigroup. In this section, we first generalize the wellknown result of Clifford, and then show that a super H+ -abundant semigroup is a normal cyber H+ -group if and only if it is a strong semilattice of completely J + -simple semigroups. This result generalizes of the result of Petrich in the class of semiabundant semigroups.

Lemma 3.1 Let S be an H+ -abundant semigroup. Then H+ is a congruence if and only if for any a, b ∈ S, (ab)0 = (a0 b0 )0 .

Proof: (Necessity). For any a, b ∈ S, then aH+ a0 and bH+ b0 . Since H+ is a congruence, we have abH+ a0 b0 . But abH+ (ab)0 , and so that (ab)0 = (a0 b0 )0 since every H+ -class contains a unique idempotent. 10

(Sufficiency). Since H+ is an equivalent relation on the semigroup S, we only need to show that H+ is compatible with the semigroup multiplication on S. Let (a, b) ∈ H+ and c ∈ S. Then (ca)0 = (c0 a0 )0 = (c0 b0 )0 = (cb)0 so that H+ is left compatible with the semigroup multiplication. Dually, H+ is also right compatible with the semigroup multiplication. Thus, H+ is indeed a congruence on S. 2 Lemma 3.2 A super abundant completely J + -simple S is primitive for idempotents. Proof: Let e, f be idempotents in S with e 6 f . Since S is J + -simple and H+ -abundant semigroup, it follows from Lemma 2.10 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 by g 6 f, we have a0 = ga0 (gf )a0 = g(f a0 ) = gf = g. Now we have g 6 f and gLf so that f = f g = g. But since g 6 e,we have e = f and so that all idempotents of S are primitive. 2 Theorem 3.3 Let S be super H+ -abundant semigroup. Then S is a semilattice Y of completely J + -simple semigroups Sα (α ∈ Y ) such that for α ∈ Y + + + and a ∈ Sα , L+ a (S) = La (Sα ), Ra (S) = La (Rα ). Proof: If a ∈ S, then aH+ a2 so that certainly J + (a) = J + (a2 ).Now for a, b ∈ S,(ab)2 ∈ SbaS,and so J + (ab) = J + ((ab)2 ) ⊆ J + (ba) and by symmetry, J + (ab) = J + (ba). By Lemma 2.10,we can deduce that T J + (a) = Sa0 S, J + (b) = Sb0 S so that if c ∈ J + (a) J + (b), then c = xa0 y = 11

zb0 t for some x, y, z, t ∈ S. Now c2 = zb0 txa0 y ∈ Sb0 txa0 S ⊆ J + (b0 txa0 ) so that by using the arguments in Lemma 3.2,we have J + (b0 txa0 ) = J + (a0 b0 tx). This leads to c2 ∈ J + (a0 b0 ) and since c H+ c2 , we have c ∈ J + (a0 b0 ). Since a H+ a0 , b H+ b0 and H+ is a congruence on S so that ab H+ a0 b0 . T Hence c ∈ J + (ab),and consequently, J + (a) J + (b) ⊆ J + (ab). Because the T opposite inclusion is clear, we have J + (a) J + (b) = J + (ab). We have already observed that the set Y of principal ideals+ of S forms a semilattice under set intersection and that the map a 7→ J + (a) is a homomorphism from S onto Y . The inverse image of J + (a) is just the J + -class Ja+ which is a subsemigroup of S. Hence S is a semilattice Y of the semigroups Ja+ . Now let a, b be elements of the J + -class J + and suppose that (a, b) ∈ L+ (J + ). Then we have a0 , b0 ∈ J + so that we have (a0 , b0 ) ∈ L+ (J + ).This leads to a0 b0 = a0 , b0 a0 = b0 and (a0 , b0 ) ∈ L+ (S). It follows that (a, b) ∈ + + + + L+ (S) and consequently,by L+ a (S) ⊆ J , we have La (S) = La (J ). By using a similar argument as before,we can show that Ra+ (S) = Ra+ (J + ).

From the last paragraph, we have Ha+ (J + ) = Ha+ (S) so that J + is H+ abundant semigroup. Furthermore, if a, b ∈ J + , then by Lemma 2.11, we have (a, b) ∈ D+ (S) so that, by Corollary 2.8, there is an element c in T + T + + + L+ Rb (S) = L+ Rb (J ). Thus a, b are D+ -related in J + and a (S) a (J ) hence,the semigroup J + is J + -simple. 2

Lemma 3.4 In a completely J + -simple semigroup S, the set of all regular elements of S forms a regular subsemigroup of S.

Proof: Let a, b be regular elements of S. Since S consists of a single D+ class(by Lemma 2.11), it follows from Corollary 2.8 that there is an element 12

c ∈ S with aL+ cR+ b. Hence we have aL+ c0 R+ b. Since a is regular, we have c0 b = b and aLc0 . This leads to abLb and the regularity of ab follows from the regularity of b. 2 Lemma 3.5 Let S = (Y ; Sα ) be a super H+ -abundant semigroup. Then the following statements hold: (i) Let a ∈ Sα and α > β. Then there exists b ∈ Sβ with a > b; (ii) Let a, b, c ∈ S with bH+ c and a > b, c. Then we have b = c; (iii) Suppose that a ∈ E(S) and b ∈ S. If a > b,then b ∈ E(S).

Proof: (i) Let b ∈ Sβ .Then, by Lemma 3.1, a(aba)0 , (aba)0 a and (aba)0 are all in the same H+ -class so that a(aba)0 = (aba)0 a(aba)0 = (aba)0 a. Let b = a(aba)0 .Then b ∈ Sβ and hence a > b. (ii) By the definition “>”, there exist e, f, g, h ∈ E(S) such that b = ea = af , c = ga = ah. From eb = b and bH+ b0 , we have eb0 = b0 . Similarly, c0 h = c0 . Thus, ec = ec0 c = eb0 c = b0 c = c. Similarly, bh = b so that b = bh = eah = ec = c, as required. (iii) We have b = ea = af, for some e, f ∈ E(S), and whence b2 = (ea)(af ) = ea2 f = b. 2 The following lemma is a crucial lemma for normal cyber H+ -groups. 13

Lemma 3.6 Let S = (Y ; Sα ) be a normal cyber H+ -group.Then the following statements hold: (i) For any α > β on Y and a ∈ Sβ , there exists a unique element aβ ∈ Sβ such that a > aβ . (ii) For any α > β on Y and a ∈ Sα , x ∈ Sβ , if a0 > e for some idempotent e ∈ Sβ then eax = ax, xae = xa, ea = ae and (ea)0 = e. Proof: (i) By Lemma 3.5 (i), there exists aβ = a(aca)0 = (aca)0 a ∈ Sβ for any c ∈ Sβ such that a > aβ . If there is another b ∈ Sβ such that a > b, then there are idempotents g, h ∈ E(S) such that b = ga = ah. Hence aβ H+ = (aca)0 H+ aH+ = aH+ (aca)0 H+ and bH+ = gH+ aH+ = aH+ hH+ , that is, aβ H+ 6 aH+ and bH+ 6 aH+ . It can be easily seen that S/H+ = (Y ; Sα /H+ ).Hence, aβ H+ = bH+ since S/H+ is a normal band. By Lemma 3.5 (ii),we have aβ = b. (ii) Since (a0 (ax)0 a0 )a0 = a0 (ax)0 a0 = a0 (a0 (ax)0 a0 ) and a0 (ax)0 a0 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 ∈ Sβ , we have ax ∈ Sβ 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. Because x is an arbitrary element in Sβ , we can particularly choose x = e. In this case, we obtain that ea = ae and consequently, by Lemma 3.1, we have (ea)0 = (ea0 )0 = e. 2 We now formulate the following theorem. Theorem 3.7 ( Main theorem) Let S be a super H+ -abundant semigroup.Then S is a normal cyber H+ -group if and only if it is a strong semilattice of completely J + -simple semigroups.

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Proof: (Necessity). By Theorem 3.3, we can represent the super H+ − abundant semigroup S by S = (Y ; Sα ), where Y is a semilattice and each Sα (α ∈ Y ) is a completely J + -simple semigroup. Now, for α > β on Y and a ∈ Sα , by Lemma 3.6, we can define a map φα,β : a 7→ aβ from Sα into Sβ , where aβ is the unique element in Sβ such that aβ 6 a. We proceed to show that S = [Y ; Sα , φα,β ] is a strong semilattice of completely J + -simple semigroups Sα . (i) For every α ∈ Y , a, b ∈ Sα and a > b. Then there exist idempotents e, f ∈ S such that b = ea = af so that bH+ = eH+ aH+ = aH+ f H+ . This implies that aH+ = bH+ since S/H+ = (Y ; Sα /H+ ) is a normal band and so by Lemma 3.5,we have a = b since a > a. (ii) Let α > β > γ on Y .Then by the transitivity of 6, we have φα,β φβ,γ = φα,γ . (iii) Let α, β ∈ Y with a ∈ Sα , b ∈ Sβ . Then,by the proof of Lemma 3.6 (i), we can let aφα,αβ = a1 = a(aca)0 = (aca)0 a and bφβ,αβ = b1 = b(bcb)0 = (bcb)0 b, for any c ∈ Sαβ . Thus by Lemma 3.6 (ii), a1 b1 = (aca)0 ab(bcb)0 = (aca)0 ab = ab(bcb)0 so that a1 b1 6 ab. Now, we have ab = a1 b1 = (aφα,αβ )(bφβ,αβ ) by using the same proof of (i). (Sufficiency). Let S = [Y ; Sα , φα,β ] be a strong semilattice of completely J -simple semigroups Sα . We only need to show that H+ is a congruence and S/H+ is a normal band. Let a ∈ Sα and b ∈ Sβ , Then we have ab = (aφα,αβ )(bφβ,αβ ) and so (ab)0 = [(a0 φα,αβ )(b0 φβ,αβ )]0 = (a0 b0 )0 . Thus by Lemma 3.1, S is a H+ is a congruence on the semigroup S. +

For a ∈ Sα , x ∈ Sβ and y ∈ Sγ , let δ = αβγ. Then by direct computation, we can verify that (axya)H+ = aφα,δ H+ = (ayxa)H+ . Thus S/H+ is indeed a normal band. Hence, the proof is completed. 2

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References [1] A. H. Clifford and G. B. Preston, The Algebraic Theorey of Semigroups, Mathematical Surveys 7, Vols 1 and 2, American Mathematical Society, Providence, R. I. 1967. [2] A.H.Clifford, Semigroups admitting relative inverses, Ann. of Math.,Vol. 42, 1941, pp.1037-1049. [3] J.B., Fountain, Abundant semigroups, Proc. London Math. Soc., Vol. 44(3), 1982, pp.103-129. and K.P.Shum,On [4] X.J.Guo Journal,Vol,2004,No.8,pp.705-717.

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[5] J.M. Howie, Fundamental of Semigroup Theory, Clarendon Press, Oxford,1995. [6] F. Pastijn, A representation of a semigroup by a semigroup of matrices over a group with zero, Semigroup Forum, Vol. 10, 1975, pp.238-249. [7] M. Petrich, The structure of completely regular semigroups, Trans. Amer. Math. Soc., Vol. 189, 1974, pp.211-236. [8] M. Petrich, N.R.Reilly, Completely Regular Semigroups, John Wiley & Sons,1999. [9] X.M.Ren and K.P.Shum,Strucurue of superabundant semigroups,Science in China,2004. [10] X. M. Ren and K.P.Shum,Superabundant semigroups whose set of idempotents forms a subsemigroup,( to appear in Algebra Colloqium).

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