Some Properties of Regular Semigroups. Possess a Medial Idempotent

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1 International Journal of Algebra, Vol. 4, 2010, no. 9, Some Properties of Regular Semigroups Possess a Medial Idempotent S. Hussain 1, T. Anwer 1, H. Chien 2 1 Higher Education Department, Pakistan fbsaso@gmail.com 2 Department of Mathematics, Hangzhou Teacher,s College, Zhejiang China Abstract In this paper we study the properties of all such semigroups particularly all regular semigroups that contains a medial idempotent z that is, an idempotent such that Ē is the subsemigroup of S generated by the idempotents, then ( for all a belongs to Ē) a = aza which is normal. Mathematics Subject Classification: 20M10, 20M99 Keywords: Medial idempotent, Normal medial idempotent 1 Preliminaries Definition 1 By [3] a semigroup is an algebraic structure consisting of a nonempty set S together with an associative binary operation. Definition 2 By [1] an inverse semigroup S is a semigroup in which every element x in S has a unique inverse y in S in the sense that x = xyx and y = yxy. Definition 3 By [4] let S be a regular semigroup Ē be the subsemigroup that is generated by the set E of idempotents of S. An idempotent z belongs to S will be called medial if it is such that ( for all a belong to Ē) a = aza. Example 4 Let B be a band, if z belong to B is a middle unit ( for all a,b in B) azb = ab, then z is medial.

2 434 S. Hussain, T. Anwer, H. Chien Definition 5 By [2] a medial idempotent z will be called normal if the sub-band zēz is commutative. Example 6 If S is a regular semigroup with an identity 1 then S is an inverse semigroup if and only if the medial idempotent is normal. We shall denote by F z and z F the equivalences associated with the translation α z : x zx and β z : x xz. 2 Some Results on Regular Semigroups Possess a Medial Idempotent Theorem 7 If S contains a medial idempotent z, then the subsemigroup Ē is periodic. Proof. For every a belongs to Ē we have a2 = a 2 za 2 = a.aza.a = a.a.a = a 3, and so Ē is periodic. Theorem 8 Let B be a band with right identity z, let B be a band with left identity z, and suppose that B,B contain the common sub-band Ê = zbz = zb z. Let B B Ê, described by (a, x) a x, be a mapping such that for all a,b in B, for all x, y in B (a x) b y = (a x)(b y) =a x(b y) (1) for all a in B, for all x in B (a z) a = a and x (z x) =x (2) Let B z B = {(x, a) in B B ; z x = a z} and on B z B define (x, a)(y, b) = (x (a y), (a y) b). Then B z B is an idempotent generated a regular semigroup containing a medial idempotent. Proof. First we show that B z B is well defined. By (1) and (2) z x(a y) = (z x)(a y) = (a z)(a y) = (a z)a y = a y and, similarly,(a y)b z =(a y)(b z) =(a y)(z y) =a y(z y) =a y This multiplication is associative; which is obvious. In the resulting semigroup B z B, the element (z, z) is idempotent; for, using (2) we have (z, z)(z, z) =(z(z z), (z z)z) =(z, z). Observe that (x, a)(z, z)(x, a) =(x, a) by (2). This shows that B z B is regular and is idempotent generated. Since (x, a) =(x, a)(z, z) (z, z)(x, a); moreover (z, z) is a medial idempotent.

3 Properties of regular semigroups 435 Theorem 9 Let S be a regular semigroup and let z belongs to S be medial idempotent then the following are equivalent. (1) z is normal (2) Green,s relation RL on Ē are given by R = z F and L = F z Proof. (1) implies (2).We have xrxz and xlzx for every x in S. It follows that z F is subset of R and F z is a subset of L. To obtain the reverse inclusions, suppose now that z is normal, and that x, y belongs to Ē are such that xry. Then xzryz and so, since xz, yz are idempotents, we have xz yz = yz and yz xz = xz. Consequently, zxz zyz = zyz, and zyz zxz = zxz. Since zēz is a semilattice, it follows that zxz = zyz. We then have xz = xzxz = xzyz and yz = yzyz = yzxz, that is xzlyz. Thus we see that xzhyz, since xz and yz are idempotents, xz = yz, so that R is subset of z F. Similarly, we have L is subset of F z. (2) implies (1) Suppose now that R = z F and L = F z on Ē. Given that x, y belongs to Ē we observe that xzyyzx. Since m = mzm for every m belongs to Ē we have that zmz belongs to V (m), and hence, zyzxz belongs to V (yzx). Now, since zē, Ēz and zēz are bands. xzy zyzxz xzy = xzy z xzy = xzy, zyzxz xzy zyzxz = zyz xzx z yzy zxz = zyzxz zyzxz = zyzxz, and consequently zyzxz belongs to V (xzy). Since xzy and yzx thus have an inverse in common, it follows that xzyyzx. Since R = z F and L = F z, there exists t belongs to Ē such that zxzy = zt and tz = yzxz. Consequently, zxz zyz = zxzyz = ztz = zyzxz = zyz zxz and so the band zēz is commutative, hence a semilattice. Remark 10 The method described in theorem 8 of constructing all idempotentgenerated regular semigroup that have a medial idempotent can be adopted to yield a construction of all idempotent-generated regular semigroup that contains a normal medial idempotent. To find conditions under with a medial idempotent is normal, we proceed as follows. Using the notation of theorem 8, we see from properties (1) and (2) of that result,and the fact that z is the identity element of Ê, that for all x, y belongs to B we have (z x) (z y) =[(z x) z](z y) =(z x)(z y). Similarly, we have (a x) (b z) for all a, b belongs to B. It follows that if z x = a z, and z y = b z. Then (z x, a z)(z y, b z) =((z x)[(a z) (z y)], [(a z) (z y)](b z)) = ((z x)(z y), (a z)(b z)). Now for all (x, a) belongs to B z B, (z, z)(x, a)(z, z) =(z(z x)(a z), (z x)(a z)z) =(z x, a z). We deduce that the medial idempotent (z, z) is normal if and only if for all x, y belongs to B (z x) (z y) = (z y) (z x) (3) or, equivalently, for all a, b belongs to B (a z) (b z) = (b z) (a z) (4)

4 436 S. Hussain, T. Anwer, H. Chien. Now for every x belongs to S, and for every x belongs to V (x), we have xx zx = x = xzx x. It follows from this that zx z belongs to V (x), and that xzx = xx,and x zx = x. It follows that zsz is an inverse semigroup whose semilattice of idempotent is Ê = zēz. Now we have the following results. Theorem 11 (for all x belongs to S)(for all x belongs to V (x))zx z belongs to V (zxz). Proof. zx z zxz zx z = z x z xzx z = z x xx z = zx z, and, similarly, zxz zx z zxz = zxz. Theorem 12 (for all x belongs to S) z V (x) z =1 Proof. Since zsz is an inverse semigroup, it follows, by Theorem 11 that, for every x belongs to V (x),zx z belongs to V (zxz) zsz = {(zxz) 1 }. In what follows, we shall denote the unique element of z V (x) z by x. Thus, we have x = zx z for every x belongs to V (x). Note that since x belongs to V (x ), we have x = zxz, and (x ) = x =(x ). Now from[see,for example,6] that if e, f are idempotents of S, then the sandwich set S(e, f) is defined by S(e, f) ={k belongs to S, k 2 = k = ke = fk,ekf = ef}. In the case where S is locally inverse ( that is ese is an inverse semigroup for every idempotent e), it is know that sandwich set are singletons. As the following result shows, this happens in particular whenever S contains a normal medial idempotent. Theorem 13 Let S be a regular semigroup that contains a normal medial idempotent then S is locally inverse. Proof. Let z be a normal medial idempotent. Then, if e, f, g are idempotents, with f,g belongs to ese, we have fg = efe ege = eze f eze g eze = e zefez zegez e = e zegez zefez e = gf, thus idempotent in ese commute. Theorem 14 S(e, f) =(fze) 2. Proof. That (fze) 2 is idempotent follows from the fact that, since zēz is a semilattice, fze fze fze fze = f zefz zefz zefz e = f zefz e. It is clear that e is a left identity, and that for e is a left identity for (fze) 2. Finally, we have e (fze) 2 f = efzefzef = efzef zef = efzef = ef. Lemma 15 (for all x, y belongs to S)S(x x, yy ) =(yy x x) 2.

5 Properties of regular semigroups 437 Proof. Since x belongs to zsz, we have zx = x = x z, so yy zx x = yy x x, The property of sandwich sets is that if x, y belongs to S, and x belongs to V (x), y belongs to V (y). Then l belongs to S(x x, yy ) implies y lx belongs to V (xy). Theorem 16 ( for all x, y belongs to S) (xy) = y x xyy x. Proof. By Lemma 15, we have y (yy x x) 2 x belongs to V (xy). Consequently, we have (xy) = zy (yy x x)x z = zy yy x xyy x xx z = y x xyy x. Lemma 17 ( for all x, y belongs to S)(xy) xy = y x xy and xy(xy) = xyy x. Proof. (xy) xy = y x xyy x xy = y zx xyy z x xy = y zx xyy z zx xyy z y = y zx xyy z y = y x xy, and by similar way, xy(xy) xyy x. Now we can obtain a decomposition theorem for a regular semigroup S that contains a normal medial idempotent z. Moreover, if we denote by I the relation on such a semigroup that is given by xiy if and only if x = y if and only if zxz = zyz, then we have the following observation. Theorem 18 If z is a normal medial idempotent then F z z F = z F F z = I. Proof. Suppose that x(f z z F )y. Then there exists m belongs to S, such that zx = zm, and mz = yz. Then zxz = zmz = zyz, and so xiy. Conversely, suppose that xiy. Then x = y, and so x = x = y = y. Now let s = yy x = yx x then sz = yy xz = yzy zxz = yy zyz = yz, zs = zyx x = zyzx zx = zxzx x = zx, and hence x(f z z F )y. Thus F z z F = I and similarly, we can show that z F F z = I. Using the above result, we can prove that S Sz I zs, where Sz I zs = {(xz, za); xia}. References [1] A. H. CLIFFORD AND G. B. PRESTON, The algebraic theory of semigroups,vols. l and ll, Math. surveys of the Amer. Math. Soc. 7, Providence, Rhode Island, [2] H.CHIEN, Construction of a kind of abundant semigroups, Mathematical Communication 11(2006), [3] J. M. HOWIE, An introduction to semigroup theory, Academic Press, New York, 1976.

6 438 S. Hussain, T. Anwer, H. Chien [4] M. L. Than, Regular semigroup with a medial idempotent, Semigroup Forum 36(1987) Received: September, 2009

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