In his paper, Hall T. R, proved some local properties for semigroups.In this paper we provide alternative proofs for some local properties of semigroups studied by Hall.In addition we introduce some new local properties of semigroups and generalize some of the Hall's results. Aslo we give some examples to illustrate the results

Edmond PISHA

Osman HYSA

Epoka University

“1st International Symposium on Computing in Informatics and Mathematics (ISCIM 2011)”in Collabaration between EPOKA University and “Aleksandër Moisiu” University of Durrëson June  2-4 2011, Tirana-Durres, ALBANIA.473Some Properties of SemigroupsEdmond PISHA1), Osman HYSA2)1)Department of Mathematics, Faculty of Natural Sciences,Universityof Tirana, Tirana,Albania.Email: pishamondi@yahoo.com – Phone: +3556925324762)Department of Mathematics, “Aleksandër Moisiu” University, Durrës.Email: o.hysa@yahoo.com – Phone: +355692040155ABSTRACTIn his paper, Hall T. R, proved some local propertiesfor semigroups.In this paper weprovide alternative proofs for some local properties of semigroups studied by Hall.In addition we introducesome new local properties of semigroups and generalizesome of the Hall’s results. Aslo we give some examples to illustrate the results.Keywords: Local subsemigroup, regular elements, idempotent elements,weaklycommutative semigroup, Green’s relations, Archimedian semigroup, localproperties.INTRODUCTIONDefinition 1.1 [4] A Local subsemigroup of a semigroup S will be called everysubsemigroup of form eSe, where e is an idempotent element of S.Nambooripad in Decalb’s Conferenca 1979 raised the problem: If S is a regularsemigroup such that for every idempotent e S, idempotents of eSe form aband, then idempotents of eS (or Se) form a band too.The answer of this question was given in affirmative form Hall [3]. Beforecontinuing with presenting the results of Hall we will explain some concepts.We assume that the reader is familiar with Green’s relations L, R, H, D in theform they are defined by Clifford A. H. and Preston G. B. [1].Definition 1.2. [1] By Reg(S) we denote the set of all regular elements of semigroupS.Definition 1.3. By V(x) we denote the set of all inverse elements of x in S.Definition 1.4. The (p, q, r)-regular element of semigroup S we call an element a ofS such that aapSaqSar (p, q, r are non negative integers).“1st International Symposium on Computing in Informatics and Mathematics (ISCIM 2011)”in Collabaration between EPOKA University and “Aleksandër Moisiu” University of Durrëson June  2-4 2011, Tirana-Durres, ALBANIA.474Definition 1.5. By (p, q, r) – Reg(S) we denote the set of all (p, q, r)-regularelements of semigroup S.Definition 1.6. The element a of semigroup S will be called right (left) divisiblefrom element b, if and only if a = xb, xS (a = by, yS). The element a will becalled divisible from element b if it is left and right divisible from b.Definition 1.7. A semigroup S is called right archimedian if for two elements a, bSthere is a power of a such that is divided by b and there is a power of b that isdivided by a.Definition 1.8. A semigroup S is called weakly commutative if for a, b  S hold(ab)k = xa = by for any x, y  S and k positive integer number.Definition 1.9. A semigroup S is called bi-simple if it contains only one D-class.For other concept we use here the reader may refer the book of Clifford and Preston.We will use these results:Proposition 1.1. (Lemma 2.14 [1]) Every element e of a semigroup S is right neutralelement for Le, left neutral element for Re and neutral element for He.Proposition 1.2. (Theorem 2.17 [1]) If a and b are elements of semigroup S thenabRaLb RbLa has idempotent element. In this case aHb = Hab = HaHb = Hab =RaLb.Proposition 1.3. (Lemma 2.13 [1]) If a is a regular element of semigroup S, then aS1= aS and S1a = Sa.Proposition 1.4. (Theorem 2.16 [1]) If a, b and ab are elements of a H-class H ofsemigroup S, then H is subgroup of S. In particular, every H-class that contain anidempotent element is a subgroup of S.Finally, we note that if xV(x), then xLxx and xR xx.Now we give some alternative proofs of T. E. Hall results.Theorem 1.5. (i) If Reg(eSe) is a subsemigroup generated by idempotents,then Ref(eS) is also a subsemigroup generated by idempotents.“1st International Symposium on Computing in Informatics and Mathematics (ISCIM 2011)”in Collabaration between EPOKA University and “Aleksandër Moisiu” University of Durrëson June  2-4 2011, Tirana-Durres, ALBANIA.475(ii) If Reg(eSe) is subsemigroup in which H is congruence, thenReg(eS) is subsemigroup in which H is congruence.(iii) If Reg(eSe) is union of groups (non necessarily a subsemigroup),then and Reg(eS) is union of groups.(iv) If eSe has at most one idempotent for every L-class, then eS has atmost one idempotent for everyL-class.Proof.(i) If Reg(eSe) is subsemigroup, then, as it is showed by Hall [3], Reg(eS) issubsemigroup also. Now, it is clear Reg(eS)  Reg(eSe)E(eS). Since g(eSe) isgenerated by idempotents, then Reg(eS) is generated by idempotents also.(ii) Since Reg(eSe) is subsemigroup, then, as is showed by Hall [3], Reg(eS) issubsemigroup also. Let us prove that H is congruence in  Reg(eS). Let p and qbe elements from Reg(eS) such that p H q. Let us show that c, d Reg(eS),pcHqc and dpH dq. From pHq  pLq  pRq. Since L is right congruencepcLqc, so peLqe. It is easy to show that from pR q we have pe R p R qR qe(since it is easy to show pe R p and q R qe ). Now it is clear that pe R qe,which combined with pe L qe implies pe H qe. From conditions we have pce Hqce (for ce Reg(eSe)). Thus pce L qce, pce R qce. So pce R pc, qce R qctherefore pc R qc.From pc L qc and pc R qc we have pc H qc.Similarly p H q imply p L q and p R q and since Ris left congruencewe willhave pd R dq. Let’s show now pd L dq. We have Reg(eSe) = Reg(eS)eso p L qimply Reg(eS)p = Reg(eS)qand then Reg(eSe)pe = Reg(eSe)qe. NowReg(eSe)depe = Reg(eSe)deqe thus Reg(eS)dpe = Reg(eS)dqe; So Reg(eS)dp =Reg(eS)dq (becauseq Reg(eS)p). Finally,dp L dq in Reg(eS).From dp L dq and dp R dq results dp H dq. Finally we have shown that H is acongruence in Reg(eS).(iii) Let us prove that Reg(eS) is union of groups. Since union of subgroups ofReg(eS) is subset of Reg(eS), then it is sufficient to show the inclusion ofReg(eS) in a union of subgroups of Reg(eS).For a Reg(eS) we prove that Ha = RaaLaa has an idempotent. From ae Reg(eSe) and conditions we have Hae = RaaeLaae has an idempotent. FromProposition 1.2, aa2ae RaaeLaae. Now we have aa2a R aa2ae R aae Raa. From aa2a L aae and aa2a L aa. So aa2a RaaLaa. From“1st International Symposium on Computing in Informatics and Mathematics (ISCIM 2011)”in Collabaration between EPOKA University and “Aleksandër Moisiu” University of Durrëson June  2-4 2011, Tirana-Durres, ALBANIA.476Proposition 1.2 Ha = RaaLaa has an idempotent and from Proposition 1.4 wehave that Ha is subgruop; so we showed that a is an element of the subgroupHa.(iv) Let f, g  E(eS) such that f L g. We must show f = g. From f L git follwosthat fe L ge. Since fe, ge  E(eSe) we have that fe = ge. Now we have f = f f =fef = gef = gf. From f L g and Proposition 1.1 we get gf = g and consequently f= g.Now Theorem 1 (iv) of [3] states that if Reg(eSe) is semigroup, then Reg(eS)is subsemigroup or equivalently: If <E(eSe)> is regular subsemigroup, then and<E(eS)> is a regular semigroup.In an attempt to generalize this result concerning to (p, q, r)-regularity weintroduce this problem:Problem: If (p, q, r) – Reg(eSe) is subsemigroup, then is (p, q, r) – Reg(eS)subsemigroup?Answer to this question is partially positive.We distinct these cases concerning to triples (p, q, r):First case: (1) (p, q, r) = (0, 0, 0).Second case: p  1; q, r  0(2) (1, 0, 0) (3) (1, 0, 1) (4) (1, 1, 0)(5) (1, 0, r), r > 1 (6) (1, 1, 1) (7) (1, 1, r) r > 1(8) (1, q, 0), q > 1 (9) (1, q, 1), q > 1 (10) (1, q, r), q, r > 1(11) (p, 0, 0), p > 1 (12) (p, 0, 1), p > 1 (13) (p, 0, r), p, r > 1(14) (p, 1, 0), p > 1 (15) (p, 1, r), p, r > 1 (16) (p, q, 0), p, q > 1(17) (p, 1, 1), p > 1 (12) (p, q, 1), p, q > 1 (19) (p, q, r), p, q, r >1Third case: p = 0; q, r  0(20) (0, 0, 1) (21) (0, 0, r), r > 1 (22) (0, 1, 0)(23) (0, q, 0), q > 1 (24) (0, 1, 1) (25) (0, 1, r), r > 1(26) (0, q, 1), q > 1 (27) (0, q, r), q, r > 1.“1st International Symposium on Computing in Informatics and Mathematics (ISCIM 2011)”in Collabaration between EPOKA University and “Aleksandër Moisiu” University of Durrëson June  2-4 2011, Tirana-Durres, ALBANIA.477Theorem 1.6. If (0, 0, 0) – Reg(eSe) is subsemigroup, then so is and (0, 0, 0)– Reg(eS).Proof.Let supposse a, b (0, 0, 0) – Reg(eS), then a = x1x2 for x1, x2eS and b= y1y2 for y1, y2  eS. Now ab = (x1x2) (y1y2) = (ab)(x1x2) (ab)(y1y2) (ab)(ab)(eS) (ab)(eS) (ab).So ab  (0, 0, 0) – Reg(eS).Thus for the first case the response of the question is positive.The response for second case will give thisTheorem 1.7. If (p, q, r) – Reg(eSe) is subsemigrup, then so is and (p, q, r) –Reg(eS) for p  1 and q, r  0.Proof. Let a, b (p, q, r) – Reg(eS). If p  1, then b  b(eS) and b = bec foranyc  S. We will have ae, be  (p, q, r) – Reg(eSe). So abe = (ae)(be)  (p, q,r)– Reg(eSe).Thus abe  Te, where T = (ab)p (eS) (ab)q (eS) (ab)r and therefore ab = abec  Tec T.Also the response of the question is positive for the second case.For the third case we will give an example, which shows that the answer isnegative for this case.Example 1.8. Let F be a free monoid over the alphabet {0, 1 and  the emptyword. For x, y  Fwill have xy  u 010 vy = u 1v for u, v  F and xy, if andonly if there exists a string x = x1, x2, ..., xn = y (n 1),where xixi+1 orxi+1xi fori = 1, 2, ..., n – 1 (n 2). It is clear  is congruence in F.Let us denote by T = F/ the factor monoid with unity  = {}. For X, YT wehave XY=  X =  = Y. Let us supose that B, CT, where 1 B and 0 C. Itis evidente that B = CBCand B2 = XB2Y  X =  = Y.Let us denote by E the zero right semigroup which contain two elements ,and consider the semigroup ExT (where the operation is coordinatewise). ExTsatisfy the condition:“1st International Symposium on Computing in Informatics and Mathematics (ISCIM 2011)”in Collabaration between EPOKA University and “Aleksandër Moisiu” University of Durrëson June  2-4 2011, Tirana-Durres, ALBANIA.478(u, X)(u, Y) = (, )  (u, Y) = (, ).Let us denote by S = (EXT) \ {(, )} and e = (, ), b = (, B), c = (, C). Wethen havee2= e, eb = b and ec = c. It is evident that the semigroup S is generated by e, b,c and eS = S, eSe = Se. Now, for the element x  Se we have x = exe  SexSeand so(0, 1, 0) – Reg(eSe) = Se is a semigroup of S. It is clear that b = cbc and so b(0, 1, 0) – Reg(eS). Let supposse that b2 (0, 1, 0) – Reg(eS). Then b2  Sb2Sand b2 = xb2y for any x, y  S. Thus we have (, B2) = (u, X) (, B2) (v, Y),where x = (u, X) and y = (v, Y). So v =  and B2 = XB2Y henceY =  and y = (,) S.Consequently b2Sb2S and b2 (0, 1, 0) – Reg(eS). As a result (0, 1, 0) –Reg(eS) is not a subsemigroup of S.The following theorems give some other local properties of semigroups.Theorem 1.8.(i) If eSe is a bisimple subsemigroup, then eS is also a bisimplesubsemigroup.(ii) If Reg(eSe) is a zero right semigroup, then Reg(eS) is also a zero rightsemigroup.Proof. (i) Let Da and Db be two D-classes of eS, where a, b  eS. We mustprove Da = Db.Let p  Da, then p D a or p L◦Ra and p L z, z R a for any zeS. We will havepe L ze and ze R ae (since z R ze and a R ae); finally we have pe D ae. Fromthe assumption we have that Dae = Dbe, so pe D be which shows that pe L k andk R be for any k  eSe. It is clear that be R b. From pe L k and k R b it followsthat pe D b. Further, from pe R p we have that pe D p and then p D pe D b, p Db, bDb. So, we have showed that Da  Db. Similarly we prove Db  Da andthen Da = Db.(ii) Since Reg(eSe) is subsemigroup, then, as it is shown by Hall [3] Theorem1-(iv), Reg(eS) is a subsemigroup. If p, q Reg(eS), then we havepq = pqqq = pqeqq = qeqq (from the assumption) = qqq = q.Thus, pq = q for all p, q Reg(eS), which means that Reg(eS) is zero rightsemigroup.“1st International Symposium on Computing in Informatics and Mathematics (ISCIM 2011)”in Collabaration between EPOKA University and “Aleksandër Moisiu” University of Durrëson June  2-4 2011, Tirana-Durres, ALBANIA.479Theorem 1.9. In weakly commutative semigroup S, if eSe is a right (left)archimedian subsemigroup, then eS is aright (left) archimediansubsemigroup.Proof. Let eSe be a right archimedian subsemigroup of semigroup S and let usshow that eS is also a right archimedian subsemigroup of S. If a, bS, then ae,be  eSe. Since eSe is right archimedian, then (ae)n = x(be) and (be)n = y(ae)for any n, m positive integers andx, y  eSe. Since an+1 = ana = (ae)aa (as aeS) = x(be)a = xba, so an+1 = xba.Thus, a(n+1)p = (xba)p = u(xb) for any u  eSe and p – positive integer.Similarly, bm+1 = bmb = (be)mb = (yae)b = yab and b(m+1)q = (yab)q = v(ya) forany veSe and q – positive integer.Thus we have proved that eSis right archimedian. In a similar fashion we provethe case for left archimedian semigroups.REFERENCES[1] Clifford AH and Preston GB, (19??) The algebraic theory of semigroups,Math. Surveys, No. 7, Amer. Math. Soc, Providence, R1 Vol. 1.[2] Hall TE(1973), On regular semigroups, J. Algebra, 24, 1 – 24.[3] Hall TE (1982) Some properties of local subsemigroups inherited by largersubsemigroups, Semigroup Forum, Vol. 25, 35 – 49.[4] Mc Alister DB, (1981) Regular Rees matrix semigroups and regular Dubreil –Jacotin semigroups, J. Austral. Math. Soc. (Series A), 31, 325 – 336.

Some Properties of Semigroups

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Some Properties of Semigroups

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