# Abstract. semigroups which are the disjoint union of

Abstract. We prove that in the ideal case, up to isomorphism, there are only one type ofsemigroups which are the union of two copies of the free monogenic semigroup. Similarly, thereare only five types of semigroups which are the union of three copies of the free monogenicsemigroup. And there are only two types of semigroups which are the union of two copies of thefree semigroup in two generators. We provide finite presentations for each of these semigroups. Introduction and Preliminaries There are several well-known examples of structural theorems for semigroups, which involvedecomposing a semigroup into a disjoint union of subsemigroups.

For example, up to isomorphism,the Rees Theorem states that every completely simple semigroup is a Rees matrix semigroup overa group G, and is thus a disjoint union of copies of G, see 8, Theorem 3.3.1; every Cliffordsemigroup is a strong semilattice of groups and as such it is a disjoint union of its maximalsubgroups, see 8, Theorem 4.

2.1; every commutative semigroup is a semilattice of archimedeansemigroups, see 7, Theorem 2.2. If S is a semigroup which can be decomposed into a disjoint union of subsemigroups, thenit is natural to ask how the properties of the subsemigroups influence S. For example, if thesubsemigroups are finitely generated, then so is S. There are several further examples in theliterature where such questions are addressed: Arau ?jo et al. 4 consider the finite presentability ofsemigroups which are the disjoint union of finitely presented subsemigroups; Golubov 5 showedthat a semigroup which is the disjoint union of residually finite subsemigroups is residually finite;in 3 the authors proved that every semigroup which is a disjoint union of finitely many copies of Nis finitely presented, and such a semigroup has linear growth which implies that the correspondingsemigroup algebra is a PI algebra, see 1, Theorem 2.

10 and Corollary 2.11; further references are6, 10; in 2 we completely classify those semigroups which are the disjoint union of two or threecopies of the free monogenic semigroup in general. In this paper we classify those semigroups but in a special case when it has an ideal, which isone of the copies or is a disjoint union of two copies. And we classify also the semigroup whichis a disjoint union of two copies of the free semigroup in two generators in which one of the twocopies is an ideal. The main theorems of this paper are the following. Theorem 0.1.

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Let S be a semigroup. Let ?a? and ?b? be two copies of the free monogenic semi-group. Then S is a disjoint union of ?a? and ?b?, where ?a? is an ideal in S, if and only if S isisomorphic to the semigroup defined by the following presentation: ? c,d ? cd=dc=ck ? for some k?1 . Theorem 0.2.

Let S be a semigroup. Let ?a?, ?b? and ?c? be three copies of the free monogenicsemigroup. Then S is a disjoint union of ?a?, ?b? and ?c?, where the subsemigroup ?a? or ?a? ? ?b?is an ideal in S, if and only if S is isomorphic to the semigroup defined by one of the followingpresentations: (i) ?d,f,g?df=di,fd=di,dg=dj,gd=dj,fg=dk,gf=dk ?wherei+j=k+2andi,j,k?N; Date: 01.11.2017.1991 Mathematics Subject Classification.

20M05.Key words and phrases. Free Semigroup, Ideal, Union, finitely presented . 1 2 UNIONS OF SEMIGROUPS (ii)?d,f,g?df=di,fd=di,dg=dj,gd=dj,fg=fk,gf=fk ?wherei+j+k?ik=2andi,j,k ? N; (iii) ? d,f,g ? df =di,fd=di,dg=di,gd=di,fg=g2,gf =f2 ? where i?N;(iv) ? d,f,g ? df =di,fd=di,dg=g2,gd=d2,fg=gi,gf =di ? where i?N;(v) ? d,f,g ? df =di,fd=di,dg=g2,gd=d2,fg=gi,gf =gi ? where i?N; Theorem 0.

3. Let S be an Equitable semigroup. Let ?a1,a2? and ?b1,b2? be two copies of thefree semigroup in two generators. Then S is a disjoint union of ?a1,a2? and ?b1,b2?, where thesubsemigroup ?b1,b2? is an ideal in S, if and only if S is isomorphic to the semigroup defined byone of the following presentations: (i)  ? c1, c2, d1, d2 ? c1d1 = d21, d1c1 = d21, c2d1 = d2d1, d1c2 = d1d2, c1d2 = d1d2, d2c1 = d2d1,c2d2 = d2, d2c2 = d2 ?; (ii)  ? c1,c2,d1,d2 ? c1d1 = d21, d1c1 = d21, c2d1 = d21, d1c2 = d21, c1d2 = d1d2, d2c1 = d2d1,c2d2 = d1d2, d2c2 = d2d1?. We prove Theorems 0.1 and 0.2 in Section 1, and Theorem 0.

3 in Section 2 respectively. Let A be a set, and let S be any semigroup. Then we denote by A+ the free semigroup on A,which consists of the non-empty words over A. Any mapping ? ? A ? S can be extended in aunique way to a homomorphism ? ? A+ ? S, and A+ is determined up to isomorphism by theseproperties. If A is a generating set for S, then the identity mapping on A induces an epimorphism? ? A+ ? S. The kernel ker(?) is a congruence on S; if R ? A+ × A+ generates this congruencewe say that ?A ? R? is a presentation for S. We say that S satisfies a relation (u, v) ? A+ × A+ if?(u)=?(v); we write u=v in this case. Suppose we are given a set R?A+ ×A+ and two wordsu,v?A+.

Wewriteu?vifuandvareequalaselementsofA+. Wesaythattherelationu=vis a consequence of R if there exist words u ? w1,w2,…,wk?1,wk ? v (k ? 1) such that for eachi=1,…,k?1wecanwritewi ??iui?i andwi+1 ??ivi?i where(ui,vi)?Ror(vi,ui)?R.

Wesay that ?A ? R? is a presentation for S if and only if S satisfies all relations from R, and everyrelation that S satisfies is a consequence of R: see 9, Proposition 1.4.2. If A and R are finite,then S is finitely presented. Let ? be a congruence on a semigroup S, and let ? ? S ? T be a homomorphism such that? ? ker ?. Then there is a unique homomorphism ? ? S/? ? T defined by s/? ? ?(s) and such thatim ? = im ?; 8, Theorem 1.5.3.

Proposition 0.4 (9). Let S be the semigroup defined by the presentation ?A ? R?. If T is anysemigroup satisfying the relations R, then T is a homomorphic image of S. Let ?a? be the free monogenic semigroup.

Then any two non-empty subsemigroups S and T of?a? have non-empty intersection, since ai ? S and aj ? T implies aij ? S ? T . Let S be a semigroupwhich is the disjoint union of m ? N copies of the free monogenic semigroup, and let a1 , . . . , am ? Sbe the generators of these copies.

Suppose that S is also the disjoint union of n ? N copies of thefree monogenic semigroup. Then there exist b1 , . . . , bn ? S such that ?b1 ?, . . .

, ?bn ? are free, disjoint,and S = ?a1? ? ? ? ?am? = ?b1? ? ? ? ?bn?. If n > m, say, then there exist i,j such that bi,bj ? ?ak? for some k. But then ?bi? ? ?bj? =/ ?,a contradiction. Hence a semigroup cannot be the disjoint union of m and n copies of the freemonogenic semigroup when n =/ m. Lemma 0.5. 2, Lemma 1.

4 Let S and T be semigroups which are the disjoint union of m ? Ncopies of the free monogenic semigroup, and let A = {a1,…,am} and B = {b1,…,bm} be thegenerators of these copies in S and T , respectively.

Then every homomorphism ? ? T ? S suchthat ?(ai) = bi for all i is an isomorphism. Proof. Since ? is surjective, it follows that the function f ? {1,..

.,m} ? {1,…,m} defined by?(ai) ? ?bf(i)? is a bijection. Suppose that there exist x,y ? S such that ?(x) = ?(y).

Then there exist a,b ? A such thatx = ai and y = bj for some i,j ? N. It follows that ?(a)i = ?(b)j, which implies that ?(a),?(b) ? ?c?for some c?B. Hence a=b, since f is a bijection, and so x=y.

UNIONS OF SEMIGROUPS 3 Since the free monogenic semigroup is anti-isomorphic to itself, it follows that a semigroup Sis the disjoint union of m copies of the free monogenic semigroups if and only if any semigroupanti-isomorphic to S has this property. Lemma0.6. LetAbeaset,let?beacongruenceonA+,let??A?T?{0}beanymappingwhereT is a free semigroup in two generators, and let ? ? A+ ? T ? {0} be the unique homomorphismextending ?. If ? ? ker? and a,b ? A such that ?(a) =? 0 and ?(b) =? 0, then ?a/????b/?? is infinitesubsemigroups of A+/?.

Proof. Since ? ? ker?, it follows that ? ? A+/? ? T ? {0} defined by ?(w/?) = ?(w) is a homo-morphism. Homomorphisms map elements of finite order to elements of finite order, and since?(a) =? 0 and ?(b) =? 0 do not have finite order, a/? ? b/? must have infinite order in A+/?. 1.

Two and three copies of the free monogenic semigroupIn this section we prove Theorems 0.1 and 0.2. Proof of Theorem 0.1. (?) We proved in 2, Theorem1.1 that the semigroup S? which definedby the presentation ?c,d?cd=dc=ck ? forsomek?1,is a disjoint union of two copies of the free monogenic semigroup ?c? and ?d?.

It is clear from the presentation that ?c? ? S? ? ?c? and S? ? ?c? ? ?c?, which implies that ?c? is an ideal in S?.(?) Let S be a semigroup which is the disjoint union of the free semigroups ?a? and ?b?. Clearly one of the following must hold: (a) ab,ba??a?,(b) ab,ba??b?, (c) ab??a? and ba??b?,(d) ab??b? and ba??a?. In case (b), S is isomorphic to a semigroup satisfying (a) and in case (c) and (d) neither ?a? nor?b? is an ideal which is a contradiction. Hence we may assume without loss of generality that just(a) holds. Case (a) There exist m,n ? N such that ab = am and ba = an. Henceam+1 =ama=(ab)a=a(ba)=aan =an+1 and so m = n. So, in this case, S is a homomorphic image of the semigroup T defined by thepresentation ? a, b ? ab = ba = am ?.

It follows from Lemma 0.5 that S is isomorphic to T . Proof of Theorem 0.2.

(?) We proved that the semigroup defined by one of the presentations in2, Theorem1.2 is a disjoint union of three copies of the free monogenic semigroup.Case (i): ?d? is an ideal in S? which implies that ?a? is an ideal in S.

Case (ii):?d? is an ideal in S? which implies that ?a? is an ideal in S. Case (iii): ?d? is an ideal in S? which implies that ?a? is an ideal in S.Case (iv): ?d? ? ?g? is an ideal in S? which implies that ?a? ? ?c? is an ideal in S.

Case (v): ?d? ? ?g? is an ideal in S? which implies that ?a? ? ?c? is an ideal in S.Cases (vi,vii,viii,ix)): There is no ideal which is a disjoint union of copies of the free monogenicsemigroup in these cases.Thus, we just have 5 types of semigroups satisfy the ideal condition. (?) As we have seen in 2, Theorem 1.2 that the semigroup which is a disjoint union ofthree copies of the free monogenic semigroup is isomorphic to a semigroup defined by one of thepresentations in that theorem . Therefore, the semigroup which is a disjoint union of three copies 4 UNIONS OF SEMIGROUPS of the free monogenic semigroup which contains an ideal ?a? or ?a? ? ?b? must be defined by oneof the presentations (i,ii,iii,iv,v) in 2, Theorem 1.

2. 2. Two copies of the free semigroup in two generatorsDefinition 2.1. An Equitable semigroup defines by a presentation of the following form ?a,b,c,d ? ac=wac, ca=wca, bc=wbc, cb=wcb, ad=wad, da=wda,bd=wbd, db=wdb ? where wxy ? {a,b,c,d}+, ?wxy? = 2. Classifying all possible Equitable semigroups which are disjoint unions of two copies of the freesemigroup in two generators is still quite complicated.

In this paper, our aim is to classify Equitable semigroups in the case that one of the two copiesis an ideal in S. Lemma 2.2. Let S be an Equitable semigroup which is the disjoint union of the free semigroup?a,b?, ?c,d? where ?c,d? is an ideal in S. Then one of the following must hold: (i) ac=c2 and then ca=c2, bc=cb=c2, ad=cd, da=dc, bd=cd,db=dc.(ii) ac=dc and then ca=cd, bc=dc, cb=cd, ad=da=bd=db=d2. Proof.

As ?c,d? is an ideal then ac ? {c2,dc,d2,cd}. But ac =? d2 and ac =? cd because if ac = d2 orac = cd then c(ac) =? (ca)c in the both cases. Thus ac ? {c2, dc}. Now if ac = c2 then by the associativity on c(ac) = c3 and (ca)c = wcac which implies thatwca = ca = c2. And c(ad) = cwa,d, (ca)d = c2d which implies wad = ad = cd. Therefore da = dc bythe associativity on (dad). Also, d(bc) = dwbc and (db)c = wdbc, which means that the word wbcends with the letter c and the word wdb starts with the letter d. So there are two possibilities,wbc =c2 or dc and wdb =db=d2 or dc and then if wbc =c2 that gives us wdb =dc by associativity,and if wbc = dc that implies wdb = d2 and after that we can continue to get cb and bd.

So whenwbc =c2 thismeanscb=c2 andbd=cdandwhenwbc =dcthatmeanscb=cdandbd=d2.Therefore, if ac = c2 then we have two types of semigroups: (1) The semigroup with relations ac=ca=bc=cb=c2,ad=bd=cd,da=db=dc;(2) The semigroup with relations ac=ca=c2,ad=cb=cd,da=bc=dc,db=bd=d2. The second value is when ac = dc and then by the associativity on (cac) we get ca = cd. Thenc(ad) = cwa,d and (ca)d = cd2 which implies that wa,d = ad = d2 and similarly da = d2 by theassociativity on (dad). Also by the associativity on (cbc) we have two possibilities bc ? {c2,dc}and cb ? {c2, cd} and we get that: (i) cb=bc=c2 and,(ii) cb=cd, bc=dc. In case (i) we have db = dc, bd = cd by the associativity on (dbc) and (cbd) respectively.In case (ii) we have bd = d2, db = d2 by the associativity on (cbd) and (dbc) respectively. Therefore there are two types of semigroups.

(1′) The semigroup with the relations: ac=bc=dc,ca=cb=cd,ad=da=bd=db=d2; (2′) The semigroup with the relations: ac = db = dc,ca = bd = cd,ad = da = d2,bc = cb = c2; Notice that the semigroup in (1) is isomorphic to the semigroup in (1′) by just replacing c by dand d by c in (1), and similarly the semigroup in (2) is isomorphic to the semigroup in (2′) by, asabove, replacing c by d and d by c in (2). As a result, up to isomorphism, we just have two typesof semigroups with relations (1) and (2). Proof of Theorem 0.3.

(?) To prove the converse implication, it suffices to show that thesemigroups mentioned in Theorem 0.3 are disjoint unions of two copies of the free semigroup intwo generators. Let S be the semigroup defined by the presentation?c1,c2,d1,d2?c1d1 =d21,d1c1 =d21,c2d1 =d2d1,d1c2 =d1d2,c1d2 =d1d2,d2c1 =d2d1,c2d2 =d2,d2c2 =d2?. UNIONS OF SEMIGROUPS 5 It is clear that every word from S is a product of letters in {c1,c2} or in {d1,d2}, and so S =?c1, c2? ? ?d1, d2?. Since there is no relation in the presentation that can be applied to a wordfrom{c1,c2}+,itfollowsthat?c1,c2???d1,d2?=?and?c1,c2?isinfinite. Weshowthat?d1,d2?isinfinite using Lemma 0.

6. Let ? be the congruence on {c1,c2,d1,d2}+ generated by the relations c1d1 = d21, d1c1 = d21, c2d1 = d2d1, d1c2 = d1d2, c1d2 = d1d2, d2c1 = d2d1, c2d2 = d2, d2c2 = d2, let??{c1,c2,d1,d2}?T bedefinedby?(c1)=f,?(c2)=g,?(d1)=f,?(d2)=g,whereT isafreesemigroup in two generators f and g and let ? ? {c1, c2, d1, d2}+ ? T be the unique homomorphismextending ?. Then and, similarly, ?(c1d1) = ?(c1)?(d1) = ff = f2 = ?(d21) ?(d1c1) = ?(d1)?(c1) = ff = f2 = ?(d21);?(c2d1) = ?(c2)?(d1) = gf = ?(d2d1);?(d1c2) = ?(d1)?(c2) = fg = ?(d1d2);?(c1d2) = ?(c1)?(d2) = fg = ?(d1d2);?(d2c1) = ?(d2)?(c1) = gf = ?(d2d1);?(c2d2) = ?(c2)?(d2) = gg = g2 = ?(d2);?(d2c2) = ?(d2)?(c2) = gg = g2 = ?(d2). Hence ? ? ker ? and so ?d1, d2? is infinite in S, by Lemma 0.6.Let G be the semigroup defined by the presentation ? c1,c2,d1,d2 ? c1d1 = d21, d1c1 = d21, c2d1 = d21, d1c2 = d21, c1d2 = d1d2, d2c1 = d2d1,c2d2 = d1d2, d2c2 = d2d1?. Then as above G = ?c1, c2? ? ?d1, d2?.

It is clear that every word from S is a product of letters in{c1, c2} or in {d1, d2}, and so S = ?c1, c2???d1, d2?. Since there is no relation in the presentation thatcan be applied to a word from {c1, c2}+, it follows that ?c1, c2???d1, d2? = ?. The proof that ?c1, c2?and ?d1, d2? are infinite follows using a similar argument as above but where ? ? {c1, c2, d1, d2} ? Tis defined by ?(c1) = ?(c2) = ?(d1) = f and ?(d2) = g, where T is a free semigroup in twogenerators f and g. Let ? ? {c1, c2, d1, d2}+ ? T be the unique homomorphism extending ?. Then and, similarly, ?(c1d1) = ?(c1)?(d1) = ff = f2 = ?(d21) ?(d1c1) = ?(d1)?(c1) = ff = f2 = ?(d21);?(c2d1) = ?(c2)?(d1) = ff = f2 = ?(d21);?(d1c2) = ?(d1)?(c2) = ff = f2 = ?(d21);?(c1d2) = ?(c1)?(d2) = fg = ?(d1d2);?(d2c1) = ?(d2)?(c1) = gf = ?(d2d1);?(c2d2) = ?(c2)?(d2) = fg = ?(d1d2);?(d2c2) = ?(d2)?(c2) = gf = ?(d2d1).

Hence ? ? ker ? and so ?d1, d2? is infinite in S, by Lemma 0.6.(?) Let S? be a semigroup defined by one of the following presentations: (a)  ? c1, c2, d1, d2 ? c1d1 = d21, d1c1 = d21, c2d1 = d2d1, d1c2 = d1d2, c1d2 = d1d2, d2c1 = d2d1,c2d2 = d2, d2c2 = d2 ?; (b)  ? c1,c2,d1,d2 ? c1d1 = d21, d1c1 = d21, c2d1 = d21, d1c2 = d21, c1d2 = d1d2, d2c1 = d2d1,c2d2 = d1d2, d2c2 = d2d1?. Let S be a semigroup which is the disjoint union of the free semigroups ?a, b? and ?c, d?.

Thus oneof the following must hold by Lemma 2.2: (i) ac=ca=bc=cb=c2,ad=bd=cd,da=db=dc;(ii) ac=ca=c2,ad=cb=cd,da=bc=dc,db=bd=d2. 6 UNIONS OF SEMIGROUPS Clearly S with the relations in case (i) and case (ii) satisfies the relations in the presentation(a) and (b) respectively and then S is a homomorphic image of S? by Proposition 0.4. If it is aproper homomorphic image then there is without loss of generality u1 and u2 ? ?c1, c2? or u andv in ?c1, c2? and ?d1, d2? respectively such that u1 = u2 or u = v which contradicts with the factthat there is no element in the free semigroup of rank two is of finite order or contradicts with?c1,c2???d1,d2?=?.

ThusS? ?S.

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