Let I be a semilattice, and Si (i ∈ I) be a family

of disjoint semigroups. Then we prove that the strong semilattice

S = S[I, Si

, φj,i] of semigroups Si with homomorphisms φj,i : Sj →

Si (j ≥ i) is finitely presented if and only if I is finite and each

Si (i ∈ I) is finitely presented. Moreover, for a finite semilattice

I, S has a soluble word problem if and only if each Si (i ∈ I)

has a soluble word problem. Finally, we give an example of nonautomatic semigroup which has a soluble word problem

Ayık, G.

Ayık, H.

Unlu, Y.

Algebra and Discrete Mathematics

Наукова електронна бібліотека періодичних видань НАН України (Vernadsky National Library of Ukraine)

Journal Algebra Discrete Math.Algebra and Discrete Mathematics RESEARCH ARTICLENumber 4. (2005). pp. 28 – 35c© Journal “Algebra and Discrete Mathematics”Presentations and word problem for strongsemilattices of semigroupsGonca Ayık, Hayrullah Ayık and Yusuf U¨nlu¨Communicated by V. I. SushchanskyAbstract. Let I be a semilattice, and Si (i ∈ I) be a familyof disjoint semigroups. Then we prove that the strong semilatticeS = S[I, Si, φj,i] of semigroups Si with homomorphisms φj,i : Sj →Si (j ≥ i) is finitely presented if and only if I is finite and eachSi (i ∈ I) is finitely presented. Moreover, for a finite semilatticeI, S has a soluble word problem if and only if each Si (i ∈ I)has a soluble word problem. Finally, we give an example of non-automatic semigroup which has a soluble word problem.IntroductionFinite presentability of semigroup constructions has been widely studiedin recent years (see, for example, [1, 2, 8, 9, 10]). One of semigroupstructures is a strong semilattice of semigroups which is an importantstructure for completely regular semigroups (see [7]).Let I be a semilattice, and let Si (i ∈ I) be a family of disjointsemigroups. Suppose that, for any two elements i, j ∈ I, with i ≤ j(i ≤ j if and only if ij = i), there is a homomorphism φj,i : Sj −→ Si,and that these homomorphisms satisfy the following conditions:1. for each i ∈ I, φi,i is the identity map 1Si ;2. for all i, j, k ∈ I, such that i ≤ j ≤ k, φk,jφj,i = φk,i.2000 Mathematics Subject Classification: 20M05.Key words and phrases: Semigroup presentations, strong semilattices of semi-groups, word problems..Journal Algebra Discrete Math.G. Ayık, H. Ayık, Y. U¨nlu¨ 29Define a multiplication on S = ∪i∈ISi in terms of the multiplications inthe components Si and the homomorphisms φj,i, by the rule that, foreach x ∈ Si and y ∈ Sj ,xy = (xφi,ij)(yφj,ij).Respect to this multiplication, S is a semigroup, called a strong semi-lattice of semigroups, and denoted by S[I;Si, φj,i]. Finite presentabilityof strong semilattice of semigroups was investigated in [1]. The authorsof [1] first proved that a band of semigroups Si (i ∈ I) is finitely pre-sented if I is finite and each Si is finitely presented (see [1, Corollary4.5]). Moreover, they proved that a band of monoids Si (i ∈ I) is fi-nitely presented if and only if I is finite and each Si is finitely presented(see [1, Corollary 5.6]). Then, from Theorem 1.3 in [11] they concludedthat a strong semilattice of semigroups S[I;Si, φj,i] is finitely presentedif each Si (i ∈ I) is finitely presented and if I is finite. In this paper weconstruct more specific presentations for strong semilattice of semigroupsfrom given presentations for Si (i ∈ I) and a presentation for each Si froma given presentation for S[I;Si, φj,i]. Moreover, if S[I;Si, φj,i] is finitelygenerated then we prove that S[I;Si, φj,i] has a soluble word problem ifand only if each Si has a soluble word problem.1. The presentationsFor a given alphabet A, let A+ be the free semigroup on A (i.e. theset of all non-empty words over A) and let A∗ be the free monoid onA (i.e. A+ together with the empty word, denoted by ε). A semigrouppresentation is a pair 〈A | R〉, with R ⊆ A+ × A+. A semigroup S isdefined by the presentation 〈A | R〉 if S is isomorphic to the semigroupA+/ρ, where ρ is the congruence on A+ generated by R. For any twowords w1, w2 ∈ A+ we write w1 ≡ w2 if they are identical words and writew1 = w2 if w1ρ = w2ρ (i.e. if they represent the same element in S). Itis well known that w1 = w2 if and only if this relation is a consequence ofR, that is there is a finite sequence w1 ≡ η1, η2, ..., ηk ≡ w2 of words fromA+, in which every term ηi (1 < i ≤ k) is obtained from the previousone by applying one relation from R. Let P = 〈A | R〉 be a presentationfor S and let W ⊆ A+. If W contains exactly one element representingeach element of S then W is called a canonical form for S respect to P.The semigroup S is finitely presented if S has a presentation 〈A | R〉 suchthat both A and R are finite. A (finite) monoid presentation is definedsimilarly with respect to the free monoid A∗. Notice that a monoid Mis finitely presented as a monoid if and only if it is finitely presented asJournal Algebra Discrete Math.30 Presentations and word problem for strong ...a semigroup. Furthermore, if 〈A | R〉 is a semigroup presentation for amonoid M then 〈A | R,w = 1〉 is a monoid presentation for M, wherew ∈ A+ is a word representing the identity of M .Now we construct a presentation for S = S[I;Si, φj,i] from givenpresentations 〈Ai | Ri〉 for Si (i ∈ I). Since S is a disjoint union ofsemigroups Si (i ∈ I), it is clear that A = ∪i∈IAi generates S. Forsimplicity of notation we assume that Ai ⊆ Si.Consider the following presentationP = 〈A | R, aiaj = (aiφi,ij)(ajφj,ij) (ai ∈ Ai, aj ∈ Aj)〉where A = ∪i∈IAi and R = ∪i∈IRi.Proposition 1. With above notation if a ∈ Ai and u ∈ A+j with i 6= j,then there exists v ∈ A+ij such that the relation ua = v is a consequenceof the relations aja = (ajφj,ij)(aφi,ij) (a ∈ Ai, aj ∈ Aj).Proof. We use induction on the length of u. If | u |= 1 then take v ≡(uφj,ij)(aφi,ij). Next we suppose that u ≡ a1 · · · an (a1, . . . , an ∈ Aj)with n ≥ 2 and that the result is true for any word of length n − 1.By the inductive hypothesis, there exists bv1 with b ∈ Aij and v1 ∈ A∗ijsuch that the relation (a2 · · · an)a = bv1 is a consequences of the relationsaja = (ajφj,ij)(aφi,ij) (a ∈ Ai, aj ∈ Aj). Since jij = ij in I and φij,ij isthe identity map on Sij , it follows thatua ≡ a1((a2 · · · an)a) = (a1b)v1 = ((a1φj,jij)(bφij,jij)v1 ≡ ((a1φj,ij)b)v1.Since (a1φj,ij) ∈ A+ij , taking v ≡ (a1φj,ij)bv1 completes the proof.Proposition 2. If u ≡ ai1 · · · aim (ai1, . . . , aim ∈ A ) and k = i1 · · · im,then there exists u ∈ A+k such that the relation u = u is a consequence ofthe relations aiaj = (aiφi,ij)(ajφj,ij) (ai ∈ Ai, aj ∈ Aj).Proof. We use induction on m. If m = 1 then we take u ≡ u ≡ ai1 .Now we suppose that m ≥ 2 and that the result is true for m − 1. Letk1 = i1 · · · im−1. By inductive hypothesis there exists u1 ∈ A+k1suchthat the relation u1 ≡ ai1 · · · aim−1 = u1 is a consequence of the relationsaiaj = (aiφi,ij)(ajφj,ij) (ai ∈ Ai, aj ∈ Aj). By the previous propositionthere exists u ∈ A+k1im such that u1aim = u. Since k1im = k, for u ∈ A+kwe have u ≡ (ai1 · · · aim−1)aim = u1aim = u, as required.Theorem 1. Let S[I;Si, φj,i] be a strong semilattice of semigroups and letPi = 〈 Ai | Ri 〉, with Ai ∩Aj = ∅ for i 6= j, be a semigroup presentationfor Si (i ∈ I). If we take A = ∪i∈IAi and R = ∪i∈IRi thenP = 〈A | R, aiaj = (aiφi,ij)(ajφj,ij) (ai ∈ Ai, aj ∈ Aj with i 6= j)〉is a presentation for S[I;Si, φj,i].Journal Algebra Discrete Math.G. Ayık, H. Ayık, Y. U¨nlu¨ 31Proof. It is clear that A generates S[I;Si, φj,i] and that all the relationsin P hold in S[I;Si, φj,i]. Notice that since φi,i is the identity map on Siand i2 = i (i ∈ I), we can consider P as 〈A | R, aiaj = (aiφi,ij)(ajφj,ij)(ai ∈ Ai, aj ∈ Aj)〉 (i.e. without conditions i 6= j).Let u and v be any two words in A+ such that u = v holds inS[I;Si, φj,i], that is they represent the same element of S[I;Si, φj,i]. ByProposition 2, there exist k, l ∈ I, u ∈ A+k and v ∈ A+l such that u = uand v = v are consequences of the relations (aiφi,ij)(ajφj,ij) (ai ∈ Ai,aj ∈ Aj). Since the relation u = v holds in both Sk and Sl, we have k = land u = v as a consequence of Rk. Therefore u = v is a consequence ofrelations in P, as required.Notice that if Wi ⊆ A+i (i ∈ I) is a set of canonical forms for Si(i ∈ I) with respect to 〈 Ai | Ri 〉, then W = ∪i∈IWi is a set of canonicalforms for S = S[I;Si, φj,i] with respect to P.Now we find a presentation for each Si (i ∈ I) from a presentationfor S = S[I;Si, φj,i].Proposition 3. Let A be a generating set for S = S[I;Si, φj,i] and Bj ={a ∈ A | a ∈ Sj} (j ∈ I). Then, for each i ∈ I,Ai =⋃j≥i{aj,i ∈ Si | aj ∈ Bj and ajφj,i = aj,i}generates Si.Proof. Since Ai ⊂ Si, it is enough to show that Si ⊆ 〈Ai〉.For s ∈ Si, there exist ai(1), . . . , ai(m) ∈ A such that s = ai(1) · · · ai(m).By the multiplication defined on S, we must have i(1), . . . , i(m) ≥ i andthere exists at least one k ∈ {1, . . . ,m} such that i(k) = i, and so s =ai(1) · · · ai(m) = (ai(1)φi(1),i) · · · (ai(m)φi(m),i) ≡ ai(1),i · · · ai(m),i ∈ 〈Ai〉, asrequired.Notice that aiφi,i = ai,i ≡ ai. Moreover, for aj , a′j ∈ Bj , aj 6= a′j itis possible to obtain ajφj,i = a′jφj,i. For every aj,i ∈ Ai, we fix aj ∈ Bjsuch that ajφj,i = aj,i and denote this fixed aj by aj,i.Let 〈A | R〉 be a finite presentation for S = S[I;Si, φj,i]. Next weconstruct a presentation for Si (i ∈ I) in terms of Ai. Let Φi be the uniquehomomorphism from(⋃j≥iBj)+to A+i such that, for each a ∈ Bj , j ≥ i,aΦi = aφj,i, and let Wi =(⋃j≥iBj)∗Bi(⋃j≥iBj)∗. Notice that, forw ∈ A+, w represent an element of Si if and only if w ∈Wi.Journal Algebra Discrete Math.32 Presentations and word problem for strong ...Theorem 2. Let Q = 〈A | R〉 be a presentation for S = S[I;Si, φj,i].With above notation,Qi = 〈Ai | {(rΦi, sΦi) | (r, s) ∈ R ∩ (⋃j≥iBj∗×⋃j≥iBj∗)}〉is a presentation for Si (i ∈ I).Proof. By Proposition 3, Ai is a generating set for Si. LetRi = {(rΦi, sΦi) | (r, s) ∈ R ∩ (⋃j≥iBj∗×⋃j≥iBj∗)}.It is clear that all the relation in Ri hold in Si. Foru ≡ aj(1),i · · · aj(m),i and v ≡ aλ(1),i · · · aλ(n),iwhere aj(1),i, . . . , aj(m),i, aλ(1),i, . . . , aλ(n),i ∈ Ai, let the relation u = vholds in Si. To complete the proof we have to show that u = v is aconsequence of Ri.Let u,v ∈ A+ denote the words A+ obtained from u and v by replacingaj,i by aj,i, respectively. It is clear that the relation u = v holds in S,and so there is a finite sequence u ≡ α1, α2, ..., αm ≡ v of words fromA+, in which every term αk+1 (1 ≤ k < m) is obtained from αk byapplying one relation from R. If αk ≡ βkrkγk and αk+1 ≡ βkskγk whereβk, γk ∈ A∗ and (rk, sk) ∈ R (or equivalently (sk, rk) ∈ R) then βk, γk, rkand sk must be in(⋃j≥iBj)∗. Thus αkΦi ≡ (βkΦi)(rkΦi)(γkΦi) andαk+1Φi ≡ (βkΦi)(skΦi)(γkΦi) where βkΦi, γkΦi ∈ A∗i and (rkΦi, skΦi) ∈Ri (or equivalently (skΦi, rkΦi) ∈ Ri), and so we have a finite sequenceu ≡ α1Φi, α2Φi, ..., αmΦi ≡ v of words from A+i . Hence u = v is aconsequence of Ri, as required.With above notation notice that if Wi =(⋃j≥iBj)∗Bi(⋃j≥iBj)∗then, for w ∈ A+, w represent an element of Si if and only if w ∈ Wi.Since there exist finitely many Bi, there exist finitely manyWi so that I isfinite and every Qi (i ∈ I) in Theorem 2 is a finite presentation wheneverQ is a finite presentation. Moreover, the presentation P in Theorem 1 isfinite if I is finite and every Pi is finite. Therefore, we have the followingcorollary.Corollary 1. The strong semilattice S[I;Si, φj,i] of disjoint semigroupsSi (i ∈ I) is finitely presented if and only if I is finite and every Si isfinitely presented.Journal Algebra Discrete Math.G. Ayık, H. Ayık, Y. U¨nlu¨ 332. Word ProblemA finitely generated semigroup S is said to have a soluble word problemwith respect to a finite generating set A if there exists an algorithm which,for any two words u, v ∈ A+, decides whether or not the relation u = vholds in S (in finite steps). It is easy to see that the solubility of the wordproblem does not depend on the choice of the finite generating set for afinitely generated semigroup ([11]).Theorem 3. Let I be a finite semilattice, and let Si (i ∈ I) be a fam-ily of disjoint finitely generated semigroups. The strong semilattice S =S[I;Si, φj,i] has a soluble word problem if and only if, for each i ∈ I, Sihas a soluble word problem.Proof. (⇒) Suppose that S = S[I;Si, φj,i] has a soluble word problem.Let Ai be a finite generating set for Si. We know that A =⋃i∈I Ai isa finite generating set for S. Then S has a soluble word problem withrespect to generating set A. Since, for each i ∈ I, if u, v ∈ A+i , thenu, v ∈ A+ and there exists an algorithm which decides whether or notu = v holds in S and so in Si. Thus Si has a soluble word problem.(⇐) Suppose that, for each i ∈ I, Si has a soluble word problem andBi is a finite generating set for Si. Let Ai =⋃j≥iBjφj,i for each i ∈ I.It is clear that, for each i ∈ I, Ai is a finite generating set for Si, andmoreover, Si has a soluble word problem with respect to the generatingset Ai.Then we show that S = S[I;Si, φj,i] has a soluble word problemwith respect to the generating set A =⋃i∈I Ai. Let u, v ∈ A+ beany two words. Then we have u ≡ ai(1) · · · ai(m) and v ≡ aj(1) · · · aj(n)where ai(k) ∈ Ai(k) and aj(l) ∈ Aj(l) (1 ≤ k ≤ m, 1 ≤ l ≤ n). Takei = i(1) · · · i(n) and j = j(1) · · · j(n). Since u represents an elementof Si and v represents an element of Sj , if i 6= j, then by the mul-tiplication defined on S, u = v does not hold in S. Now supposethat i = j and take u ≡ (ai(1)φi(1),i) · · · (ai(m)φi(m),i) ∈ A+i and v ≡(aj(1)φj(1),j) · · · (aj(m)φj(m),j) ∈ A+i . Since u = u and v = v holds in S,u = v holds in S if and only if u = v holds in Si. Since, for each i ∈ I, Sihas a soluble word problem with respect to Ai, there exists an algorithmwhich decides whether u = v holds in Si in finite steps. Therefore S hasa soluble word problem.Automatic semigroups were first introduced in [5] and they have beenwidely studied for semigroup structures, such as direct product of semi-groups, Rees matrix semigroups, etc. (see [4, 6]). It is shown that ifa semigroup S is automatic then S has a soluble word problem (see [5,Journal Algebra Discrete Math.34 Presentations and word problem for strong ...Corollary 3.7]). In general the converse of Corollary 3.7 in [5] is not true.For this, consider the free group G1 = 〈a, b | 〉 of rank two and the freeproduct G2 = 〈c, d | c2 = 1 d2 = 1〉 of two cyclic groups of order two. Itis a well-known fact that free groups of finite rank, and finite groups hassoluble word problem. In addition, the free product of two groups whichhave soluble word problems has also soluble word problem. Therefore,G1 and G2 have soluble word problems. Moreover the groups G1 and G2are automatic (see [3]).Let φ1,2 : G1 → G2 be the homomorphism defined by aφ1,2 = c andbφ1,2 = d, φ1,1 be the identity map of G1, φ2,2 be the identity map ofG2. Then consider the strong semilattice G = S[I;Gi, φj,i] of the groupsG1 and G2 where I = {1, 2} is the semilattice with the multiplicationi ·j = max{i, j}. It follows from Theorem 3 that the strong semilattice Gof the groups G1 and G2 has a soluble word problem however it is shownthat G is not automatic (see [6]).References[1] I.M. Araujo, M.J.J. Branco, V.H. Fernandes, G.M.S. Gomes, N. Ruskuc, OnGenerators and Relations of Unions of Semigroups, Semigroup Forum, 63, 2001,pp. 49-62.[2] H. Ayık, N. Ruskuc, Generators and Relations of Rees Matrix Semigroups, Proc.Edinburgh Math. Soc., 42, 1999, pp. 482-495.[3] G. Baumslag, S.M. Gersten, M. Shapiro, H. Short, Automatic Groups and Amal-gams, J. Pure Appl. Algebra 76, 1991, pp. 229-316.[4] C.M. Campbell, E.F. Robertson, N. Ruskuc, R.M. Thomas, Direct Products ofAutomatic Semigroups, J. Austral. Math. Soc., 69, 2000, pp. 19-24.[5] C.M. Campbell, E.F. Robertson, N. Ruskuc, R.M. Thomas, Automatic Semi-groups, Theoret. Comput. Sci., 250, 2001, pp. 365-391.[6] C.M. Campbell, E.F. Robertson, N. Ruskuc, R.M. Thomas, Automatic Com-pletely Simple Semigroups, Acta Math. Hungar., 95, 2002, pp. 201-215.[7] J.M. Howie, Fundamentals of Semigroup Theory, Clarendon Press, 1995.[8] J.M. Howie, N. Ruskuc, Constructions and Presentations for Monoids, Comm.Algebra, 22, 1994, pp. 6209-6224.[9] T.G. Lavers, Presentations of General Products of Monoids, J. Algebra, 204,1998, pp. 733-741.[10] E.F. Robertson, N. Ruskuc, J. Wiegold, Generators and Relations of Direct Prod-uct of Semigroups, Trans. Amer. Math. Soc., 350, 1998, pp. 2665–2685.[11] N. Ruskuc, On Large Subsemigroups and Finiteness Conditions of Semigroups,Proc. London Math. Soc., 76, 1998, pp. 383–405.Journal Algebra Discrete Math.G. Ayık, H. Ayık, Y. U¨nlu¨ 35Contact informationG. Ayık C¸ukurova University, Department of Math-ematics, 01330-Adana, TurkeyE-Mail: agonca@cu.edu.trH. Ayık C¸ukurova University, Department of Math-ematics, 01330-Adana, TurkeyE-Mail: hayik@cu.edu.trY. U¨nlu¨ C¸ukurova University, Department of Math-ematics, 01330-Adana, TurkeyE-Mail: yusuf@cu.edu.trReceived by the editors: 12.09.2005and final form in 15.12.2005.

Presentations and word problem for strong semilattices of semigroups

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Presentations and word problem for strong semilattices of semigroups

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