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On Subsemigroups of Finitely Presented Semigroups
JOURNAL OF ALGEBRA ARTICLE NO.
180, 1]21 Ž1996.
0049
On Subsemigroups of Finitely Presented Semigroups C. M. Campbell, E. F. Robertson, and N. Ruskuc ˇ Mathematical Institute, Uni¨ ersity of St Andrews, St Andrews KY16 9SS, Scotland
and R. M. Thomas* Department of Mathematics and Computer Science, Uni¨ ersity of Leicester, Leicester LE1 7RH, England Communicated by Peter M. Neumann Received June 8, 1994
1. INTRODUCTION The purpose of this paper is to investigate properties of subsemigroups of finitely presented semigroups, particularly with respect to the property of being finitely presented. More precisely, we will consider various instances of the following: Main Problem. Let S be a finitely presented semigroup. Ži. Is every subsemigroup Žideal, one-sided ideal. of S finitely generated? Žii. Is every subsemigroup Žideal, one-sided ideal. of S which is finitely generated as a semigroup finitely presented? Žiii. Is every subsemigroup Žideal, one-sided ideal. of finite index in S finitely presented? ŽFor the definition of index see below.. For example, if S is finite all these questions have an affirmative answer. Less trivially, if S is a commutative semigroup then Žii. has an affirmative answer; see w12x. Also, in w2x it has been proved that a two-sided ideal of finite index in a finitely presented semigroup is always finitely presented. * The authors acknowledge the support from the Edinburgh Mathematical Society Centenary Fund and the European Community Grant ERBCHRXCT930418, which helped to make possible the visit of the fourth author to the University of St Andrews while this work was being undertaken. 1 0021-8693r96 $18.00 Copyright Q 1996 by Academic Press, Inc. All rights of reproduction in any form reserved.
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TABLE I Subsemigroups
Ideals
One-sided ideals
Ži. f.g.
No ŽExample 5.3.
No ŽExample 5.3.
No ŽExample 5.3.
Žii. f.g.« f.p.
No ŽExample 4.5.
No Žfuture paper.
No Žfuture paper.
Žiii. f.i.« f.p.
Not known ŽOpen Problem 5.2.
Yes Žw2, Corollary 4.2x.
Yes ŽTheorem 5.1.
Note. f.g. denotes finitely generated, f.p. denotes finitely presented, and f.i. denotes finite index.
Group versions of questions Ži., Žii., and Žiii. have been investigated extensively. The main tool for these investigations is the Reidemeister] Schreier rewriting process, which gives a presentation for a subgroup of a group defined by a presentation; see w8x. Probably the best known application of this rewriting process is the Reidemeister]Schreier theorem which states that a subgroup of finite index in a finitely presented group is itself finitely presented; see w6x and w8x. A general theory of Reidemeister]Schreier-type rewriting for semigroups has been developed in w2x. This theory was then applied to two-sided ideals of finite index, as well as 0-minimal ideals; see also w1x. Here we continue this investigation and prove that a one-sided ideal of finite index in a finitely presented semigroup is itself finitely presented; see Theorems 5.1 and 5.6 and Corollary 5.7. Also, in a future paper we give an example of an ideal of a finitely presented semigroup, which is finitely generated as a semigroup but is not finitely presented. The present status of the Main Problem for general finitely presented semigroups is given in Table I. Another important application of the Reidemeister]Schreier rewriting for groups is the Nielsen]Schreier theorem, which states that every subgroup of a free group is itself free. A straightforward generalisation of this result does not hold: subsemigroups and one-sided ideals may or may not be free, while proper two-sided ideals are never free; see Propositions 3.1 and 3.3 and Example 3.2. Nevertheless, one can ask whether subsemigroups and ideals of a free semigroup have some other, weaker properties. For example, Spehner w14x has shown that every finitely generated subsemigroup admits a finite Malcev presentation. Here we consider free semigroups in the context of the Main Problem. We prove, among other things, that, in a free semigroup, two-sided and one-sided ideals which are finitely generated as semigroups, as well as subsemigroups of finite index,
SUBSEMIGROUPS OF F.P. SEMIGROUPS
3
TABLE II Subsemigroups
Ideals
One-sided ideals
free
No ŽWell known.
No ŽProposition 3.1.
No ŽProposition 3.3.
Ži. f.g.
No ŽExample 5.3.
No ŽExample 5.3.
No ŽExample 5.3.
Žii. f.g.« f.p.
No ŽExample 4.5.
Yes ŽCorollary 3.6.
Yes ŽTheorem 4.3.
Žiii. f.i.« f.p.
Yes ŽTheorem 3.8.
Yes ŽTable 1.
Yes ŽTable 1.
Note. The notation is the same as that used in Table I.
are finitely presented; see Corollary 3.6 and Theorems 3.8 and 4.3. The information obtained about substructures of free semigroups is summarised in Table II. It is also worth mentioning that Margolis and Meakin w9x have obtained results of Nielsen]Schreier type for free inverse semigroups; see also w10x. In this paper we shall use the standard semigroup theory notation and definitions as presented in w3x. For a semigroup T of S we define the Rees equi¨ alence determined by T to be r T s T = T j Ž s, s .: s g S4 . If T is a left Žright. ideal then r T is a left Žright. congruence, and if T is a two-sided ideal then r T is a congruence; the corresponding factor semigroup is denoted by SrT. We define the index of T in S to be < S y T < q 1, which is the number of equivalence classes of r T . Since we frequently consider ideals as subsemigroups, there is a certain ambiguity in the use of terms such as ‘‘generated by’’ and ‘‘finitely generated.’’ We adopt the convention that ‘‘generated’’ will always mean ‘‘generated as a subsemigroup,’’ except when stated otherwise. Thus, for example, the principal right ideal aS 1 is not necessarily finitely generated although it is finitely generated as a right ideal. For an alphabet A, the free semigroup and the free monoid on A are denoted by Aq and A* respectively. The empty word is denoted by e , and the length of a word w g A* is denoted by < w <. A semigroup presentation is an ordered pair ² A N R :, where R : Aq= Aq. A semigroup S is said to be defined by the presentation ² A N R : if S ( Aqrh , where h is the smallest congruence on Aq containing R. Usually we identify a word from Aq with the element of S it represents. However, to avoid confusion, we write w 1 ' w 2 if the words w 1 and w 2 are identical, and w 1 s w 2 if they represent the same element of S. For a subset T of S we use the notation LŽ A, T . for the set of all words from Aq which represent
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elements of T. For a systematic introduction to presentations of semigroups we refer the reader to w5x.
2. PRELIMINARY RESULT In this section, we prove the following preliminary result which we will need later. This is probably well known, but, for completeness, we include a proof here. THEOREM 2.1. If S is a semigroup and I is a finitely presented ideal of S, and if the Rees quotient SrI is finitely presented, then S is finitely presented. Proof. Suppose that I and SrI are defined by presentations ² X N R : and ² Y N G : respectively. Suppose for the moment that no generator from Y represents the zero of the semigroup SrI. In this case S is obviously generated by the set X j Y. Each word w g Yq which represents the zero of SrI represents an element of I in S, and so there exists a word a w g Xq such that w s a w in S. Also, for each x g X and each y g Y the words xy and yx represent elements from I in S; let b x y , b y x be words from Xq such that xy s b x y and yx s b y x hold in S. Let us fix a word z g Yq representing the zero of SrI. We define the following new sets of relations: G 1 s Ž u s ¨ . g G : u / z in SrI 4 , G 2 s Ž u s ¨ . g G : u s z in SrI 4 , G 2 s u s a u , ¨ s a¨ : Ž u s ¨ . g G 2 4 j z s a z 4 , G 3 s xy s b x y , yx s b y x : x g X , y g Y 4 , and let T denote the semigroup defined by the presentation ² X j Y N R j G1 j G 2 j G 3:.
Ž 1.
We claim that S ( T. It is obvious that S satisfies all the relations from R j G 1 j G 2 j G 3 , and hence there is a natural epimorphism f : T ª S. Assume now that two words w 1 , w 2 g Ž X j Y .q represent the same element of S. If both w 1 and w 2 contain a letter from X then we can use relations from G 3 to find words w 1 , w 2 g Xq such that w 1 s w 1 and w 2 s w 2 in T. Since the relation w 1 s w 2 holds in I, it can be deduced from R, thus giving w 1 s w 2 in T. If w 1 g Yq and w 1 represents an element of I in S then w 1 s z holds in SrI, and so z can be obtained from w 1 by applying relations from G. If no relations from G 2 are needed
SUBSEMIGROUPS OF F.P. SEMIGROUPS
5
in this deduction then w 1 s z s a z holds in T. Otherwise, instead of the first application of a relation from G 2 we can use the corresponding relation from G 2 and thus obtain a word w 1 containing a letter from X such that w 1 s w 1 holds in T. Applying a similar argument to w 2 , if necessary, we reduce this case to the case where both w 1 and w 2 contain a letter from X. Finally, if w 1 does not represent an element of I in S then w 1 s w 2 can be deduced by just using relations from G 1 , and again w 1 s w 2 holds in T. Therefore, f is an isomorphism and T ( S as required. Let us now suppose that Y contains generators representing the zero in SrI, and let Y0 be the set of all non-zero generators from Y. If the zero of SrI is a product of two non-zero elements from SrI then Y0 generates SrI and the argument above can be repeated. Otherwise a similar argument would show that ² X j Y0 N R j G 1 j G 3 :
Ž 2.
is a presentation for S. Finally, note that if both presentations ² X N R : and ² Y N G : are finite so are the presentations Ž1. and Ž2..
3. FREE SEMIGROUPS As we indicated in the introduction, a straightforward generalisation of the Nielsen]Schreier theorem for groups does not hold for semigroups. It is well known that subsemigroups of a free semigroup are not necessarily free; see w7x. We now show that the situation for ideals is even worse in that they are almost never free. PROPOSITION 3.1. Let F s Aq be a free semigroup and let R / F be a proper right ideal. If R is finitely generated then it is not free. Proof. Since F / R there exists a g A such that a f R. Suppose that a i f R for all i G 1. Let w be an element of R of minimal length. Then wa i g R, i G 1, since R is a right ideal, but wa i is not a product of two elements of R. Therefore, each generating set of R contains all the words wa i, i G 1, and R is not finitely generated, a contradiction. Thus R contains some power of a. Let i be the minimal such power; obviously i ) 1. The word a iq1 belongs to R since R is a right ideal, but a iq1 is not a product of two elements of R since i ) 1; hence each generating set for R contains both a i and a iq1. Since a i and a iq1 satisfy the non-trivial relation a ia iq1 s a iq1 a i, R cannot be free.
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EXAMPLE 3.2. Let F be the free semigroup on two generators a, b4 , and let R s aF 1 be the principal right ideal generated Žas a right ideal. by a. The set ab i : i G 04 is a unique minimal generating set for R, and it is easy to see that R is free on that generating set. PROPOSITION 3.3. Let F s Aq be a free semigroup and let I / F be a proper two-sided ideal of F. Then I is not free. Proof. The proposition is an immediate corollary of w5, Proposition 2.2x. For the sake of completeness, we give an alternative direct proof. Let a g A y I, and let w be an element of I of minimal length. Both words aw and wa belong to I since I is a two-sided ideal, and neither of them is a product of two elements of I; hence each generating set for I contains both these words. But then w Ž aw . s Ž wa. w is a non-trivial relation holding in I, and I is not free. Now we shall consider free semigroups in the context of our Main Problem. The following example gives a negative answer to Ži.. EXAMPLE 3.4. Let F be the free semigroup on two generators a, b4 , and let I be the set of all words containing both a and b. I is obviously a two-sided ideal Žand hence a right ideal, a left ideal, and a subsemigroup.. However, the words ab i, i G 1, are not products of words from I, and I is not finitely generated. On the other hand, as we mentioned in the Introduction, question Žiii. always has a positive answer for ideals of finite index. In particular, an ideal of finite index of a finitely generated free semigroup is finitely presented. We shall now use this result to answer question Žii. for two-sided ideals and question Žiii. for subsemigroups of finite index of a free semigroup. THEOREM 3.5. If T is a finitely generated ideal in a free semigroup S, then T has finite index in S. Proof. We show that if T has infinite index in S, then T is not finitely generated. Suppose that T has infinite index in S, and let w 1 , w 2 , . . . be distinct elements of S y T. Let x be an element of T with x of minimal length among all elements of T, so that xw1 , xw 2 , . . . are elements of T. If xwi s u¨ with u, ¨ g T, then x is a prefix of u Žby the minimality of < x <., so that ¨ is a suffix of wi ; since ¨ g T, we have that wi g T, a contradiction. So xwi cannot be expressed as a non-trivial product of elements of T, and hence any generating set for T must contain xw1 , xw 2 , . . . . So T is not finitely generated.
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COROLLARY 3.6. If T is a finitely generated ideal in a finitely generated free semigroup S, then T is finitely presented. Proof. The corollary is an immediate consequence of w2, Corollary 4.2x and Theorem 3.5. Now we turn our attention to the subsemigroups of finite index in a free semigroup. PROPOSITION 3.7. If S is a finitely generated free semigroup and T is a subsemigroup of S of finite index, then there is an ideal I of finite index in S with I : T. Proof. Let S y T s a 1 , a 2 , . . . , a k 4 , and let p s max< a i <: 1 F i F k 4 . Let I be the ideal of all words in S of length at least p q 1. We see that I : T and that I has finite index in S as required. Given this, we can give an affirmative answer to question Žiii. for subsemigroups of finite index in a free semigroup. THEOREM 3.8. If S is a finitely generated free semigroup and T is a subsemigroup of S of finite index, then T is finitely presented. Proof. By Proposition 3.7, there is an ideal I of finite index in S with I : T. By w2x, I is finitely presented. The Rees quotient TrI is a finite semigroup, and hence is also finitely presented; the result follows from Theorem 2.1.
4. IF FINITELY GENERATED, THEN FINITELY PRESENTED? In this section, we continue our investigation of free semigroups in the context of the Main Problem. We prove that finitely generated one-sided ideals of a free semigroup are finitely presented, although they are not necessarily of finite index. We also give an example of a finitely generated subsemigroup of a free semigroup which is not finitely presented. First, however, we give a necessary and sufficient condition for an ideal to be finitely generated. PROPOSITION 4.1. Let S be a free semigroup, and let R be a right ideal of S. Then R is finitely generated if and only if there exist words a 1 , a 2 , . . . , a m of S and a constant N such that R s a 1 S 1 j a 2 S 1 j ??? j a m S 1 and such that e¨ ery word in S of length greater then N contains at least one of the a i as a substring.
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Proof. Suppose R is generated by a 1 , a 2 , . . . , a m 4 . Then, since a i g R for all i and R is a right ideal, we have that
a 1 S 1 j a 2 S 1 j ??? j a m S 1 : R. On the other hand, any word in R is a product of the a i , and hence lies in a k S 1 for some a k ; so R : a 1 S 1 j a 2 S 1 j ??? j a m S 1 , and hence R s a 1 S 1 j a 2 S 1 j ??? j a m S 1. Now suppose that R is Ža right ideal. of the form a 1 S 1 j a 2 S 1 j ??? j a m S 1 for some a i . ŽNote that we are not assuming now that R is generated by the a i as a semigroup.. We want to show that R is finitely generated if and only if there is a constant N such that every word in S of length greater than N contains at least one of the a i as a substring. We may assume, without loss of generality, that no a i is a prefix of any other a j , since, if a i is a prefix of a j , then a j S 1 : a i S 1, and a j S 1 can be omitted. For any word h in S 1 and any i, a ih lies in R, and any word in R is of this form. If R is finitely generated, then there is an upper bound on the length of a generator, and so R is generated by the set S s a ih : 1 F i F m, h g S 1 ,
Ž 3.
for some N. Let z be any element of S, and consider the element a 1 z of R. Since S is a generating set for R, we may write
a 1 z s a i1h1 a i 2h 2 ??? a i khk for some i1 , i 2 , . . . , i k and h1 , h 2 , . . . , hk with
z s h1 a i 2h 2 . . . a i khk . If < z < ) N, then, since
SUBSEMIGROUPS OF F.P. SEMIGROUPS
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i 2 , and so we have b s a i1h1 a i2 j for some h1 and j with
b s a i1h1 a i 2h 2 ??? a i khk , a product of elements in S, as required. Proposition 4.1 has an obvious dual for left ideals. Therefore we have the following corollary, which shows a surprising connection between finitely generated right ideals and finitely generated left ideals. COROLLARY 4.2. If S is a free semigroup and a 1 , a 2 , . . . , a m g S, then a 1 S 1 j a 2 S 1 j ??? j a m S 1 is a finitely generated right ideal of S if and only if S 1a 1 j S 1a 2 j ??? j S 1a m is a finitely generated left ideal of S. We can now prove the main result of this section. THEOREM 4.3. If R is a finitely generated right ideal of a free semigroup S, then R is finitely presented. Proof. Let S s Aq be the free semigroup on A. Since R is finitely generated and each generator of R only involves finitely many elements from A, and since Ra : R for each a g A, we see that A is finite. Let A s a1 , a2 , . . . , a n4 . Let R be generated by a 1 , a 2 , . . . , a m 4 , where a i g Aq for each i, and let B be the alphabet b1 , b 2 , . . . , bm 4 , where bi represents the generator a i of R. Let M s max< a i <: 1 F i F m4 . We have a relation
g Ž b1 , b 2 , . . . , bm . s d Ž b1 , b 2 , . . . , bm . in R if and only if g Ž a 1 , a 2 , . . . , a m . ' d Ž a 1 , a 2 , . . . , a m . in Aq. We define the weight of the relation g Ž b1 , b 2 , . . . , bm . s d Ž b1 , b 2 , . . . , bm . to be the length of the word g Ž a 1 , a 2 , . . . , a m . in Aq. Let R denote the set of all relations g Ž b1 , b 2 , . . . , bm . s d Ž b1 , b 2 , . . . , bm . of weight at most 3 M that hold in R. We claim that ² B N R : is a presentation for R Žand the proof will then be completed, since R is finite.. We need to show that any relation g Ž b1 , b 2 , . . . , bm . s d Ž b1 , b 2 , . . . , bm . holding in R is a consequence of R. So suppose we have a relation bi1 bi 2 bi 3 ??? bi r s bj1 bj 2 bj3 ??? bj s
Ž 4.
holding in R, so that
a i1 a i 2 a i 3 ??? a i r ' a j1 a j 2 a j3 ??? a j s in Aq. We argue by induction on the weight of the relation Ž4., the case of weight 1 being a relation in R. If a i1 ' a j1, we have bi 2 bi 3 ??? bi r s
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bj 2 bj3 ??? bj s, and the result follows by the inductive hypothesis. So suppose that < a i1 < - < a j1 <, so that a i1 is a proper prefix of a j1, say a j1 ' a i1 z . We now have
a i1 a i 2 a i 3 ??? a i r ' a i1 za j 2 a j3 ??? a j s , so that a i 2 a i 3 ??? a i r ' za j 2 a j3 ??? a j s. If a i 2 is a prefix of z then z g R, and so z ' a k 1 a k 2 ??? a k t. Both relations bi 2 bi 3 ??? bi r s bk 1 ??? bk t bj 2 ??? bj s and bj1 s bi1 bk 1 ??? bk t are of lower weight than Ž4., and hence are consequences of R by the inductive hypothesis. Now we have bi1 bi 2 ??? bi r s bi1 bk 2 ??? bk t bj 2 ??? bj s s bj1 bj 2 ??? bj s , which shows that Ž4. is a consequence of R as well. So let us assume now that z is a proper prefix of a i 2 . Then we have a i 2 ' za j 2 a j3 ??? a j qy 1 j for some q and j , where j is a prefix of a j q. Since za j 2 a j3 ??? a j qy 1 j represents an element of R and za j 2 a j3 ??? a j qy 1 j is a prefix of za j 2 a j3 ??? a j q, we see, since R is a right ideal, that za j 2 a j3 ??? a j q represents an element of R. So
za j 2 a j3 ??? a j q ' a k 1 a k 2 ??? a k t for some a k 1, a k 2 , . . . , a k t. Now < za j 2 a j3 ??? a j qy 1 j < s < a i 2 < F M, and so < a j a j a j ??? a j a j < s < a i za j a j ??? a j a j < 1 2 3 qy 1 q 1 2 3 qy 1 q F < a i1 < q < za j 2 a j3 ??? a j qy 1 j < q < a j q < F 3 M. Now
a j1 a j 2 a j3 ??? a j qy 1 a j q ' a i1 za j 2 a j3 ??? a j qy 1 a j q ' a i1 a k 1 a k 2 ??? a k t , and the corresponding relation bj1 bj 2 ??? bj qy 1 bj q s bi1 bk 1 bk 2 ??? bk t is a relation of weight at most 3 M, and so is in R by definition. So we may use this relation from R to transform the relation bi1 bi 2 bi 3 ??? bi r s bj1 bj 2 bj3 ??? bj s to bi1 bi 2 bi 3 ??? bi r s bi1 bk 1 bk 2 ??? bk t bj qq 1 bj qq 2 ??? bj s. So we only need to deduce the relation bi 2 bi 3 ??? bi r s bk 1 bk 2 ??? bk t bj qq 1 bj qq 2 ??? bj s , which is of lower weight than our original relation, and the result follows by induction.
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Theorem 4.3 is a generalisation of Corollary 3.6. The following example shows that a generalisation of Theorem 3.5 does not hold for right ideals. EXAMPLE 4.4. Let F be the free semigroup on a and b, and let R s a2 S 1 j abS 1 j b 2 S 1 . Since any word of length at least 3 must contain a2 , ab, or b 2 as a subword, R is a finitely generated right ideal of F by Proposition 4.1. However, F y R s a, b4 j baS 1 is not finite; in other words, R does not have finite index in F. The hypothesis in Theorem 4.3 that R is a right ideal cannot be omitted, in that a finitely generated subsemigroup of a finitely generated free semigroup need not be finitely presented, as our next example shows. EXAMPLE 4.5. Let F be the free semigroup on three generators a, b, and c, and let I be the subsemigroup generated by ¨ s ba, w s ba2 , x s a3, y s ac, and z s a2 c. I is clearly finitely generated, and we claim that I is not finitely presented. First note that, if a semigroup is finitely presented with respect to one generating set, then it is finitely presented with respect to any finite generating set, since we can pass from a presentation on the first set to a presentation on the second by means of Tietze transformations. ŽFor the semigroup version of Tietze transformations see w13x or w11x.. So it is sufficient to show that I is not finitely presented with respect to the particular generating set ¨ , w, x, y, z 4 . Decomposing the word a s ba3Ž nq1. c in two different ways, we see that the relations ¨ x n z s wx n y Ž n G 0. hold in I, and so any set of defining relations must imply these. Since any proper subword of a representing an element of I can be expressed as an element of ¨ , w, x, y, z 4 q in only one way, there is no non-trivial relation holding in I which we can apply to a proper subword of ¨ x n z or wx n y. So any set of defining relations for I must include all the relations ¨ x n z s wx n y, and so I is not finitely presented.
5. RIGHT IDEALS OF FINITE INDEX Now we turn our attention to a general finitely presented semigroup, and we give an affirmative answer to question Žiii. of the Main Problem in the case of one-sided ideals. In other words, we prove THEOREM 5.1. If S is a finitely presented semigroup and R is a right Ž or left . ideal of finite index in S, then R is finitely presented. The case of subsemigroups still remains open:
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Open Problem 5.2. If T is a subsemigroup of finite index in a finitely presented semigroup S, is T necessarily finitely presented? As mentioned in the Introduction, Theorem 5.1 is a generalisation of w2, Corollary 4.2x. It also strengthens the analogy between the Reidemeister] Schreier type theorems for semigroups and groups: both subgroups and one-sided ideals give rise to one-sided congruences Ži.e., the group and the semigroup respectively act on the corresponding factor sets by pre- or post-multiplication.. However, as the following example shows, the finite index condition in Theorem 5.1 cannot be replaced by the weaker condition that R is an equivalence class of a congruence with finitely many equivalence classes. EXAMPLE 5.3. Let F be the free semigroup on two generators a, b4 , and let S be the semigroup defined by the presentation ² A, B: A2 s A, B 2 s B, AB s BA: . We have a natural homomorphism u from F to S defined by au s A and bu s B. The kernel h of this homomorphism has three equivalence classes. Let I be the pre-image of AB in F. We see that I is the ideal of F consisting of all words involving both a and b. In Example 3.4 we have shown that I is not even finitely generated, let alone finitely presented. Our strategy in the proof of Theorem 5.1 is to use w2, Theorem 2.1x which gives a presentation ² B N I : for R. Although B is finite when R is of finite index, the set I is always infinite, and we aim to find a finite set of relations equivalent to I. First, however, we shall recall the necessary notation from w2x. Let S be defined by a Žfinite. presentation ² A N R :. Every element of S is represented by a word in Aq, and, as before, we let LŽ A, R . denote the set of all words in Aq representing elements of R. A set of representati¨ es of S y R is a set V of words in A* such that: Ž1. V contains the empty word e ; Ž2. each non-empty word in V represents an element of S y R; Ž3. each element of S y R is represented by one, and only one, word in V. It is worth noting that the empty word e is included in V to simplify notation in what follows; e does not represent an element of S. We choose V here in such a way that, for each non-empty word v in V, v is a word of minimal length representing the corresponding element of S y R.
SUBSEMIGROUPS OF F.P. SEMIGROUPS
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It was shown in w2, Theorem 3.1x that R is generated by the set X s r a s : r , s g V , a g A, r a g L Ž A, R . 4 . Note that, if we have r a g LŽ A, R ., then we also have r a s g LŽ A, R ., as R is a right ideal. We define a new alphabet B s br , a, s : r , s g V , a g A, r a g L Ž A, R . 4 in one]one correspondence with X. This correspondence can be extended to an epimorphism c : Bqª R in a natural way. The representati¨ e function w ¬ w associates to every element w g S y R its representative w g V Žand associates the empty word to the empty word.. We note the following immediate consequence of this definition: LEMMA 5.4. The representati¨ e function has the following properties. Ži. x ' x for each x f R; Žii. if y s z g S y R, then y ' z. For a word w g LŽ A, R . let w9a, where w9 g A* and a g A, be the shortest initial segment of w in LŽ A, R ., and let w0 be the rest of w. Define
f : L Ž A, R . ª Bq by
f Ž w. s
½
bw 9, a, w 0
if w0 f L Ž A, R .
bw 9, a, e f Ž w0 .
if w0 g L Ž A, R . .
Ž 5.
It is proved in w2, Lemma 3.4x that f is a rewriting mapping, in the sense that
c Ž f Ž w. . s w
in S.
Ž 6.
ŽIntuitively, f ‘‘rewrites’’ any word from Aq which represents an element of R into a product of generators from X.. Now we can easily deduce the following presentation for R from w2, Theorem 2.1x. PROPOSITION 5.5.
R is defined by the presentation
² B N f Ž w 1w 2 . s f Ž w 1 . f Ž w 2 . : w 1 , w 2 g L Ž A, R . 4 j f Ž w 3 uw4 . s f Ž w 3¨ w4 . : w 3 uw4 g L Ž A, R . , Ž u s ¨ . g R 4 : .
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CAMPBELL ET. AL.
Proof. By w2, Theorem 2.1x we have the following presentation for R: ² B N br , a, s s f Ž r a s . : br , a, s g B 4 j f Ž w 1w 2 . s f Ž w 1 . f Ž w 2 . : w 1 , w 2 g L Ž A, R . 4 j f Ž w 3 uw4 . s f Ž w 3¨ w4 . : w 3 uw4 g L Ž A, R . , Ž u s ¨ . g R 4 : . Note that the relations br , a, s s f Ž r a s . are identically true for our particular choice of the rewriting mapping, and the proposition follows. We want to prove that R is finitely presented; in fact, we will give an explicit finite set of defining relations for R with respect to the generating set B, namely the union of the following six Žfinite. sets: J1 s f Ž r 1 a1 r 2 . f Ž r 3 a2 r4 . s f Ž r 1 a1 . f Ž r 2 r 3 a2 r4 . : a i g A, r i g V ,
r 1 a1 , r 2 r 3 a2 , r 3 a2 g L Ž A, R . 4 ; J2 s f Ž r 1 a1 r 2 . f Ž r 3 a2 r4 . s f Ž r 1 a1 r 2 r 3 a2 r4 . : a i g A, r i g V ,
r 1 a1 , r 3 a2 g L Ž A, R . , r 2 r 3 a2 r4 f L Ž A, R . 4 ; J3 s f Ž r 1 r 2 . f Ž r 3 a r4 r 5 . s f Ž r 1 r 2 r 3 a r4 r 5 . : a g A, r i g V ,
r 1 r 2 , r 3 a g L Ž A, R . , r 2 r 3 a r4 f L Ž A, R . 4 ; J4 s f Ž r 1 r 2 . f Ž r 3 r4 . s f Ž r 1 r 2 r 3 r4 . : r i g V ,
r 1 r 2 , r 3 r4 g L Ž A, R . , r 2 r 3 f L Ž A, R . 4 ; J5 s f Ž r 1 r 2 a r 3 r4 . s f Ž r 1 r 2 a r 3 r4 . : a g A, r i g V ,
r 1 r 2 a g L Ž A, R . , r 2 a r 3 f L Ž A, R . 4 ; J6 s f Ž r 1 u r 2 . s f Ž r 1¨ r 2 . : Ž u s ¨ . g R , r i g V , r 1 u r 2 g L Ž A, R . 4 . Now we have THEOREM 5.6. If S is a semigroup with finite presentation ² A N R :, R is a right ideal of finite index in S, V is a set of representati¨ es of S y R, and f is the corresponding rewriting mapping, and if J1 , . . . , J6 are defined as abo¨ e, then R has presentation ² B N J1 j ??? j J6 :. Theorem 5.1 is clearly an immediate consequence of Theorem 5.6, since the sets J1 , . . . , J6 are all finite. It has to be admitted that the nature of our relations is a little complex. A more natural finite Žthough larger. set of relations is given in the following
SUBSEMIGROUPS OF F.P. SEMIGROUPS
15
COROLLARY 5.7. Let S be a semigroup with finite presentation ² A N R :, let R be a right ideal of finite index, let V be a set of representati¨ es of S y R, and let f be the corresponding rewriting mapping. If m s max < r < , < u < , < ¨ < : r g V , Ž u s ¨ . g R 4 then R is defined by the presentation ² B N f Ž w 1w 2 . s f Ž w 1 . f Ž w 2 . : < w 1w 2 < F 5m q 2 4 j f Ž w 3 . s f Ž w4 . : w 3 s w4 in S, < w 3 < F 5m q 2, < w4 < F 5m q 2 4 : . Proof. All the relations from the given presentation hold in R, since f is a rewriting mapping, and they include all the relations from the presentation given in Theorem 5.6. We now embark on the proof of Theorem 5.6. We need to show that all the relations in J1 j ??? j J6 hold in R and that all the relations in the presentation from Proposition 5.5 are consequences of the relations in J1 j ??? j J6 . The former is a straightforward consequence of the fact that f is a rewriting mapping; we therefore concentrate on the latter. The first step in the proof of Theorem 5.6 is the following technical lemma, which gives certain rules for applying the rewriting mapping f . LEMMA 5.8.
Each word b g LŽ A, R . can be written as
b ' b 1 a1 b 2 a2 ??? b k a k b kq1 , where k G 1, b 1 , . . . , b kq1 g A*, a1 , . . . , a k g A, b 1 , . . . , b kq1 f LŽ A, R ., b 1 a1 , . . . , b k a k g LŽ A, R ., and then
f Ž b . ' f Ž b 1 a1 . ??? f Ž b ky1 a ky1 . f Ž b k a k b kq1 . . If biX g Aq is such that bi s biX is S then
f Ž b . ' f Ž b 1 a1 . ??? f Ž biX a i . ??? f Ž b ky1 a ky1 . f Ž b k a k b kq1 . . In particular,
f Ž b . ' f Ž b 1 a1 . ??? f Ž bi a i . ??? f Ž b ky1 a ky1 . f Ž b k a k b kq1 . . Proof. The lemma is a straightforward consequence of the definition Ž5. of f and the fact that R is a right deal. The main work in the proof of Theorem 5.6 is contained in the following result.
16
CAMPBELL ET. AL.
PROPOSITION 5.9. If S is a semigroup with finite presentation ² A N R :, R is a right ideal of finite index in S, V is a set of representati¨ es of S y R, and f is the corresponding rewriting mapping, and if J1 , . . . , J6 are defined as abo¨ e, then the following relations are consequences of J1 j ??? j J6 : Ži. f Ž w 1w 2 . s f Ž w 1 . f Ž w 2 ., w 1 , w 2 g LŽ A, R .; Žii. f Ž abg . s f Ž abg ., a , g g A*, b g Aq, abg g LŽ A, R ., b f LŽ A, R .. Proof. We proceed by simultaneous induction on Ži. and Žii.. First we prove Ža. f Ž w 1w 2 . s f Ž w 1 . f Ž w 2 . if < w 1w 2 < s n, assuming that f Ž u1 u 2 . s f Ž u1 . f Ž u 2 . if < u1 u 2 < - n, and that f Ž abg . s f Ž abg . if < abg < - n; and then we prove Žb. f Ž abg . s f Ž abg . if < abg < s n, assuming that f Ž u1 u 2 . s Ž f u1 . f Ž u 2 . if < u1 u 2 < F n, and that f Ž dzu . s f Ž dzu . if < dzu < - n. For n s 1 there is nothing to prove in Ža., while in Žb. we have a ' g ' e Žthe empty word., and b g A, and the relation f Ž abg . s f Ž abg . belongs to J5 . Now we prove Ža. for a general n. Let w 1 ' a 1 a1 a 2 a2 ??? a k a k a kq1 and w 2 ' b 1 b1 b 2 b 2 ??? b l bl b lq1 as in Lemma 5.8. If k ) 1 then
f Ž w 1w 2 . ' f Ž a 1 a1 a 2 a2 ??? a k a k a kq1w 2 . ' f Ž a 1 a1 . f Ž a 2 a2 ??? a k a k a kq1w 2 .
Ž by Ž 5. .
s f Ž a 1 a1 . f Ž a 2 a2 ??? a k a k a kq1 . f Ž w 2 .
Ž by induction .
s f Ž a 1 a1 a 2 a2 ??? a k a k a kq1 . f Ž w 2 .
Ž by induction .
' f Ž w1 . f Ž w 2 . . So we may assume that k s 1, i.e., that w 1 ' a 1 a1 a 2 . Suppose that a 2 w 2 g LŽ A, R ., so that
f Ž w 1w 2 . ' f Ž a 1 a1 . f Ž a 2 w 2 . by definition of f . If a 2 b 1 b1 . . . b ly1 bly1 g LŽ A, R ., then f Ž w 1w 2 . ' f Ž a 1 a1 . f Ž a 2 b 1 b1 ??? b ly1 bly1 b l bl b lq1 . s f Ž a 1 a1 . f Ž a 2 b 1 b1 ??? b ly1 bly1 . f Ž b l bl b lq1 . Ž by induction . ' f Ž a 1 a1 a 2 b 1 b1 ??? b ly1 bly1 . f Ž b l bl b lq1 .
Ž by Ž 5. .
s f Ž a 1 a1 a 2 . f Ž b 1 b1 ??? b ly1 bly1 . f Ž b l bl b lq1 . Ž by induction . ' f Ž a 1 a1 a 2 . f Ž b 1 b1 ??? b ly1 bly1 b l bl b lq1 . ' f Ž w1 . f Ž w 2 . .
Ž by Ž 5. .
SUBSEMIGROUPS OF F.P. SEMIGROUPS
17
On the other hand, if a 2 b 1 b1 . . . b ly1 bly1 f LŽ A, R ., then
f Ž w 1w 2 . ' f Ž a 1 a1 . f Ž a 2 b 1 b1 ??? b ly1 bly1 b l bl b lq1 . s f Ž a 1 a1 . f Ž a 2 b 1 b1 ??? b ly1 bly1 ? b lbl b lq1 .
Ž by induction .
s f Ž a 1 a1 a 2 b 1 b1 ??? b ly1 bly1 . f Ž b lbl b lq1 .
Ž relations J1 .
' f Ž a 1 a1 a 2 b 1 b1 ??? b ly1 bly1 . f Ž b l bl b lq1 .
Ž by Lemma 5.8.
s f Ž a 1 a1 a 2 b 1 b1 ??? b ly1 bly1 . f Ž b l bl b lq1 .
Ž by induction . .
If l s 1, we have finished; otherwise
f Ž w 1w 2 . ' f Ž a 1 a1 a 2 . f Ž b 1 b1 ??? b ly1 bly1 . f Ž b l bl b lq1 . ' f Ž a 1 a1 a 2 . f Ž b 1 b1 ??? b ly1 bly1 b l bl b lq1 . ' f Ž w 1 . f Ž w 2 . by induction. So we may assume that a 2 w 2 f LŽ A, R ., and hence that no initial segment of a 2 w 2 is in LŽ A, R .. Now
f Ž w 1w 2 . ' f Ž a 1 a1 a 2 b 1 b1 ??? b l bl b lq1 . ' f Ž a 1 a1 a 2 b 1 b1 ??? b l bl b lq1 . , by Lemma 5.8, and
f Ž w 1 . f Ž w 2 . ' f Ž a 1 a1 a 2 . f Ž b 1 b1 ??? b l bl b lq1 . ' f Ž a 1 a1 a 2 . f Ž b 1 b1 . ??? f Ž b ly1 bly1 . f Ž b lbl b lq1 .
Ž by Lemma 5.8. s f Ž a 1 a1 a 2 b 1 b1 . ??? f Ž b ly1 bly1 . f Ž b lbl b lq1 .
Ž relations J2 . s ??? s f Ž a 1 a1 a 2 b 1 b1 ??? b ly1 bly1 . f Ž b lbl b lq1 . Ž relations J2 . s f Ž a 1 a1 a 2 b 1 b1 ??? b ly1 bly1 b l bl b lq1 .
Ž relations J2 .
' f Ž a 1 a1 a 2 b 1 b1 ??? b ly1 bly1 b l bl b lq1 .
Ž by Lemma 5.8.
' f Ž w 1w 2 . . This completes the proof of Ža..
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CAMPBELL ET. AL.
We now start on the proof of Žb.. First suppose that a g LŽ A, R ., so that a ' a 1 a1 ??? a k a k a kq1 as in Lemma 5.8. If a kq1 bg f LŽ A, R ., then
f Ž abg . ' f Ž a 1 a1 ??? a k a k a kq1 bg . ' f Ž a 1 a1 ??? a k a k a kq1 bg . ' f Ž abg . by Lemma 5.8, and we are done. If a kq 1 bg g LŽ A, R ., then
f Ž abg . ' f Ž a 1 a1 ??? a k a k . f Ž a kq1 bg . s f Ž a 1 a1 ??? a k a k . f Ž a kq1 bg . ' f Ž a 1 a1 ??? a k a k a kq1 bg . ' f Ž abg . by Lemma 5.8 and induction; so suppose that a f LŽ A, R .. If ab f LŽ A, R ., the result follows immediately from Lemma 5.8; so suppose that ab g LŽ A, R .. If g g LŽ A, R . Žin particular, g / e ., then
f Ž abg . s f Ž ab . f Ž g . s f Ž ab . f Ž g . s f Ž abg . by induction; so we may assume that g f LŽ A, R .. Since a f LŽ A, R ., we may write
ab ' ab 1 b1 b 2 b 2 ??? b k bk b kq1 , where ab 1 b1 is the shortest prefix of ab in LŽ A, R .. If k ) 1, then
f Ž abg . ' f Ž ab 1 b1 b 2 b 2 ??? b k bk b kq1 g . ' f Ž ab 1 b1 b 2 b 2 ??? b ky1 bky1 . f Ž b k bk b kq1 g . Ž by Lemma 5.8. ' f Ž ab 1 b1 b 2 b 2 ??? b ky1 bky1 . f Ž b k bk b kq1g .
Ž by Lemma 5.8.
s f Ž ab 1 b1 b 2 b 2 ??? b ky1 bky1 . f Ž b k bk b kq1g .
Ž by induction .
s f Ž ab 1 b1 b 2 b 2 ??? b ky1 bky1 b k bk b kq1g .
Ž relations J3 .
' f Ž abg . . Since < b < F < b <, a similar argument shows that f Ž abg . s f Ž abg .. But b ' b by Lemma 5.4, and thus f Ž abg . s f Ž abg .. Hence we may assume that k s 1, so that ab ' ab 1 b1 b 2 . If b 2 g g LŽ A, R ., then the definition of f gives that
f Ž abg . ' f Ž ab 1 b1 b 2 g . ' f Ž ab 1 b1 . f Ž b 2 g . ' f Ž ab 1 b1 . f Ž b 2 g . .
SUBSEMIGROUPS OF F.P. SEMIGROUPS
19
Since b 1 b1 , g f LŽ A, R ., we may use induction and relations J4 to get
f Ž abg . s f Ž ab 1 b1 . f Ž b 2g . s f Ž ab 1 b1 b 2g . s f Ž abg . . As above we obtain f Ž abg . s f Ž abg .. So we may assume that b 2 g f LŽ A, R .. Given this, Lemma 5.8 gives that
f Ž abg . ' f Ž ab 1 b1 b 2 g . ' f Ž ab 1 b1 b 2g . , and relations J5 then give that
f Ž abg . s f Ž ab 1 b1 b 2g . s f Ž abg . . Again we have that
f Ž abg . s f abg ' f Ž abg . s f Ž abg . ,
ž
/
and this concludes the proof. We complete the proof of Theorem 5.6 by proving PROPOSITION 5.10. If S is a semigroup with finite presentation ² A N R :, R is a right ideal of finite index in S, V is a set of representati¨ es of S y R, and f is the corresponding rewriting mapping, and if J1 , . . . , J6 are defined as abo¨ e, then the following relations are consequences of J1 j ??? j J6 :
f Ž w 3 uw4 . s f Ž w 3¨ w4 . ,
Ž u s ¨ . g R , w 3 , w4 g A*, w 3 uw4 g L Ž A, R . .
Proof. If w 3 u f LŽ A, R ., then w 3¨ f LŽ A, R ., and
f Ž w 3 uw4 . ' f Ž w 3 uw4 . ' f Ž w 3¨ w4 . ' f Ž w 3¨ w4 . by Lemma 5.8; so we may assume that w 3 u g LŽ A, R .. If w4 g LŽ A, R . Žin particular, w4 / e ., then
f Ž w 3 uw4 . s f Ž w 3 u . f Ž w4 . s f Ž w 3¨ . f Ž w4 . s f Ž w 3¨ w4 . by Proposition 5.9Ži. and induction, so we may assume that w4 f LŽ A, R .. Assume that w 3 g LŽ A, R .; then w 3 ' a 1 a1 ??? a k a k a kq1 as in Lemma 5.8. If a kq 1 uw4 g LŽ A, R . then
f Ž w 3 uw4 . ' f Ž a 1 a1 ??? a k a k a kq1 uw4 . s f Ž a 1 a1 ??? a k a k . f Ž a kq1 uw4 . s f Ž a 1 a1 ??? a k a k . f Ž a kq1¨ w4 . ' f Ž w 3¨ w4 .
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CAMPBELL ET. AL.
by Lemma 5.8, Proposition 5.9, and induction, while if a kq 1 uw4 f LŽ A, R . we have
f Ž w 3 uw4 . ' f Ž a 1 a1 . . . a k a k a kq1 uw4 . ' f Ž a 1 a1 . . . a k a k a kq1¨ w4 . ' f Ž w 3¨ w4 . by Lemma 5.8. Given that w 3 f LŽ A, R . and that w4 f LŽ A, R ., we have
f Ž w 3 uw4 . s f Ž w 3 uw4 . s f Ž w 3¨ w4 . s f Ž w 3¨ w4 . by Proposition 5.9Žii. and relations J6 as required. This completes the proof of Theorem 5.6.
ACKNOWLEDGMENTS The authors are grateful to the referee for prompting them to summarise the present state of the Main Problem; this has led to some new results which will be published in a future paper. The fourth author thanks Hilary Craig for all her help and encouragement.
REFERENCES 1. C. M. Campbell, E. F. Robertson, N. Ruskuc, and R. M. Thomas, Rewriting a semigroup ˘ presentation, Internat. J. Algebra Comput., 5 Ž1995., 81]103. 2. C. M. Campbell, E. F. Robertson, N. Ruskuc, and R. M. Thomas, Reidemeister]Schreier ˘ type rewriting for semigroups, Semigroup Forum, 51 Ž1995., 47]62. 3. J. M. Howie, ‘‘An Introduction to Semigroup Theory,’’ Academic Press, London, 1976. 4. A. Jura, Determining ideals of finite index in a finitely presented semigroup, Demonstratio Math. 11 Ž1978., 813]827. 5. G. Lallement, ‘‘Semigroups and Combinatorial Applications,’’ Wiley]Interscience, New York, 1979. 6. R. C. Lyndon and P. E. Schupp, ‘‘Combinatorial Group Theory,’’ Springer-Verlag, Berlin, 1977. 7. M. Lothaire, ‘‘Combinatorics on Words,’’ Addison]Wesley, Reading, MA, 1983. 8. W. Magnus, A. Karrass, and D. Solitar, ‘‘Combinatorial Group Theory,’’ Interscience, New York, 1966. 9. S. W. Margolis and J. C. Meakin, Free inverse semigroups and graph immersions, Internat. J. Algebra Comput. 3 Ž1993., 79]99. 10. J. C. Meakin, An invitation to inverse semigroup theory, in ‘‘Proceedings of the Conference on Ordered Structures and Algebra of Computer Languages’’ ŽK. P. Shum and P. C. Yuen, Eds.., pp. 91]115, World Scientific, Singapore, 1993. 11. S. J. Pride, Geometric methods in combinatorial semigroup theory, in Proceedings of International Conference on Groups, Semigroups and Formal Languages, York, 1993; also in ‘‘Semigroups, Formal Languages, and Groups’’ ŽJ. Fountain, Ed.., pp. 215]232, Kluwer, Dordrecht, 1995.
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21
12. L. Redei, ‘‘The Theory of Finitely Generated Commutative Semigroups,’’ Pergamon ´ Press, Oxford, 1965. 13. N. Ruskuc, Matrix semigroups}Generators and relations, Semigroup Forum, 51 Ž1995., ˘ 319]333. 14. J. C. Spehner, Every finitely generated submonoid of a free monoid has a finite Malcev’s presentation, J. Pure Appl. Algebra 58 Ž1989., 279]287.
180, 1]21 Ž1996.
0049
On Subsemigroups of Finitely Presented Semigroups C. M. Campbell, E. F. Robertson, and N. Ruskuc ˇ Mathematical Institute, Uni¨ ersity of St Andrews, St Andrews KY16 9SS, Scotland
and R. M. Thomas* Department of Mathematics and Computer Science, Uni¨ ersity of Leicester, Leicester LE1 7RH, England Communicated by Peter M. Neumann Received June 8, 1994
1. INTRODUCTION The purpose of this paper is to investigate properties of subsemigroups of finitely presented semigroups, particularly with respect to the property of being finitely presented. More precisely, we will consider various instances of the following: Main Problem. Let S be a finitely presented semigroup. Ži. Is every subsemigroup Žideal, one-sided ideal. of S finitely generated? Žii. Is every subsemigroup Žideal, one-sided ideal. of S which is finitely generated as a semigroup finitely presented? Žiii. Is every subsemigroup Žideal, one-sided ideal. of finite index in S finitely presented? ŽFor the definition of index see below.. For example, if S is finite all these questions have an affirmative answer. Less trivially, if S is a commutative semigroup then Žii. has an affirmative answer; see w12x. Also, in w2x it has been proved that a two-sided ideal of finite index in a finitely presented semigroup is always finitely presented. * The authors acknowledge the support from the Edinburgh Mathematical Society Centenary Fund and the European Community Grant ERBCHRXCT930418, which helped to make possible the visit of the fourth author to the University of St Andrews while this work was being undertaken. 1 0021-8693r96 $18.00 Copyright Q 1996 by Academic Press, Inc. All rights of reproduction in any form reserved.
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TABLE I Subsemigroups
Ideals
One-sided ideals
Ži. f.g.
No ŽExample 5.3.
No ŽExample 5.3.
No ŽExample 5.3.
Žii. f.g.« f.p.
No ŽExample 4.5.
No Žfuture paper.
No Žfuture paper.
Žiii. f.i.« f.p.
Not known ŽOpen Problem 5.2.
Yes Žw2, Corollary 4.2x.
Yes ŽTheorem 5.1.
Note. f.g. denotes finitely generated, f.p. denotes finitely presented, and f.i. denotes finite index.
Group versions of questions Ži., Žii., and Žiii. have been investigated extensively. The main tool for these investigations is the Reidemeister] Schreier rewriting process, which gives a presentation for a subgroup of a group defined by a presentation; see w8x. Probably the best known application of this rewriting process is the Reidemeister]Schreier theorem which states that a subgroup of finite index in a finitely presented group is itself finitely presented; see w6x and w8x. A general theory of Reidemeister]Schreier-type rewriting for semigroups has been developed in w2x. This theory was then applied to two-sided ideals of finite index, as well as 0-minimal ideals; see also w1x. Here we continue this investigation and prove that a one-sided ideal of finite index in a finitely presented semigroup is itself finitely presented; see Theorems 5.1 and 5.6 and Corollary 5.7. Also, in a future paper we give an example of an ideal of a finitely presented semigroup, which is finitely generated as a semigroup but is not finitely presented. The present status of the Main Problem for general finitely presented semigroups is given in Table I. Another important application of the Reidemeister]Schreier rewriting for groups is the Nielsen]Schreier theorem, which states that every subgroup of a free group is itself free. A straightforward generalisation of this result does not hold: subsemigroups and one-sided ideals may or may not be free, while proper two-sided ideals are never free; see Propositions 3.1 and 3.3 and Example 3.2. Nevertheless, one can ask whether subsemigroups and ideals of a free semigroup have some other, weaker properties. For example, Spehner w14x has shown that every finitely generated subsemigroup admits a finite Malcev presentation. Here we consider free semigroups in the context of the Main Problem. We prove, among other things, that, in a free semigroup, two-sided and one-sided ideals which are finitely generated as semigroups, as well as subsemigroups of finite index,
SUBSEMIGROUPS OF F.P. SEMIGROUPS
3
TABLE II Subsemigroups
Ideals
One-sided ideals
free
No ŽWell known.
No ŽProposition 3.1.
No ŽProposition 3.3.
Ži. f.g.
No ŽExample 5.3.
No ŽExample 5.3.
No ŽExample 5.3.
Žii. f.g.« f.p.
No ŽExample 4.5.
Yes ŽCorollary 3.6.
Yes ŽTheorem 4.3.
Žiii. f.i.« f.p.
Yes ŽTheorem 3.8.
Yes ŽTable 1.
Yes ŽTable 1.
Note. The notation is the same as that used in Table I.
are finitely presented; see Corollary 3.6 and Theorems 3.8 and 4.3. The information obtained about substructures of free semigroups is summarised in Table II. It is also worth mentioning that Margolis and Meakin w9x have obtained results of Nielsen]Schreier type for free inverse semigroups; see also w10x. In this paper we shall use the standard semigroup theory notation and definitions as presented in w3x. For a semigroup T of S we define the Rees equi¨ alence determined by T to be r T s T = T j Ž s, s .: s g S4 . If T is a left Žright. ideal then r T is a left Žright. congruence, and if T is a two-sided ideal then r T is a congruence; the corresponding factor semigroup is denoted by SrT. We define the index of T in S to be < S y T < q 1, which is the number of equivalence classes of r T . Since we frequently consider ideals as subsemigroups, there is a certain ambiguity in the use of terms such as ‘‘generated by’’ and ‘‘finitely generated.’’ We adopt the convention that ‘‘generated’’ will always mean ‘‘generated as a subsemigroup,’’ except when stated otherwise. Thus, for example, the principal right ideal aS 1 is not necessarily finitely generated although it is finitely generated as a right ideal. For an alphabet A, the free semigroup and the free monoid on A are denoted by Aq and A* respectively. The empty word is denoted by e , and the length of a word w g A* is denoted by < w <. A semigroup presentation is an ordered pair ² A N R :, where R : Aq= Aq. A semigroup S is said to be defined by the presentation ² A N R : if S ( Aqrh , where h is the smallest congruence on Aq containing R. Usually we identify a word from Aq with the element of S it represents. However, to avoid confusion, we write w 1 ' w 2 if the words w 1 and w 2 are identical, and w 1 s w 2 if they represent the same element of S. For a subset T of S we use the notation LŽ A, T . for the set of all words from Aq which represent
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CAMPBELL ET. AL.
elements of T. For a systematic introduction to presentations of semigroups we refer the reader to w5x.
2. PRELIMINARY RESULT In this section, we prove the following preliminary result which we will need later. This is probably well known, but, for completeness, we include a proof here. THEOREM 2.1. If S is a semigroup and I is a finitely presented ideal of S, and if the Rees quotient SrI is finitely presented, then S is finitely presented. Proof. Suppose that I and SrI are defined by presentations ² X N R : and ² Y N G : respectively. Suppose for the moment that no generator from Y represents the zero of the semigroup SrI. In this case S is obviously generated by the set X j Y. Each word w g Yq which represents the zero of SrI represents an element of I in S, and so there exists a word a w g Xq such that w s a w in S. Also, for each x g X and each y g Y the words xy and yx represent elements from I in S; let b x y , b y x be words from Xq such that xy s b x y and yx s b y x hold in S. Let us fix a word z g Yq representing the zero of SrI. We define the following new sets of relations: G 1 s Ž u s ¨ . g G : u / z in SrI 4 , G 2 s Ž u s ¨ . g G : u s z in SrI 4 , G 2 s u s a u , ¨ s a¨ : Ž u s ¨ . g G 2 4 j z s a z 4 , G 3 s xy s b x y , yx s b y x : x g X , y g Y 4 , and let T denote the semigroup defined by the presentation ² X j Y N R j G1 j G 2 j G 3:.
Ž 1.
We claim that S ( T. It is obvious that S satisfies all the relations from R j G 1 j G 2 j G 3 , and hence there is a natural epimorphism f : T ª S. Assume now that two words w 1 , w 2 g Ž X j Y .q represent the same element of S. If both w 1 and w 2 contain a letter from X then we can use relations from G 3 to find words w 1 , w 2 g Xq such that w 1 s w 1 and w 2 s w 2 in T. Since the relation w 1 s w 2 holds in I, it can be deduced from R, thus giving w 1 s w 2 in T. If w 1 g Yq and w 1 represents an element of I in S then w 1 s z holds in SrI, and so z can be obtained from w 1 by applying relations from G. If no relations from G 2 are needed
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5
in this deduction then w 1 s z s a z holds in T. Otherwise, instead of the first application of a relation from G 2 we can use the corresponding relation from G 2 and thus obtain a word w 1 containing a letter from X such that w 1 s w 1 holds in T. Applying a similar argument to w 2 , if necessary, we reduce this case to the case where both w 1 and w 2 contain a letter from X. Finally, if w 1 does not represent an element of I in S then w 1 s w 2 can be deduced by just using relations from G 1 , and again w 1 s w 2 holds in T. Therefore, f is an isomorphism and T ( S as required. Let us now suppose that Y contains generators representing the zero in SrI, and let Y0 be the set of all non-zero generators from Y. If the zero of SrI is a product of two non-zero elements from SrI then Y0 generates SrI and the argument above can be repeated. Otherwise a similar argument would show that ² X j Y0 N R j G 1 j G 3 :
Ž 2.
is a presentation for S. Finally, note that if both presentations ² X N R : and ² Y N G : are finite so are the presentations Ž1. and Ž2..
3. FREE SEMIGROUPS As we indicated in the introduction, a straightforward generalisation of the Nielsen]Schreier theorem for groups does not hold for semigroups. It is well known that subsemigroups of a free semigroup are not necessarily free; see w7x. We now show that the situation for ideals is even worse in that they are almost never free. PROPOSITION 3.1. Let F s Aq be a free semigroup and let R / F be a proper right ideal. If R is finitely generated then it is not free. Proof. Since F / R there exists a g A such that a f R. Suppose that a i f R for all i G 1. Let w be an element of R of minimal length. Then wa i g R, i G 1, since R is a right ideal, but wa i is not a product of two elements of R. Therefore, each generating set of R contains all the words wa i, i G 1, and R is not finitely generated, a contradiction. Thus R contains some power of a. Let i be the minimal such power; obviously i ) 1. The word a iq1 belongs to R since R is a right ideal, but a iq1 is not a product of two elements of R since i ) 1; hence each generating set for R contains both a i and a iq1. Since a i and a iq1 satisfy the non-trivial relation a ia iq1 s a iq1 a i, R cannot be free.
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EXAMPLE 3.2. Let F be the free semigroup on two generators a, b4 , and let R s aF 1 be the principal right ideal generated Žas a right ideal. by a. The set ab i : i G 04 is a unique minimal generating set for R, and it is easy to see that R is free on that generating set. PROPOSITION 3.3. Let F s Aq be a free semigroup and let I / F be a proper two-sided ideal of F. Then I is not free. Proof. The proposition is an immediate corollary of w5, Proposition 2.2x. For the sake of completeness, we give an alternative direct proof. Let a g A y I, and let w be an element of I of minimal length. Both words aw and wa belong to I since I is a two-sided ideal, and neither of them is a product of two elements of I; hence each generating set for I contains both these words. But then w Ž aw . s Ž wa. w is a non-trivial relation holding in I, and I is not free. Now we shall consider free semigroups in the context of our Main Problem. The following example gives a negative answer to Ži.. EXAMPLE 3.4. Let F be the free semigroup on two generators a, b4 , and let I be the set of all words containing both a and b. I is obviously a two-sided ideal Žand hence a right ideal, a left ideal, and a subsemigroup.. However, the words ab i, i G 1, are not products of words from I, and I is not finitely generated. On the other hand, as we mentioned in the Introduction, question Žiii. always has a positive answer for ideals of finite index. In particular, an ideal of finite index of a finitely generated free semigroup is finitely presented. We shall now use this result to answer question Žii. for two-sided ideals and question Žiii. for subsemigroups of finite index of a free semigroup. THEOREM 3.5. If T is a finitely generated ideal in a free semigroup S, then T has finite index in S. Proof. We show that if T has infinite index in S, then T is not finitely generated. Suppose that T has infinite index in S, and let w 1 , w 2 , . . . be distinct elements of S y T. Let x be an element of T with x of minimal length among all elements of T, so that xw1 , xw 2 , . . . are elements of T. If xwi s u¨ with u, ¨ g T, then x is a prefix of u Žby the minimality of < x <., so that ¨ is a suffix of wi ; since ¨ g T, we have that wi g T, a contradiction. So xwi cannot be expressed as a non-trivial product of elements of T, and hence any generating set for T must contain xw1 , xw 2 , . . . . So T is not finitely generated.
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COROLLARY 3.6. If T is a finitely generated ideal in a finitely generated free semigroup S, then T is finitely presented. Proof. The corollary is an immediate consequence of w2, Corollary 4.2x and Theorem 3.5. Now we turn our attention to the subsemigroups of finite index in a free semigroup. PROPOSITION 3.7. If S is a finitely generated free semigroup and T is a subsemigroup of S of finite index, then there is an ideal I of finite index in S with I : T. Proof. Let S y T s a 1 , a 2 , . . . , a k 4 , and let p s max< a i <: 1 F i F k 4 . Let I be the ideal of all words in S of length at least p q 1. We see that I : T and that I has finite index in S as required. Given this, we can give an affirmative answer to question Žiii. for subsemigroups of finite index in a free semigroup. THEOREM 3.8. If S is a finitely generated free semigroup and T is a subsemigroup of S of finite index, then T is finitely presented. Proof. By Proposition 3.7, there is an ideal I of finite index in S with I : T. By w2x, I is finitely presented. The Rees quotient TrI is a finite semigroup, and hence is also finitely presented; the result follows from Theorem 2.1.
4. IF FINITELY GENERATED, THEN FINITELY PRESENTED? In this section, we continue our investigation of free semigroups in the context of the Main Problem. We prove that finitely generated one-sided ideals of a free semigroup are finitely presented, although they are not necessarily of finite index. We also give an example of a finitely generated subsemigroup of a free semigroup which is not finitely presented. First, however, we give a necessary and sufficient condition for an ideal to be finitely generated. PROPOSITION 4.1. Let S be a free semigroup, and let R be a right ideal of S. Then R is finitely generated if and only if there exist words a 1 , a 2 , . . . , a m of S and a constant N such that R s a 1 S 1 j a 2 S 1 j ??? j a m S 1 and such that e¨ ery word in S of length greater then N contains at least one of the a i as a substring.
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Proof. Suppose R is generated by a 1 , a 2 , . . . , a m 4 . Then, since a i g R for all i and R is a right ideal, we have that
a 1 S 1 j a 2 S 1 j ??? j a m S 1 : R. On the other hand, any word in R is a product of the a i , and hence lies in a k S 1 for some a k ; so R : a 1 S 1 j a 2 S 1 j ??? j a m S 1 , and hence R s a 1 S 1 j a 2 S 1 j ??? j a m S 1. Now suppose that R is Ža right ideal. of the form a 1 S 1 j a 2 S 1 j ??? j a m S 1 for some a i . ŽNote that we are not assuming now that R is generated by the a i as a semigroup.. We want to show that R is finitely generated if and only if there is a constant N such that every word in S of length greater than N contains at least one of the a i as a substring. We may assume, without loss of generality, that no a i is a prefix of any other a j , since, if a i is a prefix of a j , then a j S 1 : a i S 1, and a j S 1 can be omitted. For any word h in S 1 and any i, a ih lies in R, and any word in R is of this form. If R is finitely generated, then there is an upper bound on the length of a generator, and so R is generated by the set S s a ih : 1 F i F m, h g S 1 ,
Ž 3.
for some N. Let z be any element of S, and consider the element a 1 z of R. Since S is a generating set for R, we may write
a 1 z s a i1h1 a i 2h 2 ??? a i khk for some i1 , i 2 , . . . , i k and h1 , h 2 , . . . , hk with
z s h1 a i 2h 2 . . . a i khk . If < z < ) N, then, since
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i 2 , and so we have b s a i1h1 a i2 j for some h1 and j with
b s a i1h1 a i 2h 2 ??? a i khk , a product of elements in S, as required. Proposition 4.1 has an obvious dual for left ideals. Therefore we have the following corollary, which shows a surprising connection between finitely generated right ideals and finitely generated left ideals. COROLLARY 4.2. If S is a free semigroup and a 1 , a 2 , . . . , a m g S, then a 1 S 1 j a 2 S 1 j ??? j a m S 1 is a finitely generated right ideal of S if and only if S 1a 1 j S 1a 2 j ??? j S 1a m is a finitely generated left ideal of S. We can now prove the main result of this section. THEOREM 4.3. If R is a finitely generated right ideal of a free semigroup S, then R is finitely presented. Proof. Let S s Aq be the free semigroup on A. Since R is finitely generated and each generator of R only involves finitely many elements from A, and since Ra : R for each a g A, we see that A is finite. Let A s a1 , a2 , . . . , a n4 . Let R be generated by a 1 , a 2 , . . . , a m 4 , where a i g Aq for each i, and let B be the alphabet b1 , b 2 , . . . , bm 4 , where bi represents the generator a i of R. Let M s max< a i <: 1 F i F m4 . We have a relation
g Ž b1 , b 2 , . . . , bm . s d Ž b1 , b 2 , . . . , bm . in R if and only if g Ž a 1 , a 2 , . . . , a m . ' d Ž a 1 , a 2 , . . . , a m . in Aq. We define the weight of the relation g Ž b1 , b 2 , . . . , bm . s d Ž b1 , b 2 , . . . , bm . to be the length of the word g Ž a 1 , a 2 , . . . , a m . in Aq. Let R denote the set of all relations g Ž b1 , b 2 , . . . , bm . s d Ž b1 , b 2 , . . . , bm . of weight at most 3 M that hold in R. We claim that ² B N R : is a presentation for R Žand the proof will then be completed, since R is finite.. We need to show that any relation g Ž b1 , b 2 , . . . , bm . s d Ž b1 , b 2 , . . . , bm . holding in R is a consequence of R. So suppose we have a relation bi1 bi 2 bi 3 ??? bi r s bj1 bj 2 bj3 ??? bj s
Ž 4.
holding in R, so that
a i1 a i 2 a i 3 ??? a i r ' a j1 a j 2 a j3 ??? a j s in Aq. We argue by induction on the weight of the relation Ž4., the case of weight 1 being a relation in R. If a i1 ' a j1, we have bi 2 bi 3 ??? bi r s
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bj 2 bj3 ??? bj s, and the result follows by the inductive hypothesis. So suppose that < a i1 < - < a j1 <, so that a i1 is a proper prefix of a j1, say a j1 ' a i1 z . We now have
a i1 a i 2 a i 3 ??? a i r ' a i1 za j 2 a j3 ??? a j s , so that a i 2 a i 3 ??? a i r ' za j 2 a j3 ??? a j s. If a i 2 is a prefix of z then z g R, and so z ' a k 1 a k 2 ??? a k t. Both relations bi 2 bi 3 ??? bi r s bk 1 ??? bk t bj 2 ??? bj s and bj1 s bi1 bk 1 ??? bk t are of lower weight than Ž4., and hence are consequences of R by the inductive hypothesis. Now we have bi1 bi 2 ??? bi r s bi1 bk 2 ??? bk t bj 2 ??? bj s s bj1 bj 2 ??? bj s , which shows that Ž4. is a consequence of R as well. So let us assume now that z is a proper prefix of a i 2 . Then we have a i 2 ' za j 2 a j3 ??? a j qy 1 j for some q and j , where j is a prefix of a j q. Since za j 2 a j3 ??? a j qy 1 j represents an element of R and za j 2 a j3 ??? a j qy 1 j is a prefix of za j 2 a j3 ??? a j q, we see, since R is a right ideal, that za j 2 a j3 ??? a j q represents an element of R. So
za j 2 a j3 ??? a j q ' a k 1 a k 2 ??? a k t for some a k 1, a k 2 , . . . , a k t. Now < za j 2 a j3 ??? a j qy 1 j < s < a i 2 < F M, and so < a j a j a j ??? a j a j < s < a i za j a j ??? a j a j < 1 2 3 qy 1 q 1 2 3 qy 1 q F < a i1 < q < za j 2 a j3 ??? a j qy 1 j < q < a j q < F 3 M. Now
a j1 a j 2 a j3 ??? a j qy 1 a j q ' a i1 za j 2 a j3 ??? a j qy 1 a j q ' a i1 a k 1 a k 2 ??? a k t , and the corresponding relation bj1 bj 2 ??? bj qy 1 bj q s bi1 bk 1 bk 2 ??? bk t is a relation of weight at most 3 M, and so is in R by definition. So we may use this relation from R to transform the relation bi1 bi 2 bi 3 ??? bi r s bj1 bj 2 bj3 ??? bj s to bi1 bi 2 bi 3 ??? bi r s bi1 bk 1 bk 2 ??? bk t bj qq 1 bj qq 2 ??? bj s. So we only need to deduce the relation bi 2 bi 3 ??? bi r s bk 1 bk 2 ??? bk t bj qq 1 bj qq 2 ??? bj s , which is of lower weight than our original relation, and the result follows by induction.
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Theorem 4.3 is a generalisation of Corollary 3.6. The following example shows that a generalisation of Theorem 3.5 does not hold for right ideals. EXAMPLE 4.4. Let F be the free semigroup on a and b, and let R s a2 S 1 j abS 1 j b 2 S 1 . Since any word of length at least 3 must contain a2 , ab, or b 2 as a subword, R is a finitely generated right ideal of F by Proposition 4.1. However, F y R s a, b4 j baS 1 is not finite; in other words, R does not have finite index in F. The hypothesis in Theorem 4.3 that R is a right ideal cannot be omitted, in that a finitely generated subsemigroup of a finitely generated free semigroup need not be finitely presented, as our next example shows. EXAMPLE 4.5. Let F be the free semigroup on three generators a, b, and c, and let I be the subsemigroup generated by ¨ s ba, w s ba2 , x s a3, y s ac, and z s a2 c. I is clearly finitely generated, and we claim that I is not finitely presented. First note that, if a semigroup is finitely presented with respect to one generating set, then it is finitely presented with respect to any finite generating set, since we can pass from a presentation on the first set to a presentation on the second by means of Tietze transformations. ŽFor the semigroup version of Tietze transformations see w13x or w11x.. So it is sufficient to show that I is not finitely presented with respect to the particular generating set ¨ , w, x, y, z 4 . Decomposing the word a s ba3Ž nq1. c in two different ways, we see that the relations ¨ x n z s wx n y Ž n G 0. hold in I, and so any set of defining relations must imply these. Since any proper subword of a representing an element of I can be expressed as an element of ¨ , w, x, y, z 4 q in only one way, there is no non-trivial relation holding in I which we can apply to a proper subword of ¨ x n z or wx n y. So any set of defining relations for I must include all the relations ¨ x n z s wx n y, and so I is not finitely presented.
5. RIGHT IDEALS OF FINITE INDEX Now we turn our attention to a general finitely presented semigroup, and we give an affirmative answer to question Žiii. of the Main Problem in the case of one-sided ideals. In other words, we prove THEOREM 5.1. If S is a finitely presented semigroup and R is a right Ž or left . ideal of finite index in S, then R is finitely presented. The case of subsemigroups still remains open:
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Open Problem 5.2. If T is a subsemigroup of finite index in a finitely presented semigroup S, is T necessarily finitely presented? As mentioned in the Introduction, Theorem 5.1 is a generalisation of w2, Corollary 4.2x. It also strengthens the analogy between the Reidemeister] Schreier type theorems for semigroups and groups: both subgroups and one-sided ideals give rise to one-sided congruences Ži.e., the group and the semigroup respectively act on the corresponding factor sets by pre- or post-multiplication.. However, as the following example shows, the finite index condition in Theorem 5.1 cannot be replaced by the weaker condition that R is an equivalence class of a congruence with finitely many equivalence classes. EXAMPLE 5.3. Let F be the free semigroup on two generators a, b4 , and let S be the semigroup defined by the presentation ² A, B: A2 s A, B 2 s B, AB s BA: . We have a natural homomorphism u from F to S defined by au s A and bu s B. The kernel h of this homomorphism has three equivalence classes. Let I be the pre-image of AB in F. We see that I is the ideal of F consisting of all words involving both a and b. In Example 3.4 we have shown that I is not even finitely generated, let alone finitely presented. Our strategy in the proof of Theorem 5.1 is to use w2, Theorem 2.1x which gives a presentation ² B N I : for R. Although B is finite when R is of finite index, the set I is always infinite, and we aim to find a finite set of relations equivalent to I. First, however, we shall recall the necessary notation from w2x. Let S be defined by a Žfinite. presentation ² A N R :. Every element of S is represented by a word in Aq, and, as before, we let LŽ A, R . denote the set of all words in Aq representing elements of R. A set of representati¨ es of S y R is a set V of words in A* such that: Ž1. V contains the empty word e ; Ž2. each non-empty word in V represents an element of S y R; Ž3. each element of S y R is represented by one, and only one, word in V. It is worth noting that the empty word e is included in V to simplify notation in what follows; e does not represent an element of S. We choose V here in such a way that, for each non-empty word v in V, v is a word of minimal length representing the corresponding element of S y R.
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It was shown in w2, Theorem 3.1x that R is generated by the set X s r a s : r , s g V , a g A, r a g L Ž A, R . 4 . Note that, if we have r a g LŽ A, R ., then we also have r a s g LŽ A, R ., as R is a right ideal. We define a new alphabet B s br , a, s : r , s g V , a g A, r a g L Ž A, R . 4 in one]one correspondence with X. This correspondence can be extended to an epimorphism c : Bqª R in a natural way. The representati¨ e function w ¬ w associates to every element w g S y R its representative w g V Žand associates the empty word to the empty word.. We note the following immediate consequence of this definition: LEMMA 5.4. The representati¨ e function has the following properties. Ži. x ' x for each x f R; Žii. if y s z g S y R, then y ' z. For a word w g LŽ A, R . let w9a, where w9 g A* and a g A, be the shortest initial segment of w in LŽ A, R ., and let w0 be the rest of w. Define
f : L Ž A, R . ª Bq by
f Ž w. s
½
bw 9, a, w 0
if w0 f L Ž A, R .
bw 9, a, e f Ž w0 .
if w0 g L Ž A, R . .
Ž 5.
It is proved in w2, Lemma 3.4x that f is a rewriting mapping, in the sense that
c Ž f Ž w. . s w
in S.
Ž 6.
ŽIntuitively, f ‘‘rewrites’’ any word from Aq which represents an element of R into a product of generators from X.. Now we can easily deduce the following presentation for R from w2, Theorem 2.1x. PROPOSITION 5.5.
R is defined by the presentation
² B N f Ž w 1w 2 . s f Ž w 1 . f Ž w 2 . : w 1 , w 2 g L Ž A, R . 4 j f Ž w 3 uw4 . s f Ž w 3¨ w4 . : w 3 uw4 g L Ž A, R . , Ž u s ¨ . g R 4 : .
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Proof. By w2, Theorem 2.1x we have the following presentation for R: ² B N br , a, s s f Ž r a s . : br , a, s g B 4 j f Ž w 1w 2 . s f Ž w 1 . f Ž w 2 . : w 1 , w 2 g L Ž A, R . 4 j f Ž w 3 uw4 . s f Ž w 3¨ w4 . : w 3 uw4 g L Ž A, R . , Ž u s ¨ . g R 4 : . Note that the relations br , a, s s f Ž r a s . are identically true for our particular choice of the rewriting mapping, and the proposition follows. We want to prove that R is finitely presented; in fact, we will give an explicit finite set of defining relations for R with respect to the generating set B, namely the union of the following six Žfinite. sets: J1 s f Ž r 1 a1 r 2 . f Ž r 3 a2 r4 . s f Ž r 1 a1 . f Ž r 2 r 3 a2 r4 . : a i g A, r i g V ,
r 1 a1 , r 2 r 3 a2 , r 3 a2 g L Ž A, R . 4 ; J2 s f Ž r 1 a1 r 2 . f Ž r 3 a2 r4 . s f Ž r 1 a1 r 2 r 3 a2 r4 . : a i g A, r i g V ,
r 1 a1 , r 3 a2 g L Ž A, R . , r 2 r 3 a2 r4 f L Ž A, R . 4 ; J3 s f Ž r 1 r 2 . f Ž r 3 a r4 r 5 . s f Ž r 1 r 2 r 3 a r4 r 5 . : a g A, r i g V ,
r 1 r 2 , r 3 a g L Ž A, R . , r 2 r 3 a r4 f L Ž A, R . 4 ; J4 s f Ž r 1 r 2 . f Ž r 3 r4 . s f Ž r 1 r 2 r 3 r4 . : r i g V ,
r 1 r 2 , r 3 r4 g L Ž A, R . , r 2 r 3 f L Ž A, R . 4 ; J5 s f Ž r 1 r 2 a r 3 r4 . s f Ž r 1 r 2 a r 3 r4 . : a g A, r i g V ,
r 1 r 2 a g L Ž A, R . , r 2 a r 3 f L Ž A, R . 4 ; J6 s f Ž r 1 u r 2 . s f Ž r 1¨ r 2 . : Ž u s ¨ . g R , r i g V , r 1 u r 2 g L Ž A, R . 4 . Now we have THEOREM 5.6. If S is a semigroup with finite presentation ² A N R :, R is a right ideal of finite index in S, V is a set of representati¨ es of S y R, and f is the corresponding rewriting mapping, and if J1 , . . . , J6 are defined as abo¨ e, then R has presentation ² B N J1 j ??? j J6 :. Theorem 5.1 is clearly an immediate consequence of Theorem 5.6, since the sets J1 , . . . , J6 are all finite. It has to be admitted that the nature of our relations is a little complex. A more natural finite Žthough larger. set of relations is given in the following
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COROLLARY 5.7. Let S be a semigroup with finite presentation ² A N R :, let R be a right ideal of finite index, let V be a set of representati¨ es of S y R, and let f be the corresponding rewriting mapping. If m s max < r < , < u < , < ¨ < : r g V , Ž u s ¨ . g R 4 then R is defined by the presentation ² B N f Ž w 1w 2 . s f Ž w 1 . f Ž w 2 . : < w 1w 2 < F 5m q 2 4 j f Ž w 3 . s f Ž w4 . : w 3 s w4 in S, < w 3 < F 5m q 2, < w4 < F 5m q 2 4 : . Proof. All the relations from the given presentation hold in R, since f is a rewriting mapping, and they include all the relations from the presentation given in Theorem 5.6. We now embark on the proof of Theorem 5.6. We need to show that all the relations in J1 j ??? j J6 hold in R and that all the relations in the presentation from Proposition 5.5 are consequences of the relations in J1 j ??? j J6 . The former is a straightforward consequence of the fact that f is a rewriting mapping; we therefore concentrate on the latter. The first step in the proof of Theorem 5.6 is the following technical lemma, which gives certain rules for applying the rewriting mapping f . LEMMA 5.8.
Each word b g LŽ A, R . can be written as
b ' b 1 a1 b 2 a2 ??? b k a k b kq1 , where k G 1, b 1 , . . . , b kq1 g A*, a1 , . . . , a k g A, b 1 , . . . , b kq1 f LŽ A, R ., b 1 a1 , . . . , b k a k g LŽ A, R ., and then
f Ž b . ' f Ž b 1 a1 . ??? f Ž b ky1 a ky1 . f Ž b k a k b kq1 . . If biX g Aq is such that bi s biX is S then
f Ž b . ' f Ž b 1 a1 . ??? f Ž biX a i . ??? f Ž b ky1 a ky1 . f Ž b k a k b kq1 . . In particular,
f Ž b . ' f Ž b 1 a1 . ??? f Ž bi a i . ??? f Ž b ky1 a ky1 . f Ž b k a k b kq1 . . Proof. The lemma is a straightforward consequence of the definition Ž5. of f and the fact that R is a right deal. The main work in the proof of Theorem 5.6 is contained in the following result.
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PROPOSITION 5.9. If S is a semigroup with finite presentation ² A N R :, R is a right ideal of finite index in S, V is a set of representati¨ es of S y R, and f is the corresponding rewriting mapping, and if J1 , . . . , J6 are defined as abo¨ e, then the following relations are consequences of J1 j ??? j J6 : Ži. f Ž w 1w 2 . s f Ž w 1 . f Ž w 2 ., w 1 , w 2 g LŽ A, R .; Žii. f Ž abg . s f Ž abg ., a , g g A*, b g Aq, abg g LŽ A, R ., b f LŽ A, R .. Proof. We proceed by simultaneous induction on Ži. and Žii.. First we prove Ža. f Ž w 1w 2 . s f Ž w 1 . f Ž w 2 . if < w 1w 2 < s n, assuming that f Ž u1 u 2 . s f Ž u1 . f Ž u 2 . if < u1 u 2 < - n, and that f Ž abg . s f Ž abg . if < abg < - n; and then we prove Žb. f Ž abg . s f Ž abg . if < abg < s n, assuming that f Ž u1 u 2 . s Ž f u1 . f Ž u 2 . if < u1 u 2 < F n, and that f Ž dzu . s f Ž dzu . if < dzu < - n. For n s 1 there is nothing to prove in Ža., while in Žb. we have a ' g ' e Žthe empty word., and b g A, and the relation f Ž abg . s f Ž abg . belongs to J5 . Now we prove Ža. for a general n. Let w 1 ' a 1 a1 a 2 a2 ??? a k a k a kq1 and w 2 ' b 1 b1 b 2 b 2 ??? b l bl b lq1 as in Lemma 5.8. If k ) 1 then
f Ž w 1w 2 . ' f Ž a 1 a1 a 2 a2 ??? a k a k a kq1w 2 . ' f Ž a 1 a1 . f Ž a 2 a2 ??? a k a k a kq1w 2 .
Ž by Ž 5. .
s f Ž a 1 a1 . f Ž a 2 a2 ??? a k a k a kq1 . f Ž w 2 .
Ž by induction .
s f Ž a 1 a1 a 2 a2 ??? a k a k a kq1 . f Ž w 2 .
Ž by induction .
' f Ž w1 . f Ž w 2 . . So we may assume that k s 1, i.e., that w 1 ' a 1 a1 a 2 . Suppose that a 2 w 2 g LŽ A, R ., so that
f Ž w 1w 2 . ' f Ž a 1 a1 . f Ž a 2 w 2 . by definition of f . If a 2 b 1 b1 . . . b ly1 bly1 g LŽ A, R ., then f Ž w 1w 2 . ' f Ž a 1 a1 . f Ž a 2 b 1 b1 ??? b ly1 bly1 b l bl b lq1 . s f Ž a 1 a1 . f Ž a 2 b 1 b1 ??? b ly1 bly1 . f Ž b l bl b lq1 . Ž by induction . ' f Ž a 1 a1 a 2 b 1 b1 ??? b ly1 bly1 . f Ž b l bl b lq1 .
Ž by Ž 5. .
s f Ž a 1 a1 a 2 . f Ž b 1 b1 ??? b ly1 bly1 . f Ž b l bl b lq1 . Ž by induction . ' f Ž a 1 a1 a 2 . f Ž b 1 b1 ??? b ly1 bly1 b l bl b lq1 . ' f Ž w1 . f Ž w 2 . .
Ž by Ž 5. .
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On the other hand, if a 2 b 1 b1 . . . b ly1 bly1 f LŽ A, R ., then
f Ž w 1w 2 . ' f Ž a 1 a1 . f Ž a 2 b 1 b1 ??? b ly1 bly1 b l bl b lq1 . s f Ž a 1 a1 . f Ž a 2 b 1 b1 ??? b ly1 bly1 ? b lbl b lq1 .
Ž by induction .
s f Ž a 1 a1 a 2 b 1 b1 ??? b ly1 bly1 . f Ž b lbl b lq1 .
Ž relations J1 .
' f Ž a 1 a1 a 2 b 1 b1 ??? b ly1 bly1 . f Ž b l bl b lq1 .
Ž by Lemma 5.8.
s f Ž a 1 a1 a 2 b 1 b1 ??? b ly1 bly1 . f Ž b l bl b lq1 .
Ž by induction . .
If l s 1, we have finished; otherwise
f Ž w 1w 2 . ' f Ž a 1 a1 a 2 . f Ž b 1 b1 ??? b ly1 bly1 . f Ž b l bl b lq1 . ' f Ž a 1 a1 a 2 . f Ž b 1 b1 ??? b ly1 bly1 b l bl b lq1 . ' f Ž w 1 . f Ž w 2 . by induction. So we may assume that a 2 w 2 f LŽ A, R ., and hence that no initial segment of a 2 w 2 is in LŽ A, R .. Now
f Ž w 1w 2 . ' f Ž a 1 a1 a 2 b 1 b1 ??? b l bl b lq1 . ' f Ž a 1 a1 a 2 b 1 b1 ??? b l bl b lq1 . , by Lemma 5.8, and
f Ž w 1 . f Ž w 2 . ' f Ž a 1 a1 a 2 . f Ž b 1 b1 ??? b l bl b lq1 . ' f Ž a 1 a1 a 2 . f Ž b 1 b1 . ??? f Ž b ly1 bly1 . f Ž b lbl b lq1 .
Ž by Lemma 5.8. s f Ž a 1 a1 a 2 b 1 b1 . ??? f Ž b ly1 bly1 . f Ž b lbl b lq1 .
Ž relations J2 . s ??? s f Ž a 1 a1 a 2 b 1 b1 ??? b ly1 bly1 . f Ž b lbl b lq1 . Ž relations J2 . s f Ž a 1 a1 a 2 b 1 b1 ??? b ly1 bly1 b l bl b lq1 .
Ž relations J2 .
' f Ž a 1 a1 a 2 b 1 b1 ??? b ly1 bly1 b l bl b lq1 .
Ž by Lemma 5.8.
' f Ž w 1w 2 . . This completes the proof of Ža..
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We now start on the proof of Žb.. First suppose that a g LŽ A, R ., so that a ' a 1 a1 ??? a k a k a kq1 as in Lemma 5.8. If a kq1 bg f LŽ A, R ., then
f Ž abg . ' f Ž a 1 a1 ??? a k a k a kq1 bg . ' f Ž a 1 a1 ??? a k a k a kq1 bg . ' f Ž abg . by Lemma 5.8, and we are done. If a kq 1 bg g LŽ A, R ., then
f Ž abg . ' f Ž a 1 a1 ??? a k a k . f Ž a kq1 bg . s f Ž a 1 a1 ??? a k a k . f Ž a kq1 bg . ' f Ž a 1 a1 ??? a k a k a kq1 bg . ' f Ž abg . by Lemma 5.8 and induction; so suppose that a f LŽ A, R .. If ab f LŽ A, R ., the result follows immediately from Lemma 5.8; so suppose that ab g LŽ A, R .. If g g LŽ A, R . Žin particular, g / e ., then
f Ž abg . s f Ž ab . f Ž g . s f Ž ab . f Ž g . s f Ž abg . by induction; so we may assume that g f LŽ A, R .. Since a f LŽ A, R ., we may write
ab ' ab 1 b1 b 2 b 2 ??? b k bk b kq1 , where ab 1 b1 is the shortest prefix of ab in LŽ A, R .. If k ) 1, then
f Ž abg . ' f Ž ab 1 b1 b 2 b 2 ??? b k bk b kq1 g . ' f Ž ab 1 b1 b 2 b 2 ??? b ky1 bky1 . f Ž b k bk b kq1 g . Ž by Lemma 5.8. ' f Ž ab 1 b1 b 2 b 2 ??? b ky1 bky1 . f Ž b k bk b kq1g .
Ž by Lemma 5.8.
s f Ž ab 1 b1 b 2 b 2 ??? b ky1 bky1 . f Ž b k bk b kq1g .
Ž by induction .
s f Ž ab 1 b1 b 2 b 2 ??? b ky1 bky1 b k bk b kq1g .
Ž relations J3 .
' f Ž abg . . Since < b < F < b <, a similar argument shows that f Ž abg . s f Ž abg .. But b ' b by Lemma 5.4, and thus f Ž abg . s f Ž abg .. Hence we may assume that k s 1, so that ab ' ab 1 b1 b 2 . If b 2 g g LŽ A, R ., then the definition of f gives that
f Ž abg . ' f Ž ab 1 b1 b 2 g . ' f Ž ab 1 b1 . f Ž b 2 g . ' f Ž ab 1 b1 . f Ž b 2 g . .
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Since b 1 b1 , g f LŽ A, R ., we may use induction and relations J4 to get
f Ž abg . s f Ž ab 1 b1 . f Ž b 2g . s f Ž ab 1 b1 b 2g . s f Ž abg . . As above we obtain f Ž abg . s f Ž abg .. So we may assume that b 2 g f LŽ A, R .. Given this, Lemma 5.8 gives that
f Ž abg . ' f Ž ab 1 b1 b 2 g . ' f Ž ab 1 b1 b 2g . , and relations J5 then give that
f Ž abg . s f Ž ab 1 b1 b 2g . s f Ž abg . . Again we have that
f Ž abg . s f abg ' f Ž abg . s f Ž abg . ,
ž
/
and this concludes the proof. We complete the proof of Theorem 5.6 by proving PROPOSITION 5.10. If S is a semigroup with finite presentation ² A N R :, R is a right ideal of finite index in S, V is a set of representati¨ es of S y R, and f is the corresponding rewriting mapping, and if J1 , . . . , J6 are defined as abo¨ e, then the following relations are consequences of J1 j ??? j J6 :
f Ž w 3 uw4 . s f Ž w 3¨ w4 . ,
Ž u s ¨ . g R , w 3 , w4 g A*, w 3 uw4 g L Ž A, R . .
Proof. If w 3 u f LŽ A, R ., then w 3¨ f LŽ A, R ., and
f Ž w 3 uw4 . ' f Ž w 3 uw4 . ' f Ž w 3¨ w4 . ' f Ž w 3¨ w4 . by Lemma 5.8; so we may assume that w 3 u g LŽ A, R .. If w4 g LŽ A, R . Žin particular, w4 / e ., then
f Ž w 3 uw4 . s f Ž w 3 u . f Ž w4 . s f Ž w 3¨ . f Ž w4 . s f Ž w 3¨ w4 . by Proposition 5.9Ži. and induction, so we may assume that w4 f LŽ A, R .. Assume that w 3 g LŽ A, R .; then w 3 ' a 1 a1 ??? a k a k a kq1 as in Lemma 5.8. If a kq 1 uw4 g LŽ A, R . then
f Ž w 3 uw4 . ' f Ž a 1 a1 ??? a k a k a kq1 uw4 . s f Ž a 1 a1 ??? a k a k . f Ž a kq1 uw4 . s f Ž a 1 a1 ??? a k a k . f Ž a kq1¨ w4 . ' f Ž w 3¨ w4 .
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by Lemma 5.8, Proposition 5.9, and induction, while if a kq 1 uw4 f LŽ A, R . we have
f Ž w 3 uw4 . ' f Ž a 1 a1 . . . a k a k a kq1 uw4 . ' f Ž a 1 a1 . . . a k a k a kq1¨ w4 . ' f Ž w 3¨ w4 . by Lemma 5.8. Given that w 3 f LŽ A, R . and that w4 f LŽ A, R ., we have
f Ž w 3 uw4 . s f Ž w 3 uw4 . s f Ž w 3¨ w4 . s f Ž w 3¨ w4 . by Proposition 5.9Žii. and relations J6 as required. This completes the proof of Theorem 5.6.
ACKNOWLEDGMENTS The authors are grateful to the referee for prompting them to summarise the present state of the Main Problem; this has led to some new results which will be published in a future paper. The fourth author thanks Hilary Craig for all her help and encouragement.
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