A translation of G. S. Makanin's 1966 Ph.D. thesis "On the Identity Problem for Finitely Presented Groups and Semigroups"
aa r X i v : . [ m a t h . G R ] F e b USSR Academy of Sciences
Steklov Mathematical Institute
G. S. Makanin
On The Identity Problem For FinitelyPresented Groups and Semigroups
Dissertation for the degree of Candidate of Physical and Mathematical Sciences.SupervisorsCorresponding Member of the USSR Academy of Sciences
A. A. Markov
Doctor of Physical and Mathematical Sciences
S. I. Adian
Moscow (1966) ontents
Introduction 7Chapter 1. Algorithmic problems for ( k, ℓ )-semigroups 111.1. Construction of elementary words 111.2. The identity problem for ( k, ℓ )-semigroups 231.3. The divisibility problems for ( k, ℓ )-semigroups 311.4. The maximal subgroup of a ( k, ℓ )-semigroup 33Chapter 2. The Identity Problem In Finitely Presented Groups 352.1. Representing words equal to the identity in K / -groups 352.2. Solving the identity problem for K / -groups 39Bibliography 55 otes on the Translation The many fascinating ideas in this thesis, worked out in full detail and withclose attention paid to ensuring a high standard of rigour, have made it a delightto translate. Nearly sixty years later, G. S. Makanin’s insights into the structure ofspecial monoids and small cancellation theory remain exceptional, and that withoutany of the modern machinery to approach these topics.In transcribing and translating this thesis from the original Russian, I haveattempted to capture its original minimal style, in which the mathematics itself isin focus. Perhaps the largest change made to this original style is that proof envi-ronments, which were originally omitted, have been added, to ensure proper cut-offbetween statements, their proofs, and remarks following them. Any Cyrillic lettersused for indexing, especially the convoluted nested cases in Chapter 2, have beenreplaced by Latin or Greek counterparts. While the page numbering is differentfrom the original (with the original thesis being 98 pages), the numbering of allsections, lemmas, equations, etc. remains the same. Particular attention was givento accurately typesetting the numerous involved, and originally hand-drawn, smallcancellation diagrams in Chapter 2.As for linguistic choices, a direct translation of some of the definitions has notalways been possible or feasible, as, for instance, some use terminology that wouldbe best translated by an already overloaded term in English. In these cases, I haveinstead focused on capturing the mathematical sense of the word to be translated.Finally, some typos have been fixed – but there were very few to be found indeed,as could be expected of a student working under the watchful gaze of S. I. Adian.Any typos still present in this translation are entirely my own.I wish to thank Prof L. D. Beklemishev of the Steklov Mathematical Institutefor his support related to the project.
C. F. Nyberg-Brodda
29 January 2021ennadij Semenovich Makanin (1938–2017). his page intentionally left blank. ntroduction
Let Π be a semigroup given by a set of generators(1) a , a , . . . , a n and defining relations(2) A i = B i ( i = 1 , , . . . , m )The system of generators (1) is called the alphabet of the semigroup Π. The setof words over the alphabet (1) are called the words of the semigroup Π. Theempty word will be denoted 1, and the length of the word X will be denoted [ X ∂ .Graphical equality of two words X and Y will be indicated by X ≖ Y .The rules of inference by which, based on the relations (2), we determine therelation of equality for words of Π, generate a form of calculus (see [ Mar54 ], pp.205, 206). Most of the auxiliary concepts and statements of the present work isassociated with this calculus. By considering this calculus as a means of defininga semigroup, we will not use the term calculus in our work, but rather phrase allconcepts in terms of semigroups instead.In this work we will reason about certain normal decision problems , and provethe reducibility of some problems to others. In order to accurately formulate thenotion of a normal decision problem, it is necessary to fix a way of writing the inputdata over the alphabet, and then pose the problem of finding a normal algorithmover this alphabet. For simplicity, we will not do this, and use the term normal de-cision problem without further specification, with the implicit understanding thatit is based on this properly refined formulation. The principle of normalisation(see [
Mar54 ] II. §
5) makes it possible to assert that such a refinement of a decisionproblem is uniquely defined up to equivalence. In the same way, the proofs of thereducibility of one normal decision problem to another, which would require rea-soning about normal algorithms, will be replaced for simplicity by a description ofthe processes by which the solution of a given problem, corresponding to a normaldecision problem, is reduced to solving single cases of another normal decision prob-lem. We are not presented with any particular fundamental difficulty in carryingout these proofs in full detail, but it is very cumbersome.We will call an elementary transformation of the semigroup Π the transitionof a word of the form XA i Y to the word XB i Y , or vice versa, where X and Y arearbitrary words of the semigroup Π, and A i = B i is one of the defining relations in(2). An elementary transformation is denoted by one of the two forms XA i Y → XB i Y, XB i Y → XA i Y. We also say that the tautological transition X → X is an elementary transforma-tion. The relations (2) define the equality of words in the semigroup Π, in a wayconnected to the elementary transformations of the semigroup Π in the followingway: two words X and Y of the semigroup Π are equal in Π if and only if thereexists some sequence of elementary transformations of the semigroup Π(3) X ≖ X → X → · · · → X i → X i +1 → · · · → X k ≖ Y, transforming the word X into the word Y .We will call G the group defined by the generators (1) and the defining relations(2) if it is the semigroup given by the alphabet a , a , . . . , a n , a − , a − , · · · , a − n and the system of defining relations(2) A i = B i ( i = 1 , , . . . , m )(4) a j a − j = 1 a − j a j = 1 (cid:27) ( j = 1 , , . . . , n )The alphabet (1) will be called the positive alphabet of the group G . The letter a − j will be called the inverse of the letter a j . The defining relations (4) will be called trivial relations. These relations are uniquely determined by the alphabet of thegroup, and are therefore generally note written out when defining a group.A group will be called a free group , if all defining relations are trivial. Theequality of two words X and Y in the free group will be denoted by X ≡ Y .The identity problem for words in a semigroup Π (group G ) is the followingproblem: provide an algorithm, by which one can determine for any two words X and Y whether or not X and Y are equal in the semigroup Π (in the group G ).A semigroup given over the alphabet(1) a , a , . . . , a n and the finitely many defining relations(5) A i = 1 ( i = 1 , , . . . , k )such that(6) [ A i∂ ≤ ℓ ( i = 1 , , . . . , k ) , will be called a ( k, ℓ ) -semigroup. If the ( k, ℓ )-semigroup Π is such that we have[ A i∂ = ℓ ( i = 1 , , . . . , k ), then we say that Π is a homogeneous ( k, ℓ )-semigroup.We will say, that a word X in some semigroup Π is left (right) invertible , ifthere exists some word Y in the semigroup such that Y X = 1 (resp. XY = 1) inthe semigroup Π. A word X is two-sided invertible in the semigroup Π if X is bothleft and right invertible.If in the ( k, ℓ )-semigroup Γ every word is two-side invertible, then Γ will becalled a ( k, ℓ ) -group .We will say that the word X in some semigroup Π is left (right) divisible by theword Y , if there exists some word Z such that in Π we have the equality X = Y Z (resp. X = ZY ).The left (right) divisibility problem for a given semigroup Π is defined as follows:provide an algorithm, which decides for any two words X and Y whether or not X is left (right) divisible by Y in the semigroup Π. NTRODUCTION 9
S. I. Adian, in the third chapter of [
Adi66 ], has proved: if the natural numbers k and ℓ are such that there exists an algorithm for solving the identity problem inevery ( k, ℓ )-group, then there is an algorithm, which solves the identity problem inany homogeneous ( k, ℓ )-semigroup.In Sections 1.1, 1.2, and 1.3 of the present work, it is proved that under thesame assumptions as those made by S. I. Adian, there is an algorithm which solvesthe identity problem in any ( k, ℓ )-semigroup.We will call the maximal subgroup of a semigroup the subgroup generated bythe collection of two-sided invertible words of the semigroup.S. I. Adian, in the third chapter of [ Adi66 ], has proved: if the natural numbers k and ℓ are such that there exists an algorithm for solving the identity problemin every ( k, ℓ )-group, then there is an algorithm which for any homogeneous ( k, ℓ )-semigroup Π produces a ( k, ℓ )-group, isomorphic to the maximal subgroup of Π.In Section 1.4 of the present work, it is proved that under the same assumptionsas those made by S. I. Adian, there is an algorithm which for any ( k, ℓ )-semigroupΠ produces a ( k, ℓ )-group, isomorphic to the maximal subgroup of Π.Two systems of defining relations Ω and Ω over some alphabet A will becalled equivalent , if for every choice of two words over A , the words are equal inthe semigroup with the system of defining relations Ω if and only if the words areequal in the semigroup with system of defining relations Ω .If two semigroups Π and Π are given by equivalent systems of defining re-lations, then the identity problem in Π , either of the divisibility problems in Π ,and the problem of computing a group isomorphic to the maximal subgroup of Π,can each be reduced to the respective problem for the semigroup Π .Let the group G be defined by the generators(1) a , a , . . . , a n and a finite system of defining relations(7) R i = 1 ( i = 1 , , . . . , m )The left-hand sides of the defining relations R i = 1 ( i = 1 , , . . . , m ) are wordswritten over the alphabet ( a ± , . . . , a ± n ), which we will call the defining words ofthe group G . We denote by M the set of all defining words. Deleting a word of theform a j a − j or a − j a j from some word W will be called reducing W . A word, whichcannot be reduced, will be called reduced . If Z ≖ a ε i a ε i · · · a ε v − i v − a ε v i v where ε i = ± i = 1 , , . . . , v ), then we will call the word( a ε v i v ) − ( a ε v − i v − ) − · · · ( a ε i ) − ( a ε i ) − the inverse of Z , and denote it by Z − or Z .Without loss of generality, we may assume that every defining word R i of thegroup G is reduced, and that the set M is closed under the operations of takinginverses of and cyclic permutations of the words R i (this implies that each R i iscyclically reduced). Such a set of defining words will be called symmetrized .V. A. Tartakovskij [ Tar49a, Tar49b ], and subsequently A. V. Gladkij [
Gla59,Gla61 ], M. D. Greendlinger [
Gre65, Gre60 ] and others, have studied finitelypresented groups given by a symmetrized set of defining relations. In some classesof these groups, it is possible to distinguish large parts of the defining words in every non-empty, reduced word equal to the identity of the group. In many cases,this permits a positive solution to the identity and conjugacy problems, and todescribe some of the properties of these groups.A group G , with generating set (1) and defining relations (7), is said to belongto the class K α , where 0 < α <
1, if the set of M of the group G satisfies thefollowing property: if R i and R j are arbitrary, not mutually inverse words in theset M , then reducing the word R i R j removes less than α times the number ofletters in R i .In the cited works, and also in the work by Britton [ Bri56 ] and Sheik [
Sch56 ],the identity problem is solved for various generalisations of the class of groups K / .In Chapter 2 of the present work, using a generalisation of Dehn’s algorithm[ Deh11 ], the identity problem is solved for the class of groups K / .The definitions and notation found in the introduction (which are borrowedfrom the works of S. I. Adian [ Adi66 ], M. D. Greendlinger [
Gre65 ], A. A. Markov[
Mar54 ], and G. S. Tsejtin [
Tse58 ]) will be considered as known, and will not bere-introduced.The results of Chapter 1 of the present work are published in the note [
Mak66 ].The results of the present work were presented at a seminar on algorithmic problemsin algebra at Moscow State University.I wish to express my gratitude to A. A. Markov and S. I. Adian for theirattention and help related to the work.HAPTER 1
Algorithmic problems for ( k, ℓ ) -semigroups Construction of elementary words
For any list of words X , X , . . . , X s on some alphabet A , we associate a reduced list of words T ( X , X , . . . , X s ), which is obtained from the list X , X , . . . , X s , byrepeatedly applying the following two operations until no longer possible:( α ) if X j ≖
1, then remove X j from the list;( β ) if X k ≖ X ℓ and k < ℓ , then remove X ℓ from the list.Starting with X , X , . . . , X s , the above will result in a unique list X i , X i , . . . , X i t ,which we will denote by T ( X , X , . . . , X s ).Order in some way the set of all pairs of natural numbers and, with Y , Y , . . . , Y t some list of words, say that an ordered pair of words ( Y j , Y j ) precedes the orderedpair ( Y k , Y k ) if in our chosen ordering the pair of numbers ( j , j ) precedes thepair ( k , k ). Definition 1.
We say that a pair of words X and Y on an alphabet A overlap if X ≖ M N and Y ≖ N L , where N and M L are non-empty words.Suppose we are given a list of words(1) Y , Y , . . . , Y t , Z , Z , . . . , Z s . If in the list of words Y , Y , . . . , Y t we delete the words Y v , Y v , · · · , Y v m , we willdenote the remaining words of (1) by Y , Y , . . . , Y t , Y v , Y v , . . . , Y v m , Z , Z , . . . , Z s . The division algorithm ∆.Consider an arbitrary reduced list of words Y , Y , . . . , Y t on some alphabet A .The algorithm ∆, when applied to such a list, will act in one of two ways:(i) If no pair of words from the list Y , Y , . . . , Y t overlap, then∆( Y , Y , . . . , Y t ) ⇋ Y , Y , . . . , Y t . (ii) Suppose Y v ≖ M N, Y v ≖ N L , where N and M L are non-empty, where( Y v , Y v ) is a pair of words from Y , Y , . . . , Y t , such that this pair precedesall other overlapping pairs of words from this list; then∆( Y , Y , . . . , Y t ) ⇋ ∆( T ( Y , Y , . . . , Y t , Y v , Y v , M, N, L )) . Definition 2.
Consider a list of words Y , Y , . . . , Y t such that [ Y ∂ = ℓ i , where ℓ i is some positive integer for i = 1 , , . . . , t . The parameter of the list Y , Y , . . . , Y t is defined as the number P ti =1 ( ℓ i − ω ( Y , Y , . . . , Y t ).
112 1. ALGORITHMIC PROBLEMS FOR ( k, ℓ )-SEMIGROUPS
Lemma 1.
If the pair of words ( Y v , Y v ) overlap, where the words are taken fromsome list Y , Y , . . . , Y t , and Y v ≖ M N , Y v ≖ N L , where N and M L are non-empty, then ω ( Y , Y , . . . , Y t ) > ω ( T ( Y , Y , . . . , Y t , Y v , Y v , M, N, L )) . The lemma follows immediately from the facts that [ Y v ∂ = [ M ∂ + [ N ∂ and[ Y v ∂ = [ N ∂ + [ L ∂ , and that any arbitrary list of words X , X , . . . , X s has param-eter no less than that of T ( X , X , . . . , X s ).Suppose the algorithm ∆ is applied to a reduced list Y , Y , . . . , Y t . We intro-duce the notion of a step of the algorithm. If the algorithm ∆ immediately appliesrule (i), then we say that the algorithm ∆ produces a non-overlapping list of wordsin 0 steps, or that it finishes in 0 steps. If the algorithm ∆ first applies rule (ii) k times, and then applies rule (i), then we say that ∆ produces a non-overlappinglist of words in k steps, or that it finishes in k steps. Lemma 2.
For every reduced list of words Y , Y , . . . , Y t , there exists some non-negative integer k , such that the algorithm ∆ when applied to Y , Y , . . . , Y t finishesin k steps. Proof.
By Lemma 1, at every step of the algorithm ∆, we process a list witha given parameter ω i into a list with parameter ω i +1 , where ω i > ω i +1 . As theparameter ω is defined as a positive integer, it follows that the algorithm ∆ willeventually produce a list with no overlaps, at which point it will apply rule (i). (cid:3) Definition 3.
If the algorithm ∆, when applied to a list of words Y , Y , . . . , Y t ,finishes with the list of non-overlapping words V , V , . . . , V k , then we write(2) ∆[ Y , Y , . . . , Y t ] = V , V , . . . , V k . Lemma 3. If ∆[ Y , Y , . . . , Y t ] = V , V , . . . , V k , then(1) V , V , . . . , V k is a reduced list of non-overlapping words.(2) Every Y i ( i = 1 , , . . . , t ) is graphically a product of some V i , V i , . . . , V i si ,where ≤ i , i , . . . , i s i ≤ t .(3) For every V j ( j = 1 , , . . . , k ) there exists some Y i ≖ V i V i · · · V i si suchthat V j is graphically equal to at least one of V i , V i , . . . , V i si . All parts of the lemma are easily proved by induction on the number of stepstaken by the algorithm ∆ when applied to the list Y , Y , . . . , Y t . Lemma 4.
If the words XY and Y Z are invertible in some semigroup Π , then X, Y , and Z are invertible in Π . Proof. As XY and Y Z are invertible, we can find some words U , U , V , and V , such that in Π we have the equalities(3) U XY = 1 , XY U = 1 , Y ZV = 1 , V Y Z = 1 . From the first and third equalities in the above (3) it follows that Y is invertible inΠ. Observe that ZV Y = U XY ZV Y = U XY = 1, and Y U X = Y U XY ZV = Y ZV = 1 in the semigroup Π, and hence it follows that X and Z are also invertiblein Π. (cid:3) Lemma 5. If ∆[ Y , Y , . . . , Y t ] = V , V , . . . , V k and every Y i ( i = 1 , , . . . , t ) isinvertible in the semigroup Π , then all the V j ( j = 1 , , . . . , k ) are invertible in Π . .1. CONSTRUCTION OF ELEMENTARY WORDS 13 This lemma is easily proved by induction on the number of steps taken by thealgorithm ∆ when applied to the list Y , Y , . . . , Y t , and using Lemma 4.Let Π be a ( k, ℓ )-semigroup with generating set(4) a , a , . . . , a n and defining relations(5) A i = 1 ( i = 1 , , . . . , k ) , satisfying(6) [ A i∂ ≤ ℓ ( i = 1 , , . . . , k ) . Definition 4.
A finite list of words B , B , . . . , B p over the alphabet (4) is calleda list of B -words of the ( k, ℓ ) -semigroup Π if(1) B , B , . . . , B p is a reduced list of non-overlapping words.(2) Every left-hand side A i ( i = 1 , , . . . , k ) of the ( k, ℓ )-semigroup Π is graph-ically equal to a product of some B i , B i , . . . , B i si (1 ≤ i , i , . . . , i s i ≤ p ).(3) For every B j ( j = 1 , , . . . , p ) there exists some A i ≖ B i B i · · · B i si suchthat B j is graphically equal to one of the B i , B i , . . . , B i si . Lemma 6.
There exists an algorithm which for every ( k, ℓ ) -semigroup Π constructsa list of B -words for this semigroup. Proof.
Let A , A , . . . , A k be the left-hand sides of the defining relations ofsome ( k, ℓ )-semigroup Π. Apply the algorithm ∆ to the list T ( A , A , . . . , A k ).Suppose that ∆[ T ( A , A , . . . , A k )] = B , B , . . . , B p . Then by Lemma 3, the reduced list B , B , . . . , B p is a list of B -words of the ( k, ℓ )-semigroup Π. (cid:3) Lemma 7. If A ≖ B j B j · · · B j p ≖ B ′ m B ′ m · · · B ′ m s , where B j i and B ′ m i are B -words of the ( k, ℓ ) -semigroup Π , then p = s , and B j i ≖ B ′ m i for all i . Proof.
Let A be the shortest word such that some of the conclusions ofLemma 7 do not hold. Neither of the words B j and B ′ m can be a proper prefixof the other, as B -words do not overlap. Hence B j ≖ B ′ m , and B j · · · B j p ≖ B ′ m · · · B ′ m s . Call this word A ′ . As [ A ′ ∂ < [ A ∂ , and as A is the shortest word suchthat the conclusions of Lemma 7 do not hold, we have that p − s −
1, and B j i ≖ B ′ m i for i = 1 , , . . . , p . But then the conclusions of Lemma 7 hold for A .This is a contradiction. Hence, Lemma 7 is true for all words A . (cid:3) Specifying the ( k, ℓ )-semigroup Γ from a ( k, ℓ )-semigroup Π and its B -words.Establish a one-to-one connection between the B -words B , B , . . . , B p of the( k, ℓ )-semigroup Π and a set of symbols β , β , . . . , β p . As B i ≖ B j for i = j , thiscan be done.By Lemma 7 every A i can be uniquely written as B i B i · · · B i si , where ( i =1 , , . . . , k ) and (1 ≤ i , i , . . . , i s i ≤ p ).Let Γ be the ( k, ℓ )-semigroup with generating set(7) β , β , . . . , β p k, ℓ )-SEMIGROUPS and defining relations(8) M i = 1 ( i = 1 , , . . . , k )where M i ≖ β i β i · · · β i si ( i = 1 , , . . . , k ) , (1 ≤ i , i , . . . , i s i ≤ p ). That is, M i isobtained from A i ≖ B i B i · · · B i si by replacing the B -words by their correspondingletters β i , β i , . . . , β i si .As [ M i∂ ≤ [ A i∂ ( i = 1 , , . . . , k ), it follows that Γ really is a ( k, ℓ )-semigroup.From Lemma 7, it follows that the ( k, ℓ )-semigroup Γ is uniquely defined from the( k, ℓ )-semigroup Π and its B -words. Lemma 8. ∆[ T ( M , M , . . . , M k )] = β , β , . . . , β p . The lemma follows easily by induction on the number of steps taken by thealgorithm ∆. For this, one must note that any word appearing in the process ofapplying the algorithm ∆ to T ( M , M , . . . , M k ) is either empty, or else comesfrom some B -word; and then note that ∆[ T ( A , A , . . . , A k )] = B , B , . . . , B p . Definition 5.
A finite list of words C , C , . . . , C s on the alphabet (4) is called alist of C -words of the ( k, ℓ ) -semigroup Π, if C , C , . . . , C s is a list of B -words of Πand (4) The ( k, ℓ )-semigroup Γ, defined from the ( k, ℓ )-semigroup Π and the B -words C , C , . . . , C s , is a ( k, ℓ )-group. Lemma 9.
There exists an algorithm which for any ( k, ℓ ) -semigroup Π produces alist of C -words for this semigroup. Proof.
Suppose that ∆[ T ( A , A , . . . , A k )] = B , B , . . . , B p . By Lemma 6,we have that B , B , . . . , B p are B -words of Π. By Lemmas 5 and 8 all generators β , β , . . . , β p of the ( k, ℓ )-semigroup Γ, corresponding to the ( k, ℓ )-semigroup Πand the B -words B , B , . . . , B p , are invertible. Thus the ( k, ℓ )-semigroup Γ is a( k, ℓ )-group. (cid:3) Remark 1.
If the B -words are C -words, then we may speak of specifying from the( k, ℓ )-semigroup Π and its C -words the ( k, ℓ )-group Γ. Remark 2.
When there is no risk of misunderstanding, we will sometimes simplyspeak of a semigroup rather than a ( k, ℓ )-semigroup.Generating c i -words of the ( k, ℓ )-semigroup ΠThe generation of c i -words from a given ( k, ℓ )-semigroup Π, some C -words C , C , . . . , C s of this semigroup, and a specified C i from the list C , C , . . . , C s ,will be achieved by constructing a set of words, which we will call c i -words of the( k, ℓ )-semigroup Π. For this, we will assume that there exists an algorithm forsolving the identity problem in the ( k, ℓ )-group Γ associated to the semigroup Πand its C -words C , C , . . . , C s .We will define c i -words inductively for a given ( k, ℓ )-semigroup Π, given aspecified C -word C i .(1) Suppose that C i ≖ h C i C i · · · C i p h , where p ≥ h , h are non-empty; and suppose that U ≖ h C j C j · · · C j t h , with t ≥
0; and that[ U ∂ ≤ [ C i∂ and β i β i · · · β i p = β j β j · · · β j t in the group Γ. Then U is a c i -word of the ( k, ℓ )-semigroup Π. .1. CONSTRUCTION OF ELEMENTARY WORDS 15 (2) Suppose that c ( k ) i is some c i -word of the semigroup Π; suppose furtherthat c ( k ) i ≖ g C k C k · · · C k v g , where v ≥ g , g are non-empty; andsuppose that V ≖ g C m C m · · · C m w g with w ≥
0; and that [ V ∂ ≤ [ c ( k ) i ∂ and β k β k · · · β k v = β m β m · · · β m w in the group Γ. Then V is a c i -wordof the ( k, ℓ )-semigroup Π.Note that in the above construction that we may replace C e C e · · · C e m by theempty word, if β e β e · · · β e m = 1 in the group Γ; and that C i is a c i -word, as wemay replace the empty word inside C i by the empty word. Lemma 10.
There are only finitely many c i -words. Proof.
For any C -word C i the number of graphically distinct c i -words cannotexceed n [ C i∂ , where n is the number of generators of Π, which proves the lemma. (cid:3) Thus, if there is an algorithm for solving the identity problem in the ( k, ℓ )-group Γ, we may c -words, which we will denote by c (1)1 , c (2)1 , . . . , c ( v )1 ; and all c -words, which we denote by c (1)2 , c (2)2 , . . . , c ( v )2 ; . . . ; and all c s -words, which wewill denote by c (1) s , c (2) s , . . . , c ( v s ) s . Sometimes we will enumerate all these words in asingle list c , c , . . . , c t and refer to them as the c -words of the ( k, ℓ )-semigroup Π.Furthermore, the set of all c i -words c (1) i , c (2) i , . . . , c ( v i ) i will sometimes be denotedby { c ( j ) i } without specifying the set 1 , , . . . , v i through which j runs.Hence we have proved the following theorem. Theorem 1.
If the natural numbers k and ℓ are such that there exists an algo-rithm for solving the identity problem in every ( k, ℓ ) -group, then there exists analgorithm which for any ( k, ℓ ) -semigroup produces all C -words C , C , . . . , C s ofthis semigroup, and all sets { c ( j )1 } , { c ( j )2 } , . . . , { c ( j ) s } of this semigroup. For the following arguments we will focus on those sets { c ( j )1 } , { c ( j )2 } , . . . , { c ( j ) s } ,of some ( k, ℓ )-semigroup, which satisfy the following three properties.I. From the construction of c i -words, we will always have [ C i∂ ≥ [ c ( j ) i ∂ , forall i = 1 , , . . . , s and j = 1 , , . . . , v i ; for our purposes, we will need thestronger [ C i∂ = [ c ( j ) i ∂ for all i = 1 , , . . . , s and j = 1 , , . . . , v i .II. Generally speaking, we may have that some word f is an element of theset { c ( j ) i } as well as of the set { c ( j ) k } , where i = k , but for our purposesthis is not permitted.III. Some pairs of words c ( t ) p , c ( ρ ) q may overlap. For our purposes, we need thatno pair of c -words overlap.If a set of words { c ( j )1 } , { c ( j )2 } , . . . , { c ( j ) s } of some ( k, ℓ )-semigroup Π does notsatisfy property I, II, or III, then we will say that the set lacks I, II, or III, respec-tively. Remark 3.
We will make use of the definitions of affected and unaffected lettersintroduced by P. S. Novikov. These definitions can be found in the work by S. I.Adian [
Adi66 , pp. 7-8]. Furthermore, given a list of words X , X , . . . , X m suchthat each X j +1 is obtained from X j ( j = 1 , , . . . , m −
1) by a single step of theabove generating operation. Then this sequence can be represented as a sequenceof elementary transformations X → X → · · · → X m in the semigroup, in which k, ℓ )-SEMIGROUPS all substitutions allowed by the generating operation are given by the definingrelations. In this way, we naturally extend the notion of affected and unaffectedletters to a list of words X , X , . . . , X m . Definition 6.
A ( k, ℓ )- tuple is any pair, the first element of which is some ( k, ℓ )-semigroup T , and the second of which is a reduced list of C -words of this semigroup T . Such a tuple will often be denoted by { T ; C , C , . . . , C s } .If the conditions of Theorem 1 are satisfied, then for any given ( k, ℓ )-tuple { T ; C , C , . . . , C s } we can uniquely obtain the sets { c ( j )1 } , { c ( j )2 } , . . . , { c ( j ) s } , where s depends only on the tuple { T ; C , C , . . . , C s } , and unambiguously (from theunambiguity of the generating operation) obtain, from the ( k, ℓ )-semigroup T andits C -words C , C , . . . , C s , a ( k, ℓ )-group Γ. Thus in all the following reasoning wewill assume that the conditions of Theorem 1 are satisfied, and so we can speak ofthe sets { c ( j )1 } , { c ( j )2 } , . . . , { c ( j ) s } of the tuple { T ; C , C , . . . , C s } , and of the ( k, ℓ )-group Γ of the tuple { T ; C , C , . . . , C s } . Definition 7.
A ( k, ℓ )-tuple V will be said to be distinguished , if the sets { c ( j )1 } , { c ( j )2 } , . . . , { c ( j ) s } of this tuple does not lack a single one of the three properties I,II, or III. Lemma 11.
Let Γ be a ( k, ℓ ) -group with generating alphabet (9) γ , γ , . . . , γ s and defining relations (10) N i = 1 ( i = 1 , , . . . , k ) and suppose that in this ( k, ℓ ) -group there is some equality (11) γ p γ p · · · γ p v = γ q γ q · · · γ q w . Let F , F , . . . F s be a list of words over some alphabet A . Let F denote the semi-group with generating alphabet A and defining relations obtained from (10) by re-placing each letter γ i with F i for i = 1 , , . . . , s . Then in the semigroup F we havethe equality F p F p · · · F p v = F q F q · · · F q w . The lemma follows easily from the definition of equality in the semigroup.
Definition 8.
Let { Π; C , C , . . . , C s } be a ( k, ℓ )-tuple, and let Γ be the ( k, ℓ )-group associated to this tuple. Let f be some word in the set { c ( j ) m } associatedto this tuple. If we (like in Lemma 11) replace the generators γ , γ , . . . , γ s of Γby C , C , . . . C s , then we obviously obtain the semigroup Π. If we replace thegenerators γ , γ , . . . , γ m − , γ m , γ m +1 , . . . , γ s by the list C , C , . . . , C m − , f, C m +1 , . . . , C s then we obtain a ( k, ℓ )-semigroup which, as [ C m∂ ≥ [ f ∂ , we will call derived fromthe ( k, ℓ )-semigroup of the tuple { Π; C , C , . . . , C s } . Lemma 12.
Let { Π; C , C , . . . , C s } be a ( k, ℓ ) -tuple and let T be some semigroupderived from the ( k, ℓ ) -semigroup of this tuple. Then the system of relations of thesemigroup Π is equivalent to that of T . .1. CONSTRUCTION OF ELEMENTARY WORDS 17 Proof.
Let Π be a ( k, ℓ )-semigroup with generating set (4) and defining rela-tions A i = 1 ( i = 1 , , . . . , k ), and let T be a ( k, ℓ )-semigroup with generating set(4) and defining relations G i = 1 ( i = 1 , , . . . , k ). Because f is a word from { c ( j ) m } ,we have that f = C m in the semigroup Π.(a) We will show that G i = 1 ( i = 1 , , . . . , k ) in the semigroup Π. In fact, G i is obtained from A i by replacing C m by f , and as f = C m and A i = 1 inΠ, we have shown (a).(b) We will show that A i = 1 ( i = 1 , , . . . , k ) in the semigroup T . We dividethis into two separate cases.(i) Consider the p -th defining relation of the group Γ. Suppose it doesnot include γ m . Then A p ≖ G p , and hence A p = 1 in T .(ii) Consider the j -th defining relation of the group Γ. Suppose that γ m appears t times on the left side of this relation. Then A j ≖ h C m h C m · · · C m h t +1 G j ≖ h f h f · · · f h t +1 , where each h v ( v = 1 , , . . . , t + 1) is a product of some C -words. Letus repeat the process by which we obtained the word f from C m .The word C m could be written as C m ≖ m C i (1 , C i (1 , · · · C i (1 ,k ) n ≖ p , where m , n are non-empty words and 1 ≤ i (1 , , . . . , i (1 , k ) ≤ s .The word p ≖ m C j (1 , C j (1 , · · · C j (1 ,t ) n is obtained from C m by a single application of the generating opera-tion.The graphical equalities in (12) record the µ steps of the generatingoperation by which we obtain f from C m . p ≖ m C i (1 , C i (1 , · · · C i (1 ,k ) n ≖ C m p ≖ m C j (1 , C j (1 , · · · C j (1 ,t ) n ≖ m C i (2 , C i (2 , · · · C i (2 ,k ) n p ≖ m C j (2 , C j (2 , · · · C j (2 ,t ) n ≖ m C i (3 , C i (3 , · · · C i (3 ,k ) n (12) ... p µ ≖ m µ C j ( µ, C j ( µ, · · · C j ( µ,t µ ) n µ ≖ f In (12) all p k ( k = 1 , , . . . , µ ) are c m -words, so [ p k∂ ≤ [ C m∂ for all k = 1 , , . . . , µ , and because all m k and n k ( k = 1 , , . . . , µ ) are non-empty words, C m does not appear among the C -words written out in(12). By the definition of the generating operation, in obtaining theword f from C m we used the fact that in the group Γ the followingequalities hold. k, ℓ )-SEMIGROUPS γ i (1 , γ i (1 , · · · γ i (1 ,k ) = γ j (1 , γ j (1 , · · · γ j (1 ,t ) γ i (2 , γ i (2 , · · · γ i (2 ,k ) = γ j (2 , γ j (2 , · · · γ j (2 ,t ) ...(13) γ i ( µ, γ i ( µ, · · · γ i ( µ,k µ ) = γ j ( µ, γ j ( µ, · · · γ j ( µ,t µ ) As we do not have γ m among the γ appearing in the equalities in(13), it follows that in the semigroup T we have the equalities C i (1 , C i (1 , · · · C i (1 ,k ) = C j (1 , C j (1 , · · · C j (1 ,t ) C i (2 , C i (2 , · · · C i (2 ,k ) = C j (2 , C j (2 , · · · C j (2 ,t ) ...(14) C i ( µ, C i ( µ, · · · C i ( µ,k µ ) = C j ( µ, C j ( µ, · · · C j ( µ,t µ ) . In the semigroup T the equality G j = 1 is true, as G j = 1 is a definingrelation of T . Those words f which were substituted for γ m in theconstruction of G j (but not for all f belonging to G j ) we will, usingthe equalities in (14), which are performed inside the semigroup T ,perform the inverses of the substitutions which transform C m into f . As a result, the word G j will be transformed into A j . Hence, G j = A j in T , and hence A j = 1 in the semigroup T , completing theproof of Lemma 12. (cid:3) Definition 9.
Two tuples will be said to be equivalent if the system of relations ofthe semigroup associated to the first tuple is equivalent to the system of relationsof the semigroup associated to the second tuple.
Definition 10.
The index of the tuple R is an ordered pair of natural numbers( α, β ), where α is the sum of the lengths of the left-hand sides of the definingrelations of the tuple R ; and β is the parameter ω of the reduced list of C -words ofthe tuple R .The above indices will be ordered lexicographically, i.e. ( α , β ) < ( α , β ) ifand only if one of the following hold: either α < α , or α = α and β < β . Theindex of a tuple R will be denoted as J ( R ). Lemma 13.
If the tuple R lacks property I , then we can construct a tuple R ′ ,which is equivalent to R , and such that J ( R ′ ) < J ( R ) . Proof.
Let R = { Π; C , C , . . . , C s } . As R lacks property I, there exists somepositive natural number m ≤ s such that the length of the word c ( t ) m from the set { c ( j ) m } is strictly less than that of the length of C m .In the defining relations of the group of the tuple R , we will replace the genera-tors γ , γ , . . . , γ m − , γ m , γ m +1 , . . . , γ s by the words C , C , . . . , C, c ( t ) m , C m +1 , . . . , C s , .1. CONSTRUCTION OF ELEMENTARY WORDS 19 and call the resulting semigroup Π ′ . This semigroup is derived from the semigroupof R , and is given by defining relations over the same generating alphabet as Π.Construct some C -words C ′ , C ′ , . . . , C ′ s ′ of the semigroup Π ′ , and construct theassociated tuple R ′ = { Π ′ ; C ′ , C ′ , . . . , C ′ s ′ } . By Lemma 12 the tuples R and R ′ areequivalent; and as [ c ( t ) m ∂ < [ C m∂ , also α ( R ′ ) < α ( R ), and hence J ( R ′ ) < J ( R ). (cid:3) Lemma 14.
For every tuple R one may construct a tuple R , equivalent to thetuple R , and which does not lack property I. Proof.
This lemma is proved by induction on the index of the tuple R .Consider an arbitrary tuple R = { Π ; C , C , . . . , C s } , the index of whichequals ( α, α is some arbitrary positive number. The parameter ω of thelist C , C , . . . , C s equals 0, from which it follows, that every word C , C , . . . , C s is a single letter. But this means that every set { c ( j ) i } , for i = 1 , , . . . , s , of thetuple R consists of a single letter, from which it follows, that the tuple R doesnot lack property I.Suppose that for every tuple R ′ with index less than ( α , β ) we can constructsome tuple R ′′ , which is equivalent to R ′ and which does not lack property I.Consider a tuple R with index ( α , β ). If this does not lack property I, then wemay take this tuple itself as our desired tuple. If instead R lacks property I, thenby Lemma 13 we may construct a tuple R ′ , equivalent to R , and which satisfies J ( R ′ ) < ( α , β ). By the inductive hypothesis, from R ′ we may construct a tuple R ′′ , equivalent to R ′ (and hence to R ) and which does not lack property I. (cid:3) Lemma 15.
If the tuple R does not lack property I , but lacks property II , thenthere exists some tuple R ′′ , equivalent to R , with J ( R ′′ ) < J ( R ) . Proof.
Let R = { Π; C , C , . . . , C } . As R lacks property II, it follows thatthere exist natural numbers i and µ ( i = µ ) such that there is some word f in both { c ( j ) i } and { c ( j ) µ } , i.e. c ( m ) i = c ( n ) µ . As R does not lack property I, we have that[ C i∂ = [ c ( m ) i ∂ and [ C µ∂ = [ c ( n ) µ ∂ , and hence [ C i∂ = [ C µ∂ = g (where g >
1, as thelist C , C , . . . , C s is reduced, and C i ≖ C µ for i = µ ), so { c ( j ) i } and { c ( j ) µ } havelength g . From the definition of the generating operation it follows that the sets ofwords { c ( j ) i } and { c ( j ) µ } coincide, and hence there is some t such that C µ ≖ c ( t ) i .Define the semigroup Π to have the same generating alphabet as the semigroupΠ and subject to the defining relations of the group of the tuple R , in which thegenerators γ , γ , . . . , γ i − , γ i , γ i +1 , . . . , γ s have been replaced by the words C , C , . . . , C i − , c ( t ) i , C i +1 , . . . , C s . It is easy to verify (as c ( t ) i ≖ C µ ) that C , C , . . . , C i − , C i +1 , . . . , C s are C -wordsof the semigroup Π . Define R ′′ = { Π ; C , C , . . . , C i − , C i +1 , . . . , C s } . Fromthe equality [ c ( t ) i ∂ = [ C µ∂ it follows that α ( R ) < α ( R ′′ ). As [ C i∂ >
1, we have β ( R ′′ ) < β ( R ). This proves the lemma. (cid:3) Lemma 16.
If the tuple R = { Π; C , C , . . . , C s } does not lack property I but lacksproperty III , then there exists some C -word C p and some c -word c ( v ) µ which overlap. k, ℓ )-SEMIGROUPS Proof. As R lacks property III, it follows that there exists some c ( w ) p and c ( t ) µ ,such that c ( w ) p ≖ M N , and c ( t ) µ ≖ N L , where
M L and N are non-empty words.Suppose without loss of generality that M is non-empty (if M ≖
1, then L is non-empty, and all the reasoning below is symmetrical). Suppose that the generatingoperation obtains c ( w ) p from C p by the following sequence of c p -words: C p → c (1) p → c (2) p → · · · → c ( w − p → c ( w ) p , where each transition from c ( i ) p to c ( i +1) p is a single step of the generating operation.If w = 0, then c ( w ) p ≖ C p and the lemma follows.Suppose that w >
0, and that the lemma is true for all c ( k ) p , which can beobtained from C p using fewer than w steps.Consider the step c ( w − p → c ( w ) p , and write c ( w − p ≖ h C i · · · C i t h c ( w ) p ≖ h C j · · · C j u h where h , h are non-empty and γ i · · · γ i t = γ j · · · γ j u in the group Γ of R .If some C j t overlaps with c ( t ) µ , then the lemma follows. In the other case, C j · · · C j u lies entirely within M or entirely within N . If C j · · · C j u lies in M ,then doing the reverse substitution, we see that c ( w − p overlaps with c ( µ ) t , fromwhich by the inductive hypothesis the lemma follows. If C j · · · C j u lies in N , thendoing the reverse substitution, we see that c ( w − p overlaps with a word obtainedfrom c ( µ ) t by a single substitution, which is the reverse of a single substitution inthe generating operation. As R does not lack property I, all substitutions of theoperation preserve the lengths of the words to which they are applied, from whichit follows that by applying this substitution to c ( µ ) t we get some word c ( µ ) t . Thatis, c ( w − p overlap with c ( µ ) t , and by the inductive hypothesis we have shown thatsome C -word overlaps with some c -word. (cid:3) Lemma 17.
If the tuple R = { Π; C , C , . . . , C s } does not lack property I ; andthere exists some i such that C i ≖ M N , c ( p ) i ≖ N L with N and M L non-empty;and no C j ( j = 1 , , . . . , s ) does not overlap with any c ( v ) µ for j = µ ; then there issome c i -word c ( e ) i which overlaps with itself. Proof.
As the tuple R does not lack property I, we have [ C i∂ = [ c ( p ) i ∂ , fromwhich it follows that M and L are non-empty, and furthermore that all substitutionsmade by the generating operation does not change the length of the words involved.Consider a sequence of c i -words starting with the word C i , where c ( α m ) i isobtained from c ( α m − ) i by a single step of the generating operation, and such thatevery word from { c ( j ) i } is somewhere in this sequence.(15) C i → c ( α ) i → c ( α ) i → · · · → c ( α µ − ) i → c ( α µ ) i All words in the sequence (15) have the same length. We do not exclude thepossibility that some c i -word appears more than once in the sequence (15).We must have that [ C i∂ >
2, for otherwise the set { c ( j ) i } consists only of thesingle word C i , and the conditions of the lemma cannot be met. .1. CONSTRUCTION OF ELEMENTARY WORDS 21 Suppose that C i ≖ x E y , where x and y are single letters. Necessarily, x ≖ y , forotherwise C i overlaps with itself, contradicting the definition of a C -word. Theseletters x and y are not affected during the generation of the c i -words in the sequence(15), and hence c ( k ) i ≖ xe k y for k = α , α , α , . . . , α µ .The word c ( p ) i appears in the sequence (15) as some c ( α u ) i , and c ( α u ) i ≖ xe α u y .As C i and c ( α u ) i overlap, we have that M ≖ xM , N ≖ xN y , and L ≖ L y ,from which it follows that(16) C i ≖ xM xN y and c ( α u ) i ≖ xN yL y. We will prove that the letter x emphasised in the middle of C i in (16) isunaffected by the steps of the generating operation in (15).Suppose that x is not affected in the part of the sequence (15) given by C i → c ( α ) i → c ( α ) i → · · · → c ( α w ) i . Then c ( α w ) i ≖ xM w xN w y , and [ M w∂ = [ M ∂ . Suppose that the x under consider-ation is affected in the step c ( α w ) i → c ( α w +1 ) i . Then some C -word C h , the lengthof which is less than the length of the C -word C i , overlaps with the word xN w y .But xN w yL x is some c i -word c ( β ) i , and hence this word is obtained from c ( α u ) i by length-preserving substitutions, and hence C h overlaps with c ( β ) i where h = i ,which contradicts the assumptions of the lemma.Analogously, one may show that the y which appears in the middle of c ( α u ) i in(16) is unaffected by the steps in the operation in (15).Consider the word M in (16) and the word εM ε , where ε is any arbitraryletter; we will apply the generating operation as many times as possible to thisword. From this, we obtain the following list of words, all of the same length: εM ε, εM ε, . . . , εM f ε. We write N in the following form(17) N ≖ M δ xM δ xM δ x · · · M δ e xV and V ≖ M t xV , where 1 ≤ δ , δ , . . . , δ e , and t ≤ f, e ≥
0. This factorisation is unique, as all M δ i have the same length. Thus(18) C i ≖ xM xM δ xM δ x · · · M δ e − xM δ e xV yc ( α u ) i ≖ xM δ xM δ xM δ x · · · M δ e xV yL y We now show that all the x -letters marked inside C i in (18) are unaffected inthe steps of the sequence (15). We have already shown that the first two x -lettersare unaffected in (15). Suppose that the λ leftmost x -letters are unaffected in (15),and that the ( λ + 1)th is affected. By analogous reasoning to the above, this leadsto a contradiction.In the same way, we can show that the penultimate y distinguished in (18) is notaffected in (15). In the subword M δ e xV of the word C i lies y , which is not affectedby (15), and in the subword V yL of the word c α u i lies x , which is not affected by(15), and M δ e xV is transformed into V yL by the generating operation. If x wouldprecede y , then V ≖ M δ e xV , which is impossible by virtue of (17). Consequently, y precedes x . Hence, M δ e ≖ V yW and L ≖ W xV . k, ℓ )-SEMIGROUPS Hence we have C i ≖ xM xM δ xM δ x · · · xM δ e − xM δ e xV y,c ( α u ) i ≖ xM δ xM δ xM δ x · · · xM δ e xV yW xV y. But the generating operation allows the replacement of any M δ k by any M δ s in thewords C i and c ( α u ) i . By replacing inside all C i and c α u i all occurrences of M δ i by M δ e (and noting that M δ e ≖ V yW ), we obtain a c i -word c ( e ) i which overlaps withitself, proving the lemma. (cid:3) Lemma 18.
If the tuple R = { Π; C , C , . . . , C s } does not lack property I , but lacksproperty III , then there exists a tuple R ′′′ , equivalent to the tuple R , and such that J ( R ′′′ ) < J ( R ) . Proof.
By Lemma 16, there exists some C -word C p and some c -word c ( v ) µ which overlap. We break the proof of the lemma into two parts, with the first casebeing when µ = p , and the second case being when µ = p , but no C j and c ( v ) µ , j = µ , overlap.Case a ) µ = p .In the process described in Definition 8, we replace the generators of the groupΓ of the tuple R by the words C , C , . . . , C µ − , c ( v ) µ , C µ +1 , . . . , C s and obtain a( k, ℓ )-semigroup T with a system of relations equivalent to the system of relationsof the ( k, ℓ )-semigroup Π (by Lemma 12). Consider the words∆[ T ( C , C , . . . , C µ − , c ( v ) µ , C µ +1 , . . . , C s )] = C T , C T , . . . , C T s ( T ′ ) . It is easy to see that C T , C T , . . . , C T s ( T ′ ) are C -words of the semigroup T , andhence the ( k, ℓ )-semigroup associated to the tuple R ′′′ = { T ; C T , C T , . . . , C T s ( T ′ ) } is a ( k, ℓ )-group.Furthermore, α ( R ) = α ( R ′′′ ), as [ C µ∂ = [ c ( v ) µ ∂ . However, β ( R ′′′ ) < β ( R ), as c ( v ) µ and C p overlap, and hence by Lemma 1 we have ω ( C T , C T , . . . , C T s ( T ′ ) ) < ω ( C , C , . . . , C µ − , c ( v ) µ , C µ +1 , . . . , C s ) . Case b ) C p and c ( v ) p overlap, but none of the C j and c µ , for j = µ , overlap.In this case, by Lemma 17 there exists some c p -word c ( e ) p which overlaps withitself. In the process described in Definition 8, we replace the generators of thegroup Γ of the tuple R by the words C , C , . . . , C p − , c ( e ) p , C p +1 , . . . , C s and obtaina ( k, ℓ )-semigroup T , with a system of relations equivalent to the system of relationsof the ( k, ℓ )-semigroup Π. The argument then continues in the same way as in case a ) of the lemma. Here we use the fact that c ( e ) p overlaps with itself, and thereforethe parameter ω for the C -words of the semigroup T is therefore strictly less thanthe parameter ω for the semigroup Π. (cid:3) Theorem 2.
If the natural numbers k and ℓ are such that there exists an algorithmfor solving the identity problem in every ( k, ℓ ) -group, then there exists an algorithm .2. THE IDENTITY PROBLEM FOR ( k, ℓ )-SEMIGROUPS 23 which for every ( k, ℓ ) -tuple R constructs a distinguished ( k, ℓ ) -tuple V , equivalentto R . Proof.
The theorem is true, when the index of the ( k, ℓ )-tuple R has β = 0,since in this case every C -word of the tuple R is a single letter, and consequentlythe tuple R is distinguished.Assume that for all tuples with index J < J ′ the theorem is true. Suppose thatthe tuple R has index J ′ , and is lacking one of the properties (otherwise R is alreadydistinguished). By one of Lemmas 13, 15, or 18 we may find a tuple R with indexstrictly less than J ′ and equivalent to the tuple R . By the inductive hypothesis,we may find a distinguished tuple V , equivalent to the tuple R ′ , as J ( R ) < J ′ .To complete the proof, we note that the tuples V and R are equivalent. (cid:3) The identity problem for ( k, ℓ ) -semigroups We introduce some notation and assumptions which will be limited to thissection. Let the ( k, ℓ )-semigroup V be given by the generating alphabet(1) a , a , . . . , a n and the defining relations(19) { V i = 1 ( i = 1 , , . . . , k ) . Let the list C , C , . . . , C u be a list of C -words of the semigroup V . Let the ( k, ℓ )-group Γ, associated to the semigroup V and its C -words C , C , . . . , C u be givenby the generating alphabet(20) δ , δ , . . . , δ u and the defining relations(21) { Ψ i = 1 ( i = 1 , , . . . , k ) . Assume there exists an algorithm which solves the identity problem in the ( k, ℓ )-group Γ.Let the tuple R = { V ; C , C , . . . , C u } be distinguished, i.e. the set of c -words { c ( j )1 } , { c ( j )2 } , . . . , { c ( j ) u } does not lack any of the properties from the previoussections. Definition 11.
The word A over the alphabet (1) is called integral , if A ≖ c ( j ) i c ( j ) i · · · c ( j m ) i m , where c ( j ) i , c ( j ) i , . . . , c ( j m ) i m are c -words of the tuple R . The emptyword 1 is always an integral word. Definition 12.
On the set of integral words we define the function f , which mapsan integral word to a word from the group Γ. Let A ≖ c ( j ) i c ( j ) i · · · c ( j m ) i m . Then f ( A ) ⇋ δ i δ i · · · δ i m . This function is well-defined, as in the distinguished tupleof c -words no two words overlap, and because no word belongs to { c ( j ) i } and { c ( j ) p } for i = p .We notice that f ( AB ) ≖ f ( A ) f ( B ); and that if A ≖ c ( j ) i c ( j ) i · · · c ( j m ) i m and B = c ( p ) i c ( p ) i · · · c ( p m ) i m , then f ( A ) ≖ f ( B ). Definition 13.
Every elementary transformation of the ( k, ℓ )-semigroup V is ofone of the two forms(22) XY → XAY k, ℓ )-SEMIGROUPS (23)
XAY → XY where A is one of the V i of the defining relations (19), and X and Y are arbitrarywords over the alphabet (1). We say that a transformation of type (22) is an insertion , and that a transformation of type (23) is a deletion . Two integral words A and B are called equivalent with respect to the tuple R , if f ( A ) = f ( B ) in thegroup Γ of the tuple R . If S ≖ XAY and T ≖ XBY are words over the alphabet(1), with A and B equivalent, then we say that S and T are equivalent with respectto a replacement in the tuple R .It is clear that every elementary transformation is a replacement, and that everyreplacement can be carried out using some finite number of elementary transforma-tions. If [ A ∂ ≤ [ B ∂ , then the replacement XAY → XBY is called non-decreasing ,and the replacement
XBY → XAY is called non-increasing . It is sometimes con-venient to consider an insertion as a non-decreasing replacement, and a deletionas a non-increasing replacement. We extend the notion of affected and unaffected individual letters in a natural way (see Remark 3) to sequences of elementary trans-formations and replacements.
Definition 14. If A is a word over the alphabet (1), then by T ( A ) we denote theset of all such words B over the alphabet (1), such that there exists a finite sequenceof words X , X , . . . , X α , where X ≖ A, X α ≖ B , and X i +1 is obtained from X i by a non-increasing replacement for all i = 1 , , . . . , α − B ∂ ≤ [ A ∂ for every word in the set T ( A ), there exists an algorithm whichtakes as input a word A over the alphabet (1), and produces the set of words T ( A ).The word A is final if for every B ∈ T ( A ) we have [ B ∂ = [ A ∂ . It is clear that anysubword of a final word is final, and that every c -word of a distinguished tuple R is final (as R does not lack property I and III). Definition 15.
Let A be a final word over the alphabet (1), and A ≖ y y · · · y s c i (1 , c i (1 , · · · c i (1 ,m ) y s y s +2 · · · y s c i (2 , c i (2 , · · ·· · · c i (2 ,m ) y s +1 y s +2 · · · y s · · · c i ( v − , c i ( v − , · · · (24) · · · c i ( v − ,m v − ) y s v − +1 y s v − +2 · · · y s v where every y i ( i = 1 , , . . . , s v ) is a letter; and every c -word c i ( t,µ ) appearing in(24) has the following property:Denote the prefix of the word A appearing before the word c i ( t,µ ) by X , andthe suffix of the word A appearing after the word c i ( t,µ ) by Y , i.e. A ≖ X c i ( t,µ ) Y .Then there does not exist a c -word c p such that A ≖ X c p Y , where X Y is non-empty, and c p ≖ X c i ( t,µ Y , where X Y is non-empty, and A ≖ X X c i ( t,µ ) Y Y ,where X ≖ X X , Y ≖ Y Y . Furthermore, no subword y s i +1 y s i +2 · · · y s i +1 of A is a c -word.We then say that the right-hand side of the equality in (24) is the representation of the final word A .There is an obvious algorithm which for any final word A computes its repre-sentation. It is also easy to see that the representation of a final word is unique.Consider the word Y ≖ y y · · · y s v , which we will call the stable word of the finalword A , and every letter y m of the word Y will be called the m th stable letter ofthe final word A . The letters y m and y m +1 are called adjacent stable letters . It is .2. THE IDENTITY PROBLEM FOR ( k, ℓ )-SEMIGROUPS 25 clear that between two adjacent stable letters y m and y m +1 of a final word, thereis an integral word, which will be called the m th main integral word of the word A . Definition 16.
Suppose we are given a final word T and a finite sequence of anon-decreasing replacements(25) T → T → T → · · · → T µ − → T µ ( µ ≥ . The word T µ will be called a T -word , if a final word T and a finite sequence ofnon-decreasing replacements indicating how it was obtained from T is specified forit. The rank of a T -word is the number of non-decreasing replacements by whichit was derived from the final word.Let T ≖ y y · · · y s c i (1 , c i (1 , · · · c i (1 ,m ) y s y s +2 · · · y s c i (2 , c i (2 , · · ·· · · c i (2 ,m ) y s +1 y s +2 · · · y s · · · c i ( v − , c i ( v − , · · · (26) · · · c i ( v − ,m v − ) y s v − +1 y s v − +2 · · · y s v be the representation of the word T , and suppose the µ non-decreasing replace-ments of (25) are applied to T .Suppose that the stable letters of the final word T are all unaffected by theapplication of the replacements in (25). Then(27) T µ ≖ M y M y M · · · M s v y s v M s v +1 , and the letters y , y , . . . , y s v are called the stable letters of the T -word T µ ; theletters y m and y m +1 are called adjacent stable letters of the T -word T µ . The right-hand side of the equality (27) is called the representation of the T -word T µ , andthe word M i is called the i th changing word of the T -word T µ . Notice that every M i is obtained from some (possibly empty) integral word by a finite number ofnon-decreasing replacements. Definition 17.
To every T -word T i with rank i we assign the index λ = ( γ, δ ),where γ = i , and δ = [ T i∂ . We order the indices λ lexicographically (see Defini-tion 10). Definition 18.
In the distinguished tuple R every c -word is a final word. If inDefinition 16 the final word T is a c -word; if the stable letters of this stable c -wordare unaffected by the sequence (25); and if in the representation (27) of the T -word T µ the words M and M s v +1 are empty; then the word T µ is called a d -word ofrank µ .The following lemma is proved by induction on the index λ . Lemma 19.
For any λ the following three statements are true: I. The stable letters of the final word T are not affected by the sequence (25) of replacements, if ( µ, [ T µ∂ ) ≤ λ . II.
Every changing word M j of the T -word T µ , with ( µ, [ T µ∂ ) ≤ λ , is graph-ically equal to some product of d -words M j ≖ d j d j · · · d j tj , where theindex of every d j i ( i = 1 , , . . . , t j ) is strictly less than λ . III.
If the final word T does not overlap with any c -word, then T µ does notoverlap with any c -word, when ( µ, [ T µ∂ ) ≤ λ . k, ℓ )-SEMIGROUPS Proof.
If the index λ equals (0 , λ ′ = ( µ, [ T µ∂ ), and suppose that for all λ < λ ′ the lemma is true. We willprove the lemma for λ = λ ′ .Part I. The sequence (25) begins with(28) T → T → · · · → T µ − By the inductive hypothesis of I, the stable letters of the final word T are notaffected by the sequence (25) of replacements, as ( µ − , [ T µ − ∂ ) < λ ′ . By theinductive hypothesis of II, we can write T µ − ≖ d µ − , d µ − , · · · d µ − ,m ) y d µ − , d µ − , · · ·· · · d µ − ,m ) y · · · · · · y s v d µ − s v +1 , d µ − s v +1 , · · · d µ − s v +1 ,m sv +1 ) (29)where y y · · · y s v is the stable word of the final word T ; and every d µ − i,j ) ap-pearing is a d -word, and furthermore the index of every d µ − i,j ) is less than λ ′ .By the inductive hypothesis of III, none of the d µ − i,j ) appearing in (29)overlaps with any c -word.Consider the non-decreasing replacement(30) T µ − → T µ .
1) Suppose that (30) is an insertion. In this case no letter of the word T µ − is affected, and thus property I holds.2) Suppose that (30) is a deletion. That is, T µ − ≖ XAY and T µ ≖ XBY ,where A and B are non-empty, integral, and equivalent words. The word A cannot be contained in any subword of T µ − consisting only of stableletters. This would contradict the definition of the representation of thefinal word T . The word A does not overlap with the d -words appear-ing in (29). Thus none of the stable letters of T µ − are affected by thereplacement (30).Part II. Consider the non-decreasing replacement(30) T µ − → T µ .
1) Suppose that (30) is an insertion XY → XV i Y , and suppose that noneof the d -words appearing in (29) is divided when dividing T µ − into thewords X and Y . Then the representation of the word T µ − will differfrom the representation of T µ only in that some M j ≖ d j d j · · · d j tj willbe changed into M ′ j ≖ d j d j · · · d j s c i c i · · · c i m d j s +1 · · · d j tj . That is, the resulting word will be graphically equal to a product of d -words, and moreover (as c -words are d -words of rank zero) the index ofevery d -word of M ′ j is strictly less than ( µ − , [ T µ − ∂ ), which in turn isless than λ .2) Suppose that (30) is an insertion XY → XV i Y , and that at least one ofthe d -words d j m appearing in (29) is divided when dividing T µ − into thewords X and Y . Then the representation of the word T µ − differs from .2. THE IDENTITY PROBLEM FOR ( k, ℓ )-SEMIGROUPS 27 the representation of T µ only in that d j m , with index λ < ( µ − , [ T µ − ∂ ),will be replaced by some d -word with index λ + 1 < ( µ, [ T µ∂ ).3) Suppose that (30) is not an insertion, i.e. T µ − ≖ XAY and T µ ≖ XBY ,where A and B are non-empty, integral, and equivalent words, and supposethat A is included in some d µ − i,j appearing in (29), and that it is notequal to this d µ − i,j ) . Then this d µ − i,j ) , with index λ < ( µ − , [ T µ − ∂ )is replaced in the representation of T µ by some d -word with index λ +1 < ( µ, [ T µ∂ ).4) Suppose that (30) is a non-decreasing replacement which is not an inser-tion, and let A ≖ d µ − i,j ) · · · d µ − i,j + t ) . But A is an integral word, and d µ − i,j ) · · · d µ − i,j + t ) does not, by the inductive hypothesis of III, over-lap with any c -word. Therefore, each d µ − i,j ) , . . . , d µ − i,j + t ) is a c -word,and hence the replacement (30) replaces one changing word, graphicallyequal to a product of d -words of zero rank, by another changing word withthe same property.Part III. We prove this part by contradiction. Suppose that the word T µ from thesequence (25) overlaps with some c -word c s , i.e. T µ ≖ M N and c s ≖ N L , where N and M L are non-empty words.By parts I and II we can write the T -word T µ as its representation (27). As c s is non-empty, we have that c s ≖ eK , where e is some letter.1) Suppose that e ≖ y i , where y i is some stable letter. Then every d -wordappearing in (27) to the right of e is a c -word, as the distinguished tuple R does not lack property I. From here, the end of the word T µ which startswith the e in question, coincides with the end of the word T which startswith the same e , which contradicts III.2) Suppose that e belongs to some M j in the representation (27). Now M j ≖ d j d j · · · d j tj and so e appears in some d j m . But as the index of d j m is strictly less than the index of T µ by II, it follows by the inductivehypothesis of III that d j m does not overlap with any c -word, a contradic-tion.This completes the proof of Lemma 19. (cid:3) Definition 19.
We place every T -word T µ , obtained from the final word T by µ specified non-decreasing replacements, in correspondence with a word O µ with thefollowing properties:Let the representation of T be as in (26). Suppose we are given the finitesequence of non-decreasing replacements (25). Let the representation of the T -word T µ be as in (27) (by Lemma 19.I, we can speak of the representation of the T -word T µ without the assumptions in Definition 16). By Lemma 19.II, the s thchanging word of the T -word T µ is(31) M s ≖ d j ( s, d j ( s, · · · d j ( s,k s ) , where every d j ( s,t ) is a d -word; and for every d j ( s,t ) there is a c -word c j ( s,t ) , fromwhich it is obtained by the indicated non-decreasing replacements, and the indexof every d j ( s,t ) is strictly less than the index of T µ . k, ℓ )-SEMIGROUPS Then U µ ≖ c j (1 , c j (1 , · · · c j (1 ,k ) y c j (2 , c j (2 , · · · c j (2 ,k ) y · · · · · ·· · · y s v c j ( s v +1 , c j ( s v +1 , · · · c j ( s v +1 ,k sv +1 ) (32)and moreover1) Every c j ( s,t ) (for s = 1 , , . . . , s v + 1 and t = 1 , . . . , k s ) which appears in(32) is precisely the c -word from which one obtains the d -word d j ( s,t ) inthe representation (27) , (31) of the T -word T µ .2) Every integral word(33) c j ( m, c j ( m, · · · c j ( m,k m ) in the word O µ can be obtained by a single non-decreasing replacementfrom the whole word appearing between the stable letters y m − and y m in the final word T . Lemma 20.
Suppose we are given a finite sequence of non-decreasing replacements (34) T → T → T → · · · → T µ → T µ +1 where T is a final word. Then for the T -word T µ +1 we can construct the corre-sponding O µ +1 . Proof.
For T , the corresponding word O is just T itself.Suppose that for the T -word T µ we already know the corresponding word O µ .We represent T µ of the form given in (27) , (31), and the the corresponding O µ ofthe form (32).Consider the the non-decreasing replacement(35) T µ → T µ +1 .
1) Suppose that (35) is an insertion. If the insertion is inside some d -word d j ( α,β ) appearing in (27) , (31), then we set O µ ≖ O µ +1 . Indeed, fromthe properties of O µ , it is easy to see that O µ will be the word O µ +1 corresponding to T µ +1 . If instead the insertion is between two d -words, orbetween two stable letters, or between a stable letter and a d -word, then O µ +1 is obtained from O µ by an insertion between the same words (wherewe consider c i ( α,β ) and d i ( α,β ) as the same for these purposes). It is easyto verify from the definition of O µ that this choice of O µ +1 is correct.2) Suppose that (35) is a non-decreasing replacement, but not an insertion.By Lemma 19, it follows that the replaced word either completely liesinside some d -word, or else is equal to a product of d -words of rank zero(i.e. d -words which are c -words). In the first case, we set O µ ≖ O µ +1 . Inthe second case, the replaced d -words are c -words, and O µ +1 is obtainedfrom O µ by the same replacement as the one by which T µ +1 is obtainedfrom T µ . Furthermore, if in T µ +1 the replaced word is d j ( α,β ) · · · d j ( α,δ ) ,then in O µ +1 the corresponding replaced word is c j ( α,β ) · · · c j ( α,δ ) .This completes the proof of Lemma 20. (cid:3) Remark 4.
By saying that we are given a T -word T µ , we are really saying thatwe are given a final word T and a sequence of µ non-decreasing replacements. Ifwe wish to prove that some word X is a T -word T µ , then we must prove that thereexists some final word T , to which the application of µ non-decreasing replacementsproduces the word X . .2. THE IDENTITY PROBLEM FOR ( k, ℓ )-SEMIGROUPS 29 Lemma 21.
If one performs on a T -word T µ , obtained from a final word T , adeletion (36) T µ ≖ XV S Y → XY, where V s is one of the left-hand sides of the defining relations in (19) , then XY isa T -word, obtainable from the same T . Proof.
Suppose the lemma is true for all T -words whose length is less than λ , and let [ T µ∂ = λ . From the definition of (19) and Lemma 20 it follows that T µ can be obtained from T in the following way. From the final word T with thepresentation (26) we construct via non-decreasing replacements the word O µ withthe presentation (32), and then via non-decreasing replacements the word T µ withthe presentation (27) , (31); furthermore, every d j ( α,β ) in (27) , (31) is replaced bythe corresponding c j ( α,β ) in the non-decreasing replacements of (32).When reducing (36), there are two cases.1) d j ( α,β ) ≖ X V s Y . But d j ( α,β ) is a T -word, obtained from c j ( α,β ) by non-decreasing replacements, and [ d j ( α,β ) ∂ < [ T µ∂ (if [ d j ( α,β ) ∂ = [ T µ∂ , then XY ≖
1, and 1 is a T -word). By the inductive hypothesis, X Y canbe obtained from c j ( α,β ) by non-decreasing replacements, from which it iseasy to see that X Y is a d -word.2) d j ( α ,β ) d j ( α ,β ) · · · d j ( α s ,β s ) ≖ V S . As no d -word overlaps with any c -word, and V s is an integral word, it follows that any d j ( α t ,β t ) ( t = 1 , , . . . , s )is a c -word. From here, it follows that when T µ is obtained from O µ ,that then in the c -word c j ( α ,β ) c j ( α ,β ) · · · c j ( α s ,β s ) no replacements weremade. Suppose that the word X ≖ c j ( α ,β ) c j ( α ,β ) · · · c j ( α s ,β s ) is enclosed in O µ in the word y w c j ( γ ,δ ) · · · c j ( γ n ,δ n ) Xc j ( γ n +1 ,δ n +1 ) · · · c j ( γ m ,δ m ) y w +1 . The word between y w and y w +1 is derived from some integral final word Y . It is clear that c j ( γ ,δ ) · · · c j ( γ n ,δ n ) c j ( γ n +1 ,δ n +1 ) · · · c j ( γ m ,δ m ) can be obtained from Y by a non-decreasing replacement.This completes the proof of the lemma. (cid:3) Lemma 22.
If some finite number s of deletions are carried out on the T -word T µ , then the resulting word is some T -word obtained from the same final word T as T µ is obtained from. The proof is an immediate consequence of Lemma 21.
Lemma 23. If X and Y are final words, and X is obtained from Y by a sequenceof elementary transformations, then X ∈ T ( Y ) and Y ∈ T ( X ) . Proof.
Let(37) X ≖ Z → Z → · · · → Z n ≖ Y k, ℓ )-SEMIGROUPS be the sequence of elementary transformations in the statement of the lemma. As X is final, we must have that Z → Z is an insertion. Consider the sequence ofelementary transformations Z → Z → · · · → Z k ′ consisting only of insertions, such that Z k ′ → Z k ′ +1 → · · · → Z k ′′ consists only of deletions, such that Z k ′′ → Z k ′′ +1 → · · · → Z k ′′′ consists only of insertions, etc. By Lemma 22, the sequence Z → Z → · · · → Z k ′′ can be reproduced by a sequence of non-decreasing substitutions. After a finitenumber of steps, repeating this reasoning, we find that X can be obtained from Y by a sequence of non-decreasing replacements; from this, we have that X ∈ T ( Y ).Reasoning symmetrically, we also have that Y ∈ T ( X ). (cid:3) Lemma 24.
Two words X and Y are equal in the ( k, ℓ ) -semigroup V if and onlyif the sets T ( X ) and T ( Y ) have a common element. Proof.
Suppose Z ∈ T ( X ) and Z ∈ T ( Y ). Then Z = X and Z = Y in thesemigroup V , from which we have X = Y in the semigroup V .Now suppose that X = Y in the semigroup V . We find some final words X and Y such that X ∈ T ( X ) and Y ∈ T ( Y ). This is always possible, as every word ofminimal length in T ( X ) is a final word. This means that X = Y in V , from whichit follows that X is obtained from Y by a sequence of elementary transformations.By Lemma 23, X ∈ T ( Y ) and as Y ∈ T ( Y ), we have X ∈ T ( Y ). Thus, we havefound a word X such that X ∈ T ( X ) and X ∈ T ( Y ). (cid:3) Theorem 3.
If the natural numbers k and ℓ are such that there exists an algorithmfor solving the identity problem in every ( k, ℓ ) -group, then there exists an algorithmfor solving the identity problem in the semigroup of any distinguished ( k, ℓ ) -tuple. Proof.
If we are given two words X and Y over the alphabet (1), then we canfind the sets of words T ( X ) and T ( Y ), which are finite, as for any word A over thealphabet (1) the set of words T ( A ) consists of at most n [ A ∂ elements, where n isthe number of letters in the alphabet (1). We then check whether T ( X ) and T ( Y )have any element in common or not. By Lemma 24, the answer to this question isthe same as whether X and Y are equal in the semigroup V of the tuple R . (cid:3) Theorem 4.
If the natural numbers k and ℓ are such that there exists an algorithmfor solving the identity problem in every ( k, ℓ ) -group, then there exists an algorithmfor solving the identity problem in any ( k, ℓ ) -semigroup. Proof.
By Theorem 1, for every ( k, ℓ )-semigroup Π we can find a ( k, ℓ )-tuple K such that K = { Π; C k , C k , . . . , C ks } . By Theorem 2 for every ( k, ℓ )-tuple we canfind a reduced tuple R = { V ; C , C , . . . , C u } such that the ( k, ℓ )-semigroups Π and V are equivalent. Thus to prove Theorem 4 we must only show that there existsan algorithm for solving the identity problem in the semigroup of any arbitrarydistinguished ( k, ℓ )-tuple. But this is precisely what is proved in Theorem 3. (cid:3) .3. THE DIVISIBILITY PROBLEMS FOR ( k, ℓ )-SEMIGROUPS 31 The divisibility problems for ( k, ℓ ) -semigroups We retain the same notation and assumptions as in Section 1.2.
Lemma 25.
Suppose we are given a final word T which does not have any end(beginning) which overlaps with any c -word of the tuple R . Suppose we are giventhe finite sequence of non-decreasing replacements (25) T → T → · · · → T µ − → T µ ( µ ≥ Then the T -word T µ does not have any end (beginning) overlapping with any c -wordof the tuple R . Proof.
Suppose T , given by the representation (26), does not have any endoverlapping with any c -word, and T µ , given by the representation (27), overlapswith some c -word c ( j ) i . The word c ( j ) i is a non-empty word, so c ( j ) i ≖ aX . Theletter a is a stable letter of the T -word T µ , as no c -word overlaps with any d -wordby Lemma 19.III. The changing words of T µ , located to the right of the specified a , are final words (as c ( j ) i is final), and are therefore integral words. Accordingly,the end of the word T µ located to the right of the specified a is obtained from theend of T to the right of the same a (as a is a stable letter of the sequence (25)) bythe same number of replacements. Carrying out these replacements in the reverseorder, we find that T has an end which overlaps with some c -word, obtained from c ( j ) i by these replacements. This is a contradiction. We can carry out the analogousproof for the beginnings of the T -word T µ . (cid:3) Lemma 26.
If the non-empty final word T has no end which overlaps with some c -word of the tuple R , and the word X is left divisible by T in the ( k, ℓ ) -semigroup V , then X ≖ X X , where X = T in the ( k, ℓ ) -semigroup V . Proof.
By definition of left divisibility, there exists some word Z over thealphabet (1), such that in the semigroup V we have the equality T Z = X . Inother words, there exists some sequence of elementary transformations(38) T Z → · · · → X. As T is non-empty, we have that T ≖ Y a , where a is some letter. We will consider(38 ′ ) Y aZ → · · · → X and prove, that the specified letter a in the word Y aZ is unaffected by the sequenceof elementary transformations (38 ′ ).Suppose that the letter a becomes affected by the w th transformation(38 ′′ ) Y aZ ≖ Y aZ → Y aZ → · · · → Y w − aZ w − → X w → · · · X. As in the sequence (38 ′′ ) the transformations are all elementary, the w th transfor-mation, which affects a , is a deletion. The word Y is a final word, as it is a subwordof the final word T . The word Y w − can be obtained from Y by a non-decreasingreplacement (by Lemma 23). By Lemma 25, and as Y a has no end which over-laps with some c -word, it follows that Y w − a has no end which overlaps with some c -word. This is a contradiction, and hence a is not affected by the sequence (38 ′′ ),i.e. Y aZ → · · · Y w − aZ w − → Y w aZ w → · · · → Y s aZ s ≖ X where T ≖ Y a . Thus we can take X to be Y s a , in which case T = X in thesemigroup V . (cid:3) k, ℓ )-SEMIGROUPS Lemma 27. If Y is the beginning of some c -word of the semigroup V ( Y canitself be a c -word), and if the word X is left divisible by Y , then X is left divisibleby Y Y . Proof.
Let Y Z ≖ c ( i ) j , and let f ( c ( i ) j ) = δ j . For the letter δ j in the group Γ ofthe tuple R , there exists an inverse δ j δ j · · · δ j m , i.e. such that δ j δ j δ j · · · δ j m = 1in the group Γ. From this it follows that in the semigroup V we have the equality c ( i ) j C j C j · · · C j m = 1.We have Y Z C j C j · · · C j m = 1 in the semigroup V , and also X = Y Z in thesemigroup V , as X is left divisible by Y . Thus X = Y Z = Y Y Z C j C j · · · C j m Z that is, X is left divisible by Y Y . (cid:3) Theorem 5.
If the natural numbers k and ℓ are such that there exists an algorithmfor solving the identity problem in every ( k, ℓ ) -group, then there exists an algorithmwhich solves the left (right) divisibility problem in the semigroup of any distinguished ( k, ℓ ) -tuple. Proof.
Suppose X and Y are words over the alphabet (1), and we wish todecide whether or not X is left divisible by Y in the semigroup V of the distinguishedtuple R . Construct the set of words T ( Y ). Choose from the set T ( Y ) some finalword Y . We have Y = Y in the semigroup V , and hence our problem is equivalentto deciding whether X is left divisible by Y .By direct verification, we check whether some end of Y overlaps with some c -word of the tuple R . If there is some such overlap, then Y ≖ Y H , where H isa beginning of some c -word. We next check whether some end of Y overlaps withsome c -word of the tuple R . If there is some such overlap, then Y ≖ Y H , where H is the beginning of some c -word. This process continues until it ends with theconstruction of a (possibly empty) word Y k , such that it has no end which overlapswith any c -word, and the words H , H , . . . , H k − , where k ≥
0, and every H i isthe beginning of some c -word. We have Y k H k − H k − · · · H H ≖ Y . Here Y k is afinal word, as it is a subword of the final word Y .If X is left divisible by Y , then X is left divisible by Y k . This is obvious fromthe equality X ≖ Y Z ≖ Y k H k − H k − · · · H H Z . On the other hand, if X is left divisible by Y k , then by Lemma 27 X is left divisible Y k H k − H k − · · · H H ≖ Y .Thus it remains to check whether X is left divisible by Y k . To do this, we candirectly check (as the word problem is decidable in the semigroup V by Theorem 3)whether or not Y k is equal to some beginning of X . If Y k = X in the semigroup V , then X ≖ X X , then by Lemma 25 we have that X is left divisible by Y k . If X is left divisible by Y k , then X ≖ X X , where X = Y k in the semigroup V . (cid:3) Theorem 6.
If the natural numbers k and ℓ are such that there exists an algorithmfor solving the identity problem in every ( k, ℓ ) -group, then there is also an algorithmfor solving the left (right) divisibility problem in any ( k, ℓ ) -semigroup. Proof.
By Theorem 1, for any ( k, ℓ )-semigroup Π we can construct a ( k, ℓ )-tuple K such that K = { Π; C k , C k , . . . , C ks } . By Theorem 2, for every ( k, ℓ )-tuple K we can construct some distinguished tuple R = { V ; C , C , . . . , C u } such that the .4. THE MAXIMAL SUBGROUP OF A ( k, ℓ )-SEMIGROUP 33 ( k, ℓ )-semigroups Π and V are equivalent. Thus, to prove Theorem 6 we only needto prove, that there exists an algorithm which solves the left divisibility problem inthe semigroup of any distinguished ( k, ℓ )-tuple. But this is precisely what is provedin Theorem 5. (cid:3) The maximal subgroup of a ( k, ℓ ) -semigroup We retain the same notation and assumptions as in Section 1.2.
Lemma 28.
No two d -words of a tuple R overlap. Proof.
The proof is by induction on the sum of the lengths of the two consid-ered d -words. If the sum of the two d -words is 2, then these words cannot overlapby Definition 1.Suppose that for any pair of d -words for which the sum of the lengths is lessthan λ we have that they do not overlap. Suppose d α ≖ M N , d β ≖ N L , where N and M L are non-empty words, and with [ d α∂ + [ d β∂ = λ .Consider the c -word c β , such that(39) c β ≖ X → X → · · · → X m ≖ d β (see Remark 3) and the representation (26) of the final word c β . Let N be a non-empty word such that N ≖ Y a and d β ≖ Y aL . The letter a highlighted in the T -word d β is stable, for otherwise d α intersects with the d -word d γ , which is achanging word of the T -word d β , but [ d α∂ + [ d γ∂ < λ .Every changing word of the T -word d β is changed into an integral, final wordby non-increasing substitutions, by performing the sequence of elementary trans-formations (39) in reverse. We obtain that c β has a beginning which intersects withsome word, obtained from d α by non-increasing replacements, i.e. some word d δ .The proof that d δ is a d -word follows from the statement that any replacementcan be reproduced by a finite number of elementary transformations, that c α is afinal word, and Lemma 23. (cid:3) Lemma 29.
If the word X is two-sided invertible in the semigroup V , then thereexists some integral word Z such that X = Z in the semigroup V . Proof. As X is two-sided invertible in the semigroup V , there exists someword Y such that XY = 1 and Y X = 1 in V .Consider the sequence of elementary transformations(40) 1 ≖ U → U → · · · → U e − → U e ≖ XY The word 1 is an integral word. By Lemma 23, we know that 1 can be convertedto XY by a sequence of non-decreasing replacements, i.e. U e is a T -word. Con-struct the word O e of the T -word T e , and consider the sequence of non-decreasingreplacements (Lemma 20)(41) 1 → O e ≖ c i c i · · · c i s → U ′ → U ′ → · · · → U ′ e ≖ XY, where, starting from the second replacement, every replacement happens insidesome c -word c i j . In this way, XY is obtained from a d -word. In exactly the sameway, we can prove that Y X can be obtained from a d -word. But as (by Lemma 28) d -words do not intersect, it follows that X can be obtained from a d -word, and isobtained by non-decreasing replacements from the integral word c i c i · · · c i t , where t ≤ s . It is now clear that we can take Z to be c i c i · · · c i t . (cid:3) k, ℓ )-SEMIGROUPS Lemma 30.
The group Γ of the ( k, ℓ ) -tuple R is isomorphic to the maximal sub-group F of the ( k, ℓ ) -semigroup V of the ( k, ℓ ) -tuple R . Proof.
For every two-sided invertible word X of the semigroup V , we willdefine µ ( X ) as follows. By Lemma 29, for every two-sided invertible word X thereexists some integral word Z such that Z = X in V . If we choose for every X α a Z α ,then we set µ ( X α ) ⇋ f ( Z α ), see Definition 12, and, if X α is itself integral, then µ ( X α ) ⇋ f ( X α ).Thus, we have an algorithm, which to every two-sided invertible word X of thesemigroup V associates a word f ( Z α ) of the group Γ of the tuple R . Obviously,every word of the group Γ can be obtained in this way from some word X . We willprove, that this mapping is an isomorphism between the group Γ and the group F .For this, it is sufficient to prove the following lemma. Lemma 31. If X and Y are two-sided invertible words of the semigroup V , then X = Y in V if and only if f ( µ ( X )) = f ( µ ( Y )) in the group Γ . Proof.
By Lemma 29 µ ( X ) = µ ( Y ) in V is equivalent to X = Y in V .If f ( µ ( X )) = f ( µ ( Y )) in Γ, then µ ( X ) = µ ( Y ) in V by the definition of thegroup Γ of the tuple R .Suppose µ ( X ) = µ ( Y ) in V . Without loss of generality, we can assume µ ( X )is final.By Lemmas 20 and 23 there exists some sequence of non-decreasing replace-ments(42) µ ( X ) ≖ Z → Z → · · · → Z m − → Z m ≖ µ ( Y ) , and moreover Z is an integral word, where Z and Z are equivalent words; Z isan O -word of the T -word µ ( Y ); and every replacement in the sequence(43) Z → Z → · · · → Z m − → Z m where Z ≖ c i c i · · · c i n , replaces some c -word c i t , or some part of c i t , by the word d i t . In this way, µ ( Y ) is obtained from a d -word. But by definition, µ ( Y ) is anintegral word, and hence f ( Z ) = f ( µ ( Y )). As Z and Z are equivalent words, wealso have f ( µ ( X )) = f ( µ ( Y )) in Γ. This completes the proof of Lemma 31. (cid:3) Using Lemma 31, this completes the proof of Lemma 30. (cid:3)
Theorem 7.
If the natural numbers k and ℓ are such that there exists an algo-rithm for solving the identity problem in every ( k, ℓ ) -group, then there is an algo-rithm which for any ( k, ℓ ) -semigroup Π constructs a ( k, ℓ ) -group, isomorphic to themaximal subgroup of Π . Proof.
By Theorem 1, for every ( k, ℓ )-semigroup Π we can construct a ( k, ℓ )-tuple K , such that K = { Π; C k , C k , . . . , C ks } . By Theorem 2, for every ( k, ℓ )-tuple K we can construct a distinguished tuple R = { V ; C , C , . . . , C s } such that the( k, ℓ )-semigroup Π and V are equivalent. Thus, to prove Theorem 7, it suffices toprove that there exists an algorithm which for any semigroup of a distinguished( k, ℓ )-tuple constructs some ( k, ℓ )-group, which is isomorphic to the maximal sub-group of the semigroup of this tuple. But this is precisely what is proved inLemma 30. (cid:3) HAPTER 2
The Identity Problem In Finitely PresentedGroups
Representing words equal to the identity in K / -groups Let G be a group with generating set(1) a , a , . . . , a n and defining relations(2) R i = 1 ( i = 1 , , . . . , m ) . Let M be the set of defining words of G and, for the remainder of this chapter,suppose that M is symmetrised. The class K / . The group G , given by the generators (1) and the definingrelations (2), belongs to the class K / if the set M of the group G satisfies thefollowing condition: if R i and R j are arbitrary, not mutually inverse words in M ,then when reducing the product R i R j we delete fewer than 2 /
11 of the letters ofthe word R i .If a non-empty reduced word Q is equal to the identity of the group G , thenby a theorem of Dyck there exists a natural number w , words T , T , . . . , T w , anddefining words R i , R i , . . . , R i w such that Q ≡ Q wj =1 T j R ε j i j T j , where ε j = ± j =1 , , . . . , w ). As the set M is symmetrised, we may assume that this representationof Q is of the form Q ≡ Q wj =1 T j R i j T j .We are not interested in any particular numbering of the defining words of G , and hence we will instead in the sequel write the product Q wj =1 T j R i j T j as Q wi =1 T i R i T i , where the two words R t and R z can be the same word, even if t < z .To simplify notation, the subword T p R p T p . . . T k R k T k of the word Q wi =1 T i R i T i will be denoted [ p, k ].Consider the set of finite vectors { k , k , . . . , k w } , where w >
0, and k , k , . . . , k w are non-negative integers.The parameter α of the vector ω = { k , k , . . . , k w } is defined as the naturalnumber P wi =1 k i and will be denoted by α { k , k , . . . , k w } , or simply α ( ω ).The defect of the number k j in the vector { k , k , . . . , k w } is defined as thenumber ( k i − k j ) + ( k i − k j ) + · · · + ( k i v − k j ), where k i , k i , . . . , k i v are allthe numbers from the vector { k , k , . . . , k w } such that k i z > j and i z < j , with z = 1 , , . . . , v .The parameter β of the vector ω = { k , k , . . . , k w } , denoted β ( ω ), is definedas the sum of the defects of all numbers k , k , . . . , k w of the vector ω .
356 2. THE IDENTITY PROBLEM IN FINITELY PRESENTED GROUPS
The index of the vector ω = { k , k , . . . , k w } is defined as the ordered pairof natural numbers ( α ′ , β ′ ), where α ′ = α ( ω ) and β ′ = β ( ω ). The set of indiceswill be ordered lexicographically, i.e. ( α , β ) < ( α , β ), where α , α , β , β arenon-negative natural numbers, if and only if one of the following two properties istrue: I. α < α , II. α = α and β < β . We will denote the index of the vector ω as J ( ω ).We will say that the vector ω = { k , k , . . . , k w } is less than the vector ω = { k ′ , k ′ , . . . , k ′ w ′ } if J ( ω ) < J ( ω ).The following three lemmas easily follow from the definition of our ordering onthe vectors. Lemma 1.
The vector ω = { k , k , . . . , k i − , k i , k i +1 , . . . , k w } is greater than thevector ω = { k , k , . . . , k i − , s , s , . . . , s v , k i +1 , . . . , k w } , if we have ≤ v < and s z < k i ( z = 1 , , . . . , v ) . Lemma 2.
The vector ω = { k , k , . . . , k i − , k i , . . . , k w } is greater than the vector ω = { k , k , . . . , k i , k i − , . . . , k w } , if k i − > k i . Lemma 3.
For every vector ω = { k , k , . . . , k w } there exist only finitely manyvectors ω , ω , . . . , ω t such that J ( ω t ) < · · · < J ( ω ) < J ( ω ) < J ( ω ) . We will now prove a theorem regarding the form of reduced words which areequal to the identity K / . The theorem will describe some properties of theword Q wi =1 T i R i T i . For brevity, we will not describe the dual properties (the dualproperties of the word Q wi =1 T i R i T i being the properties of (cid:0)Q wi =1 T i R i T i (cid:1) − ), butall the properties are implied, and the proof of these properties coincide with theordinary properties). Theorem 1.
Let G be a group belonging to the class K / . If a non-emptyword Q is equal to the identity in G , then there exists a natural number w , words T , T , . . . , T and reduced words R , R , . . . , R w such that Q ≡ Q wi =1 T i R i T i , wherethe following properties hold: Property 1.
For every i = 1 , , . . . , w the word T i R i T i is reduced. Property 2.
Let R k ≖ U V and [ U ∂ > [ R k∂ . Then neither T i nor T i ( i =1 , , . . . , w ) contains the word U . Property 3. If R k and R i are not mutually inverse, and R k ≖ ZXY , R i ≖ ZK , T i ≖ N Y , then [ X ∂ ≥ [ Y ∂ . Property 4. If R k ≖ BAX and R s ≖ XCD are not mutually inverse, then in T i there is a word AC , such that [ B ∂ + [ D ∂ ≥ [ A ∂ + [ C ∂ . Property 5. If R k ≖ BAX and R s ≖ XCDE are not mutually inverse, if
CXED and R i ≖ DK are not mutually inverse, and T i ≖ N AC , then [ E ∂ +[ B ∂ ≥ [ C ∂ +[ A ∂ . Property 6.
Suppose R k ≖ U V and [ U ∂ ≥ [ R k∂ . Then the word U can befound inside the word T i R i T i ( i = 1 , , . . . , w ) only as a subword of the word R i . Property 7.
Suppose that R s ≖ AXY and R k ≖ Y ZU D are not mutually inverse.Suppose that [ A ∂ < [ R s∂ , [ D ∂ < [ R k∂ , and [ U ∂ < [ R k∂ . Then U ZK is nota subword of any T i R i T i ( i = 1 , , . . . , w ) such that T i ≖ ZXT ′ i , R i ≖ R ′ i U , and[ U ∂ < [ R i∂ . .1. REPRESENTING WORDS EQUAL TO THE IDENTITY IN K / -GROUPS 37 Property 8. If R p ≖ XY Z , R k ≖ U Y V , p < k and for some way of reducing Q wi =1 T i R i T i the subword Y of the word R p is reduced by the subword Y of theword R k , then the word ZXY and
Y V U are not mutally inverse.
Property 9. If R p ≖ XY Z , T k ≖ U Y V , p < k and for some way of reducing Q wi =1 T i R i T i the subword Y of the word R p reduces with the subword Y of theword T k , then [ T p∂ ≤ [ U ∂ , and [ T p∂ < [ T k∂ . Property 10. If T p ≖ AB , T j ≖ CD , where p < j , and for some way of reducing Q wi =1 T i R i T i the word B [ p +1 , j − C is absorbed, without reducing anything, theneither [ p + 1 , j − ≡
1, or p + 1 = j , or there is some k , with p < k < j , such that[ T k∂ ≥ [ T p∂ . Proof.
Any non-empty word Q can be written of the form Q ≡ Q wi =1 T i R i T i (by the theorem of Dyck). Let the vector ([ T ∂ , [ T ∂ , . . . , [ T w∂ ) be the vector of thelengths of the words T , T , . . . , T w , and suppose its index is J .Property k ( k = 1 , , . . . ,
10) of Theorem 1 will follow from Lemma 3 andthe proof of the following disjunction: either the representation Q ≡ Q wi =1 T i R i T i satisfies Property k ; or there is some representation Q ≡ Q wi =1 T ′ i R ′ i T ′ i , such thatthe index of the lengths of the words T ′ , T ′ , . . . , T ′ t is less than the index J . Property 1.
For any i = 1 , , . . . , w the word T i R i T i is reduced.The words T i , R i , and T i are reduced. If T i R i T i is a reducible word, then T i ≖ T ′ i a ε , R i ≖ a − ε X , and T i ≖ a − ε T ′ i , from which we have T i R i T i ≡ T ′ i Xa − ε T ′ i ,where Xa − ε is a reduced word. In this way, we get in the free group the equality Q ≡ [1 , i − T ′ i Xa − ε T ′ i [ i + 1 , w ], and α { [ T ∂ , [ T ∂ , . . . , [ T w∂ } > α { [ T ∂ , [ T ∂ , . . . , [ T i − ∂ , [ T ′ i ∂ , [ T i +1 ∂ , . . . , [ T w∂ } . In this way, we can write a new representation for Q with a smaller parameter α for the vector of the lengths of the words T , T , . . . , T w , and consequently, with asmaller index. Property 2.
Let R k ≖ U V and [ U ∂ > [ R k∂ . Then neither T i , nor T i ( i =1 , , . . . , w ) contains the word U .Suppose T i ≖ XU Y and [ U ∂ > [ V ∂ . Then T i R i T i ≖ XU Y R i Y U X ≡ XU V X · XV Y R i Y V X · XV U X.
Notice that
V U is a reduced word, and [ X ∂ < [ T i∂ , [ XV Y ∂ < [ T i∂ . Hence, we canwrite a new presentation for Q with a smaller parameter α for the vector of thelengths of the words T , T , . . . , T w . Property 3. If R k and R i are not mutually inverse, and R k ≖ ZXY , R i ≖ ZK , T i ≖ N Y , then [ X ∂ ≥ [ Y ∂ .Suppose [ X ∂ < [ Y ∂ . Then T i R i T i ≖ N Y ZKY N ≡ N Y ZXN · N XKZXN · N XZY N.
Notice that
Y ZK , XZY and KZ are defining words, and [ N ∂ < [ T i∂ , [ N X ∂ < [ T i∂ . Property 4. If R k ≖ BAX and R s ≖ XCD are not mutually inverse, then in T i there is a word AC , such that [ B ∂ + [ D ∂ ≥ [ A ∂ + [ C ∂ . Suppose [ B ∂ + [ D ∂ < [ A ∂ + [ C ∂ . Then T i R i T i ≖ U ACV R i V CAU ≡ U AXV U · U BXCDBU · U BDV R i V DBU · U BDCXBU · U BXAU .
Notice that
AXB , XCD , DCX , BXA , are defining words, and [ U ∂ < [ T i∂ , [ U B ∂ < [ T i∂ , [ U BDV ∂ < [ T i∂ . Property 5. If R k ≖ BAX and R s ≖ XCDE are not mutually inverse, if
CXED and R i ≖ DK are not mutually inverse, and T i ≖ N AC , then [ E ∂ +[ B ∂ ≥ [ C ∂ +[ A ∂ .Suppose [ E ∂ + [ B ∂ < [ C ∂ + [ A ∂ . Then T i R i T i ≖ N ACDKCAN ≡ N AXBN · N BXCDEBN · N BEKDEBN · NBEDCXBN · N BXAN.
Notice that
AXB , XCDE , KD , EDCX , and
BXA are all defining words, and[ N ∂ + [ T i∂ , [ N B ∂ < [ T i∂ , and [ N BE ∂ < [ T i∂ . Property 6.
Suppose R k ≖ U V and [ U ∂ ≥ [ R k∂ . Then the word U can befound inside the word T i R i T i ( i = 1 , , . . . , w ) only as a subword of the word R i .Property 6 follows from Property 3. Property 7.
Suppose that R s ≖ AXY and R k ≖ Y ZU D are not mutually inverse.Suppose that [ A ∂ < [ R s∂ , [ D ∂ < [ R k∂ , and [ U ∂ < [ R k∂ . Then U ZK is nota subword of any T i R i T i ( i = 1 , , . . . , w ) such that T i ≖ ZXT ′ i , R i ≖ R ′ i U , and[ U ∂ < [ R i∂ .Suppose the opposite. Then [ XY ∂ > [ R s∂ , [ ZU Y ∂ > [ R k∂ , [ X ∂ > [ R s∂ ,[ ZU ∂ > [ R k∂ , and [ Z ∂ > [ R k∂ . There are two cases to consider.(1) Suppose [ R k∂ ≤ [ R s∂ . Then[ Z ∂ + [ X ∂ >
311 [ R k∂ + 511 [ R s∂ ≥
411 [ R k∂ + 411 [ R s∂ > [ A ∂ + [ D ∂ . But this contradicts Property 5.(2) Suppose [ R k∂ − [ R s∂ = δ >
0. Then[ Y ∂ <
211 [ R s∂ = 211 (cid:16) [ R k∂ − δ (cid:17) . [ Z ∂ >
311 [ R k∂ + 211 δ. [ Z ∂ + [ X ∂ >
311 [ R k∂ + 211 δ + 511 [ R s∂ = 411 [ R k∂ + 111 δ + 411 [ R s∂ > [ A ∂ + [ D ∂ . But this contradicts Property 5.
Property 8. If R p ≖ XY Z , R k ≖ U Y V , p < k and for some way of reducing Q wi =1 T i R i T i the subword Y of the word R p is reduced by the subword Y of theword R k , then the word ZXY and
Y V U are not mutally inverse.Suppose
ZXY and
Y V U are mutually inverse. Then
U ZXV ≡
1. We have[ p + 1 , k − ≡ T p ZU T k , and thus we have[ p, k ] ≡ T p XV T k ≡ T p ZU T k · T k U ZXV T k ≡ [ p + 1 , k − . .2. SOLVING THE IDENTITY PROBLEM FOR K / -GROUPS 39 Property 9. If R p ≖ XY Z , T k ≖ U Y V , p < k and for some way of reducing Q wi =1 T i R i T i the subword Y of the word R p reduces with the subword Y of theword T k , then [ T p∂ ≤ [ U ∂ , and [ T p∂ < [ T k∂ . . . . T p XY ZT p . . . U Y V R k V Y U
We have [ p + 1 , k − ≡ T p ZU , and[ p, k ] ≡ T p XV R k T k ≡ T p ZU · U ZXY U · U Y V R k T k ≡ [ p + 1 , k − U ZXY U · T k R k T k . Notice that
ZXY is a defining word, from which we have [ U ∂ ≥ [ T p∂ , and since Y is a non-empty word, thus [ T k∂ > [ T p∂ . Property 10. If T p ≖ AB , T j ≖ CD , where p < j , and for some way of reducing Q wi =1 T i R i T i the word B [ p +1 , j − C is absorbed, without reducing anything, theneither [ p + 1 , j − ≡
1, or p + 1 = j , or there is some k , with p < k < j , such that[ T k∂ ≥ [ T p∂ .We have [ p + 1 , j − ≡ BC , and that[ p, j ] ≡ BAR p ADR j DC ≡ BC · CAR p AC · CDR j DC ≡ BAR p AB · BDR j DB · BC.
If [ B ∂ > [ C ∂ , then we consider the penultimate presentation; if [ B ∂ < [ C ∂ , thenwe consider the final presentation. If [ B ∂ = [ C ∂ and [ T k∂ < [ T p∂ for all k , where p < k < j , then the presentation[ p + 1 , j − CAR p ACT j R j T j has the same parameter α as the presentation [ p, j ], but a smaller parameter β , i.e.it has smaller index. (cid:3) Solving the identity problem for K / -groups Definition 1.
Consider the word W ≖ Q wi =1 T i R i T i . The number of factors of theform T q R q T q in the word W , i.e. the number w , will be called the L - length of theword W , and will be denoted by L [ W ]. Definition 2.
Suppose that when shortening the word [ i, j ] there is some R k , where i ≤ k ≤ j , such that when shortening with the subwords of [ i, j ] this word losesfewer letters than times its length. Then the word X , which remains of R k asthe result of reducing, will be called a major trace of the word [ i, j ].Suppose that when reducing the word [ i, j ] there are words R k and R k , where i ≤ k < k ≤ j , such that the two words cancel each other. Furthermore, supposethat each of these words, when reduced with the subwords of [ i, j ] (not countingthe reductions between R k and R k ) has fewer letters than times its length removed. Then the result of reducing R k will produce a subword X , and of R k a subword X . Then the word X X will be called a minor trace of the word [ i, j ].Suppose that when reducing the word [ i, j ], where i = s < z < s < z < s < · · · < z t − < s t − < z t < s t = j, t > , we have that the word R s reduces with the word R s , the word R s reduces withthe word R s , . . . , and the word R s t − reduces with the word R s t ; and moreover,any R s k ( k = 0 , , , . . . , t ) can be reduced (but not necessarily cancelled) to theright only with R z k +1 and T s k +1 , and to the left only with R z k and T s k − . Supposefurther that when reducing the word [ i, j ] there is no major or minor trace. Thenthe word which remains after reducing [ i, j ] will be called a full trace of the word[ i, j ]. Definition 3.
When reducing some word S V S V S , we will say that the high-lighted subword V reduces with the highlighted subword V if we have that V ≖ AXB , V ≖ CXD , where X is a non-empty word, BS C ≡
1, and that whencarrying out this reduction, the word BS C is first reduced, and then X is reducedwith X . We will denote this reduction by the arc S V S V S This definition naturally defines to a larger number of reductions, i.e. we have asan example the word with two arcs S V S V S V S which denotes that when reducing the word S V S V S V S , the subword V re-duces with the highlighted word V , and the highlighted subword V reduces withthe highlighted subword V .We will consider a number of statements regarding possible ways of reducinga word of the form W ≖ Q wi =1 T i R i T i , where the product Q wi =1 T i R i T i satisfiesProperties 1 −
10, and such that L [ W ] ≤ m , where m is some positive integer.These statements will concern reductions of R i on the right. For brevity, wewill not formulate the dual statements for reducing R i on the left, but all thesestatements are implied, and the proofs of these statements are entirely symmetrical. A m . If for any way of reducing
W. . . T i R i T i . . . T j R j T j . . . T k R k T k then when reducing the word [ i + 1 , k −
1] the word R j reduces with at most one R to the left and one R to the right. B m . If for every way of reducing W. . . T i R i T i . . . T j R j T j . . . T k R k T k . . . then when reducing [ i + 1 , k − R j reduces with at most one R . C m . We cannot have the reduction . . . T i R i T i . . . T j R j T j . . . T k R k T k . . . T p R p T p . . . .2. SOLVING THE IDENTITY PROBLEM FOR K / -GROUPS 41 when reducing the word W in any way; and we cannot have the reduction . . . T i R i T i . . . T j R j T j . . . T k R k T k . . . T p R p T p . . . when reducing the word W in any way. D m . We cannot have the reduction . . . T i R i T i . . . T j R j T j . . . when reducing the word W in any way. E m . We cannot have . . . T i R i T i . . . T j R j T j . . . [ k, p ] . . . when reducing the word W in any way. F m . We cannot have . . . T i R i T i . . . T j R j T j . . . T k R k T k . . . T p R p T p . . . G m . If R i reduces with T j R j , where i < p < j , then[ T p∂ < [ T j∂ and [ T p∂ < [ T i∂ .. . . T i R i T i . . . T j R j T j . . . H m . For all ways of reducing the word W , there is a trace in the resulting word. Lemma 4. ( A m , B m , C m , . . . , H m ) = ⇒ A m +1 . Proof.
Suppose that for any way of reducing the word W , where we have L [ W ] ≤ m + 1, there are the following reductions. . . . T i R i T i . . . T j R j T j . . . T k R k T k . . . Then [ i + 1 , k −
1, and by H m for any reduction of [ i + 1 , k −
1] there is a traceof this word.
Case I.
When reducing [ i + 1 , k − Case IA.
When reducing [ i + 1 , k −
1] there is a R v which reduces with at mostone R . If v < j , then the word R v cannot reduce with T k R k and by Property 6,cannot be entirely reduced when reducing the word [ i, k ]. If v = j , then statement A m +1 holds true. If v > j , then the word R v cannot reduce with T i , and thereforecannot be entirely reduced when reducing the word [ i, k ]. Case IB.
When reducing [ i + 1 , k −
1] there will be some R v , which reduces with R s and R t , where s < t . Case IB1. v < j . The word R v cannot reduce with T k R k , and by Property 6cannot be entirely reduced when reducing the word [ i, k ]. Case IB2. v = j . If s < t < j , then the resulting reductions contradict the dualof C m . If j < s < t , then to be completely reduced in [ i, k ] (by Property 6) R j must reduce with T k R k , which contradicts C m . If s < j < t , then statement A m +1 holds. Case IB3. v > j . If s < t < v , then for the word R v to be entirely reduced whenreducing [ i, k ], the word R v needs to reduce with T k R k , which contradicts C m .Suppose s < v < t . The word R v must reduce with R i , from which by F m wehave s = j . Furthermore R v must reduce with T k and R k (the reduction of R v and R k can be empty).In this way, we have the following reduction (for convenience, some indices arechanged). . . . T i R i T i . . . T s R s T s . . . T j R j T j . . . T t R t T t . . . T k R k T k . . . A BC D E FG
Suppose that the words R i and R s reduce by the word A (this means that R i ≖ X AX , and R s ≖ Y AY and for the given reduction the subword A of R i reduceswith the subword A of R s ), and that R s and R j reduce. Case IB3 α . The word R t does not reduce when reducing the word [ j + 1 , k − j, k ], R t will be reducedwith T j (say with the word H ) and with T k (say with the word J ). Let T i and R s reduce (if they reduce) by the word K , and the words R s and T j (if they reduce)by the word L . . . . T i R i T i . . . T s R s T s . . . T j R j T j . . . T t R t T t . . . T k R k T k . . . K LA B DH J
By Property 1, we have [ H ∂ ≥ [ R t∂ .We will prove that [ L ∂ < [ R s∂ and [ L ∂ < [ R t∂ . Indeed, otherwise R s and R t have a common piece, containing at least letters of one of them, and therefore R s comes from R t by a cyclic permutation. Thus, R s ≖ X a Φ Y , R t ≖ X Φ bY , where the word Φ is the shared piece of the words R s and R t , a is the first letterof R j , and b is the final letter of R j . As Φ Y X a ≖ Φ X Y b , we have a ≖ b , andhence the word R j is cyclically reduced.By the fact that [ L ∂ < [ R s∂ , and Property 6, to be entirely reduced in [ i, j ]the word R s must be reduced via reduction with some R p , where i < p < j . .2. SOLVING THE IDENTITY PROBLEM FOR K / -GROUPS 43 Suppose p < s . Then by statement G m and Property 10 we have s + 1 = j . . . . T i R i T i . . . T p R p T p . . . T s R s T s T j R j T j . . . T t R t T t . . . T k R k T k . . . AN K H
The word R p must be reduced via reduction with T i and T s , as by H m the word R p , when reducing [ i + 1 , s − R .Suppose R p and T s reduce with the word P , and T i and R p with the word N . We will prove that [ P ∂ < [ R p∂ and [ P ∂ < [ R t∂ . Indeed, otherwise R p and R t have a common piece, containing not less than letters of one of them. R p ≖ X c Φ Y , R t ≖ X Φ dY , where c is the first letter of R s , and d is the finalletter of R s . As Φ Y X c ≖ Φ X Y d , we have c ≖ d , and the word R s is cyclicallyreduced.In this way, we have that [ N U ∂ > [ R p∂ and [ U KA ∂ > [ R s∂ , which con-tradicts Property 7.Suppose p > s . Then by statement G m and Property 10 we have i + 1 = s . . . . T i R i T i T s R s T s . . . T p R p T p . . . T j R j T j . . . T t R t T t . . . T k R k T k . . . AK U LBP N H The word R p must reduce with T s and T j , as by B m the word R p , whenreducing with [ s + 1 , j − R .Suppose T s and R p reduce with the word P , and R p and T j with the word N . We will prove that [ N ∂ < [ R p∂ and [ N ∂ < [ R t∂ . Indeed, otherwise R p and R t have a common piece, containing not less than letters of one of them. R p ≖ X s a Φ Y s , R t ≖ X Φ bY , where R s ≖ X abY . As Φ Y X a ≖ Φ X Y b , wehave a ≖ b , i.e. the word R s is reduced.In this way, [ U P ∂ > [ R p∂ , which contradicts Property 6.If when reducing [ i + 1 , j − R s reduces with two R , then we havea contradiction much the same way as in the case just considered. Case IB3 β . When reducing [ j + 1 , k −
1] the word R t reduces with R p , where p < t .By statement G m and Property 10 we have t + 1 = k . . . . T i R i T i . . . T s R s T s . . . T j R j T j . . . T p R p T p . . . T t R t T t T k R k T k . . . AK LB P H QS J
Suppose T i and R s reduce with the word K ; R s and T j with the word L ; T j and R p with the word P ; and R p and T t with the word Q . Case IB3 β When reducing [ i + 1 , j −
1] the word R s does not reduce.Then [ L ∂ > [ R s∂ , [ HS ∂ ≥ [ R t∂ , as otherwise we do not have Property 7,from which [ H ∂ > [ R t∂ , and we get that R j is cyclically reduced. Case IB3 β When reducing [ i + 1 , j −
1] the word R s reduces with R z , where z < s .By statement G m and Property 10 we have s + 1 = j . . . . T i R i T i . . . T z R z T z . . . T s R s T s T j R j T j . . . T p R p T p . . . T t R t T t T k R k T k AK X LB P H QS J [ HS ∂ ≥ [ R t∂ , as otherwise we do not have Property 7, from which we have[ H ∂ > [ R t∂ . Then [ L ∂ < [ R s∂ , [ L ∂ < [ R t∂ , as otherwise R j is cyclicallyreduced.Consider the word [ i, s ][ j + 1 , k ]. . . . T i R i T i . . . T z R z T z . . . T s R s T s . . . T p R p T p . . . T t R t T t T k R k T k . . . AK X LHP QS J
As [ H ∂ > [ L ∂ , we have that R t reduces with R s and T s . LT s ≖ T j , and hencethe word T s reduces with R p by the word P . After reducing [ i, s ][ j + 1 , k ], thereremains of R s a word B such that [ B ∂ < [ R s∂ .We will prove, that the word [ i, s ][ j + 1 , k ] satisfies all ten properties of Theo-rem 1.Properties 1-5 of the word [ i, s ][ j +1 , k ] are satisfied, as these properties concernthe subword T i R i T i , which in the word [ i, s ][ j + 1 , k ] are the same as those of[ i, k ]. Properties 6-7 of the word [ i, s ][ j + 1 , k ] are satisfied, as these properties areconsequences of Properties 3 and 5.Property 8 holds for the subwords [ i, s ] and [ j + 1 , k ], and hence it holds forthe same subwords in the word [ i, k ]. Furthermore, in the word [ i, s ][ j + 1 , k ] thereis only one reduction: the word R s reduces with the word R t by the word L , but,as we have proved, [ L ∂ < [ R s∂ and [ L ∂ < [ R t∂ .Property 9 is also enough to check for a single reduction: the reduction of R t and T s .Let T j ≖ HU and that in the word [ i, k ] T j reduces with R t by the word H . Then by Property 9 we have [ U ∂ ≥ [ T t∂ . Suppose that T s and R t intersectin [ i, s ][ j + 1 , k ] by the word H . Then T s ≖ H U , but [ U ∂ ≥ [ T t∂ , from which[ T s∂ > [ T t∂ . .2. SOLVING THE IDENTITY PROBLEM FOR K / -GROUPS 45 Property 10 follows, as s + 1 = j and the word [ k + 1 , p − T p P , both in theword [ i, k ] and in the word [ i, s ][ j + 1 , k ], is absorbed from the left by the same word U . Therefore, in the word [ i, s ][ j + 1 , k ], all properties A m , B m , . . . , H m hold.Suppose that the words R z and T s reduce by the word Y , and that R z does notreduce when reducing the word [ i + 1 , s − Y ∂ ≥ [ R z∂ , then we do nothave Property 7; if [ Y ∂ ≥ [ R z∂ , then [ P ∂ < [ R p∂ (as otherwise R t is reduced),but then [ Q ∂ > [ R p∂ , [ J ∂ > [ R t∂ , [ H ∂ < [ R t∂ , and [ P F ∂ < [ R p∂ , wherewhen reducing [ s + 1 , t −
1] the word R p is reduced by the word F .Hence, when reducing [ i +1 , s − R z reduces with some R , and we can continueour argument by splitting into a case (which essentially does not differ from IB3 ).As the length L of the word W is finite, we have our contradiction. Case IB3 β When reducing [ i + 1 , j − R s reduces with R z , where s < z .By statement G m and Property 10 we have i + 1 = s . . . . T i R i T i T s R s T s . . . T z R z T z . . . T j R j T j . . . T p R p T p . . . T t R t T t T k R k T k . . . L H
As we already know, [ H ∂ > [ R t∂ , from which we have [ L ∂ < [ R s∂ and [ L ∂ < [ R t∂ , i.e. [ L ∂ < [ H ∂ , and the words R z and T j reduce.Consider the word [ s, j − j + 1 , t ]. . . . T s R s T s . . . T z R z T z . . . T p R p T p . . . T t R t T t . . . L H
The words R t and R z reduce inside the word [ s, j − j + 1 , t ]. It is easy to show,that the word [ s, j − j + 1 , t ] satisfies all ten properties of Theorem 1, but R t reduces on the left with three R , which contradicts the dual of C m . Case IB3 β When reducing the word [ i + 1 , j −
1] the word R s reduces with R z and R q , where z < s < q . . . . T i R i T i . . . T z R z T z . . . T s R s T s . . . T q R q T q . . . T j R j T j . . . T p R p T p . . . T t R t T t T k R k T k We have [ H ∂ > [ R t∂ , whence [ L ∂ < [ R s∂ and [ L ∂ < [ R t∂ , i.e. [ L ∂ < [ H ∂ ,and the word R q and T j reduce.Consider the word [ s, j − j + 1 , t ]. . . . T s R s T s . . . T q R q T q . . . T p R p T p . . . T t R t T t . . . As the word [ s, j − j + 1 , t ] satisfies all ten properties of Theorem 1, we hencehave a contradiction to the dual of C m . Case IB3 γ . When reducing [ j + 1 , k −
1] the word R t reduces with R p , where p > t .By statement G m and Property 10, we have j + 1 = t . . . . T i R i T i . . . T s R s T s . . . T j R j T j T t R t T t . . . T p R p T p . . . T k R k T k . . . Here we reach the same contradiction as in Case
IB3 β , but already when con-sidering the word [ i, j − t + 1 , k ]. Case IC.
When reducing [ i + 1 , k −
1] there is some R v which reduces with morethan two R .In this case, we immediately get a contradiction to C m or the dual of C m , as R v ,which reduces completely in [ i, k ] (by Property 6) must reduce with R i T i and T k R k . Case ID.
When reducing [ i + 1 , k − R v which reduces with T s or T p .In this case, we immediately get a contradiction to D m or E m . Case II.
When reducing [ i + 1 , k − R v and R v which reduce with each other, and that in this reduction eachword loses fewer than letters of its length. Case IIA. v < v < j .In this case, neither R v nor R v can reduce with T k R k , and the minor tracemust be completely reduced when reducing with R i T i . But by Property 7, this isimpossible. Case IIB. v < v = j . . . . T i R i T i . . . T v R v T v . . . T j R j T j . . . T k R k T k . . . By Property 7, the word R j should be reduced with T k R k . By C m and thedual of E m R j cannot reduce on the left. By C m and E m R j can be reduced onthe right only by a single R (other than the fact that R j reduces with T k R k ), butthen when reducing [ i + 1 , k −
1] by R j there remains a major trace, and we havealready considered this case. Case IIC. j = v < v . . . . T i R i T i . . . T j R j T j . . . T v R v T v . . . T k R k T k . . . .2. SOLVING THE IDENTITY PROBLEM FOR K / -GROUPS 47 R j cannot reduce with T v [ v +1 , k − i +1 , k − R j and R v no trace would remain. By E m and the dual of E m , C m and the dual of C m , and as no major trace remains of R j , then when reducing[ i + 1 , k −
1] the word R j reduces with R s and R s , where i < s < j < s < v . . . . T i R i T i . . . T s R s T s . . . T j R j T j . . . T s R s T s . . . T v R v T v . . . T k R k T k . . . By Property 7 the word R v must reduce with T k R k , and in [ i + 1 , k −
1] theword R v (by C m , E m , and F m ) must reduce with R s and R s , where v < s < k . . . . T i R i T i . . . T s R s T s . . . T j R j T j . . . T s R s T s . . . T v R v T v . . . T s R s T s . . . T k R k T k . . .R j and R v must be completely absorbed when reducing [ i, k ], and thereforeeither R j reduces with T i (and has no reductions other than those indicated) orelse R v reduces with T k R k (and has no reductions other than those indicated).We then arrive at a contradiction in the same way as in Case IB3 . Case IID. j < v < v . . . . T i R i T i . . . T j R j T j . . . T v R v T v . . . T v R v T v . . . T k R k T k . . .R v reduces with R i , R v with T k R k . The word R v reduces with R j and R s ,where v < s < v , and therefore R v does not reduce with T k R k . We then arriveat a contradiction in the same way as in Case IIC . Case III.
When reducing [ i + 1 , k −
1] there remains a full trace, i.e. there arenumbers i + 1 = s < z < s < z < s < · · · < z t − < s t − < z t < s t = k − t >
3, and the word R s reduces with the word R s , the word R s reduces withthe word R s , . . . , and the word R s t − reduces with the word R s t ; wherein R s v ( v =0 , , , . . . , t ) can only reduce with R z v +1 , T s v +1 , R z v , and T s v − ; additionally, whenreducing [ i + 1 , k −
1] no major nor minor trace remains.Then there are R s q − , R s q , R s q +1 such that the following reductions occur (3) . . . T s q − R s q − T s q − . . . T z q R z q T z q . . . T s q R s q T s q . . . T z q +1 R z q +1 T z q +1 . . . T s q +1 R s q +1 T s q +1 . . . Indeed, R s must reduce with T s , as otherwise from R s (by statements C m , D m , E m ) there remains a major trace; and R s t must reduce with T s t − , asotherwise from R s t there remains a major trace.Consider the leftmost R s α which reduces with T s α − , and the rightmost R s β in[ s , s α − ] which reduces with T s β +1 . If β − α , then from R s β +1 there remains amajor trace. If β − α = 3, then from R s β +1 and R s β +2 there remain minor traces.If β − α >
3, and from R s β +1 , R s β +2 , . . . , R s α − neither major nor minor tracesremain, then we have the reductions in (3).As when reducing [ i, k ] the full trace in question must be completely absorbed,we have that R s q reduces either with R i T i , or with T k R k , which contradicts eitherthe dual of C m , or C m . This completes the proof of Lemma 4. (cid:3) Lemma 5. ( A m +1 , B m , C m , . . . , H m ) = ⇒ B m +1 . Proof.
Suppose that for every way of reducing the word W , where L [ W ] ≤ m + 1, there are the following reductions . . . T i R i T i . . . T j R j T j . . . T k R k T k . . . If when reducing [ i + 1 , k − R j reduces with more than one R , thenwe easily come to a contradiction by the same methods as in Lemma 4. (cid:3) Lemma 6. ( A m +1 , B m +1 , C m +1 , . . . , H m ) = ⇒ C m +1 . Proof.
Suppose that when reducing the word W , where L [ W ] ≤ m + 1, thereare the following reductions . . . T i R i T i . . . T j R j T j . . . T k R k T k . . . T p R p T p . . . or the following reductions . . . T i R i T i . . . T j R j T j . . . T k R k T k . . . T p R p T p . . . Then by A m +1 and B m +1 , R j in [ i + 1 , p −
1] cannot reduce with any morethan two R , and to be completely reduced in [ i, p ] must reduce with R i T i by morethan of the letters of its length, which is impossible by Property 6. (cid:3) Lemma 7. ( A m +1 , B m +1 , C m +1 , D m , . . . , H m ) = ⇒ D m +1 . Proof.
Suppose that when reducing the word W , where L [ W ] ≤ m + 1, wehave the following reduction. . . . T i R i T i . . . T j R j T j . . . The word R j must be completely reduced on the left by the word R i T i [ i +1 , j − T j .The maximal reduction which R j can have is its reduction with R s , T s , and R p ,where i ≤ s < p < j (this follows from C m +1 , D m , and E m ). But by Property 6these three words do not absorb more than of the letters of the word R i . (cid:3) .2. SOLVING THE IDENTITY PROBLEM FOR K / -GROUPS 49 Lemma 8. ( A m +1 , . . . , D m +1 , E m , F m , G m , H m ) = ⇒ E m +1 . Proof.
Suppose that when reducing the word W , where L [ W ] ≤ m + 1, thereis the following reduction . . . T i R i T i . . . T j R j T j . . . [ k, p ] . . . But, as we saw in the proof of the statement D m +1 , the word R j can reduceon the right by at most of its length, and besides this, R j can only reduce (onthe left) by R i . In this way, R j is absorbed completely when reducing [ i, p ], andthis reduction is impossible. (cid:3) Lemma 9. ( A m +1 , . . . , E m +1 , F m , G m , H m ) = ⇒ F m +1 . Proof.
Suppose that when reducing the word W , where L [ W ] ≤ m + 1, thereare the following reductions . . . T i R i T i . . . T j R j T j . . . T k R k T k . . . T p R p T p . . . Case I.
When reducing [ i + 1 , p −
1] there remains a major trace, i.e. there is some R v , which reduces with a subword of [ i + 1 , p −
1] with fewer letters than of itslength.In this case R v , in order to be completely reduced in [ i, p ], must reduce eitherwith R i , or with R p ; if v = j , then R v also reduces with R p ; if v = k , then R v reduces with R i ; i.e. in every case we obtain a contradiction to C m +1 or the dualof C m +1 . Case II.
When reducing [ i + 1 , p −
1] there remains a minor trace, i.e. there are R v and R v , which reduce with one another, and such that each of these wordsloses fewer than of its letters.By Property 7 the cases v < v < j and k < v < v are impossible. By C m +1 and the dual of C m +1 , the other cases are also impossible. Case III.
When reducing [ i + 1 , p −
1] there remains a full trace.In Lemma 4, Case
III , it was in particular proved that if R i reduces with R p ,then when reducing [ i + 1 , p − (cid:3) Lemma 10. ( A m +1 , . . . , F m +1 , G m , H m ) = ⇒ G m +1 . Proof.
Suppose that we have the word W , where L [ W ] ≤ m + 1, and thatthe following reductions take place in W . . . . T i R i T i . . . T j R j T j . . . Case I.
When reducing [ i + 1 , j −
1] there remains a major trace.In this case, R v reduces with T i and T j . If v = p , then statement G m +1 followsdirectly from Property 9. If v = p , then, as by Property 9 we have [ T v∂ < [ T i∂ and[ T v∂ < [ T j∂ , statement G m +1 follows from statement G m in the word [ i, v ] or theword [ v, j ]. Case II.
When reducing [ i + 1 , j −
1] there remains a minor trace.If R i reduces with some R t , where i < t < j , then by the proof of D m +1 wefind that this case is actually impossible.If each of R v and R v reduces with only one of T i and T j , then by the proofof D m +1 we find that this case is actually impossible.If R v and R v reduce with both T i and T j , then statement G m +1 is proved asin Case I . Case III.
When reducing [ i + 1 , j −
1] there remains a full trace.As a particular case of Lemma 4, Case
III , we have proved that if R i reduceswith T j , then when reducing [ i + 1 , j −
1] there cannot remain a full trace. (cid:3)
Lemma 11. ( A m +1 , B m +1 , . . . , G m +1 , H m ) = ⇒ H m +1 . Proof.
Suppose that we have the word W , where L [ W ] ≤ m + 1, and considersome particular way of reducing the word W .Consider all reductions in which R z reduces with R s , where 1 ≤ z < s ≤ w ,and choose the maximal of them, i.e. the reductions in which no subword of theword [1 , z − T z R z does not reduce with any subword of the word R s T s [ s + 1 , w ](except when reducing R z and R s ).In the same way, choose all maximal reductions of R with T and T with R .Let R i , R i , . . . , R i p are all the R i appearing in W which for the chosen reduc-tion is affected by the maximal reduction of an R with an R .If i >
1, then R either does not reduce, or the maximal reduction with T k ,where 1 < k ≤ w , then the word [ k, w ] has (by statement H m ) a trace, which willalso be a trace for W .We will also prove that either i p = w , or else G m +1 holds.If some R i q ( q = 1 , , . . . , p −
1) is affected in the maximal reduction of R z with R s only on one side (e.g. the left), then either R i q does not reduce with it on theright – in which case the word [ i q + 1 , w ] will be the trace of W ; or else R i q reducesmaximally with T k – in which case the word [ k, w ] will be the trace of W .It remains only to prove Lemma 11 in the case of the following reductionstaking place in the word W . T R T . . . T i R i T i . . . T i R i T i . . . . . . T i p − R i p − T i p − . . . T i p R i p T i p where the indicated arcs are all maximal reductions of an R with an R . Case I. p = 1.In this case, either R reduces with T w – in which case R w reduces only with R , or else R , in the given reduction, loses fewer than of its letters (by state-ments C m +1 , D m +1 , E m +1 ). Case II. p = 2 , R reduces with T i , or else there remains a major trace from R . Either the word T i p − reduces with R w , or else there remains a major tracefrom R w . But if R reduces with T i , and T i p − with R w , then either there remainsa major trace from R i , or there remains a major trace from R i , or from R i and .2. SOLVING THE IDENTITY PROBLEM FOR K / -GROUPS 51 R i there remains a minor trace, depending on what R i is reduced with on theright and what R i is reduced with on the left. Case III. p > (cid:3)
Theorem 2.
If the non-empty reduced word Q is equal to the identity in the group G , then Q contains some trace as a subword. Proof.
By Theorem 1, if the non-empty word Q is equal to the identity, thenthere exists a word of the form Q wi =1 T i R i T i which satisfies Properties 1-10, andsuch that Q ≡ Q wi =1 T i R i T i .If the word Q is reduced, then after completely reducing the word Q wi =1 T i R i T i in any way, the word Q remains. Obviously, the statements A , B , . . . , H areall true. Then by Lemmas 4-11, properties A q , B q , . . . , H q all hold for all naturalnumbers q , and in particular statement H w is true. (cid:3) Lemma 12.
Let Q ≡ Q wi =1 T i R i T i . If when reducing the product Q wi =1 T i R i T i (satisfying Properties 1-10), the words R and R k (or the words R and T k ) reduce,then [ T ∂ + 2 ≥ k . Proof.
We prove the lemma by induction on k .If k = 2, then the lemma is trivially true, as [ T ∂ ≥ k ′ < k . Case I.
Suppose that the words R and R k reduce. Case IA.
Neither R nor R k reduce with any R p , where 1 < p < k . By statement G w and Property 10, we have k = 2, and the lemma is trivially true. Case IB. R reduces with R p , where 1 < p < k . T R T . . . T p R p T p . . . T k R k T k . . . The word R p is completely reduced when reducing [1 , k ], and hence R p reduceswith T and T k . Case IB1. R p does not reduce when reducing [2 , k − G w and Prop-erty 10, we have that k = 3, [ T ∂ > [ T p∂ ≥
0, from which we have [ T ∂ + 2 ≥ k . Case IB2. R p reduces with R q , where 1 < q < p . T R T . . . T q R q T q . . . T p R p T p . . . T k R k T k . . . Then by the inductive hypothesis (of the lemma dual to Lemma 12), we havethat p ≤ [ T p∂ + 2. By G w and Properties 9 and 10, we have p + 1 = k and [ T p∂ < [ T ∂ , from which we have[ T ∂ + 2 ≥ [ T p∂ + 3 ≥ p + 1 = k. Case IB3. R p reduces with R q , where p < q < k . T R T . . . T p R p T p . . . T q R q T q . . . T k R k T k . . . By the inductive hypothesis [ T p∂ + 2 ≥ k − p + 1. By G w and Properties 9 and10, we have that p = 2 and [ T p∂ < [ T ∂ , from which we have[ T ∂ + 2 ≥ [ T p∂ + 3 ≥ k − ≥ k. Case IC. R k reduces with R p , where 1 < p < k . The proof is identical to Case IB.Case II.
Suppose that R reduces with T k . The proof is identical to Case I. This completes the proof of Lemma 12. (cid:3)
Lemma 13.
Let Q ≡ Q wi =1 T i R i T i . If when reducing the product Q wi =1 T i R i T i (satisfying Properties 1-10), the word R reduces with R k , then [ Q ∂ ≥ k . Proof.
By statements C w , D w , and E w , the word R can reduce only with R t , T t , and R s , where s < t . But these reductions remove fewer than of thenumber of letters of R . The same is true for R w . It follows that when reducing Q wi =1 T i R i T i , the word T does not reduce at all, and the words R and R w arenot completely absorbed. Hence [ Q ∂ ≥ [ T ∂ + 2, and by Lemma 12, we also have[ T ∂ + 2 ≥ k . (cid:3) Lemma 14.
Let Q ≡ Q wi =1 T i R i T i . If when reducing the product Q wi =1 T i R i T i (satisfying Properties 1-10) there remains some full trace of the words R , R s , R s , . . . , R s p − , R w , then p ≤ Q ∂ . Proof.
As we already know, the words R and R w are not completely ab-sorbed when reducing Q wi =1 T i R i T i . To prove the lemma, it suffices to show thatnone of R s m − , R s m , R s m +1 , ( m = 1 , , . . . , p −
1) are not all simultaneously com-pletely absorbed when reducing Q wi =1 T i R i T i . . . . T sm − R sm − T sm − . . . T sm − R sm − T sm − . . . T sm R sm T sm . . . T sm +1 R sm +1 T sm +1 . . . T sm +2 R sm +2 T sm +2 . . . If R s m − reduces on the right with the same subword as R s m reduces with, and R s m +1 reduces on the left with the same subword as R s m , then R s m , when reduced,cannot lose more than of its letters, i.e. is not completely absorbed. Hence R s m − reduces on the right only with R s , but on the left (by C w , D w , f E w ) it cannot losefewer than of its letters, i.e. R s m − is not completely absorbed. (cid:3) Lemma 15.
Let Q ≡ Q wi =1 T i R i T i . If when reducing the product Q wi =1 T i R i T i (satisfying Properties 1-10) the word R and R k reduce, then [ T k∂ ≤ ( d + 1)[ T ∂ ,where d is the length of the longest defining word. .2. SOLVING THE IDENTITY PROBLEM FOR K / -GROUPS 53 Proof.
By statement G w and Property 10, each word R p , where 1 < p < k ,loses with T k fewer than d letters, after which, there remain of T k fewer [ T ∂ . Butby Lemma 12, we have that k ≤ [ T ∂ + 2, i.e. k − ≤ [ T ∂ . It follows that[ T k∂ ≤ d · ( k −
2) + [ T ∂ ≤ d · [ T ∂ + [ T − ∂ = ( d + 1)[ T ∂ . (cid:3) Lemma 16.
Let Q ≡ Q wi =1 T i R i T i , where [ Q ∂ = e . If when reducing the product Q wi =1 T i R i T i (satisfying Properties 1-10) there remains a full trace of the words R , R s , R s , . . . , R s p − , R w , then [ T j∂ ≤ e · ( d + 1) e , where j = 1 , , . . . , w . Proof.
We first prove the lemma for the words R , R s , R s , . . . , R s p − , R w .By Lemma 15, [ T s ∂ ≤ ( d + 1) e , as [ T ∂ < [ Q ∂ = e . We have [ T s ∂ ≤ ( d + 1) e , . . . , [ T s p ∂ ≤ ( d + 1) e p . But by Lemma 14, p ≤ e . By statement G w , for any u such that s t − < u < s t , we have [ T u∂ ≤ [ T s t ∂ . This proves the lemma. (cid:3) Lemma 17.
Let Q ≡ Q wi =1 T i R i T i , let [ Q ∂ = e , and let d be the length of thelongest defining word. If when reducing the product Q wi =1 T i R i T i (satisfying Prop-erties 1-10) there remains a full trace, then h w Y i =1 T i R i T i ∂ ≤ e ( d + 1) e . (This upper bound is very rough, and was only chosen for ease of writing). Proof.
Indeed, we have [ T i R i T i∂ ≤ e ( d + 1) e + d by Lemma 16, and that w ≤ e { e ( d + 1) e } by Lemmas 14, 12, and 16. From this we have h w Y i =1 T i R i T i ∂ ≤ { e ( d + 1) e + d } e { e { d + 1 } e }≤ { e ( d + 1) e + d } · e ≤ { e ( d + 1) e } · e ≤ e ( d + 1) e . (cid:3) Theorem 3.
There exists an algorithm for solving the identity problem in everygroup from the class K / . Proof.
To prove the theorem we use a generalisation of M. Dehn’s algorithm[
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