One-relator quotients of Partially Commutative Groups
aa r X i v : . [ m a t h . G R ] J u l One-relator quotients of PartiallyCommutative Groups
Andrew J. Duncan Arye Juh´aszJuly 19, 2019
Abstract
We generalise a key result of one-relator group theory, namely Mag-nus’s Freiheitssatz, to partially commutative groups, under sufficientlystrong conditions on the relator. The main theorem shows that underour conditions, on an element r of a partially commutative group G ,certain Magnus subgroups embed in the quotient G = G / N ( r ) ; thatif r = s n has root s in G then the order of s in G is n , and underslightly stronger conditions that the word problem of G is decidable.We also give conditions under which the question of which Magnussubgroups of G embed in G reduces to the same question in the mini-mal parabolic subgroup of G containing r . In many cases this allows usto characterise Magnus subgroups which embed in G , via a conditionon r and the commutation graph of G , and to find further examplesof quotients G where the word and conjugacy problems are decidable.We give evidence that situations in which our main theorem appliesare not uncommon, by proving that for cycle graphs with a chord Γ,almost all cyclically reduced elements of the partially commutativegroup G ( Γ ) satisfy the conditions of the theorem. Partially commutative groups have been extensively studied in several dif-ferent guises and are variously known as semi-free groups; right-angled Artin
Keywords.
One-relator group theory; partially commutative groups, right-angledArtin groups, HNN-extensions of groups.
Mathematics Subject Classification (2010) : 20F05; 20F36, 20E06 A = V ( Γ ) be the set of vertices of Γ andlet F ( A ) be the free group on A . For elements g, h of a group we denote thecommutator g − h − gh of g and h by [ g, h ] . Let R = {[ x i , x j ] ∈ F ( A ) ∣ x i , x j ∈ A and there is an edge from x i to x j in Γ } . We define the free partially commutative group with (commutation) graph
Γto be the group G = G ( Γ ) with free presentation ⟨ A ∣ R ⟩ . We shall refer tofinitely generated free partially commutative groups as partially commuta-tive groups. (Strictly speaking we have defined the class of finitely generated partially commutative groups.) The class of partially commutative groupscontains finitely generated free, and free Abelian groups; has provided severalcrucial examples in the theory of finitely presented groups and has applica-tions both in mathematics and computer science. For an introduction andsurvey of the literature see [7] or [12]. It emerges, from the work of Sageev,Haglund, Wise, Agol and others, that many well-known families of groupsvirtually embed into partially commutative groups: among these are Coxetergroups, certain one-relator groups with torsion, limit groups, and fundamen-tal groups of closed 3-manifold groups (see for example [25] for details andreferences). It is therefore natural to consider related classes, such as theirone-relator quotients as we do here.There are generalisations of one-relator group theory in several direc-tions; for instance to two-relator groups, to one-relator quotients of surfacegroups [18, 2] and very successfully to one-relator quotients of free productsof groups, see e.g. [6, 16, 17] and see [13] for details and fuller references. Akey result of one-relator group theory is Magnus’s Freiheitssatz which statesthat if F is the free group of rank n , and r is an element of F , involvingevery generator of F , then the subgroup generated by n − F / N ( r ) . (We use N ( r ) to denote the nor-mal closure of r in a given group.) The Freiheitssatz has several immediateconsequences: for instance if r = s m , where s is not a proper power in F then s has order precisely m in F / N ; if m = F / N is torsion-free; if m > F / N is conjugate to a power of s ; andone-relator groups have solvable word problem (see for example [13]).Turning to one-relator quotients of partially commutative groups, An-tolin and Kar [1] proved that a Freiheitssatz holds for one-relator quotientsof starred partially commutative groups: G ( Γ ) is starred if Γ has no full sub-graph isomorphic to the four cycle or the path graph on four vertices. More2enerally they prove versions of the Freiheitssatz for one-relator quotientsof starred partially commutative products and show that the word problemis solvable, both for one-relator quotients of starred partially commutativegroups and of starred partially commutative products of polycyclic groups.Here, with appropriate restrictions on the relator, we show that a Frei-heitssatz holds for one-relator quotients of any partially commutative groupand that, with slightly stronger restrictions, the word problem is solvable. Asummary of the situations in which a Freiheitssatz is known to hold, combin-ing our results and those of [1], is given in the Examples and final summaryof this section.In order to state our main results we need some definitions. If w is anelement of the free monoid ( A ∪ A − ) ∗ then we denote by supp ( w ) the setof elements x ∈ A such that x or x − occurs in w . For an element g ∈ G wedefine supp ( g ) to be equal to supp ( w ) , where w is an element of ( A ∪ A − ) ∗ of minimal length amongst all those elements representing g . For more detail(including the fact that supp ( g ) is well-defined) see Section 2. An element g of G is said to be cyclically minimal if it is represented by a word w ∈ ( A ∪ A − ) ∗ of length no greater than any other word representing an elementof the conjugacy class of g . (See Section 2, Lemma 2.1.)A subset Y of A is called a clique if the full subgraph Γ Y of Γ generatedby Y is a complete graph and independent if Γ Y is a null graph. For x ∈ A define the link of x to be lk ( x ) = { y ∈ A ∣ d ( x, y ) = } (edges of the graphhave length 1) and the star of x to be st ( x ) = lk ( x ) ∪ { x } . The definitionsof t -thick and t -root appear in Section 3.1, Definition 3.2 and Section 3.3,Definition 3.7, respectively.To illustrate some of the possibilities we give some examples. Example 1.1.
1. If Γ is a complete or null graph then G = G ( Γ ) is freeAbelian or free, respectively. In this case, if s ∈ G is not a properpower, t ∈ supp ( s ) and r = s n , for some positive integer n , then ⟨ A /{ t }⟩ embeds in G = G / N ( r ) and s has order n in G ; using standard resultsfrom the theory of finitely generated Abelian groups, or of one-relatorgroups, as appropriate.2. Let A = { a, b, c } , R = {[ a, b ] , [ b, c ]} and G = ⟨ A ∣ R ⟩ ≅ F × Z . Let s = abc , let N be the normal closure of s n in G , for some n >
0, and let G = G / N . Here b ∈ supp ( s ) and r = ( ac ) n b n , so [( ac ) n , a ] = G , but [( ac ) n , a ] ≠ G (as C G ( a ) = ⟨ a, b ⟩ ). Hence ⟨ A /{ b }⟩ = ⟨ a, c ⟩ does notembed in G . 3 heorem 1.2. Let G = ⟨ A ∣ R ⟩ and let s ∈ G be a cyclically minimal elementsuch that, for some t ∈ supp ( s ) ,1. lk ( t ) is a clique (or empty),2. s is t -thick,3. s ∉ ⟨ st ( t )⟩ and4. s is a t -root.Then, for n ≥ ,(a) ⟨ A /{ t }⟩ embeds in G / N , where N denotes the normal closure of s n in G , and(b) the order of sN in G / N is n .Moreover if n ≥ then the word problem is solvable in G / N . Given a cyclically reduced element r of the free group F ( A ) on A , a Mag-nus subgroup is a subgroup generated by a subset B of A such that r ∉ ⟨ B ⟩ .As given above, Magnus’s Freiheitssatz for one-relator groups applies when r involves every generator of the free group in question so every proper subsetof A generates a Magnus subgroup; which embeds in the quotient by thenormal closure of r . In the free group setting the general case reduces to thisspecial case: indeed, if r is an arbitrary element of F ( A ) and Y = supp ( r ) ⊆ A with X = A / Y non-empty then we have F ( A )/ N ( r ) ≅ [ F ( Y )/ M ( r )] ∗ F ( X ) ,where N ( r ) and M ( r ) are the normal closures of r in F ( A ) and F ( Y ) , re-spectively. From the Freiheitssatz as stated above, if Y ′ is a proper subsetof Y then the subgroup ⟨ Y ′ ⟩ of F ( Y ) embeds in F ( Y )/ M ( r ) , so ⟨ X ∪ Y ′ ⟩ embeds in F ( A )/ N ( r ) . Thus the general Freiheitssatz for Magnus subgroupsof F ( A ) reduces to the special case of Magnus subgroups for F ( Y ) .To obtain an analogous reduction for one-relator quotients of partiallycommutative groups we first note that it follows from work of B. Baumslagand S.J. Pride [4], together with the fact, proved by Diekert and Mushcoll[8], that equations are decidable over partially commutative groups; that if G = G ∗ ⋯ ∗ G k is a free product (so Γ has k connected components) andsupp ( r ) contains vertices in at least two components of Γ, then each of the G i embeds in G . In general, we consider the case where the group G decomposesas a free product with amalgamation, and one factor is generated by supp ( r ) .More precisely, we make the following definition.4 efinition 1.3. For a subset B of A define lk ( B ) = ∩ b ∈ B lk ( b ) and for g ∈ G define lk ( g ) = lk ( supp ( g )) . A subset Y of A is called synchronised if, for allvertices v of Y , the star st ( v ) of v is a subset of Y ∪ lk ( Y ) .If Y is synchronised then, writing X = A /{ Y ∪ lk ( Y )} , it follows alsothat, for all vertices v of X , the star st ( v ) is a subset of X ∪ lk ( Y ) = A / Y .Hence, if Y is synchronised, setting A = ⟨ Y ∪ lk ( Y )⟩ , A = ⟨ X ∪ lk ( Y )⟩ and U = ⟨ lk ( Y )⟩ , it follows that G = A ∗ U A . Theorem 1.4.
Let G = ⟨ A ∣ R ⟩ and let s ∈ G be a cyclically minimal el-ement such that supp ( s ) is synchronised. Let A = ⟨ supp ( s ) ∪ lk ( s )⟩ , A = ⟨ A / supp ( s )⟩ and U = ⟨ lk ( s )⟩ ; so G = A ∗ U A . Let r = s n , for some n ≥ , let K = ⟨ supp ( s )⟩ , denote by N the normal closure of r in G and by M the normal closure of r in K , and let G = G / N . Then the following hold.(a) If Y ′ is a subset of supp ( s ) such that the subgroup ⟨ Y ′ ⟩ of K embedsin K / M then ⟨ Y ′ ∪ ( A / supp ( s ))⟩ embeds in G ;(b) if s has order n in K / M then s has order n in G ,(c) if the word problem is solvable in K / M then the word problem is solvablein G and(d) if the conjugacy problem is solvable in K / M then the conjugacy problemis solvable in G . In terms of Magnus subgroups of G (defined exactly as for free groups)with the notation, and under the hypotheses, of the theorem, a Magnussubgroup generated by a subset B of A embeds in G if and only if theMagnus subgroup ⟨ B ∩ supp ( s )⟩ of K embeds in K / M .In cases where supp ( s ) is a clique we obtain a complete characterisationof the conditions under which all Magnus subgroups of G embed in G . Corollary 1.5.
Let G = ⟨ A ∣ R ⟩ and let s ∈ G be a cyclically minimal elementsuch that supp ( s ) is a clique. Let r = s n , for some n ≥ , denote by N thenormal closure of r in G and let G = G / N . Then the following are equivalent.1. G t = ⟨ A /{ t }⟩ embeds in G , for all t ∈ supp ( s ) .2. supp ( s ) is synchronised. oreover, if supp ( s ) is synchronised then s has order n in G , and the wordand conjugacy problem are decidable in G .Proof. Suppose supp ( s ) is a clique and not synchronised. This means thereexists t ∈ supp ( s ) and x ∈ A /( supp ( s ) ∪ lk ( s )) such that [ x, t ] =
1. Since x ∉ lk ( s ) , there is a ∈ supp ( s ) such that [ x, a ] ≠
1. As K = ⟨ supp ( s )⟩ is Abelianwe have r = s n = t m a n w , for some w such that supp ( w ) = supp ( s )/{ a, t } .Then x ( a n w ) = G xt − m = t − m x = G ( a n w ) x. Hence [ x, a n w ] = G [ x, a ] ≠ [ x, a n w ] ≠ G . As [ x, a n w ] ∈ ⟨ A /{ t }⟩ , this implies that ⟨ A /{ t }⟩ does not embed in G .The converse follows from Theorem 1.4, since when K is Abelian ⟨ supp ( s )/{ t }⟩ embeds in K / M , for all t ∈ supp ( s ) , and s has order n in K / M ; as in Example 1.1.1. As K / M is a finitely generated Abelian group,it has solvable conjugacy problem, so the remaining statement follows fromTheorem 1.4 (d).In the dual case, where supp ( s ) is independent, we obtain a sufficient,but not necessary, condition for all Magnus subgroups of G to embed in G . Corollary 1.6.
Let G = ⟨ A ∣ R ⟩ and let s ∈ G be a cyclically minimal elementsuch that supp ( s ) is independent. Let r = s n , for some n ≥ , denote by N the normal closure of r in G and let G = G / N . If supp ( s ) is synchronisedthen G t = ⟨ A /{ t }⟩ embeds in G , for all t ∈ supp ( s ) , s has order n in G andthe word problem is solvable in G . Moreover, if n ≥ then the conjugacyproblem is solvable in G .Proof. This follows from Theorem 1.4, since when K is free ⟨ supp ( s )/{ t }⟩ embeds in K / M , for all t ∈ supp ( s ) and s has order n in K / M ; as in Example1.1.1. As K / M is a one-relator group it has solvable word problem, and if n ≥ Example 1.7.
1. Corollaries 1.5 and 1.6 generalise Example 1.1.1. InExample 1.1.2, supp ( s ) = A so, although it is synchronised, Theorem1.4 gives no new information. In this example lk ( b ) is not a clique butlk ( c ) = lk ( a ) = { b } , a clique. However (taking c = t in Theorem 1.2) wehave s = abc and ab ∉ Maln (⟨ b ⟩) , so s is not c -thick, and Theorem 1.2does not apply. 6. Let A = { a, b, c, t } and R = {[ t, a ] , [ a, b ] , [ b, c ]} so G = ⟨ A ∣ R ⟩ has com-mutation graph the path graph P of length 3 (and 4 vertices). If s = ct then lk ( t ) is a clique and s satisfies the hypothesis of Theorem 1.2. Set-ting r = s it follows that ⟨ a, b, c ⟩ embeds in G / N ( r ) . Note that, since G = G / N ( s ) is isomorphic to Z , the subgroup ⟨ a, c ⟩ of G does notembed in G ; showing that the condition n ≥ t be a leaf of Γand let a be the vertex of Γ to which t is adjacent; so lk ( t ) = { a } . Let w be any word in ⟨ A / st ( a )⟩ . Then s = wt satisfies all the hypotheses ofTheorem 1.2 so ⟨ A /{ t }⟩ embeds in G / N ( s n ) , when n ≥
3. Moreover, if w i ∈ ⟨ A / st ( a )⟩ ∪ ⟨ a ⟩ and ε i ∈ { ± } are chosen for i = , . . . , m , such that w i ∉ ⟨ a ⟩ , for at least one i , and s = w t ε w ⋯ w m − t e m w m is cyclicallyminimal and not a proper power in G , then again Theorem 1.2 impliesthat ⟨ A /{ t }⟩ embeds in G / N ( s n ) , for n ≥ C ′ on the left of Figure 1. This is a starred graph,therefore Theorem A of [1] applies. Every subgraph of a connectedstarred graph contains a central vertex; that is a vertex incident to allother vertices. In this example the set of central vertices is B = { a, c } .From [1, Theorem A] it follows that if r ∉ ⟨ a, b, c ⟩ then ⟨ a, b, c ⟩ embedsin G = G / N ( r ) . Moreover ( loc. cit. ), if r ∉ ⟨ a, c, d ⟩ then ⟨ a, c, d ⟩ embedsin G and if r ∉ ⟨ a, c ⟩ then ⟨ a, c ⟩ embeds in G . On the other hand, { a, c } is a synchronised clique, so from our Corollary 1.5 it follows thatif r ∈ ⟨ a, c ⟩ and t ∈ supp ( r ) then ⟨ A /{ t }⟩ embeds in G . Again, { b, d } is a synchronised and independent, so from Corollary 1.6, if r ∈ ⟨ b, d ⟩ and t ∈ supp ( r ) then ⟨ A /{ t }⟩ embeds in G . The same applies whenΓ = C , as shown on the right hand side of Figure 1. In this case, ifsupp ( r ) = { b, d } and t = b or d , then Corollary 1.6 implies that ⟨ A /{ t }⟩ embeds in G ( C )/ N ( r ) .As the set of central vertices of a starred graph always forms a syn-chronised clique, for starred graphs Corollary 1.5 always complements[1, Theorem A], as in this example.5. The previous examples of a four cycle and a four cycle with a chordmay be extended to n -cycles where n ≥
5. Let C ′ n be the n -cycle witha chord, n ≥
5, as shown on the left of Figure 2, let G ′ = G ( C ′ n ) and let7 bcd C ′ abcd C Figure 1: Example 1.7.4 s ′ = a a n − t ∈ G ′ . Then G ′ , t and s ′ satisfy the conditions of Theorem1.2, so setting r ′ = s ′ the subgroup ⟨ a , a , . . . , a n − , a n − ⟩ of G ′ embedsin G ′ = G ′ / N ( r ′ ) .Next let C n be n -cycle on the right of Figure 2, let G = G ( C n ) , let s = a a n − t ∈ G and r = s . In this case lk ( t ) is not a clique so Theorem1.2 does not apply. There is a natural surjection π ′ from G to G ′ andcomposing this with the canonical map ρ ′ from G ′ to G ′ gives a surjec-tion ρ ′ π ′ of G onto G ′ . Since π ′ maps the subgroup H = ⟨ a , . . . , a n − ⟩ of G isomorphically to the subgroup H ′ = ⟨ a , . . . , a n − ⟩ of G ′ and, fromthe above, ρ ′ embeds H ′ into G ′ , it follows that ρ ′ π ′ restricts to anembedding of H into G ′ . On the other hand, denoting the canonicalmap from G to G = G / N ( r ) by ρ , and the canonical map of G to G ′ by π we have πρ = ρ ′ π ′ . Therefore ρ restricts to an embedding of H into G . This gives a restricted Freiheitssatz in the case where lk ( t ) is not aclique.To put the conditions and conclusions of Theorem 1.2 in context we con-sider circumstances under which the Theorem fails. As in Example 1.7.5, wemay salvage some form of Freiheitssatz in the case where lk ( t ) is not a clique;so we assume that, for some vertex t of supp ( s ) , the link of t is a clique. Tokeep matters simple we also assume that s = wt , where supp ( w ) = A /{ t } . Inthis case the conditions of Theorem 1.2 fail if either(i) supp ( s ) ⊂ st ( t ) , if and only if supp ( w ) ⊂ lk ( t ) ; or(ii) s is not a t -root; or(iii) s is not t -thick. 8 C ′ n a a a n − a n − tC n a a a n − a n − Figure 2: Example 1.7.5Since we have assumed that supp ( s ) = A the first of these possibilities, to-gether with the assumption that lk ( t ) is a clique, implies that G is a freeAbelian group. In this case we defer to standard results for finitely gen-erated free Abelian groups; and no longer need our Theorems. Moreover,the form s = wt that we have chosen implies that s is not a t -root. (In-deed, from the definitions in Section 3, this holds for every s with prime t -length.) This leaves the case where s is not t -thick. When s = wt is not t -thick it follows from Lemma 3.3 that there is a element b which belongsto both Y = { y ∈ A ∶ st ( y ) = A } (the set of central vertices of Γ) and tosupp ( w )/ lk ( t ) . The centre Z of G is equal to ⟨ Y ⟩ and so, setting A = A / Y and G = ⟨ A ⟩ , we have G = G × Z and w = w w z , where w ∈ G , w z ∈ Z .As s ∉ st ( t ) , we have t ∈ A and, setting U = ⟨ lk ( t )/ Y ⟩ , Lemma 3.3 impliesthat w is in Maln G ( U ) so s = w t is t -thick. Moreover, from the definitionin Section 3.3, it follows that s is not a t -root and certainly s is not inthe star of t (in the full subgraph of Γ generated by A ). The hypothesesof Theorem 1.2 therefore hold for the element s of G and the letter t ofsupp ( s ) . Thus ⟨ A /{ t }⟩ embeds in G = G / N , where N is the normal clo-sure of s n in G , for some n ≥
3. In G we have r = s n = w nz ( w t ) n = w nz s n andso G = G / N ( r ) = G × Z / W , where W = ⟨ w z ⟩ . It follows that the subgroup ⟨ A /{ t }⟩ of G embeds in G . Also, using Theorem 1.2 (b), sN ( r ) has order n in G . 9rom the results of [1], the examples above and this analysis we concludethat a version of the Freiheitssatz holds for many one-relator quotients ofpartially commutative groups. A precise classification requires a more deli-cate analysis of the ways in which s may fail to be t -thick or a t -root (or aproof which avoids some of these conditions). To add some credence to ourclaim however, we show in Section 5 that Theorem 1.2 applies for almost allcyclically reduced words in G ( C ′ n ) (as defined in Example 1.7.5).In Section 2 we review the parts of the theory of partially commutativegroups necessary for the paper. In Section 3, we describe small cancellationtheory over the HNN-presentation of G , in situations where the there isan element t in supp ( s ) such that lk ( t ) is a clique. In Section 4 we proveTheorem 1.2 and Theorem 1.4. Finally, in Section 5 we show that Theorem1.2 is almost always applicable in the situation of Example 1.7.5. First we recall some of the notation and definitions of [3] and [12]. As above,if w ∈ ( A ∪ A − ) ∗ then supp ( w ) is the set of elements x ∈ A such that x or x − occurs in w . (In [12] α ( x ) is used instead of supp ( x ) .) For u, v ∈ ( A ∪ A − ) ∗ we use u = v to mean u and v are equal as elements of G . Equality of words u, v in ( A ∪ A − ) ∗ is denoted by u ≡ v . If g ∈ G and w ∈ ( A ∪ A − ) ∗ is a wordof minimal length representing g then we say that w is a minimal form of g (or just that w is minimal ).The Cancellation Lemma, [3, Lemma 4], asserts that if w is a non-minimalword in ( A ∪ A − ) ∗ then w has a subword xux − , where x ∈ A ∪ A − , u ∈ ( A ∪ A − ) ∗ and x commutes with every letter occurring in u . The TransformationLemma, [3, Lemma 5.5.1] (see also [12, Lemma 2.3]) asserts that, if u and v are minimal and u = v then we can transform the word u into the word v usingonly commutation relations from R (that is, without insertion or deletion ofany subwords of the form x ε x − ε , x ∈ A ). From the Transformation Lemma itfollows that if u and v are minimal and u = v then supp ( u ) = supp ( v ) and ∣ u ∣ = ∣ v ∣ (where ∣ w ∣ denotes the length of the word w ). Therefore, for anyelement g ∈ G we may define supp ( g ) = supp ( w ) , and the length l ( g ) of g as l ( g ) = ∣ w ∣ , where w is a minimal form of g . For a subset S of G we definesupp ( S ) = ∪ s ∈ S supp ( s ) .If g, h ∈ G such that l ( gh ) = l ( g ) + l ( h ) then we write gh = g ⋅ h . (Written g ○ h in [12].) It follows that gh = g ⋅ h if and only if, for all minimal forms10 and v of g and h , respectively, uv is a minimal form for gh . Clearly if w ∈ ( A ∪ A − ) ∗ is minimal and w ≡ uv then, in G , w = u ⋅ v . If k = g ⋅ h thenwe say that g is a left divisor of k (and h is a right divisor of k ).We say that h ∈ G is cyclically minimal if l ( h ) ≤ l ( h g ) , for all g ∈ G . If w ∈ ( A ∪ A − ) ∗ and w ≡ uv then we call vu ∈ ( A ∪ A − ) ∗ a cyclic permutation of w . Lemma 2.1 ([11, Lemma 2.2]) . Let w ∈ ( A ∪ A − ) ∗ be a minimal form foran element h of Γ . Then the following are equivalent.(i) h is cyclically minimal.(ii) If y ∈ A ∪ A − is a left divisor of h then y − is not a right divisor of h .(iii) All cyclic permutations of w are minimal forms.Moreover, if w is cyclically minimal then supp ( w g ) ⊇ supp ( w ) , for all g ∈ G . If w is a minimal form of a cyclically minimal element g ∈ G then we saythat w is a cyclically minimal form. From, for example, Proposition 3.9 of[9], if g ∈ G then there exist u, w ∈ ( A ∪ A − ) ∗ , with w a cyclically minimalform, such that g = u − ⋅ w ⋅ u . Thus g is cyclically minimal if and only if u =
1. Observe that if g is cyclically minimal then l ( g n ) = nl ( g ) . Thereforepartially commutative groups are torsion free. Lemma 2.2.
Let U be a subgroup of G and let D be a set of double cosetrepresentatives of U in G . The following are equivalent.(i) If d ∈ D and g ∈ U dU then l ( g ) ≥ l ( d ) .(ii) If d ∈ D then d has no left or right divisor in U .There exists a unique set D of double coset representatives of U in G suchthat (i) holds and this set satisfies(iii) ∈ D and(iv) if d ∈ D then d − ∈ D .In particular, if U dU = U d − U , for some d ∈ D , then d = . roof. For each g ∈ G we may choose a minimal form d ∈ G such that l ( d ) ≤ l ( w ) , for all w ∈ U gU . Thus there exists a set D of double cosetrepresentatives having property (i). Suppose D is such a set and d ∈ D .Then d satisfies (ii), since it has minimal length amongst elements of U dU .If d ∈ D , g ∈ U dU and l ( g ) = l ( d ) then g = udv , for some u, v ∈ U . In fact,as d has no left or right divisor in U we may choose such u and v so that g = u ⋅ d ⋅ v . Unless u = v = l ( g ) > l ( d ) , and if u = v = g = d . Therefore d is the unique element of minimal length in U dU ;and D is uniquely determined by condition (i). On the other hand, if D ′ isa set of double coset representatives satisfying (ii) a similar argument showsthat D ′ satisfies (i).Assume now that D satisfies (i). Then (iii), a special case of (i), alsoholds. If d ∈ D and U d − U contains an element g such that l ( g ) ≤ l ( d − ) then g − ∈ U dU and l ( g − ) ≤ l ( d ) ; so from the above g = d − . It follows thatif d ∈ D then so is d − , and so (iv) holds. Moreover, since d ≠ d − , we have U dU ∩ U d − U = ∅ , for all d ∈ D /{ } .The elements of A are termed the canonical generators of G ( Γ ) and asubgroup of G generated by a subset Y of G is called a canonical parabolic subgroup. Let Y ⊆ A and denote by Γ Y the full subgraph of Γ with vertexset Y . Let G ( Γ Y ) be the partially commutative group with commutationgraph Γ Y . It follows from the Transformation Lemma, [3, Lemma 5.5.1],that G ( Γ Y ) = ⟨ Y ⟩ , the canonical parabolic subgroup of G generated by Y .For cyclically minimal elements g, h ∈ G we define g ∼ h if g = u ⋅ v and h = v ⋅ u , for some u, v ∈ G . Then let ∼ be the transitive closure of ∼ anddenote by [ g ] the equivalence class of g under the equivalence relation ∼ on cyclically minimal elements. The following appears in [11, Corollary 2.4](where the set [ g ] is incorrectly defined). Corollary 2.3.
Let w, g be (minimal forms of ) elements of G and let w = u − ⋅ v ⋅ u , where v is cyclically minimal. Then there exist minimal forms a , b , c , d , d and e such that g = a ⋅ b ⋅ c ⋅ d , u = d ⋅ a − , d = d ⋅ d , w g = d − ⋅ e ⋅ d , [ e ] = [ v ] , e = v b , supp ( b ) ⊆ supp ( v ) and [ supp ( b ⋅ c ) , supp ( d )] = [ supp ( c ) , supp ( v )] = . The next result is a direct consequence of this lemma.
Corollary 2.4.
Let Y be a subset of A and let w, g be (minimal forms of )elements of G such that w and w g belong to ⟨ Y ⟩ and g has no right or leftdivisor in ⟨ Y ⟩ . Then [ supp ( g ) , supp ( w )] = . (That is [ x, y ] = , for all x ∈ supp ( g ) and y ∈ supp ( w ) .) roof. In the notation of Corollary 2.3 we have u, v, d ∈ ⟨ Y ⟩ since w and w g are in ⟨ Y ⟩ . Hence a, d , d and b are in ⟨ Y ⟩ . As g has no left or right divisorsin ⟨ Y ⟩ it follows that g = c and u = d . To complete the proof we use the factthat [ supp ( b ⋅ c ) , supp ( d )] = [ supp ( c ) , supp ( v )] = Corollary 2.5.
Let Y ⊆ A be a clique and let w, g be (minimal forms of )elements of G such that w and w g belong to ⟨ Y ⟩ . Then [ supp ( g ) , supp ( w )] = .Proof. If g ∈ ⟨ Y ⟩ , the result holds as Y is a clique. Otherwise we maywrite g = a ⋅ b ⋅ c , where a, c ∈ ⟨ Y ⟩ and b has no left or right divisor in ⟨ Y ⟩ . Then w g = w b ⋅ c ∈ ⟨ Y ⟩ . Again, as Y is a clique and c ∈ ⟨ Y ⟩ , we have w b ⋅ c = w b . From Corollary 2.4, [ supp ( b ) , supp ( w )] =
1, and as Y is a clique, [ supp ( g ) , supp ( w )] =
1, as claimed.For a subset U of A denote by Γ U the full subgraph of Γ generated by U and for a cyclically minimal word w over G set Γ w = Γ supp ( w ) . For a subset Y ⊂ A define lk ( Y ) = ∩ y ∈ Y lk ( y ) and st ( Y ) = ∩ y ∈ Y st ( y ) . For an element w ∈ G define lk ( w ) = lk ( supp ( w )) and st ( w ) = st ( supp ( w )) . From [3, Korollar 3],for a ∈ A we have C G ( a ) = ⟨ a ⟩ × ⟨ lk ( a )⟩ .As necessary we shall use a normal form for elements of the partiallycommutative group G ( Γ ) , which we now define. Let ∆ be the complementof Γ (∆ has the same vertex set as Γ and { u, v } is an edge of ∆ if and onlyif { u, v } is not an edge of Γ). For any element w of G define ∆ ( w ) to be thefull subgraph of ∆ with vertices supp ( w ) . If ∆ ( w ) is connected we call w a block . If ∆ ( w ) has connected components ∆ , . . . , ∆ k then it follows that w = w ⋅ ⋯ ⋅ w k , where ∆ ( w i ) = ∆ i and [ supp ( w i ) , supp ( w j )] =
1, for i ≠ j .We call this factorisation of w the block decomposition of w .In the sequel, we shall use the notation and results of this section withoutfurther mention. In this section we review the special case of diagrams over HNN-extensionswe need in the proof of Theorem 1.2. We refer the reader to [21, Pages 291–292] for basic results on diagrams over HNN-extensions and to [19] for a moregeneral version of the definitions given here. As above, let G be a partially13ommutative group with commutation graph Γ and canonical presentation ⟨ A ∣ R ⟩ . To realise G as an HNN-extension, given any t ∈ A , set G t = ⟨ A /{ t }⟩ and define the HNN-extensionHNN ( t ) = ⟨ G t , t ∣ t − xt = x, ∀ x ∈ lk ( t )⟩ . (3.1)By definition HNN ( t ) is the group with presentation ⟨ A ∣ R t ∪ { t − ut = u, u ∈ ⟨ lk ( t )⟩}⟩ , where R t = {[ x, y ] ∈ R ∣ x ≠ t and y ≠ t } . We may perform Tietzetransformations on the latter to replace it with the presentation ⟨ A ∣ R t ∪ { t − xt = x, x ∈ lk ( t )}⟩ . As R t ∪ { t − xt = x, x ∈ lk ( t )} = R , it follows thatHNN ( t ) = G . We call HNN ( t ) the HNN-presentation of G with respect to t .For notational simplicity, write • F for the group G = HNN ( t ) with the HNN-presentation with respectto t , • H = G t and U = ⟨ lk ( t )⟩ , and in addition • let D be a set of double coset representatives for U in H , satisfyingthe properties of Corollary 2.2, and let σ be the function from H to D given by σ ( g ) = d , where U gU = U dU .(We assume that all elements of F are represented as reduced words ofthe free product H ∗ ⟨ t ⟩ , unless an explicit exception is made.) Given p ∈ F ,the factorisation p = g t ε ⋯ t ε n g n , (3.2)where g i ∈ H and ε i ∈ { ± } , is called reduced , over F , if either n ≤ n > ≤ i ≤ n −
1, if g i ∈ U then ε i ε i + =
1. In this case we say that p has t -length ∣ p ∣ t = n ∑ i = ∣ ε i ∣ . Every element of F has a reduced factorisation, and two reduced factorisa-tions which represent the same element have the same t -length. We explicitlyallow reduced words to contain sub-words of the form t ε ut ε , where u ∈ U and ε = ± u and v are elements of F such that ∣ uv ∣ t = ∣ u ∣ t + ∣ v ∣ t , then we say theproduct uv is reduced (if and only if, given reduced factorisations u = g t ε ⋯ t ε m g m and v = h t δ ⋯ t δ n h n , m + n ≤ g m h ∉ U or ε m = δ ). More generally, if u , . . . , u n arereduced factorisations of elements of F such that ∣ u ⋯ u n ∣ t = n ∑ i = ∣ u i ∣ t (3.3)we say that the product u ⋯ u n is reduced. An element p of F is said tobe cyclically reduced if the product pp is reduced. (That is, given a reducedfactorisation p = g t ε ⋯ t δ n g n , either n ≤ g n g ∉ U or ε n = ε .) Again, for u , . . . , u n as above, the product u ⋯ u n is said to be cyclically reduced if it isreduced and u n u is reduced.The aim of the definitions in the remainder of this sub-section is to allowus to avoid products which split elements of H in a non-trivial fashion. Forexample, if a, b and g are elements of H / U with g = ab , and we have s = tgt then the factorisation pq , where p = ta and q = bt , splits the factor g = ab .With this in mind, let the element p ∈ F have reduced factorisation (3.2). If g = p is said to begin with a t -letter and if g n = p is said to end with a t -letter. Now if u , . . . , u k are reduced factorisations of elements of F then the product u ⋯ u k is said to be right integral at i , where 1 ≤ i ≤ k , if itis cyclically reduced and u i ends with a t -letter and left integral at i if u i + begins with a t -letter (subscripts modulo k ). The product u ⋯ u k is integral at i if it is either left or right integral at i . (Note that this is not quite the sameas the definition in [19].) The product u ⋯ u k is said to be (right) integral if itis (right) integral at i , for all i ∈ { , . . . , k } . We say that u is a (right) integral subword of w if w = u ⋅ v , where uv is a (right) integral factorisation. [Notethat the integral property depends on the choice of reduced factorisation offactors. For example, w = tuht , where u ∈ U , h ∈ H / U , can be written asthe product pq , where p = tu and q = ht . This product is not integral, butwe may write p = ut in F and using this factorisation of p the product pq isintegral.] ( t ) Let G be a group and K a non-trivial subgroup of G . Define Maln G ( K ) = { x ∈ G ∶ x − Kx ∩ K = { }} . It follows that x ∈ Maln G ( K ) if and onlyif x − ∈ Maln G ( K ) . Also, x ∈ Maln G ( K ) if and only if KxK ⊆ Maln G ( K ) .Indeed, if x ∈ Maln G ( K ) and u, v ∈ K then ( uxv ) − K ( uxv ) ∩ K = v − x − Kxv ∩ K = v − x − Kxv ∩ v − Kv = v − ( x − Kx ∩ K ) v = { } .15 emma 3.1. Let V be a subgroup of the group G and let D V be a set of rep-resentatives of double cosets V gV in G , satisfying the conditions of Corollary2.2. If a, b ∈ Maln G ( V ) and u, v ∈ V are such that aub − = v then V aV = V bV and there exist d ∈ D V , u ′ , v ′ ∈ V such that a = ( vv ′ ) ⋅ d ⋅ u ′ and b = v ′ ⋅ d ⋅ ( u ′ u ) .Proof. Note first that x ∈ Maln G ( V ) if and only if V xV ⊆ Maln G ( V ) . As aub − ∈ V we have V aV = V bV . Let d the element of D V such that V aV = V dV = V bV . Then there exist a l , b r , u ′ , v ′ ∈ V such that a = a l ⋅ d ⋅ u ′ and b = v ′ ⋅ d ⋅ b r . By assumption a l du ′ ub − r d − v ′− = aub − = v, and since a, b ∈ Maln G ( V ) we have d ∈ Maln G ( V ) , so u ′ ub − r = a − l vv ′ =
1, fromwhich the final statement follows.
Definition 3.2.
Let w = g t ε ⋯ g k − t ε k g k be a reduced factorisation of anelement of HN N ( t ) , where g i ∈ G t and ε i = ±
1, and let U = ⟨ lk ( t )⟩ ≤ G t . Wesay w is t -thick if g i ∈ U ∪ Maln G t ( U ) , for i = , . . . , k . We say w is cyclically t -thick if w is cyclically reduced, t -thick and g k g ∈ U ∪ Maln G t ( U ) .The notion of t -thickness is well-defined since, if w has another reducedfactorisation w = h t δ ⋯ h m − t δ m h m , where δ i = ±
1, then m = k , ε i = δ i andthere are elements u , . . . , u m − ∈ U such that g = h u − , g m = u m − h m and g i = u i − h i u − i , for 1 ≤ i ≤ m −
1. Hence, from the comment preceding Lemma3.1, h i ∈ U ∪ Maln G t ( U ) if and only if g i ∈ U ∪ Maln G t ( U ) . Lemma 3.3.
Let G = ⟨ A ∣ R ⟩ be a partially commutative group, let B be asubset of A and let K = ⟨ B ⟩ . If B is a clique and w is an element of G / K then the following are equivalent.(i) w belongs to Maln G ( K ) .(ii) For all b ∈ B , there exists x ∈ supp ( w )/ B such that [ x, b ] ≠ .Proof. To see that (i) implies (ii), assume w ∈ Maln G ( K ) and let b ∈ B . Then w − bw ≠ b , so w ∉ C G ( b ) and it follows that there exists x ∈ supp ( w )/ B suchthat [ x, b ] ≠ w ∈ G / K such that w ∉ Maln G ( K ) . As-sume, in order to obtain a contradiction, that w is of minimal length amongstall elements not in K ∪ Maln G ( K ) which satisfy (ii). As w ∉ Maln G ( K ) , thereexists 1 ≠ v ∈ K such that w − vw ∈ K . We may write w = w ⋅ w , where w ∈ K w is non-trivial and has no (non-trivial) left divisor in K . By minimal-ity of w , we conclude that w =
1, and so we may assume w has no leftdivisor in K . As w has no left divisor in K , (from the Cancellation Lemma) w − vw = w − ⋅ v ⋅ w , unless w has a left divisor x ∈ A ± such that [ x, v ] = x must commute with all elements of supp ( v ) . We maywrite w = x ⋅ w and, by the assumption on w , supp ( w ) must contain anelement which does not commute with an element of supp ( v ) ; so w ∉ K .Moreover w − vw = w − vw ∈ K . It follows that w is shorter than w andsatisfies all the same properties, contrary to the choice of w . We concludethat w − vw = w − ⋅ v ⋅ w , so that w ∈ K , a contradiction. Therefore, for all w not in K ∪ Maln G ( K ) property (ii) fails, as required. In this section we follow [21, Chapter V, Section 11]. Assume that G isexpressed as the HNN-extension F = HNN ( t ) , and set H = G t and U = ⟨ lk ( t )⟩ ,as above. A set R of elements of F is said to be symmetrised if every elementof R is cyclically reduced and, for all r ∈ R , all cyclically reduced conjugatesof r and r − are in R . The symmetrised closure ̃ S of a set S of cyclicallyreduced elements of F is the smallest symmetrised subset containing S ; andconsists of all cyclically reduced conjugates of s , for all elements s ∈ S ∪ S − .From Collins’ Lemma (see [21, Theorem 2.5, p. 185]) if r ∈ F is cyclicallyreduced and ends in t ± then a cyclically reduced element s ∈ F , ending in t ± , is a conjugate of r if and only if it may be obtained by taking a cyclicpermutation of r and then conjugating by an element of U . This means thateven though R may be finite its symmetrised closure ̃ R is, in general, infinite. Definition 3.4.
Let ̃ R be a symmetrised subset of F . An element p ∈ F is a piece ( over ̃ R ) if there exist distinct elements r , r ∈ ̃ R , and elements u , u of F , such that r i factors as a reduced product r i = F pu i , for i = Definition 3.5.
Let ̃ R be a symmetrised subset of F and let m be a positiveinteger. If w ∈ ̃ R has a reduced factorisation w = p ⋯ p k , where p i is a pieceover ̃ R , then w is said to have a k -piece factorisation. If no element of ̃ R has a k -piece factorisation, where k < m , then ̃ R is said to satisfy smallcancellation condition C ( m ) .For details of disc diagrams over HNN-extensions we refer the reader to[21, pp. 291–294], and the references therein. We outline here only what17e need below, in particular defining diagrams on a disk. Let ̃ R be a sym-metrised subset of F and w a reduced factorisation of an element of F . An ̃ R diagram with boundary label w over F consists of the following. A fi-nite 2-complex M with underlying space Σ a compact, connected, simplyconnected, subset of the real plane; and a distinguished vertex O of M on ∂ Σ. (0-cells, 1-cells and 2-cells of M are called vertices, edges and regions,respectively.) A labelling function φ from oriented edges of M to H ∪ { t ± } .(Strictly speaking φ maps edges of M to freely reduced words in ( A /{ t }) ± orelements of { t ± } .) For an oriented edge a we write ¯ a for the same edge giventhe opposite orientation (if f ∶ [ , ] → a is a homeomorphism determiningthe oriented edge a then ¯ a is the edge determined by the map ¯ f mapping x ∈ [ , ] to f ( − x ) .) A boundary cycle of a region ∆ of M is a closedpath traversing the boundary ∂ ∆ of ∆ exactly once (beginning and endingat a vertex v of M ). A boundary cycle of M is a closed path traversing theboundary ∂ Σ of Σ exactly once, beginning and ending at O . In addition thefollowing conditions must be satisfied.(1) If a is an oriented edge with label w = φ ( a ) then φ ( ¯ a ) = w − .(2) If M has a boundary cycle p = a , . . . , a n , beginning and ending at O ,then the product φ ( a ) ⋯ φ ( a n ) is reduced and equal, in F , to w or w − .(3) If ∆ is a region of M and ∆ has a boundary cycle p = a , . . . , a n , thenthe product φ ( a ) ⋯ φ ( a n ) is reduced and equal in F to an element ̃ R .It follows (see for example [21, Theorem 11.5, p. 292]) that there exists adiagram D over F with boundary label w if and only if w = F / N , where N is the normal closure of R in F .If the label φ ( a ) of an edge a is in H then a is called an H -edge , andif φ ( a ) = t ± then a is called a t -edge . If v is a vertex in the boundary ∂ ∆of a region ∆ and v is incident to a t -edge of ∂ ∆, then we call v a primaryvertex , with respect to ∆. The products φ ( a ) ⋯ φ ( a n ) appearing in (2) and(3) are called boundary labels of M and ∆ respectively. In our case ̃ R is thesymmetrised closure of the single element r , which is cyclically minimal asan element of the partially commutative group G . Every element w of ̃ R istherefore equal in F to u − r ′ u , for some u ∈ U and cyclic permutation r ′ of r or r − . Let ∆ be a region with boundary label equal in F to r ′ , when readfrom an appropriate vertex o on its boundary. Attaching an edge e labelled u by identifying its initial vertex with o , we obtain a diagram with a single18 qq ∆ ∆ q Figure 3: Cancelling regions: φ ( µ ) ≡ p, φ ( ν ) ≡ φ ( ν ) ≡ q region ∆ and boundary label w . As r ′ is cyclically minimal we may replace(3) with(3’) If ∆ is a region of M and ∆ has a boundary cycle p = a , . . . , a n , then theproduct φ ( a ) ⋯ φ ( a n ) is reduced and equal in F to a cyclic permutationof r or r − .As usual we shall restrict to diagrams which do not have pairs of re-dundant regions of the following sort. Let ∆ and ∆ be distinct regionsof the diagram D with boundary cycles ρ = µν and ρ = µν (where µ, ν i are subpaths of ρ i ) meeting in the connected boundary component µ . If φ ( ν ) ≡ φ ( ν ) (as words in the free monoid ( A ∪ A − ) ∗ ) then, as in Figure 3,we may remove the interior of ∆ ∪ ∆ and identify the sub-paths ν and ν ofthe boundary cycles of ∆ and ∆ , via their (equal) labels φ ( ν ) and φ ( ν ) ,to leave a new diagram D ′ with the same reduced boundary label as D butfewer regions. The modification of D to produce D ′ is called a cancellationof regions . A diagram in which no cancellation of regions is possible is called reduced .It turns out that if a pair of distinct regions ∆ and ∆ have boundarycycles ρ = µν and ρ = µν , as above, but instead of satisfying φ ( ν ) ≡ φ ( ν ) the labels satisfy only φ ( ν ) = F φ ( ν ) then, after some minor modifications tothe diagram, which do not alter the label of its boundary, we may again cancelthe regions ∆ and ∆ . The general process of modification is described in19 x yb ay xb ∆∆ ′ ∆ ′′ yx − ∆∆ ′ ∆ ′′ Figure 4: Shuffling in the boundary label of ∆.[21, Page 292]. Here we describe modifications sufficient for our particularcase. There are two types of these; the first of which results in a new diagramin which the product of edge labels around a boundary cycle may not be freelyreduced. The second type may then be used to freely reduce labels on edgesof boundary cycles if necessary.
Shuffling labels.
Suppose that a subpath µ of the boundary of a region∆ has label with subword xy , where x, y ∈ A ± and xy = yx . Then we maymodify the diagram D (without altering the element of F represented byits boundary label) to obtain a new diagram in which this subword xy isreplaced by yx , as follows. If xy occurs as a subword of the label φ ( e ) = axyb of a single edge e of µ then we modify φ by setting φ ( e ) = ayxb . Otherwise µ contains a subpath e e , where e and e are edges, such that φ ( e ) = ax and φ ( e ) = yb . In this case e ⊆ ∆ ∩ ∆ ′ and e ⊆ ∆ ∩ ∆ ′′ , for some regions ∆ ′ and ∆ ′′ and we modify D as shown in Figure 4. There is a choice here: wemay label the new edge, in ∆ ′ ∩ ∆ ′′ with yx − , as shown, or with x − y . Inboth cases we refer to this modification as a shuffle of the label of ∆. Free reduction of labels.
Let D be an ̃ R -diagram over F and let ∆ bea region of D . Suppose a subpath µ of a boundary cycle of ∆ has labelcontaining a subword xx − , where x ∈ A ± . If xx − is a subword of thelabel φ ( e ) = axx − b of a single edge e , where a, b ∈ H , then we modify φ bysetting φ ( e ) = ab . (Such edges do not occur in D itself, but may arise aftera shuffling of labels as above.) Otherwise µ contains a subpath e e of twoedges such that φ ( e ) = ax and φ ( e ) = x − b , where a, b ∈ H (and if x = t ± then necessarily a = b =
1, since φ ( e i ) is by definition either an element of H x x − b a b ∆∆ ′ ∆ ′′ x ∆∆ ′ ∆ ′′ Figure 5: Free reduction of the boundary label of ∆or of { t ± } ). In this case we modify D as shown in Figure 5. We call this free reduction of the label of ∆. Note that this reduces the sum of lengthsof boundary labels of regions of D , where length here means length as aword over A ∪ A − , so we may repeat free reductions of labels until no suchsubpaths occur in the boundary label of any region.Note that the boundary label of D is unaffected by such modifications.Both free reduction and shuffling of labels of D result in a new diagram D ′ ,which has regions in one to one correspondence with the regions of D . If∆ and ∆ are regions of D with a common boundary component µ andboundary cycles µν and µν , respectively, such that φ ( ν ) = F φ ( ν ) thenthese modifications may be used alter the diagram so that φ ( ν ) ≡ φ ( ν ) ;and ∆ and ∆ become cancelling regions. We say a diagram D is stronglyreduced if it satisfies the condition that, whenever ∆ and ∆ are distinctregions with a common boundary component µ , of positive length, the label φ ( µ ) is a piece. From [21, Theorem 11.5] we obtain the following. Proposition 3.6.
Let R be a subset of F and ̃ R its symmetric closure. Thereexists a strongly reduced ̃ R diagram D over F with boundary label w if andonly if w = in F / N , where N is the normal closure of R in F . In this section we specialise the methods of [19] to the case in hand. Asabove, let D be a set of double coset representatives of U in H satisfying theproperties of Corollary 2.2, and let σ ∶ H → D be the function mapping h ∈ H to d ∈ D such that U hU = U dU . Partition the set D into disjoint subsets D + ,21 − and { H } such that d ∈ D + if and only if d − ∈ D − . Now define the freeproduct F D = F ( D + ) ∗ ⟨ t ⟩ , where F ( D + ) is the free group on D + .The canonical map from D + to F ( D + ) extends to an injective map from D to F ( D + ) by mapping d − in D − to d − ∈ F ( D + ) , for all d ∈ D + , andmapping 1 H to the empty word. Composing this map with the canonicalinjection from F ( D + ) into F D we have an injective map ι from D to F D . Thecomposition ι ○ σ is then a map from H to F D , which we shall now also referto as σ . We extend σ to a function from F to F D as follows. If p ∈ F hasreduced factorisation p = g t ε ⋯ t ε n g n then we define σ ( p ) = σ ( g ) t ε ⋯ t ε n σ ( g n ) . If g, h ∈ H and u ∈ U are such that h = ug then σ ( g ) = σ ( h ) , so σ is awell-defined map from F to F D . To simplify notation we may write ¯ p for σ ( p ) . Definition 3.7. If p ∈ F is such that σ ( p ) is a not a proper power in F D then we say that p is a t -root .Note that a cyclically reduced element of F of t -length at least 1, whichends in a t letter and which is a t -root; cannot be a proper power in F .Indeed, suppose that p is a cyclically reduced element of F ending in a t -letter, with ∣ p ∣ t ≥
1, and p = q n , for some element q ∈ F and positive integer n .It follows, from [21, Chapter IV, Section 2], that q is also cyclically reduced,of positive t -length and ends in a t -letter. Therefore σ ( p ) = σ ( q n ) = σ ( q ) n ,so p is not a t -root. (Note that σ is not a group homomorphism.)Let n ≥ s be a cyclically reduced wordof non-zero t -length, s = h t ε ⋯ h m − t ε m , where h i ∈ G t and ε i ∈ { ± } , for i = , . . . , m −
1. Let r = s n , let N be the normal closure of R = { r } in G andlet G = G / N . If r and r are elements of the symmetrised closure ̃ R , of R ,such that σ ( r ) = σ ( r ) we say that r and r are in periodic position . Lemma 3.8.
Assume s has t -length m ≥ , ends in a t -letter, is cyclically t -thick and a cyclically reduced t -root. Let r = s n , for some n ≥ and R = { r } .If r and r are cyclic permutations of r or r − such that r = pq and r = pq and r and r are in periodic position, then φ ( q ) = F φ ( q ) .Proof. We may assume that r = r and r is obtained from r ε , where ε = ±
1, by cyclic permutation. If ε = − σ ( r ) is a cyclic permutation of σ ( r − ) = σ ( r ) − and σ ( r ) = σ ( r ) in F D . This implies σ ( r ) =
1, so r ∈ U , a22ontradiction. Therefore ε = s is a t -root it follows that r is obtainedfrom r by a cyclic permutation of length k ∣ s ∣ = km , for some positive integer k . Hence r = F r and so φ ( q ) = F φ ( q ) . Corollary 3.9. If M is a strongly reduced diagram and µ is the commonboundary component of regions ∆ and ∆ , with boundary cycles ρ = µν and ρ = µν , respectively, then φ ( ρ ) and φ ( ρ ) are not in periodic position.In particular, φ ( µ ) is a piece.Proof. As M is strongly reduced φ ( µ ) is a piece, so from the previous lemmaand condition (3’) for diagrams, φ ( ρ ) and φ ( ρ ) are not in periodic position.For the remainder of this section assume that s and r are elements of F satisfying the hypotheses of Lemma 3.8. Let F ( X ) be the free groupon a basis X and let w be an an element of F ( X ) , written as a reducedword. We say a cyclic subword a of w is uniquely positioned if no othercyclic subword of w or w − is equal to a . As ¯ s is cyclically reduced andnot a proper power in F D , it follows from [10, Theorem 2.1] that ¯ s has acyclic permutation with reduced factorisation equal to ¯ a ¯ b , where ¯ a and ¯ b are non-empty uniquely positioned subwords of ¯ s . Therefore s has a cyclicpermutation ˜ s , with a reduced factorisation ˜ s = ab , such that a and b areintegral cyclic subwords of s , σ ( a ) = ¯ a and σ ( b ) = ¯ b . In this definition,the words a and b are not necessarily uniquely determined. Suppose s hasa cyclic permutation ˜ s which factors as ˜ s = w t ε ut ε w , where u ∈ U , then σ ( ˜ s ) = ¯ w t ε t ε ¯ w . If, in this expression, ¯ a ends with the first occurrence of t ε and ¯ b begins with the second, then a may be chosen to end t ε u or t ε , with b beginning t ε or ut ε , respectively. To avoid such ambiguity, in this situation,we always choose a to be right integral, that is a ends in t ε . Similarly, if thissituation arises with the roles of a and b interchanged, we choose b to be rightintegral. With this convention, the words a and b are uniquely determinedcyclic subwords of s .Now let ∆ be a region, of an ̃ R diagram over F , with boundary cycle ρ ,such that φ ( ρ ) = r ∈ ̃ R . By changing the base point of ρ , if necessary, wemay assume that r has reduced factorisation r = ( g t ε ⋯ g m − t ε m ) n , where s = g t ε ⋯ g m − t ε m is a cyclic permutation of s or s − . The primary verticesof ∆ form a subsequence δ , γ . . . , δ mn , γ mn of the vertex sequence of ρ , suchthat, setting γ = γ mn , φ ([ δ i + km , γ i + km ]) = t ε i and φ ([ γ i − + km , δ i + km ]) = g i − ,for 1 ≤ i ≤ m and 0 ≤ k ≤ n − [ x, y ] denotes the subpath of ρ from23ertex x to vertex y ). In this notation, if g i = γ i + km and δ i + + km are the same.Let φ = σ ○ φ , defined on subintervals of ∂ ∆ beginning and ending atprimary vertices; so φ ([ γ i − + km , δ i + km ]) = ¯ g i − , and φ ([ γ km , γ ( k + ) m ]) = s = ¯ g t ε ⋯ ¯ g m − t ε m ∈ F D , for 0 ≤ k ≤ n −
1. From the above, s has a cyclic permutation which factorisesas ab , where a and b are integral cyclic subwords of s , and s has a correspond-ing cyclic permutation which factorises as ¯ a ¯ b , where ¯ a , ¯ b are uniquely posi-tioned cyclic subwords of a cyclic permutation of ¯ s . Consequently, ¯ r = ( ¯ s ) n has a cyclic permutation which factorises as ( ¯ a ¯ b ) n . Therefore there is a subse-quence α , β , . . . α n , β n of the sequence δ , γ , . . . , δ mn , γ mn of primary verticesof ∆, such that φ ([ α i , β i ]) = a and φ ([ β i , α i + ]) = b , and ¯ φ ([ α i , β i ]) = ¯ a and¯ φ ([ β i , α i + ]) = ¯ b , for i = , . . . , n ; as illustrated in Example 3.10. Example 3.10.
Figure 6 illustrates a possible distribution of the first fourof the vertices α i , β i on ∂ ∆, assuming that s = g tg t − , g = g ′ g ′′ , g = g ′ g ′′ ,¯ a = t ¯ g t − and ¯ b = t − ¯ g . In the diagram, primary vertices are those with thelarger diameter.We define Sep ( ∆ ) to beSep ( ∆ ) = { α i , β i ∶ ≤ i ≤ n } . If µ is a simple cyclic sub-path of ρ then we may assume that µ is a pathfrom points α to β , on ∂ ∆ (when read with the same orientation as ρ ). Wedefine Sep ( µ ) = Sep ( ∆ ) ∩ [ µ /{ β }] . (The terminal point of µ never contributes to Sep ( µ ) .) Thus, if ρ has a cyclicsubpath ξ with decomposition ξ = µν , where µ and ν have disjoint interiors,then ∣ Sep ( ξ )∣ = ∣ Sep ( µ )∣ + ∣ Sep ( ν )∣ . Lemma 3.11.
Let s and r be as in Lemma 3.8. Assume ∆ is a region of astrongly reduced ̃ R diagram over F , which has boundary cycle ρ , with a sub-interval µ such that φ ( µ ) = p , where p is a piece over ̃ R . Then ∣ Sep ( µ )∣ ≤ . t g ′ g ′′ t t g ′ g ′′ t g t tα α β β s s ∆ a ab ¯ g t ¯ g ¯ g t t t ¯ g t tα α β β ∆¯ a ¯ a ¯ b Figure 6: Labelling with ¯ φ to define Sep ( ∆ ) roof. We may assume that p ∉ H and p ∉ { t ± } , since otherwise the resultholds immediately. Let p be the maximal integral subword of p (whichmust be non-empty under this assumption). Then p is a piece over ̃ R ,and as the diagram is strongly reduced, p is not in periodic position, so ¯ p is a piece over the symmetrised closure of σ ( ̃ R ) . As ¯ a and ¯ b are uniquelypositioned subwords of ¯ s , neither can be a subword of ¯ p . As a and b areintegral subwords of the boundary label of ∆, by definition of p , if a or b is a subword of p then it is a subword of p ; so ¯ a or ¯ b is a subword of¯ p , a contradiction. Thus neither a or b is a subword of p , and the Lemmafollows.Following [22] we make the following definition. Definition 3.12.
Let ∆ be a region of a diagram M over ̃ R and supposethat ∆ has a boundary cycle that factors as µ ⋯ µ k β , where µ i is the commonboundary component of ∆ and a region ∆ i , and β is a component of aboundary cycle of M . Then ∆ is called a k -shell, with boundary component β . Proposition 3.13.
Let s , r = s n and R be as in Lemma 3.8.(a) If ∆ is a k -shell of a strongly reduced diagram over ̃ R , with boundarycomponent β then l ( φ ( β )) ≥ ⎧⎪⎪⎨⎪⎪⎩( n − k + ) l ( s ) + , if k is even ( n − k + ) l ( s ) + , if k is odd . (3.4) (b) ̃ R satisfies small cancellation condition C ( n ) . We remark that this theorem applies to diagrams over ̃ R , and not todiagrams over the free presentation for G . Indeed, G may contain subgroupsisomorphic to F × F , in which case, as shown by Bigdely and Wise [5], nofree presentation for G satisfies C ( ) . Proof. (a) Let ∆ have boundary cycle ρ with decomposition ρ = µ ⋯ µ k β , asin the Definition of k -shell. From Corollary 3.9, φ ( µ i ) = p i is a piece over ̃ R , for 1 ≤ i ≤ k . Therefore Lemma 3.11 implies that Sep ( µ ⋯ µ k ) ≤ k .As Sep ( ∆ ) = n , this gives Sep ( β ) ≥ n − k . We claim that if ν is asubpath of ρ with Sep ( ν ) ≥ K then l ( φ ( ν )) ≥ (( K − )/ ) l ( s ) +
2, if26 is even, and that l ( φ ( ν )) ≥ (( K − )/ ) l ( s ) +
1, if K is odd. Thiscertainly holds if K = , a , b may have length 1.An elementary induction then shows that the claim holds for all K ≥ K = n − k , the first part of the Lemma follows.(b) Let w be an element of ̃ R with a k -piece factorisation w = p ⋯ p k . Firstwe note that, if p is a piece over ̃ R , and v = pq ∈ ̃ R , where v = u − v u , forsome cyclically minimal element v and u ∈ U , then v = upu − ( uqu − ) and it follows that u − pu is also a piece over ̃ R . Hence we may assumethat w is cyclically minimal and we may form a diagram with a singlecell ∆ and boundary cycle ρ such that ρ = µ ⋯ µ k , where φ ( µ i ) = p i .Then 2 n = Sep ( ∆ ) = ∑ ki = Sep ( µ i ) ≤ k , proving the second part of thelemma. Proof of Theorem 1.2.
Let r = s n , R = { r } and G t = ⟨ A /{ t }⟩ . If lk ( t ) isempty then G = G t ∗ ⟨ t ⟩ and Theorem 1.2 follows from well known results onone-relator products. We may therefore assume lk ( t ) is a non-empty clique.(a) Let w ∈ (( A /{ t }) ± ) ∗ be a freely reduced word such that w = G . From Proposition 3.6, there exists a strongly reduced ̃ R -diagram M over F , with boundary label w . From Proposition 3.13, ̃ R satis-fies C ( n ) , where n ≥
3. Therefore Greendlinger’s Lemma for C ( ) diagrams (see for example [22, Theorem 9.4]) implies that M has a k -shell, for k ≤
3. Hence φ ( β ) has length at least ( n − ) l ( s ) ≥ l ( s ) , sois a word of positive t -length, contrary to the assumption that w ∈ G t .(b) Let M be a strongly reduced ̃ R -diagram, with boundary label s k , forsome positive integer k < n . As ∂M has length kl ( s ) , M must containat least 2 regions and since s k is freely reduced, from Greendlinger’sLemma for C ( ) diagrams again, M must contain at least three d -shells, where d ≤
3, so ∣ ∂M ∣ ≥ ( n − ) l ( s ) . As n ≥
3, 3 ( n − ) ≥ n , sothis implies kl ( s ) = ∣ ∂M ∣ ≥ nl ( s ) , a contradiction.For the final statement let w ∈ F ( A ) . We may assume without loss of gen-erality that w is a minimal form for a non-trivial element of F . If w = / N then there exist elements y , . . . y k ∈ F ( A ) and ε , . . . , e k ∈ { ± } , suchthat w = k ∏ i = y − i r ε i y i . (4.1)From this expression we construct, in the usual way, a diagram over ̃ R , withboundary label w and regions ∆ i , i = , . . . , k . After deleting cancellingregions if necessary we obtain a reduced diagram M with at most k regionsand boundary label w . From Proposition 3.6 we may assume M is stronglyreduced. As n ≥
4, from Proposition 3.13, ̃ R satisfies C ( ) and so fromstandard small cancellation arguments (e.g. [15, page 246, Proposition 27])the number of regions of M is at most 8 l ( w ) . To decide whether or not w = F / N therefore amounts to deciding whether or not one of equations (4.1),in variables y i , with k ≤ l ( w ) has solution. From [8] equations are decidableover partially commutative groups, so the latter problem is decidable. Proof of Theorem 1.4.
As above, we may decompose G as a free productwith amalgamation, G = A ∗ U A . By definition of lk ( s ) , we have A = U × K so G = G / N ≅ ( U × K / M ) ∗ U A . If ⟨ Y ′ ⟩ embeds in K / M it follows that ⟨ lk ( s ) ∪ Y ′ ⟩ embeds in U × K / M , so ⟨( A / supp ( s )) ∪ Y ′ ⟩ embeds in ( U × K / M ) ∗ U A , as required. Moreover, if s has order n in K / M , it follows that s also has order n in G .Given an element w ∈ F ( A ) we may write w in the form w = g u h ⋯ g k u k h k u k + = [∏ ki = ( g i u i h i )] u k + , where k ≥ u i ∈ U , g i ∈ K and h i ∈ A ; using the Transformation Lemma (see Section 2), and the factthat [ K, U ] =
1. Moreover, we may assume w is a minimal length wordin ( A ∪ A − ) ∗ representing its class as an element of G , and that h i has noleft or right divisor in U , for all i . Then w ∈ N if and only if g i ∈ M , for i = , . . . , k , and u h ⋯ u k h k u k + = A . If the word problem is solvablein K / M we may decide whether or not g i ∈ M ; and so the word problem isdecidable in G .Given elements v, w ∈ F ( A ) , to decide whether or not v and w are conju-gate in G , we apply [23, Theorem 4.6, page 212]. We may first replace v and w by elements of F ( A ) representing cyclically minimal elements of G . To28implify notation let A = U × K / M . If v ∈ U , and v is conjugate to w , then w represents an element of A ∪ A and there exists a sequence v, v , . . . v l = w ,where v i ∈ U , for i < l and consecutive terms are conjugate in A j , for j = A = U × K / M , this implies that v is conjugate to w in A , and thismay be effectively verified, in the partially commutative group A . If v ∉ U but v ∈ A ∪ A then v and w are conjugate only if they belong to the samefactor and are conjugate in that factor. If v and w belong to A we mayverify if they are conjugate or not, as in the previous case. If both belong to A then we use the solvability of the conjugacy problem in K / M to decidewhether or not they are conjugate.Otherwise v and w are cyclically minimal as elements of A ∗ U A which wemay write (after a cyclic permutation if necessary) w = g u h ⋯ g k u k h k u k + and v = g ′ u ′ h ′ ⋯ g ′ l u ′ l h ′ l u ′ m + , where k > u i , u ′ j ∈ U , g i , g ′ j ∈ K , h i , h ′ j ∈ A and h i , h ′ j have no left or right divisors in U , for all i, j . We may identifyany i such that g i = G w , replacing g i by 1, using the factthat the conjugacy (so word) problem is solvable in K / M . After repeatingthis process sufficiently often we may assume that g i ∉ M , for all i ; withoutaltering the conjugacy class of w in G . Similarly we may assume no g ′ i ∈ M . Note that w is cyclically reduced as an element of ( U × K / M ) ∗ U A ifand only the first and last letters come from distinct factors (that is g ≠ h k ≠
1) and we may assume w begins with an element of K / M , bycyclically permuting if necessary. We may therefore decide whether or not w is cyclically reduced and, if it is not, replace it by a cyclic permutation withfewer factors. Hence we may assume that w and v have been rewritten asrepresentatives of cyclically reduced forms (in the sense of free products withamalgamation) of elements of A ∗ U A , both beginning with an element of K . Then v is conjugate to w if and only if a cyclic permutation of v , followedby conjugation by an element of U , results in a representative of w . We mayassume that v as written above is an appropriate cyclic permutation and itremains to decide if u − vu = w , for some u ∈ U .First consider the case where u = v = w . Then we have g u h ⋯ g k u k h k u k + = g ′ u ′ h ′ ⋯ g ′ l u ′ l h ′ l u ′ l + . (4.2)We claim that this holds if and only if k = l , g i = g ′ i , h i = h ′ i , [ h i , u ′ i + ⋯ u ′ k + u − k + ⋯ u − i + ] =
1, for i = , . . . , k and u ′ ⋯ u ′ k + = u ⋯ u k + . Tosee this, first observe that if (4.2) holds then we have h ′ l u ′ l + u − k + h − k ∈ U , andsince h k and h ′ l have no left or right divisors in U , this forces h ′ l = h k and29 h k , u ′ l + u − k + ] =
1. Therefore g u h ⋯ h k − u k u k + g k = g ′ u ′ h ′ ⋯ h l − u ′ l u ′ l + g ′ l , from which we have g ′ l g − k ∈ U , so g ′ l = g k . Continuing this way, it follows that k = l , g i = g ′ i and [ h i , u ′ i + ⋯ u ′ k + u − k + ⋯ u − i + ] =
1, for i = , . . . , k . This in turn,together with (4.2), implies that u ′ ⋯ u ′ k + = u ⋯ u k + . Conversely, if all theseconditions hold, then so does (4.2), by direct computation.In the general case, u − vu = w if and only if g u h ⋯ g k u k h k u k + = g ′ u − u ′ h ′ ⋯ g ′ l u ′ l h ′ l u ′ l + u. (4.3)From the claim above this holds if and only if k = l , g i = g ′ i , h i = h ′ i , [ h i , u ′ i + ⋯ u ′ k + uu − k + ⋯ u − i + ] = , for i = , . . . , k and u − u ′ ⋯ u ′ k + u = u ⋯ u k + . That is, u − vu = w if and only if k = l , g i = g ′ i , h i = h ′ i and the system of k + x − ( u ′ i + ⋯ u ′ k + ) − h i u ′ i + ⋯ u ′ k + x = ( u i + ⋯ u k + ) − h i u i + ⋯ u k + ,x − u ′ ⋯ u ′ k + x = u ⋯ u k + , (4.4)in the variable x , where 1 ≤ i ≤ k , has a solution in the group U . In theterminology of [8], U is a normalised rational subset of the partially commu-tative group A . Thus (4.4) is a system of equations over A with normalisedrational constraints. From [8, Corollary 1], the system (4.4) is decidable.Therefore we may decide whether or not v is conjugate to w by an elementof U . As v has only finitely many cyclic permutations, combining the abovewe have a solution to the conjugacy problem in G .The final part of this proof is unsatisfactory, in that it uses the decidabil-ity of all systems of equations over partially commutative groups, to lift todecidability of the conjugacy problem in G . We therefore give an alternativeargument, which depends only on decidability of the conjugacy problem. First let a ∈ A ∩ K , such that a ∉ M . Then, replacing g i and g ′ i by a , for all i in (4.3), the argument above shows that au h ⋯ au k h k u k + = u − au ′ h ′ ⋯ au ′ l h ′ l u ′ l + u, (4.5) The authors are grateful to Armin Weiss for pointing out this approach. u ∈ U , if and only if k = l , h i = h ′ i and the system of equations(4.4) has a solution x ∈ U . Since a ∉ U and h i has no left or right divi-sor in U the subgraph of the non-commutation graph ∆ of G with vertices { a } ∪ ⋃ ki = supp ( h i ) is connected. Therefore ˆ w = au h ⋯ au k h k u k + has blockdecomposition ˆ w = b b ⋯ b p , where the b i are blocks, b = av h ⋯ av k h k v k + ,for some v i ∈ U , and b i ∈ U , for i ≥
1. If (4.5) holds, then it follows from [12,Proposition 5.7] that its right hand side, ˆ v = au ′ h ⋯ au ′ k h k u ′ k + , has blockdecomposition ˆ v = b ′ b ′ ⋯ b ′ p , where the b ′ i are blocks, b ′ = av ′ h ⋯ av ′ k h k v ′ k + ,for some v ′ i ∈ U and b ′ i ∈ U , for i ≥
1. Moreover, in this case, after reorderingthe b ′ i s , i ≥
1, if necessary, the blocks b i and b ′ i are cyclically minimal andconjugate. From the Transformation Lemma, b is conjugate to b ′ if and onlyif there is an element z ∈ U ∩ ⟨ supp ( b )⟩ such that z − b ′ z = b , and for i ≥ b i is conjugate to b ′ i if and only if there exist z i such that z i ∈ ⟨ supp ( b i )⟩ ,and z − i b i z i = b ′ i , for i = , . . . , p . As [ supp ( b i ) , supp ( b j )] =
1, for all i ≠ j ,it follows that there exists u ∈ U such that u − ˆ wu = ˆ v if and only if thereexist such z i . For i ≥ z i exists isthe conjugacy problem in the partially commutative group ⟨ supp ( b i )⟩ , so isdecidable. For i =
0, the question of existence of such a z is decidable by [12,Proposition 5.8]. Therefore, we may decide whether or not ˆ w is conjugate toˆ v . This means that we may also decide whether or not (4.4) has a solution x ∈ U ; and applying this argument to all cyclic permutations of v , the resultfollows. In this section we argue that Theorem 1.2 almost always applies, and aFreiheitssatz holds, in the situation of Example 1.7.5. We conjecture that asimilar statement holds for arbitrary graphs. Here though we use only naivecounting arguments; a study of the conjecture over arbitrary graphs is likelyto require a more systematic approach.As in Example 1.7.5, G ′ = G ( C ′ n ) , where C ′ n is the graph on the left ofFigure 2, and n ≥
5. The subgroup H of G ′ generated by A = { a , . . . , a n − } is isomorphic to G ( C n − ) , lk ( t ) = { a , a n − } , U = ⟨ lk ( t )⟩ and HNN ( t ) = ⟨ H, t ∣ t − at = a, ∀ a ∈ lk ( t )⟩ . As input to the question “for which elementsof G ′ does Theorem 1.2 hold?” we choose a particular set L (see Section5.1) of words over ( A ∪ A − ) ∗ with the property that every cyclically mini-mal element of H is represented by a unique element of L , every cyclically31inimal element of G ′ ending in a t -letter is represented by a unique elementof L and every element of L represents a cyclically minimal element of G ′ .Every element of L is reduced with respect to HNN ( t ) and is written as w = g t α g ⋯ g m t α m , where m ≥ g i ∈ H , ε i = ±
1, and α i ∈ Z /{ } . The set ofelements w ∈ L for which ∣ α ∣ + ⋯ + ∣ α m ∣ ≤ k and l ( g i ) ≤ d , for all i , is denotedby L ( d, k ) . We show that almost all words of L satisfy the hypotheses ofTheorem 1.2, in a sense which we now make precise. For a subset S of L wedefine the asymptotic density of S to be ρ ( S ) = lim d,k → ∞ ∣ S ∩ L ( d, k )∣∣ L ( d, k )∣ . (In this limit, both d and k must become, simultaneously, sufficiently large.The asymptotic density of S is undefined if the limit does not exist.) Forgeneral discussion of asymptotic density and it properties we refer, for ex-ample, to [24, 20, 14]: the 2-parameter “bidimensional” asymptotic densitydefined in [14] plays a role equivalent to the limits as both d and k approachinfinity, defined here.As subset S ⊆ L is said to be generic if ρ ( S ) = negligible if ρ ( S ) = Proposition 5.1.
In the above terminology, let L Y be the subset of L con-sisting of words which satisfy the hypotheses of Theorem 1.2. Then L Y isgeneric. In the remainder of this section we define the set L and prove this propo-sition. The definition of L depends on a choice of normal forms for elements of H . The subgroup H is G ( C n − ) where C n − is the cycle graph with vertices A . We may define a set L H of unique normal forms for all n ≥
5, as wedo below, but observe that the case n = H = F ( a , a ) × F ( a , a ) , and an alternative normal form may be chosen usingthis decomposition. It turns out that some of the bounds we derive forour general normal form involve division by n −
5, so would need specialtreatment when n =
5. Moreover, the alternative normal form for the case n = H in the case n = n ≥
6. Thesedefinitions result in different choices of sets L in the two cases.In the case n = w over ( A ∪ A − ) ∗ is a square normalform if w = w w , where w is a reduced word in F ( a , a ) and w is a reducedword in F ( a , a ) . (The graph C is a “square”.) Then every square normalform is a minimal form and every element of H is represented by a uniquesquare normal form.For n ≥
5, we say a word w over ( A ∪ A − ) ∗ is a normal form if(i) w is freely reduced and(ii) contains no subword of the form a εi + a βi − a δi , (subscripts modulo n − ≤ i ≤ n − ε, δ ∈ { ± } and β ∈ Z ( β may be zero).Note that if w is a word in normal form then it is minimal, because thecentraliser of a i in H is C H ( a i ) = ⟨ a i − , a i , a i + ⟩ , and w can contain no subwordof the form a − δi ua δi , with u ∈ ⟨ a i − , a i + ⟩ . We claim that every element of H is equal to a unique word in normal form. To see this, define a word to be prohibited if it has the form ( r ∏ j = a α j i + a β j i − ) a δi , where r ≥ δ ∈ { ± } , α j ≠
0, for all j and β j ≠
0, for j < r (where subscriptsare written modulo n − v be a minimal length word representing anelement of H and suppose that v ≡ v v v , where v is prohibited and is theleftmost prohibited subword of v : that is, if v ≡ u u u , where u is prohib-ited, then ∣ v ∣ ≤ ∣ u ∣ . If v = (∏ rj = a α j i + a β j i − ) a δi then let v ′ = a δi (∏ rj = a α j i + a β j i − ) ,so v = H v v ′ v and if v v ′ v ≡ u u u , where u is prohibited, then ∣ u ∣ > ∣ v ∣ (as the first letter a δi of v ′ cannot be part of a prohibited subword). Contin-uing this way we may eventually write v to a normal form u with v = H u .Hence every element of H is represented by a normal form.It remains to show that normal forms are unique, for which we shall usethe fact that, if w is a normal form and w ≡ w a γk w , where [ a k , supp ( w )] =
1, then w w is also a normal form. Indeed, if w w contains a subword u = a εi + a βi − a δi , with ε, δ ∈ { ± } , then the first letter a εi + of u occurs in w and the last letter a δi occurs in w . Thus w = w ′ a εi + a β i − and w = a β i − a δi w ′ ,where β + β = β and w ′ , w ′ are initial and terminal subwords of w and w ,respectively. As [ a k , supp ( w )] =
1, we have [ a k , a i ] = k ∈ { i − , i, i + } .33f k = i − w has a subword a εi + a β + γi − a δi . If k = i then w has a subword a εi + a β i − a γi and if k = i + w has a subword a γi + a β i − a δi . None of these arepossible as w is a normal form, so w w is a normal form, as claimed.We now use induction on length to show each element of H has exactlyone normal form. This is certainly true for the element of length 0, and weassume that it is true for all elements of length at most m , for some m ≥ w and w ′ are normal forms with w = H w ′ , l ( w ) = m + l ( w ′ ) = m +
1) and w ≢ w ′ . Let w ≡ w a εi , where ε ∈ { ± } . Then w ′ ≡ w ′ a δj ,for some δ ∈ { ± } , where j ≠ i ; otherwise w and w ′ are distinct normalforms of length m for the same element of H . Thus w has right divisors a εi and a δj , so [ a i , a j ] = w ≡ w a δj w and w ′ ≡ w ′ a εi w ′ , where [ a j , supp ( w )] = [ a i , supp ( w ′ )] = a j ∉ supp ( w ) , a i ∉ supp ( w ′ ) and j = i ± w ≡ w a δj w a εi and w ′ ≡ w ′ a εi w ′ a δj and, using the fact above, both w w a εi and w ′ a εi w ′ are normal forms. As w w a εi = H w ′ a εi w ′ , the inductiveassumption implies w w a εi ≡ w ′ a εi w ′ and, as a i ∉ supp ( w ′ ) it follows that w ′ ≡
1. Similarly w ≡
1, and now we have w ≡ w ′ . Therefore w ≡ w a δj a εi and w ′ ≡ w a εi a δj , with j = i ±
1. This is a contradiction, since w and w ′ cannotboth be normal forms, and the result follows.Therefore we have unique normal forms for elements of H , for n ≥
5. Nextwe define subsets of the set of normal forms needed for analysis of Theorem1.2. • For n =
5, let L H be the set of square normal forms for elements of H ,and for n ≥ L H be the set of normal forms for elements of H . Inboth cases let L H ( d ) be the set of elements of L H of length at most d ,and set l H ( d ) = ∣ L H ( d )∣ . • Let L UH ⊆ L H be the set of normal forms for elements of H whichhave no non-trivial left divisor in U . Let L UH ( d ) = L UH ∩ L H ( d ) and let l UH ( d ) = ∣ L UH ( d )∣ . • Let L U be the subset { a αn − a β ∶ α, β ∈ Z } of L H . Then L U is the subsetof L H consisting of normal forms for elements of U (for all n ≥ L U ( d ) be the set of elements of L U of length at most d and set l U ( d ) = ∣ L U ( d )∣ . • Let L H,S ( d ) , L UH,S ( d ) and L U,S ( d ) be the subsets L H , L UH and L U ,respectively, of elements of length exactly d . Set l H,S ( d ) = ∣ L H,S ( d )∣ , l UH,S ( d ) = ∣ L UH,S ( d )∣ and l U,S ( d ) = ∣ L U,S ( d )∣ .34 Let L t ( k ) be the set of cyclically reduced words of ( A ∪ A − ) ∗ rep-resenting elements of G ′ which end in a t -letter and have the form ug t ε ⋯ g k t ε k , where g i ∈ L UH and u ∈ L U .As the unique element of L U ∩ L UH is 1, the condition that elements of L t ( k ) are cyclically reduced amounts to the condition that, for all i ; either g i ≠ ε i − = ε i (subscripts modulo n ).Fix k ≥ w be an element of L t ( k ) . We say that w is composed if either(i) w = ut k or w = ut − k , for some u ∈ U , or(ii) w = ug t α ⋯ g r t α r , where g i ∈ L UH /{ } , u ∈ L U , ∣ α i ∣ ≥ (∣ α ∣ , . . . , ∣ α r ∣) is a composition of k into r parts; for some r such that 1 ≤ r ≤ k . • Let L ( k ) the set of composed elements of L t ( k ) and let L ( i ) ( k ) and L ( ii ) ( k ) be the sets of elements of L ( k ) of types (i) and (ii), respectively. • For d ≥ L ( d, ) the set of normal forms in L H representingcyclically minimal elements of H and set l ( d, ) = ∣ L ( d, )∣ . • For d ≥ k ≥ L ( i ) S ( d, k ) the set of elements of L ( i ) ( k ) such that u ∈ L U ( d ) . Set l ( i ) S ( d, k ) = ∣ L ( i ) S ( d, k )∣ . • For d ≥ k ≥ L ( ii ) S ( d, k ) the set of elements of L ( ii ) ( k ) such that g i ∈ L UH ( d ) , for all i , u ∈ L U and l ( ug ) ≤ d . Set l ( ii ) S ( d, k ) = ∣ L ( ii ) S ( d, k )∣ . • For d ≥ k ≥ L S ( d, k ) = L ( i ) S ( d, k ) ∪ L ( ii ) S ( d, k ) and let l S ( d, k ) = ∣ L S ( d, k )∣ . • For d ≥ k ≥ L ( i ) ( d, k ) = ∪ kl = L ( i ) S ( d, l ) , L ( ii ) ( d, k ) = ∪ kl = L ( ii ) S ( d, l ) , L ( d, k ) = L ( d, ) ∪ L ( i ) ( d, k ) ∪ L ( ii ) ( d, k ) and l ( i ) ( d, k ) = ∣ L ( i ) ( d, k )∣ , l ( ii ) ( d, k ) = ∣ L ( ii ) ( d, k )∣ and l ( d, k ) = ∣ L ( d, k )∣ . • For each integer r such that 1 ≤ r ≤ k and each composition P of k into r parts, set L ( ii ) S ( d, k, r, P ) = { ug t α ⋯ g r t α r ∈ L ( ii ) S ( d, k ) ∶ (∣ α ∣ , . . . , ∣ α r ∣) = P } . Set l ( ii ) S ( d, k, r, P ) = ∣ L ( ii ) S ( d, k, r, P )∣ . • Let L = ⋃ d ≥ ,k ≥ L ( d, k ) . 35 .2 Counting normal forms To find bounds on the size of L ( d, k ) we first compute bounds on the sizesof L H ( d ) , L U ( d ) and L UH ( d ) . We consider cases n = n ≥ n ≥ n =
5, except incases where division by n − n =
5. All wordsof length 0 and 1 are in L H , so l H ( ) = l H,S ( ) = ( n − ) =
8. For k ≥ w w such that l ( w ) = p and l ( w ) = k − p ,where 1 ≤ p ≤ k −
1, is 4 k − , and there are 8 ⋅ k − normal forms w ⋅ w oflength k with either w or w trivial. Hence l H,S ( k ) = ⋅ k − ( k + ) , for k ≥
1. Summing over k from 0 to d gives l H ( d ) = + d d − , when n = . (5.1)Now consider the case n ≥
6. Again all words of length 0 and 1 are in L H ,so l H ( ) = l H,S ( ) = ( n − ) . A word of length 2 is in L H if and onlyif it is of the form a εi a δj , where ε, δ ∈ { ± } , and satisfies the conditions that j ≠ i − i = j then ε = δ . Therefore l H,S ( ) = ( n − )( n − ) . Weclaim that, setting α = n − γ = n − ( n − ) γ m − ≤ l H,S ( m ) ≤ ( n − ) α m − , for all m ≥ . (5.2)To verify this, assume that it holds for some m ≥
2. If w is a normal formof length m + w = ua δj , for some u of length m , which is necessarilyin L H , a j ∈ A and δ = ±
1. From the argument used to establish the valueof l H,S ( ) ; for each such u , there are at most 2 n − a δj . Hence l H,S ( m + ) ≤ αl H,S ( m ) ≤ ( n − ) α m . Some of these choices do not give rise tonormal forms, as they result in prohibited subwords of w . On the other handif u = va εi , and i ∉ { j − , j + } then (as u ∈ L H ) ua δj contains no prohibitedsubword, so is in L H (as long as it is reduced). Hence there are at least 2 n − a εi which do result in ua εi being a normal form. Therefore thereare at least γl H,S ( m ) possibilities for w . That is, l H,S ( m + ) ≥ γl H,S ( m ) ≥ ( n − ) αγ m − . Hence, as α ≥ γ , (5.2) follows by induction on m . Summingover m from 0 to d we obtain1 + n − n − [ γ d − ] ≤ l H ( d ) ≤ + n − n − [ α d − ] , (5.3)for all d ≥
1, in the case n ≥
6. 36ormal forms of elements of U all have the form a an − a b , for some integers a, b , and for all n ≥
5. Therefore the number of elements of U of lengthexactly d is 4 d , and l U ( d ) = + d ( d + ) , for all d ≥ n ≥ . (5.4)Next we consider L UH , the set of normal forms for elements of H whichhave no non-trivial left divisor in U . When n = w w belongsto L UH if and only if w does not begin with a ± and w does not begin with a ± . There are 2 ⋅ l − elements of F ( a , a ) of length l ≥
1, beginning with a ± ;and similarly for F ( a , a ) and a ± . It follows that the number of elementsof L UH of length exactly k is 4 ⋅ k − ( + k ) . Therefore l UH ( d ) = d − ( + d ) , for all d ≥ n = . (5.5)In the case n ≥ L UH ( d ) . We partition L UH into three subsets: (a) normal forms beginning with elements a ± i , where3 ≤ i ≤ n −
3; (b) normal forms beginning a ± and (c) normal forms beginning a ± n − .(a) Normal forms beginning a ± i , where 3 ≤ i ≤ n −
3, have no left divisor in U . There are 2 ( n − ) choices for the first letter a εi , where 3 ≤ i ≤ n − n − n − n − d ≥ L UH contains at least 2 ( n − ) αγ d − ≥ ( n − ) γ d − and at most 2 ( n − ) α d − elements of type (a), with length exactly d .Therefore, the number a ( d ) of elements of type (a) in L UH ( d ) satisfies n − n − ( γ d − ) ≤ a ( d ) ≤ n − n − ( α d − ) . (5.6)(b) To bound the number b ( d ) of normal forms of length at most d whichbegin with a ± and have no left divisor in U , we begin by finding boundson the number b S ( d ) of such normal forms of length exactly d . We have b S ( ) = b S ( ) =
2, as the possibilities are a ± . To estimate b S ( d ) ,for d >
1, it is convenient to consider normal forms beginning withelements of the set M of reduced words in { a ± }({ a ± , a ± n − }) ∗ . To thisend suppose w is a normal form of type (b) and w ≡ gh , where g is amaximal prefix of w belonging to M and g has length t ≥
1. There are37 possibilities for the first letter of g and 3 for each subsequent letter;so 2 . t − such g altogether. To find those gh with no left divisor in U we distinguish between those g ending a ± and those ending a ± n − . Let N be the set of all elements of M ending in a ± and let P = M / N . Let n ( t ) and p ( t ) be the number of elements of N and P , respectively, oflength t ≥
1. It follows (by induction) that n ( t ) = ⎧⎪⎪⎨⎪⎪⎩ t − − t is even3 t − + t is odd , and so p ( t ) = ⎧⎪⎪⎨⎪⎪⎩ t − + t is even3 t − − t is odd , for all t ≥ w ≡ gh , where g ∈ P so g ends in a ± n − , then h cannot begin witha letter in { a ± , a ± n − } , by maximality of g , cannot begin with a ± n − ,by definition of normal form, and cannot begin with a ± as w has noleft divisor in U . Thus the first letter of h belongs to { a ± , . . . , a ± n − } .As neither a or a n − commutes with any element of this set, eachsubsequent letter of h may be any member of A ± which results in anormal form. As before, the the number of possibilities for such an h of length s is at most 2 ( n − ) α s − and at least 2 ( n − ) γ s − .On the other hand, if w ≡ gh , where g ∈ N so g ends in a ± , then h cannot begin with a letter in { a ± , a ± n − } and cannot begin with a ± .As a n − does not commute with a and a does not commute with thefirst letter of h , each subsequent letter of h may again be any elementof A ± which results in a normal form. The number of possibilitiesfor h of length s in this case is then at most 2 ( n − ) α s − and at least2 ( n − ) γ s − .Now, for k ≥
2, let b ( k, t ) be the number of normal forms w ∈ L UH oflength k , of type (b), where w ≡ gh , as above, and g has length t ≥ h is of length s = k − t . When t < k (so s ≥ b ( k, t ) ≤ n ( t )( ( n − ) α s − ) + p ( t )( ( n − ) α s − ) = α s − [( n − ) n ( t ) + ( n − ) p ( t )] . (5.7)Similarly, b ( k, t ) ≥ γ s − [( n − ) n ( t ) + ( n − ) p ( t )] , (5.8)where s = k − t . For the cases where t = k and h is trivial, b ( k, k ) = n ( k ) + p ( k ) = ⋅ k − . We leave consideration of b ( d ) at this point38nd turn to normal forms of type (c), as this makes later computationsimpler.(c) Again, to bound the number c ( d ) of normal forms of length at most d ,which begin with a ± n − and have no non-trivial left divisor in U , we shallfirst bound the number c S ( d ) of such normal forms of length exactly d . As before, c S ( ) = c S ( ) =
2. To find bounds on c S ( k ) for k ≥ w which factor as gh , where g is amaximal prefix in the set of reduced words in { a ± n − }({ a ± n − , a ± }) ∗ . Letthe number of such g of length t ≥ a n − be n ′ ( t ) andthe number ending in a be p ′ ( t ) . Then n ′ ( t ) = n ( t ) and p ′ ( t ) = p ( t ) ,where n ( t ) and p ( t ) are as in case (b) above. Following through theargument of case (b), we find that the number c ( k, t ) of normal forms w ∈ L UH of length k , of type (c), where w ≡ gh , as above, and g haslength t ≥
1, satisfies c ( k, t ) ≤ α s − [( n − ) n ( t ) + ( n − ) p ( t )] and (5.9) c ( k, t ) ≥ γ s − [( n − ) n ( t ) + ( n − ) p ( t )] , (5.10)where k ≥ s = k − t and α and γ are as above.As [( n − ) n ( t ) + ( n − ) p ( t )] + [( n − ) n ( t ) + ( n − ) p ( t )] = ( n − ) t − combining (5.7), (5.8), (5.9) and (5.10) and setting β = n −
9, gives4 βγ s − t − ≤ b ( k, t ) + c ( k, t ) ≤ βα s − t − , when k ≥ t ≥
1. Also, c ( k, k ) = b ( k, k ) = ⋅ k − , when k ≥ n ≥ k ≥ b S ( k ) + c S ( k ) ≤ [ k − + β k − ∑ t = α k − t − t − ] = ( k − + β α k − − k − α − ) = n − ( k − + βα k − ) ≤ n − ( β + ) α k − , as 3 ≤ α, = α k − . (5.11)39imilarly, for n ≥ k ≥ b S ( k ) + c S ( k ) ≥ [ k − + β k − ∑ t = γ k − t − t − ] = ( k − + β γ k − − k − γ − ) = n − ( − k − + βγ k − ) ≥ n − ( β − ) γ k − , as 3 ≤ γ, = γ k − . (5.12)Note that in fact (5.11) and (5.12) hold for all k ≥
1. Then b ( d ) + c ( d ) = d ∑ k = b S ( k ) + c S ( k ) ≤ d ∑ k = α k − = ( α d − α − ) = ( α d − n − ) . (5.13)Similarly, b ( d ) + c ( d ) ≥ ( γ d − n − ) . (5.14)As l UH ( d ) = a ( d ) + b ( d ) + c ( d ) , from (5.6), (5.13) and (5.14), γ d − ≤ l UH ( d ) ≤ α d − , when n ≥ . (5.15)In what follows we shall need to know how many elements of L UH ( d ) donot belong to U ∪ Maln H ( U ) . The elements of H which are not in Maln H ( U ) or U are those which belong either to • ⟨ a n − ⟩⟨ a n − , a ⟩ and have support containing a n − , or to • ⟨ a ⟩⟨ a n − , a ⟩ and have support containing a .As elements of L UH have no non-trivial left divisor in U , elements of L UH whichdo not belong to U ∪ Maln H ( U ) are those in ⟨ a n − , a ⟩ which begin with a ± n − ,or those in ⟨ a , a n − ⟩ which begin with a ± , all of which are in normal form,as long as they are freely reduced. There are 4 ⋅ p − elements of length p of40hese forms, so the number e ( d ) of elements of L UH ( d ) which do not belongto U ∪ Maln H ( U ) is e ( d ) = ( d − ) , for all n ≥ . (5.16)To bound the size of L ( d, ) we simply note that1 ≤ l ( d, ) ≤ l H ( d ) , for all d ≥ . (5.17)Bounds on l ( d, k ) , for k >
0, are found by considering L ( i ) ( d, k ) and L ( ii ) ( d, k ) separately. From (5.4), there are l ( i ) S ( d, k ) = [ + d ( d + )] elements ut ± k of type (i) in L ( i ) S ( d, k ) . Therefore l ( i ) ( d, k ) = k [ + d ( d + )] , (5.18)for all n ≥ L ( ii ) S ( d, k, r, P ) have the form ug t α ⋯ g r t α r subject to theconditions of the definition on page 35. Here ug ∈ H and l ( ug ) ≤ d and,as every element of H may be uniquely expressed as vg , where v ∈ U and g has no left divisor in U , there are l H ( d ) possibilities for ug . For i > g i ∈ L UH , so there are l UH ( d ) r − possibilities for ( g , . . . , g r ) . Given the fixedcomposition P = ( a , . . . , a r ) of k , there are 2 r choices for ( t α , . . . , t α r ) , with (∣ α ∣ , . . . , ∣ α r ∣) = ( a , . . . , a r ) . Therefore l ( ii ) S ( d, k, r, P ) = r l H ( d )( l UH ( d )) r − . (5.19)There are ( k − r − ) compositions of k into r parts so the number l ( ii ) S ( d, k, r ) of elements of L ( ii ) S ( d, k ) which involve a partition of k into r parts is l ( ii ) S ( d, k, r ) = ( k − r − ) l ( ii ) S ( d, k, r, P ) . As l ( ii ) S ( d, k ) = ∑ kr = l ( ii ) S ( d, k, r ) , (5.19) gives l ( ii ) S ( d, k ) = k ∑ r = ( k − r − ) r l H ( d )( l UH ( d )) r − = l H ( d )( l UH ( d ) + ) k − . l ( ii ) ( d, k ) = k ∑ p = l ( ii ) S ( d, p ) , = k ∑ p = l H ( d )( l UH ( d ) + ) p − = l H ( d ) ( l UH ( d ) + ) k − l UH ( d ) = l H ( d ) l UH ( d ) [( l UH ( d ) + ) k − ] . (5.20)We are now ready to calculate the asymptotic density of subsets of L which satisfy conditions 1 – 4 of Theorem 1.2. For each i, from 1 to 4, let Z i be the subset of L which satisfies conditioni of Theorem 1.2. We shall show that ρ ( Z i ) =
1, for 1 ≤ i ≤
4. As a fi-nite intersection of generic sets is generic, and L Y = ∩ i = Z i , this proves theproposition. Theorem 1.2, condition 1.
An element s ∈ L satisfies this condition ifand only if it does not belong to H . Thus the intersection of L ( d, k ) withthe set Z c of these failing elements is L ( d, ) . From (5.17), the asymptoticdensity of the set Z c is ρ ( Z c ) ≤ lim d,k → ∞ l H ( d ) l ( d, k ) , and from (5.1), (5.3), (5.20) and the definition of L ( d, k ) , it follows that ρ ( Z c ) =
0. Therefore, ρ ( Z ) =
1; that is the set of elements of L satisfyingcondition 1 has asymptotic density 1. Theorem 1.2, condition 2.
There are no t -thick elements in L ( d, ) andelements of L ( i ) , are by definition, all t -thick. Thus the number of t -thick ele-ments in L ( d, k ) is z ( d, k ) = l ( i ) ( d, k ) + z ( ii ) ( d, k ) , where z ( ii ) ( d, k ) is the num-ber of t -thick elements in L ( ii ) ( d, k ) . To find the number of t -thick elementsin L ( ii ) ( d, k ) , begin by considering the t -thick elements of L ( ii ) S ( d, k, r, P ) . An42lement w = ug t α ⋯ g r t α r ∈ L ( ii ) S ( d, k, r, P ) fails to be t -thick only if at leastone of the g i is not in Maln H ( U ) . For i ≥ g i is an element of L UH ( d ) and,from (5.16), there are e ( d ) = ( d − ) elements of L UH ( d ) which are not inMaln H ( U ) . Therefore, there are t UH ( d ) = l UH ( d ) − ( d − ) possible choices foreach g i in the t -thick element w .Now consider ug . By definition, l ( ug ) ≤ d and supposing that u haslength s , where 1 ≤ s ≤ d −
1, from (5.16) and the fact that there are 4 s elements of U of length s , there are 4 s ⋅ e ( d − s ) possible ug , with g ∉ Maln H ( U ) . Summing over s , such that 0 ≤ s ≤ d −
1, the total possiblenumber of such ug is e ′ ( d ) where e ′ ( d ) = e ( d ) + d − ∑ s = s ⋅ e ( d − s ) = e ( d ) + d − ∑ s = s ( d − s − ) = ( d − ) + ( ( d − ) − d ) − d ( d − ) = ( d − ) − d ( + d ) . The element ug is an arbitrary element of L H ( d ) , so there are t H ( d ) = l H ( d ) − e ′ ( d ) possible choices for ug in the t -thick element w of L ( ii ) S ( d, k, r, P ) .Combining these facts, as in the calculation of l ( ii ) ( d, k ) , we see that thenumber z ( ii ) ( d, k ) of t -thick elements in l ( ii ) ( d, k ) is z ( ii ) ( d, k ) = t H ( d ) t UH ( d ) [( t UH ( d ) + ) k − ] . As the number of t -thick elements in L ( d, k ) is z ( d, k ) = l ( i ) ( d, k ) + z ( ii ) ( d, k ) , ρ ( Z ) = lim d,k → ∞ z ( d, k ) l ( d, k ) = l ( i ) ( d, k ) + z ( ii ) ( d, k ) l ( d, k ) ≥ l ( i ) ( d, k ) + z ( ii ) ( d, k ) l H ( d ) + l ( i ) ( d, k ) + l ( ii ) ( d, k ) , from (5.17) = [ l ( i ) ( d, k ) ⋅ t UH ( d ) + t H ( d )(( t UH ( d ) + ) k − )] l UH ( d )[( l H ( d ) + l ( i ) ( d, k )) l UH ( d ) + l H ( d )(( l UH ( d ) + ) k − )] t UH ( d ) . First note that l UH ( d ) t UH ( d ) = l UH ( d ) l UH ( d ) − e ( d ) = + e ( d ) l UH ( d ) − e ( d ) . n = e ( d ) l UH ( d ) − e ( d ) = ( d − ) d − ( + d ) − ( d − ) , which has limit 0 as d, k → ∞ , solim d,k → ∞ l UH ( d ) t UH ( d ) = . (5.21)If n ≥ γ d − ≤ l UH ( d ) , and γ ≥
5, from (5.15), so e ( d )/( l UH ( d ) − e ( d )) has limit 0 as d, k → ∞ , and (5.21) holds again.Next,lim d,k → ∞ l ( i ) ( d, k ) ⋅ t UH ( d )( l H ( d ) + l ( i ) ( d, k )) l UH ( d ) + l H ( d )(( l UH ( d ) + ) k − ) = , for all n ≥
5. Finally, aslim d,k → ∞ ( l H ( d ) + l ( i ) ( d, k )) l UH ( d )( l UH ( d ) + ) k − = , lim d,k → ∞ t H ( d )(( t UH ( d ) + ) k − )( l H ( d ) + l ( i ) ( d, k )) l UH ( d ) + l H ( d )(( l UH ( d ) + ) k − )] = lim d,k → ∞ t H ( d )(( t UH ( d ) + ) k − ) l H ( d )(( l UH ( d ) + ) k − )] . We have t H ( d ) l H ( d ) = l H ( d ) − e ′ ( d ) l H ( d ) = − e ′ ( d ) l H ( d ) . If n = e ′ ( d ) l H ( d ) = ( d − ) − d ( + d ) + d ⋅ d − , which has limit 0 as d, k → ∞ .If n ≥
6, then from (5.3), l H ( d ) ≥ ( n − )( γ d − )/( n − ) so again e ′ ( d )/ l H ( d ) has limit 0. Therefore in all caseslim d,k → ∞ t H ( d ) l H ( d ) = . d,k → ∞ ( t UH ( d ) + ) k − ( l UH ( d ) + ) k − = . Combining all the above we see that ρ ( Z ) =
1: that is the set of elements of L which satisfy Theorem 1.2.2 has asymptotic density 1. Theorem 1.2, condition 3.
A word in L represents an element of ⟨ st ( t )⟩ if and only if it belongs to L U or L ( i ) . Therefore, denoting by Z c the set ofelements of L which do not satisfy Theorem 1.2, condition 3, ρ ( Z c ) is thelimit of [ l U ( d ) + l ( i ) ( d, k )]/ l ( d, k ) as d, k → ∞ . As this limit is 0, it followsthat ρ ( Z ) = Theorem 1.2, condition 4 If w in G ′ is not a t -root we say it is a t -power .An element g ∈ L ( d, ) has σ ( g ) ∈ D , so σ ( g ) is a generator, or the inverseof a generator, of the factor F ( D + ) of F D ; so σ ( g ) is not a proper power.Consequently there are no t -powers in L ( d, ) . We shall count the number ofelements of L ( d, k ) which are t -powers and are in L ( i ) and in L ( ii ) separately,and show that both have asymptotic density zero, from which it follows thatthe set of elements Z c of L which do not satisfy condition 4 has asymptoticdensity zero. Elements ut ± k of L ( i ) are all t -powers. There are 2 [ + d ( d + )] of these in L ( d, k ) , from (5.18), and it follows from (5.20) and the definitionof L ( d, k ) , that the set of t -powers that are in L ( i ) has asymptotic densityzero.Let p ( ii ) ( d, k ) denote the number of t -powers in L ( ii ) ( d, k ) . If w = ug t α r ⋯ g r t α r ∈ L ( ii ) S ( d, k, r, P ) is a t -power then there are elements d i ∈ D and u i ∈ U and an integer p , such that 1 ≤ p ≤ ⌊ r / ⌋ , r = pq , w = F ud u t α ⋯ d r u r t α r and σ ( w ) = d t α ⋯ d r t α r = ( d t α ⋯ d p t α p ) q . (The d i are fixed generatorsof the free group F ( D + ) (or their inverses).) In this case it follows that P = (∣ α ∣ , . . . , ∣ α r ∣) = P qp , where P p = (∣ α ∣ , . . . , ∣ α p ∣) is a composition of q = r / p into p parts. Conversely, if there exist positive integers p , q such that p = r / q ,2 ≤ q ≤ r and P = P qp for some composition P p of q into p parts, then for all • elements d i ∈ D and u, u i,j ∈ U such that 1 ≤ i ≤ p and 1 ≤ j ≤ q , with • l ( ud u , ) ≤ d , l ( d i u i,j ) ≤ d and [ u i,j , d i ] ≠
1, and • all ( α , . . . , α p ) such that P p = (∣ α ∣ , . . . , ∣ α p ∣) ;45 = ud u , t α d u , t α ⋯ d p u ,p t α p ⋯ t α p d u q, t α ⋯ d p u q,p t α p is a t -power in L ( d, k, r, P ) . Here w has initial subword w p = ud u , t α d u , t α ⋯ d p u ,p t α p with normal form in L ( ii ) S ( d, q, p, P p ) and σ ( w p ) is the q th root of σ ( w ) . Therefore the t -power w is determined by the el-ement w p ∈ L ( ii ) S ( d, q, p, P p ) and the sequence ( u , , . . . , u ,p , . . . u q, , . . . , u q,p ) of length p ( q − ) = k − p , of elements of U (each of length at most d ).There are ( d ) k − p sequences of length k − p , of such elements of U , so writing p ( ii ) ( d, k, r, P ) for the number of t -powers in L ( ii ) S ( d, k, r, P ) , and assuming anappropriate composition P p of r / p , exists for each p such that 1 ≤ p ≤ ⌊ r / ⌋ ,we have p ( ii ) ( d, k, r, P ) ≤ ⌊ r / ⌋ ∑ p = ( d ) k − p l ( ii ) S ( d, q, p, P p ) = k l H ( d ) ⌊ r / ⌋ ∑ p = ( d ) k − p ( l UH ( d )) p − from (5.19) , ≤ k ( d ) k −⌊ r / ⌋ l H ( d ) ( l UH ( d )) ⌊ r / ⌋ l UH ( d ) − d , Summing over all compositions of k into r parts and over r from 1 to k , andsetting a = l UH ( d ) , b = l H ( d ) and c = d , the number p ( ii ) S ( d, k ) of t -powers in L ( ii ) S ( d, k ) satisfies p ( ii ) S ( d, k ) ≤ k ba − c k ∑ r = ( k − r − ) c k −⌊ r / ⌋ a ⌊ r / ⌋ ≤ k ba − c k ∑ r = ( k − r − ) c ( k − r + )/ a r / = k a / bc ( k + )/ a − c k ∑ r = ( k − r − ) c ( k − r )/ a ( r − )/ = k a / bc ( k + )/ a − c ( a / + c / ) k − . p ( ii ) ( d, k ) of t -powers in L ( ii ) ( d, k ) , satisfies p ( ii ) ( d, k ) ≤ a / ba − c k ∑ p = p c ( p + )/ ( a / + c / ) p − = a / bca − c k ∑ p = [ c / ( a / + c / )] p − ≤ a / bc [ c / ( a / + c / )] k ( a − c )[ c / ( a / + c / ) − ] . In current notation, from (5.20), l ( d, k ) ≥ l ( ii ) ( d, k ) = b [( a + ) k − ]/ a , so p ( ii ) ( d, k ) l ( d, k ) ≤ a / bc [ c / ( a / + c / )] k a ( a − c )[ c / ( a / + c / ) − ] b [( a + ) k − ] = k + a / c ( k + )/ [ a / + c / ] k ( a − c )[ c / ( a / + c / ) − ][( a + ) k − ] ≤ a / c [( ca ) / + c ] k ( a − c )[ c / ( a / + c / ) − / ] a k = ( c / a )[ − ( c / a )][ c / ( + ( c / a ) / ) − ( / a / )] [( c / a / ) + ( c / a )] k . For n = n ≥ c / a / tends to 0 as d → ∞ and so, in both cases, it canbe seen that lim d,k → ∞ p ( ii ) ( d,k ) l ( d,k ) =
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