The following lemma is Theorem VI.3.2 from [2] (or Theorem 19.5 from [30]).
Lemma 4.18. — The centralizer of every non-trivial element of B(m,n)is cyclic of order n.
We are going to use the following Lemma repeatedly in Section 6.
Lemma 4.19. — If a,b,c ∈ B(m,n), n 1 odd, a = 1, c = 1, [a,c] = 1, [bab−1,c] =1 then [b,c] = [b,a] =1.
Proof. — Indeed, suppose that [a,c] = 1, [bab−1,c] = 1. By Lemma 4.18, the centralizer of every non-trivial element ofB(m,n)is cyclic of order n. Hencea,bab−1,c belong to the same cyclic subgroup C of B(m,n). Therefore bCb−1 and C are non-trivially intersecting cyclic subgroups of order n. By Lemma 4.18, bCb−1=C. Hence b,C is a finite subgroup of B(m,n) (or order at most n2). By Theorem 19.6 from [30], b,C is cyclic of order n, so b ∈C. Hence [b,c] =1 as required.
In order to prove Proposition 6.17 below, we will need some properties of free products. Let P and Q be two free groups. Elements of the free product P∗Q are represented by words of the form p1q1p2...qsps+1 wherepi ∈P, i=1, ...,s+1, is called a P-syllable, qi∈Q, i =1, ...,s, is called a Q-syllable. By definition, a P-syllable (resp.
Q-syllable) is a maximal subword over the set of generators of P (resp. Q). Elements ofP and Q are called 1-syllable elements.
A word p1q1...psqsps+1 fromP∗Q, where pi,qj are syllables, is called *-reduced in P∗Q if none of its syllables except possibly for p1,ps+1 is equal to 1. If a product is
*-reduced, it is not equal in P∗Q to such a product with smaller number of syllables.
The P-syllable ps+1 is called the last syllable of p1q1...psqsps+1 (even if ps+1 is empty).
If in p1q1...psqsps+1, a syllable, say, pi, is freely equal to 1, we can remove it and glue qi−1 withqi to form a newQ-syllable. Applying this operation several times, we get a
*-reduced in P∗Q form of the product which is unique. Notice that a word which is
*-reduced inP∗Q may not be freely reduced (as a word written in the free generators ofP and Q).
Let ψ :P→ ¯P be a homomorphism of P onto a group P¯ satisfying the identity xn =1. We say that a word h of P∗Q is ψ-trivial if the ψ-image of every syllable of h is 1 in P.¯
Lemma 4.20. — Let h1,h2 be words in P∗Q which are equal modulo Burnside relations.
Then the following properties hold.
(i) If h1 is ψ-trivial, and h2 is of minimal length among all words which are equal to h1
modulo Burnside relations then h2 is ψ-trivial.
(ii) Suppose thath1has exactly twoP-syllablesp,p with non-trivial ψ-images,h1≡xpypz, and y is not equal to a word in P modulo Burnside relations. Then h2 is not ψ-trivial.
(iii) If h1 is ψ-trivial, h1 = 1 modulo Burnside relations, h2 ∈ P, and modulo Burnside relations h1=xh2x−1 for some x∈P∗Q, then there exists a word x1∈P∗Q such that x1h2x1−1 =h1 modulo Burnside relations and the word x1h2x−11 is ψ-trivial.
Proof. — (i) Obviously we can assume thath1 is *-reduced. Consider a g-reduced diagram ∆ over the graded presentation of the free Burnside quotient of P∗Q from Section 18 of [30] corresponding to the equality h1 =h2. If the rank of ∆ is 0 then h1=h2 inP∗Q, and each syllable of the word h2 (of minimal length) must be a prod-uct of several syllables of h1. Hence h2 is ψ-trivial.
Suppose that the rank of ∆ is > 0, and ∂(∆)=s1s2 where φ(s1)=h1, φ(s−21)= h2. Since s2 is geodesic, s2 is a smooth section of rank |s2|. By Lemma 2.4, part a), and Lemma 2.3 there exists a cell Π of rank > 0 and its contiguity subdiagram Γ to s1 of contiguity degree > 1/2−γ −α > 3ε. By Lemma 2.6, ∆ contains a cell π and a rank 0 contiguity subdiagram Γ of π to s1, such that the contiguity degree of Γ is > ε. Let t be the contiguity arc of Γ on ∂(π), |t| > ε|∂(π)|. There are two possibilities.
Case 1. The period of φ(∂(π)) is a 1-syllable word. Then φ(∂(π)) is a 1-syllable word, and one can cut the subdiagram π ∪ Γ from ∆ preserving the ψ-triviality ofφ(s1). Here we use the fact that P¯ satisfies the identity xn =1.
Case 2. The period A of φ(∂(π)) contains at least 2 syllables. Since Γ is a dia-gram of rank 0, andεn > 2, every syllable ofA is a syllable ofφ(s1)±1 =h±11. HenceA is ψ-trivial (since h1 is ψ-trivial). Therefore, again, one can cut the subdiagram Γ∪π from ∆ preserving the ψ-triviality of φ(s1): we replace subwordA of φ(s1)±1 by A1−n and reduce the resulting word.
Thus in each of these cases we replace ∆ by a diagram of smaller type, and complete the proof by induction on the type of ∆.
(ii) By contradiction assume that h2 is ψ-trivial. We have that pyp = x−1h2z−1. By part (i) of this lemma we can replace x−1h2z−1 by a ψ-trivial word h of minimal
possible length. Consider a g-reduced diagram∆with∂(∆)=s1t1s2t2 whereφ(s1)=p, φ(t1)=y, φ(s2)=p, φ(t2−1)=h. If the rank of ∆ is 0 then pyp =h in P∗Q. Since y is not equal to a word from P and is ψ-trivial, and ψ(p) = 1, we have that the
*-reduced in P∗Q form of pyp is not ψ-trivial, a contradiction with the ψ-triviality ofh.
If the rank of ∆ is > 0 then as in part (i), we get a cell π and a contiguity subdiagram Γ of π to one of the subpaths s1,t1,s2 with contiguity degree ≥ ε, such that Γ has rank 0. Since φ(s1) and φ(s2) are 1-syllable words, the period A of π is a 1-syllable word if Γ is a contiguity subdiagram to s1 or s2. If Γ is a contiguity subdiagram to t1, A is a ψ-trivial word. As in part (i) we can remove Γ∪π from ∆ preserving the structure of the boundary of ∆ and the condition that φ(y) is a ψ -trivial, and does not belong toP modulo Burnside relations, φ(s1), φ(s2)are 1-syllable words fromP, ψ(φ(s1)), ψ(φ(s2))=1. We can complete the proof by induction of the type of ∆.
(iii) Consider a g-reduced annular diagram∆ with pointed contourss1,s2, where φ(s1)≡h1, φ(s2)≡ h2. If the rank of ∆ is 0 then h1 = xh2x−1 in P∗Q and xh2x−1 is ψ-trivial.
Suppose that the rank of ∆ is > 0. As in parts (i), (ii), we can remove a cell of rank > 0 from ∆ and change s1 or s2 preserving φ(s1), φ(s2) modulo Burnside re-lations. The labels of the contours s1, s2 of the new annular diagram satisfy all the conditions of part (iii) of the lemma, and again we can complete the proof by
induc-tion on the type of∆.
Lemma 4.21. — Suppose that h,g ∈ P∗Q, and there is no x ∈ P∗Q such that g = xpx−1,h = xpx−1 modulo Burnside relations for some p,p ∈ P. Suppose also that h, the
*-reduced forms of ghg−1, and a word b which is equal to g−1hg modulo Burnside relations are ψ-trivial. Then g is ψ-trivial.
Proof. — Suppose that g is not ψ-trivial. Let g ≡ g1pg2 where g2 is ψ-trivial, p is a P-syllable, ψ(p)= 1. Consider the *-reduced form of ghg−1 ≡ g1pg2hg2−1p−1g1−1. Since this word is ψ-trivial, and g2 is also ψ-trivial, the syllable p is glued with a P-syllable ph ofh and with the P-syllable p−1 of g−1. This means that h=g2−1phg2 in the free group P∗Q, and the *-reduced form of ghg−1 is g1pphp−1g−11 . This implies that g1
is ψ-trivial, so g has only one P-syllable with non-trivial ψ-image. Now consider the
*-reduced form b of the word g−1hg=g2−1p−1g1−1g2−1phg2g1pg2.
Since b is equal to b modulo Burnside relations and b is ψ-trivial, we can con-clude by Lemma 4.20(ii) that g1−1g2−1phg2g1 is equal to a word p ∈ P modulo Burn-side relations. Taking the projection of the equality (g2g1)−1phg2g1 = p onto the sub-group P, we get P(g2g1)−1phP(g2g1) = p modulo Burnside relations, where P(g2g1) is the P-projection of g2g1. Therefore P(g2g1)−1(g2g1) centralizes ph modulo Burnside re-lations. Since the centralizers of ph in the free Burnside quotient ofP and P∗Q have
order n (by Lemma 4.18), these centralizers coincide. Hence P(g2g1)−1g2g1 belongs to Pmodulo Burnside relations, so g2g1 ∈P modulo Burnside relations. Then g2gg2−1∈P, g2hg2−1 =g2g2−1phg2g2−1 =ph ∈P modulo Burnside relations, which is impossible by our
assumption.
5. The presentation of the group
In this section, we define an S-machine consisting of a set of (positive) S-rules M+. Using this S-machine, we define our groups H (Theorem 1.3).
Fix some numbers first. We use the same odd number n 1 as in the previous sections. Let N > n, and let m=m(n) be the number which exists by Theorem 4.4.