A just-nonsolvable torsion-free group defined on the binary treeA. M. Brunner, Said Sidki, Ana Cristina Vieira

AbstractA two generator torsion-free subgroup of the group of finite state

automorphisms of the binary tree is constructed having the propertiesof being just-nonsolvable and residually ”torsion-free solvable”. Apresentation is produced for this subgroup in two generators, and tworelations together with their images under the iterated application ofa certain simple substitution.

1 Introduction

The group of automorphisms of rooted trees has attracted interest for beinga source of examples of new phenomena in combinatorial group theory, aswell as for its connections with other areas such as automata theory [3], [4],and dynamical systems [2]. This group together with many of its finitelygenerated subgroups exhibit rich recursive structures which reflect the recur-siveness of the trees themselves [12]. One form the recursiveness takes is inthe closure of the subgroup structure under wreath products by cyclic groupshaving order equal to the valency of the tree. Thus, if L is a group definedon the binary tree then the wreath product of L by the cyclic group of ordertwo C2 is also defined in a canonical manner on the tree.

Of particular interest are the automorphisms of n-ary trees which cor-respond to finite state automata defined on an alphabet of size n. Theseconstitute the enumerable group of finite state automorphisms. We haveshown in [4] that the integral linear group GL(m,Z) has a faithful represen-tation into the group of finite state automorphims of a one-rooted regularn-ary tree for some n. Besides representing integral linear groups, finite stateautomorphisms also represent finitely generated groups which are not linear,such as the finitely generated infinite Burnside groups constructed in [1], [13],[5], [8] and which enjoy diverse additional remarkable properties such as be-ing just-infinite [9], or having intermediate growth [6]. A fractal-like featurecommon to many of these groups is that they contain subgroups of finiteindex which are direct products of two or more copies of the same groups.

This paper grew out of investigating torsion-free subgroups generatedby finite state automorphisms of the binary tree. One special finite state

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automorphism is the so called ”binary adding machine” which correspondsto addition by 1 in the binary numbers; in notation to be explained below,this automorphism is τ = (e, τ)σ. It was proven in [4] that the centralizerof τ in the group of automorphisms of the binary tree is isomorphic to thedyadic integers under addition, and moreover, any automorphism of the treewith an abelian centralizer subgroup is conjugate to τ . We observe that inthe recent Lecture Notes by Bass et al. ([2], page 116), there is a treatmentof the more general q-adic adding machine.

The torsion-free group H which we describe in this paper is generatedby τ = (e, τ)σ, and by one of its conjugates µ = (e, µ−1)σ. The generatorsencapsulate in their definitions the permutations they induce on the differentlevels of the tree. In general, it is a very difficult task fathoming the structureof a group merely from a recursive definition of how its generators act on thetree. Consequently, it is perhaps surprising to see how much detail about thegroup can be revealed and understood on this basis. We prove

Theorem 1. H be the subgroup of the automorphisms of the binary treegenerated by µ = (e, µ−1)σ, τ = (e, τ)σ. Then the following properties hold:(i) H is residually a ”torsion-free solvable group”, and is just-nonsolvable;(ii) H is residually a ”finite 2-group”, and every finite quotient H of Hfactors as H = O(H).S, where O(H) is a nilpotent Hall 2′-subgroup of Hhaving class at most two and S is a Sylow 2-subgroup of H.

The quotients of the lower central series γi(H) of H are studied usingtechniques developed by Vieira in [14]. Let O2′(H) denote the odd orderresidual ∩{K � H | H/K has odd order}.

Theorem 2.(i) The cyclic subgroups 〈µ〉 , 〈τ〉 are self centralizing in H;(ii) the direct product group γ2(H) × γ2(H) is a normal subgroup of γ2(H)and the quotient γ2(H)/ (γ2(H)× γ2(H)) is infinite cyclic;(iii) the quotient group H/γ3(H) is torsion-free;(iv) γ3(H) = O2′(H);(v) the central quotients γi(H)/γi+1(H) have exponents divisors of 8, for alli ≥ 3.

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The group H admits an elegant presentation with two generators, andtwo relators together with their images under the iterated application of anendomorphism of the free group of rank two.

Theorem 3. Let F be the free group of rank 2 generated by a, b, r =[b, ba], r

′= [b, ba3

]. Furthermore let ε be the endomorphism of F determinedby ε : a → a2, b → a2b−1a2. Define the group L =< a, b | εk(r), εk(r′), k ≥0 >. Then the map ϕ : a → τ, b → λ(= τµ−1) extends to an isomorphismfrom L onto H.

The above presentation is in the same spirit as that given by Lysenok in[10] for the Grigorchuk 2-group. The Gupta-Sidki 3-group also has such apresentation, however the substitutions are more involved [11].

2 Preliminaries

2.1 Notation

The one-rooted binary tree T2 is labelled by the free monoid Y , freely gener-ated by the set Y = {0, 1}, with identity element Ø, ordered by the relation:m ≤ m′ if, and only if, m′ is a prefix of m. There is a level function onT2 arising from |m| , the number of syllables in m ∈ M ; the root vertex Øhas level 0. Let A denote the automorphism group of T2, and let σ be theautomorphism of T2 interchanging 0m and 1m for any m in the monoid Y .

An automorphism α which fixes the vertices labelled by 0, 1 is repre-sented as an ordered pair (α0, α1) where αi is the automorphism of the sub-tree headed by i ∈ {0, 1}. Since these subtrees are (standardly) isomorphicto T2, we identify the αi’s with automorphisms of T2. Therefore, a generalautomorphism α may be represented inductively as α = (α0, α1) .σiφ , whereiφ = 0, 1 modulo 2, and similarly α0 = (α00, α01)σ

i0 , α1 = (α10, α11)σi1 ,

and so on. Multiplication is determined by σ. (α0, α1) .σ = (α1, α0) and(α0, α1) . (β0, β1) = (α0β0, α1β1). Successive developments of α produce forevery u ∈ Y an automorphism αu = αu(α) of the tree, together with apermutation σu = σu(α) of Y .

Define the following involutory automorphisms of T2 : σ0 = σ, σ1 =(σ, e) , σ2 = (σ1, e) = ((σ, e), (e, e)) , and inductively, σi+1 = (σi, e). Then the

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group generated by {σi | 0 ≤ i ≤ k} is the wreath product okC2 = (... o C2)oC2

of the cyclic group C2, iterated k times, while the {σi | i ≥ 0} is the infinitelyiterated restricted wreath product o∞C2. We observe that A induces on thek-th level vertices a permutation group isomorphic to ok−1C2 for all k ≥ 1.Let Ak denote k-th level stabilizer subgroup of A; that is, Ak is the kernelof the action of A on the k-th level vertices. Then A/Ak

∼= ok−1C2, and∩{Ak | k ≥ 0} is trivial. In particular, A is residually ”a finite 2-group”.

Given α ∈ A, the set Q (α) = {αu : u ∈ Y } is called the set of states of α.A state αu is said to be inactive if iu = 0, otherwise it is active. It is possibleto interpret the automorphism α as an automaton with alphabet Y : the setof states is Q (α); when the automaton is in state β, the output function isgiven by y → z = (y)σφ(β); the state transition function is β → βz. We callα a finite state automorphism if the set of states Q (α) is finite.

For an automorphism α of the binary tree, we let α(1) denote (α, α) andinductively, α(k+1) = (α(k))(1). Then, α(k) ∈ Ak.

2.2 First facts about the group H

(i) Define λ = τµ−1. Then, as τ = (e, τ)σ, µ = (e, µ−1)σ, λ = (e, τ)σ.(µ, e)σ =(e, τ)(e, µ) = (e, τµ) = (e, τλ−1τ). Therefore,

λ = (e, τλ−1τ)

Clearly, H =< τ, λ >. Then H1 = H ∩ A1 contains the elements λ =(e, τλ−1τ), τ 2 = (τ, τ) = τ (1). Therefore, H = H1 < τ >, [H : H1] = 2,and H1 projects in its first and second coordinates onto H. That is, H1 is asub-direct product of X

2H = H ×H seen as a subgroup of A. Our group H

itself is a subgroup of(X2H

)< σ > and the embedding can be continued to

produce the chain of subgroups

H ≤(X2H

). < σ >≤ ... ≤

(X2k

H). < σi | 0 ≤ i ≤ k >

(ii) Let λi denote the conjugates λτ iof λ, for i any integer. Also define

r = [λ, λ1], r′ = [λ, λ3]. Then λ1 = (λ−1τ 2, e), λ3 = (λ−1τ 2, e)τ2

= (λ−ττ 2, e),and r = e = r′ follow. Indeed we observe that [λi, λj] = e whenever i, j haveopposite parities. Let Λ be the normal closure of λ in H, Λ1 =< λi | i isodd>, and Λ2 =< λi | i is even>. Therefore,

H1 = Λ < τ 2 >, Λ = Λ1 + Λ2

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(iii) We calculate τ−2λτ−2 = (τ−2, λ−1), thus separating λ from τ in thesecond coordinate. We can separate λ at successively lower levels: τ 2λτ 2 =(τ 2, (τ 2, λ)), τ−6λτ−6 = (τ−6, (τ−2, (τ−2, λ−1))), and so on.(iv) The relation r can be produced at the first level as follows:

τ−2λτ−2 = (τ−2, λ−1),(τ−2λτ−2

)τ2

= τ−4λ = (τ−2, λ−11 ),

r′′

= [τ−2λτ−2,(τ−2λτ−2

)τ2

] = [τ−2λτ−2, τ−4λ] = (e, [λ−1, λ−11 ]) = e

Therefore,

r′′

= (e, [λ, λ1]λ−11 λ−1

) = (e, r)τ−4λτ−2λτ−2

(v) Let F be the free group freely generated by a, b. A word w ∈ F iswritten as w = w(a, b) = ai1bj1ai2bj2 ...aisbjs for some integer s ≥ 0, andsome integers it, jt, (1 ≤ t ≤ s). Let w be a reduced word. We will use theb-length |w|b =

∑{|jh| | 1 ≤ h ≤ s} in analyzing word combinatorics. Thislength function is especially suitable since w(τ, λ) = τ i1λj1τ i2λj2 ...τ isλjs =(w0(τ, λ), w1(τ, λ))σi where i =

∑{|ih| | 1 ≤ h ≤ s} modulo 2, and wherew0(a, b), w1(a, b) are words such that |w0|b + |w1|b ≤ |w|b.(vi) The group Λ is six-generated. We note that

r′′

= [λ2τ−4, λ4τ

−4] =(λ2τ

−4)−1 (

λ4τ−4

)−1λ2τ

−4λ4τ−4 = λ−1

−2λ−1−4λ−6λ0 = e

On conjugating this last equation by τ 6, we obtain λ−14 λ−1

2 λ0λ6 = e, and so,

λ6 = λ−10 λ2λ4, Λ =< λi | 0 ≤ i ≤ 5 >

(vii) The generators τ and µ are conjugate by θ = θ(2)σ(1). Since θ2 =(θ2)(2), and its states θ2, (θ2)(1) are inactive it follows that θ2 = e.

We note that

θτθ = (e, σθ(1)τσθ(1))σ,

µ−1θτθ = (µ, e)σ(e, σθ(1)τσθ(1))σ = (µσθ(1)τσθ(1), e)

= (µ, e)σ(e, σθ(1)τσθ(1))σ = (µσθ(1)τσθ(1), e),

where the first coordinate is

µσ.θ(1)τσθ(1) = (e, µ−1)(θ2, θτθ) = (e, µ−1θτθ)

Thus, µ−1θτθ = ((e, µ−1θτθ), e), and µ−1θτθ = e follows.

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2.3 Decomposition of the subgroups H ′, γ3(H), H ′′

Let H ′ denote the first derived subgroup, H ′′ the second derived subgroup,and more generally H(k) the k-th derived subgroup of H. Furthermore definec = [λ, τ ] ∈ H ′ and the sequence of tree automorphisms c1 = (c, e), ci+1 =(ci, e) for i ≥ 1.

Proposition 4. The derived group H′factors as H

′= (X

2H

′) < c >, and

H′is the normal closure of < c1, c > in H.

proof From c = (λ−1τ 2, τ−1λτ−1), cτ = (τ−2λ, λ−1τ 2) we calculate

ccτ = (e, τ−1λτ−1λ−1τ 2) = (e, λτλ−τ2

) = (e, c−τ )

Therefore,ccτ = (e, c−τ )

From this we conclude that c1 ∈ H ′. For,

τ (1)τ−1τ (1)(ccτ )−1τ−(1)ττ−(1) = (e, cτ )τ−(1)ττ−(1)

= c1

As the normal closure of c1 in H1 is H ′×{e}, we conclude that H ′×{e} ≤ H ′

and H′is the normal closure of < c1, c > in H. Hence X

2H

′is a normal sub-

group of H, cτ = c−1 modulo X2H

′, and we obtain the required factorization

of H′.

Proposition 5. The third term of the lower central series γ3(H) factors as

γ3(H) =(X2γ3(H)

)< [c, τ 2] >< [c, τ ] >

and [c, τ 2] ≡ (c−1, c) modulo X2γ3(H).

proof: From c1 = (c, e) ∈ H′and τ 2 = (τ, τ), we obtain

[c1, τ2] = ([c, τ ], e) ∈ γ3(H)

The group γ3(H) is the normal closure of [c, τ ] in H, since [c, λ] = e. Weconclude that the normal closure of [c1, τ

2] in H1 is γ3(H) × {e} and thatthis is a subgroup of γ3(H). Therefore

X2γ3(H) ≤ γ3(H)

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The assertion [c, τ 2] ≡ (c−1, c) modulo X2γ3(H) follows from

[c, τ 2] = ([λ−1τ 2, τ ], [τ−1λτ−1, τ ]) ≡ ([λ, τ ]−1, [λ, τ ]) ≡ (c−1, c)

modulo X2γ3(H). We note that modulo X

2γ3(H), the element (c−1, c) of γ3(H)

is inverted by τ and is centralized by λ, c, [c, τ ].Let D(H ′) be the normal closure of (c−1, c) in H. Then

D(H ′) ≤ γ3(H),(X2γ3(H)

)D(H ′) =

(X2γ3(H)

)< (c−1, c) > .

In order to complete the description of γ3(H) we will compute the com-mutators [c, τ, λ], [c, τ, τ ] modulo X

2γ3(H). Another form of cτ = c−1(e, c−τ )

is[c, τ ] = c−2(e, c−τ )

Since [c, λ] = e, we have

[c, τ, λ] = [c−2(e, c−τ ), λ] ≡ [c−2(e, c−1), λ]

≡ [c−2, λ](e,c−1)[(e, c−1), λ] ≡ [(e, c−1), λ]

≡ (e, [c−1, τλ−1τ ]) ≡ e

modulo X2γ3(H). Hence

[c, τ ] = c−2(e, c−1) modulo X2γ3(H),

[c, τ, λ] ∈ X2γ3(H).

By commutator calculus [c, τ 2] = [c, τ ]2[c, τ, τ ][c,τ ]. Since by the previousproposition [c, τ 2] ≡ (c−1, c) moduloX

2γ3(H), we find that [c, τ, τ ][c,τ ] ≡ [c, τ ]−2(c−1, c),

and therefore[c, τ, τ ] ≡ [c, τ ]−2(c−1, c) modulo X

2γ3(H)

With this we arrive at the required factorization of γ3(H).

Proposition 6. The second term of the derived series H ′′ factors as

H ′′ = X2

{(X2γ3(H)

)< (c−1, c) >

}

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Proof: We conclude from the factorization of the first term of the derivedseries

H′= (X

2H

′) < c >

that H ′′ is the normal closure of [X2H

′, c] in H. Therefore by using Proposition

4, H ′′ is the normal closure of [c1, c], [c2, c] in H. First we calculate [c2, c] =([c1, λ

−1τ 2], e) = (([c, τ ], e), e) and conclude that H ′′ ≥ X4γ3(H). Now we

calculate [c1, c] in H modulo X4γ3(H),

[c1, c] = ([c, λ−1τ 2], e) = ([c, τ 2], e) ≡ ((c−1, c), e)

from which the factorization of H ′′ follows.

2.4 Centralizers

Lemma 7. The centralizer of the derived group H ′ in H is trivial.

Proof: Let w ∈ F have least b−length |w|b such that w(τ, λ) = (w0, w1)σi

centralizes H ′. Since c1 = (c, e) ∈ H ′, clearly, i = 0, and therefore, w0, w1

also centralize H ′. Hence, |w|b = 0, and w(τ, λ) is a power of τ . Sinceci = (ci−1, e) ∈ H ′, we conclude that w = e.

Lemma 8. The generators τ, µ are self centralizing in H.

Proof: We will prove the assertion for τ . The case for µ will be analo-gous. Suppose CH(τ) 6=< τ >. Let w ∈ F have least non-zero b−lengthsuch that w(τ, λ) = (w0, w1)σ

i centralizes τ . Since |w|b 6= 0, it follows that|w0|b , |w1|b < |w|b. If i = 1, then w(τ, λ)τ = (w0τ, w1) commutes withτ 2 = (τ, τ). Therefore w0, w1 commute with τ and so, w0 = τ j, w1 = τ k forsome integers j, k. Now conjugation of w by τ shows that k = j+1 and there-fore w = τ 2j+1; a contradiction. The proof for the case w(τ, λ) = (w0, w1)proceeds similarly.

3 Descending central series

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Proposition 9. The commutator quotient H/H ′ is torsion-free of rank 2and H ′/γ3(H) is torsion-free cyclic.

Proof: (i) H/H ′is a torsion-free group of rank 2. Let i, j be integers suchthat u = λiτ j ∈ H ′. Since u ∈ H1, it follows that j = 2j1. Therefore,u = (τ j1 , c′λ−iτ 2i+j1) for some c′ ∈ H ′. On the other hand, from thedecomposition of H ′, there exist an integer k and c′′, c′′′ ∈ H ′ such thatu = ck = (c′′λ−kτ 2k, c′′′λkτ−2k). Therefore,

(τ j1 , c′λ−iτ 2i+j1) = (c′′λ−kτ 2k, c′′′λkτ−2k)

from which we conclude

λ−kτ 2k−j1 , λk+iτ−(2i+2k+j1) ∈ H ′

On using λiτ j ∈ H ′ with the later conclusions, we produce τ 2(i+j) ∈ H ′. Thuswe have

λiτ j, τ 2(i+j), λ−kτ 2k−j1 ∈ H ′

The fact λ−kτ 2k−j1 ∈ H ′ implies that 2k − j1 is even, and therefore j1 isalso even; j = 4j2. On substituting λiτ j ∈ H ′ by τ 2(i+j) ∈ H ′ in the previousargument, we conclude that 2(i + j) is a multiple of 4, and so, i = 2i1.Therefore, on considering λ−kτ 2k−j1 ∈ H ′, we conclude that k = 2k1. Hence,λ−kτ 2k−j1 = λ−2k1τ 2(2k1−j2). Again, there exists an integer k′ = 2k′1 such thatλ−k′τ 2k′−(2k1−j2) ∈ H ′. Therefore, 2k′−(2k1−j2) is even, and so, j = 8j3. Bygoing back to τ 2(i+j) ∈ H ′ we conclude that i = 4i2. This procedure may berepeated to prove that 2s | j, and 2s−1 | i, for any s ≥ 1. Hence, j = i = 0.

(ii) H ′/γ3(H) is torsion-free. Clearly c /∈ γ3(H). We have from the proofof Proposition 4 that ccτ = (e, c−τ ), and therefore

c2 = (e, c−1) modulo γ3(H),

from which we conclude c2 = c2τ = (c−1, e) modulo γ3(H).Let k be the smallest non-negative integer such that c2k ∈ γ3(H). Then

(ck, e) ∈ γ3(H) and from the decomposition γ3(H) =(X2γ3(H)

)< (c−1, c) ><

[c, τ ] >, we obtain (ck, e) = (c′, c′′)(c−1, c)l[c, τ ]m, for some c′, c′′ ∈ γ3(H),and l,m integers. Since [c, τ ] = c−2(e, c−τ ), c = (λ−1τ 2, τ−1λτ−1) we obtain

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(τ−2λ)2m ∈ H ′, and therefore m = 0 follows. Hence, ck+l, cl ∈ γ3(H), and weconclude that ck ∈ γ3(H).

Proposition 10. The central quotient γi(H)/γi+1(H) has exponent a divisorof 8, ∀i ≥ 3.

Proof: It is sufficient to prove the assertion for γ3(H)/γ4(H).We develop the relation r′′ = [τ−2λτ−2, τ−4λ] = e in terms of commuta-

tors: e = [(τ−4λ)−1

.τ−2λτ−2, τ−4λ] = [[λ, τ−2], τ−4λ] = [λ, τ−2, λ][λ, τ−2, τ−4]λ,and so we arrive at the dependence equation in γ3(H)

[λ, τ−2, τ−4] = [λ, τ−2, λ−1]

On considering this equation modulo γ4(H) we obtain [λ, τ, τ ]8 ≡ [λ, τ, λ]2 =e. Therefore

[c, τ ]8 ≡ e modulo γ4(H),

and γ3(H)/γ4(H) is cyclic of exponent a divisor of 8.

4 Torsion-freeness and Just-nonsolvability

Proposition 11. The group H is residually ”torsion-free solvable” and isnonsolvable.

Proof: First we will prove that for all k ≥ 0, (X2k

H ′)/(X2k+1

H ′) is torsion-

free. For the case k = 0, we recall the decomposition H ′ = (H ′ × H ′) <c > and we suppose that cn ∈ H ′ × H ′ for some integer n. Then as cn =((λ−1τ 2)n, (τ−1λτ−1)n), we conclude that (λ−1τ 2)n ∈ H ′, and n = 0 followsfrom the fact that H/H ′ is torsion-free. The general case is treated similarly.

We conclude from the above argument that H/(X2k

H ′) is a torsion-free

group. Since H ≤ (X2k

H).(okC2), it follows that H/(X2k

H ′) is homomorphic

image of the solvable group (X2k

H/H ′).(okC2). Also since X2k

H ′ ≤ Hk, the

k-th stabilizer subgroup, ∩X2k

H ′ | k ≥ 0} ≤ ∩{Hk | k ≥ 0} = {e}. Hence H

is residually a ”torsion-free solvable” group.

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Clearly, H ′′ is non-trivial. Since H ′′ ≥ X2D(H ′), we have that for all k ≥ 1,

H(k+1) ≥ (X2D(H ′))(k) which is a sub-direct product of X

4H(k). By induction

on k, we conclude that H is non-solvable.

Proposition 12. The group H is just-nonsolvable.

Proof: Let w ∈ H, w 6= e. We will show that the normal closure W of anynon-trivial element w in H contains a term of the derived series of H. Thereare two cases to consider depending upon whether or not w ∈ H1.

Case 1. Let w = (w0, w1)σ. Then a direct calculation shows that

[w, λ] = (w−11 τ−1λτw1, τλ−1τ),

[w, λ, λ2] = (e, [τλ−1τ, λ−1τ 2]) = (e, λ0λ3λ−12 λ−1

1 )

Let ν = λ0λ3λ−12 λ−1

1 , and N be the normal closure of ν in H. Then, inΛ modulo N , λ3 ≡ λ−1

0 λ1λ2 = λ−10 λ2.λ1, and so, [λ3, λ1] ≡ e. Therefore,

Λ/N is abelian of rank at most 3, H/N is metabelian, and H ′′ ≤ N . Fromthe embedding H ≤ (H ×H) < σ >, we conclude that N × N ≤ W ,H ′′ ×H ′′ ≤ W and H is solvable.

Case 2. Let w = (w0, w1). We may assume w1 6= e. Let W1 be the normalclosure of w1 in H. Therefore, by Lemma 7, [w1, H

′] is non-trivial. Now,W ≥ [(w0, w1), (e, H

′)] = (e, [w1, H′]), and so W ≥ (e, [W1, H

′]). Clearly,if W1 contains a derived subgroup H(l) for some l, then W would containH(l+1) × H(l+1), and as H/

(H(l+1) ×H(l+1)

)is solvable, the case would be

done. The desired conclusion is reached as there exists a minimum levelwhere w is active.

5 Finite Quotients of H

Proposition 13.(i) The odd order residual O2′(H) is equal to γ3(H).(ii) Any finite quotient H of H decomposes as H = O(H).S where O(H) isa normal Hall 2′-group of H, which is nilpotent of class at most 2, and whereS is a 2-group.

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Proof: (i) Let n be an odd number and N be the normal closure of τn.Then, in H modulo N : λn = λ0, and therefore Λ1 = Λ2 = Λ, and Λis an abelian group generated by λ0, λ2, λ4. The equation λ−1

0 λ2λ4λ−16 = e

translates additively to λ0.(−1 + τ 2 + τ 4 − τ 6) = 0.Define the polynomials l(x) = −1+x2+x4−x6 and m(x) = −1+xn. Then

l(x), m(x) factor as l(x) = (−1+x)l1(x), where l1(x) = (1−x)(1+x)2(1+x2),and m(x) = −1+xn = (−1+x)m1(x). Let GCD denote the greatest commondivisor operator applied to polynomials with integer coefficients. Then sinceGCD(1+x, m1(x)) = 1, GCD(1+x2, m1(x)) = 1, GCD((1−x), m1(x)) = n,we conclude that GCD(l(x), m(x)) = n(−1 + x).

Let a(x), b(x) be polynomials over the integers such that

a(x)(1 + x)2(1 + x2) + b(x)m1(x) = 1

Then, on multiplying the above by (1− x)2 we obtain

−a(x)l(x)− b(x)m(x) = (1− x)2

Therefore the following equations hold

λ0.n(−1 + τ) = 0, λ0(1− τ)2 = 0

in H modulo N . Returning to multiplicative notation,

Nλn0 = Nλn

2 = Nλn4 ,

γ3(H) ≤ N

Now we reach the conclusion O2′(H) = γ3(H) since, easily, ∩{Hk | k odd} ≤γ3(H).

(ii) Let n = 2kn′, n′ odd, M the normal closure of τ 2k, and N the normal

closure of τn in H.We assert that M =

(X2k

H ′)

< τ 2k>. Since τ 2k

= τ (k), and there exist in-

tegers i, j, ...m, such that ξ = (τ i, (τ j, (..., (τm, λ)...)) ∈ H where λ is isolatedat the k-th level, we produce [τ 2k

, ξ] = (e, e, ..., e, [τ, λ]) ∈ Hk. Therefore, Mcontains the normal subgroup X

2kH ′. It is easy to see by induction on k that

[τ 2k, λ] ∈ X

2kH ′. With this, our assertion is established.

Now τn =(τ 2k

)n′

= (τn′)(k), and it follows from part (i) that N contains

X2k

γ3(H). Since M/(X2k

Xγ3(H))

=(X2k

H ′/γ3(H)). <

(X2k

γ3(H))τ (k) > is

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clearly a class 2 nilpotent group, therefore also M/N is also nilpotent of classat most 2.

Let N = Hn be the group generated by {hn | h ∈ H}, and let M = H2k.

Then H/M is a finitely generated solvable group of exponent 2k, and so, itis a finite 2-group. Thus, M is a finitely generated group. Now M/N is afinitely generated nilpotent group of class at most 2, and has odd exponentn′. Therefore M/N is a finite group of odd order, and H/N is a finite grouphaving the type of structure as had been affirmed.

Since any finite quotient group of H is a quotient of H/Hn for someinteger n, the proof follows.

6 A Presentation for H

Let K be the subgroup of H1 generated by τ 2, λ. Also, let ρ : K → H be theprojection on the second coordinate. Then as τ 2 = (τ, τ), λ = (e, τλ−1τ), wehave ρ : τ 2 → τ, λ → τλ−1τ and therefore it is an epimorphism. It is actuallyan isomorphism. However, first we need to prove the following.

Lemma 14. Let s be an integer, s ≥ 1. Then the tree automorphisms definedby (e, τ)s, (τ, e)s do not belong to H.

Proof: Since τ 2s = (τ s, τ s) ∈ H, it sufficient to prove the assertion forα = (e, τ)s. Suppose that α ∈ H. Since H/H ′ is 2-generated, there existintegers i, j such that αλiτ j ∈ H ′. Therefore, j = 2j′,

αλiτ j = (τ j′ , τ s(τλ−1τ)iτ j′),

and from the decomposition H ′ = (H ′ ×H ′) < c >, we obtain the secondexpression

αλiτ j = (c′, c′′)ck

for some c′, c′′ ∈ H ′ and some integer k. However, on substituting

ck = ((λ−1τ 2)k, (τ−1λτ−1)k)

in the above equation and on comparing first coordinates, we arrive at τ j′ =c′(λ−1τ 2)k. Since H/H ′ is freely generated by H ′λ, H ′τ , we conclude that

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k = 0 = j′ = j. Thus, αλi = (e, τ s′(τλ−1τ)i) = (c′, c′′), τ s′+2iλ−i ∈ H ′, andi = 0 follows.

Let w be a word in τ 2, λ. Then w = w(τ 2, λ) = (w(τ, e), ρ(w)). Ifρ(w) = e, then w = (w(τ, e), e) = (τ s, e), for some integer s. Thus (τ s, e)is an element of H, and s = 0 by the above lemma; hence w = e. Defineε : H → K to be the inverse map of ρ. Then, ε : τ → τ 2, λ → τ 2λ−1τ 2

determines an isomorphism from H onto K.

Proposition 15. Let r = [b, ba], r′ = [b, ba3] be words in the free group F

freely generated by a, b. Also consider the endomorphism of F determined byε : a → a2, b → a2b−1a2, and define the group

L =< a, b | εk(r), εk(r′), k ≥ 0 > .

Then, the map ϕ : a → τ, b → λ extends to an homomorphism from L ontoH.

Proof: Explicitly, εk : a → a2k, b → alkbδkalk , where δk = (−1)k, lk =

23(2k − δk). In order to write down εk(r), εk(r′), we define bs = bas

. Then,

εk(b) = b−1−lk

a2lk ,

εk(r) = b−δklk

b−δk

3.lk+2kbδk3lk

bδk

lk+2k ,

εk(r′) = b−δklk

b−δk

3(lk+2k)bδk3lk

bδk

lk+3.2k

for all k ≥ 0. We note that for k = 1, δk = −1, lk = 2, and thus,

ε(r) = b2b8b−16 b−1

4 , ε(r′) = b2b12b−16 b−1

8

Let B be the normal closure of b in L,B1 =< bk | k odd>,B2 =< bm | m even>. The relation r = e meansthat b0 commutes with b1, and by conjugation by a we conclude that b1 com-mutes with b0, b2. Now r′ = e implies that b1 commutes with b4 as well.Since ε(r) = b2b8b

−16 b−1

4 = e, we have b0b6b−14 b−1

2 = e, and therefore, b1 com-mutes with b0, b2, b4, b6. Furthermore, since b2b8b

−16 b−1

4 = e, we deduce thatb1 commutes with b0, b2, b4, b6, b8. It is clear that on repeating this argu-ment we obtain that b1 commutes with bi, for all i even. Thus, L factors asL = B < a >, and B = B1 + B2.

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An element w(a, b) of L can be written as w(a, b) = w1(a, b)w2(a, b)an

where w1(a, b) = bi1k1

bi2k2

...bisks∈ B1, kt odd for1 ≤ t ≤ s, and where w2(a, b) =

bj1m1

bj2m2

...bjqmq

∈ B2, mp even for 1 ≤ p ≤ q; note that |w1|b + |w2|b = |w|b.On applying the ϕ map we have ϕw(a, b) = (w′

2(τ, λ), w′1(τ, λ))τn. Here,

w′1(a, b) = (b−1

k′1a2)i1(b−1

k′2a2)i2 ...(b−1

k′sa2)is , where k′h = kh

2−1 for 1 ≤ h ≤ s, and

w′2(a, b) = (b−1

m′1a2)j1(b−1

m′2a2)j2 ...(b−1

m′qa2)jq , where m′

h = mh−12

for 1 ≤ h ≤ q.

Thus, |w′1|b = |w1|b , |w′

2|b = |w2|b.Suppose that w(a, b) is a nontrivial element in the kernel of ϕ, and let it be

of minimum b- length |w|b. Then, ϕw(a, b) = (w′2(τ, λ), w′

1(τ, λ))τn = e, andso, n = 2n0, and (w′

2(τ, λ)τn0 , w′1(τ, λ)τn0) = e. Hence, w′

1(a, b)an0 , w′2(a, b)an0

are also in the kernel of ϕ, and

|w′1(a, b)an0|b = |w1|b , |w′

2(a, b)an0 |b = |w2|b

If both w1, w2 have b−lengths shorter than w, then w′1(a, b)an0 = e =

w′2(a, b)an0 , w1(a, b) = e = w2(a, b), and consequently, n = 0, w = e; a

contradiction is reached. Thus, we may assume w(a, b) = w2(a, b)an, q > 1.Then, n = 2n0, and both w2(a, b)an, w′

2(a, b)an0 are in the kernel of ϕ. Nowwe replace in the above argument w by

w′2(a, b)an0 = (b−1

m′1a2)j1(b−1

m′2a2)j2 ...(b−1

m′qa2)jqan0 = w′′

2(a, b)an1

where n1 = 2j1 + ... + 2jq + n0. Thus, again, w′′2(a, b) ∈ B1, or B2. That is,

m′h has the same parity as m′

1 for all 1 ≤ h ≤ q. Hence, m′h −m′

1 = mh−12

−m1−1

2= mh−m1

2is even and so, 4 | mh −m1. If j1 > 1 then since (b−1

m′1a2)j1 =

b−1m′

1b−1m′

1−2..., the consecutive indices cannot be congruent modulo 4. Therefore

|jh| = 1 for all 1 ≤ h ≤ q. As q > 1, it follows that jh+1 = − jh, and qis even. We may assume j1 = 1. Thus, w′

2(a, b)an0 = b−1m′

1bm′

2...b−1

m′q′bm′

q′+1an0 .

Once more, we obtain that 2 | n0, 4 | m′h − m′

1. Therefore, 8 | mh − m1. Arepetition of this argument produces, mh = m1, n = 0. Hence, w2(a, b) =bj1m1

bj2m2

...bjqmq

= e, an = e, w = e; a contradiction is reached.

7 Final Comments

(i) On changing b by its conjugate b = ba2in the above presentation for H,

the definition of the endomorphism becomes ε : a → a2, b → b−1a4. The new

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relators may still be chosen to be r = [b, ba], r′ = [b, ba3] together with their

images under repeated applications of ε.

(ii) It was shown in the second section of this paper that conjugation byθ = θ(2)σ(1) is an automorphism of H of order two which interchanges τ andµ. We note that conjugation by δ = δ(1)σ inverts both τ and µ. It appearsthat the 4-group < δ, θ > is the group of outer automorphisms of H.

(iii) Define Rl to be the normal closure of < εk(r), εk(r′) | 0 ≤ k ≤ l > in thefree group F . It can be proven that {Rl | l ≥ 0} is a strictly ascending chainof subgroups, and therefore H is not finitely presentable.

(iv) The proof of Proposition 14 provides an algorithm for solving the wordproblem in H. It would be interesting to see if the approach of Wilson-Zaleskii in [15] leads to the solution of the conjugacy problem for this group.

(v) A torsion-free group which is an extension of an abelian group of infiniterank by the Grigorchuk 2-group has been defined to act on a tree with infinitevalency and was shown to have intermediate growth [7]. This raises thequestion about the growth function of our group.

References

[1] S. V. Aleshin, Finite automata and the Burnside problem for periodicgroups, Math. Notes 11 (1972), 199-203.

[2] H. Bass, M. Otero-Espinar, D. Rockmore, and C. P. L. Tresser, ”CyclicRenormalization and Automorphism Groups of Rooted Trees”, LectureNotes in Mathematics 1621, Springer, Berlin (1995).

[3] A. M. Brunner, and S. Sidki, On the automorphism group of the one-rooted binary tree, J. Algebra 195, (1997), 465-486

[4] A. M. Brunner, and S. Sidki, The generation of GL(n, Z) by finite au-tomata, Int. Journal of Algebra and Computation. To appear.

[5] R. I. Grigorchuk, On the Burnside problem for periodic groups, Func-tional Anal. Appl. 14 (1980), 41-43.

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[6] R. I. Grigorchuk, Degrees of growth of finitely generated groups and thetheory of invariant means, Izv. Akad. Nauk SSSR, Ser. Mat. 48 (1984),939-985.

[7] R. I. Grigorchuk, On degrees of growth of p-groups and torsion-freegroups, Mat. Sbornik, 126 (1985), 194–214.

[8] N. Gupta and S. Sidki, On the Burnside problem for periodic groups,Math. Z. 182 (1983), 385-388.

[9] N. Gupta and S. Sidki, Extensions of groups by tree automorphisms,Contributions to Group Theory, Contemporary Mathematics 33 ( Amer-ican Mathematical Society, Providence, 1984), 232-246.

[10] I. G. Lysenok, A system of defining relations for a Grigorchuk group,Math. Notes 25 (1985), 784-792.

[11] S. Sidki, On a 2-generated infinite 3 -group: the presentation problem,J. Algebra 110 (1987), 13-23.

[12] S. Sidki, Regular trees and their automorphisms, 1996. Notes of a Course,XIV Escola de Algebra, Rio de Janeiro, July 1996].

[13] V. I. Suschanskii, Periodic p-groups of permutations and the unrestrictedBurnside problem, Soviet Math. Dokl. 20 (1979), no.4, 776-770.

[14] A. C. Vieira, On the lower central series and the derived series of theGupta-Sidki 3-group, Communications in Algebra (24) (1998), no.4,1319-1333.

[15] J. S. Wilson and P. A. Zaleskii, Conjugacy separability of certain torsiongroups, Arch. Math. 68 (1997), 441-449.

DEPARTMENT OF MATHEMATICS,UNIVERSITY OF WISCONSIN-PARKSIDE,KENOSHA, WISCONSIN 513141-2000, USAe-mail address: brunner@uwp.edu

DEPARTAMENTO DE MATEMATICA,UNIVERSIDADE DE BRASILIA,

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70910-900 BRASILIA-DF,BRAZILe-mail address: sidki@mat.unb.br

DEPARTAMENTO DE MATEMATICA,UNIVERSIDADE FEDERAL DE MINAS GERAIS,30161-901 BELO HORIZONTE-MG,BRAZILe-mail address: anacris@mat.ufmg.br

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