Maximal subgroups of non locally finite p-groups

E.L. Pervova^{∗}

Abstract Kaplansky’s conjecture claims that the Jacobson radical J K[G] of a group algebra

K[G], where K is a field of characteristic p > 0, coincides with its augmentation ideal AK[G] if and only if G is a locally finite p-group. By a theorem of Passman, if G is finitely generated and J K[G] = AK[G] then any maximal subgroup of G is normal of index p. In the present paper, we consider three series of finitely generated infinite p-groups (hence not locally finite p-groups), Grigorchuk’s 2-groups, GGS-groups, and EGS-groups. We prove that their maximal subgroups are nonetheless normal of index p.

Contents

1 Main definitions 3 1.1 Grigorchuk’s groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.2 Periodic GGS-groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.3 EGS-groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 1.4 Scheme of the proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2 Upper companion groups of subgroups in Grigorchuk’s groups: one special case 11

3 Upper companion groups of subgroups in GGS-groups: one special case 20

4 Dense subgroups 29 4.1 Grigorchuk’s groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 4.2 GGS-groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 4.3 EGS-groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

5 Maximal subgroups 38

∗Partially supported by RFFI, grant r2001 Ural N 01-01-96404, and by Russian Ministry of Education, grant E00-1.0-150”

Introduction

Let K be a field and let G be a group. The group algebra K[G] is called semiprimitive if its Jacobson radical J K[G] is trivial.

The study of semiprimitivity of group algebras seeks to establish all groups G with semiprimitive group algebras K[G]. More generally, it is desirable to understand the structure of the Jacobson radical J K[G]. The answer is well-known in case of finite groups, due to a result of Maschke [9]. The result states that if the characteristic of K is zero, then K[G] is always semiprimitive, and if the characteristic is some p > 0 then K[G] is semiprimitive iff G has no elements of order p. However, the situation with infinite groups is much more complex; some of the main results in that direction can be found in [10].

Suppose now that K is a field of characteristic p > 0 and let G be a finitely generated group. If AK[G] is the augmentation ideal of K[G] then it is a maximal right ideal of K[G] and therefore contains the Jacobson radical J K[G]. There is the following conjecture of Kaplansky.

Conjecture. If charK = p > 0 and G is a finitely generated group, then J K[G] = AK[G] if and only if G is a finite p-group.

The “if” direction is obtained easily (see [11]), so the real concern is with the converse direction. Theorem 6.3 of [11] states that if G is a finitely generated group, K is a field of characteristic p > 0, and J K[G] = AK[G], then, in particular, G is a p-group and any maximal subgroup of G is normal of index p. Thus, should G turn out to be infinite, it would a be finitely generated, infinite p-group. Such groups are called Burnside groups for being counterexamples to the unrestricted Burnside problem.

Many examples of such groups are presently known. Among them are the groups constructed by Golod [4, 5], Grigorchuk’s 2-groups [6], and so-called GGS-groups [3, 1] generalizing the examples by Gupta and Sidki [8].

In the present paper we study maximal subgroups of three classes of Burnside groups, the Grigorchuk’s groups, GGS-groups, and EGS-groups [13]. For such groups to have a maximal subgroup of infinite index would be in accordance with Kaplansky’s conjecture. However, it turned out that all their maximal subgroups are normal of index p; in particular, this answers question (1) of [16].

This main result is stated in Theorem 5.1. Some intermediate results given in Theorems 4.1, 4.2, and 4.3 may also be of independent interest.

The groups studied in the paper all act as automorphism groups of regular p-trees; in fact, they are just infinite (meaning all their proper quotients are finite) and belong to the class of so-called branch groups. The class of just infinite branch groups is interesting in itself. Indeed, it is known ([7]) that every finitely generated infinite group can be mapped onto a just infinite group. In its turn, an abstract just infinite group either is a branch group or contains a normal subgroup of finite index which is isomorphic to the direct product of a finite number of copies of a group L, where L is either simple or hereditarily just infinite. There is also a similar result for profinite groups; refer to [7, 18] for more details.

Acknowledgments The crucial part of this research has been carried out in the supportive environment of Max Planck Institute for Mathematics in Bonn whose staff I would like to thank for their kind support and hospitality. I also would like to thank A.V. Rozhkov for his attention to this work, R.I. Grigorchuk for attracting my interest to the problem, Donald Passman for pointing out the connections to the group representation theory, and A.V. Ershov for productive discussions, which were very helpful in clarifying the topological aspects of this research.

1 Main definitions

Let us start with introducing some notions from the theory of branch groups, which provides a language very suitable for dealing with the groups we intend to consider.

Definition 1.1. A group G is said to be a branch group, if there exist two decreasing sequences of subgroups (L_{i})_{i∈N}, (M_{i})_{i∈N} and a sequence of integers (k_{i})_{i∈N} such that

L_{0} = M_{0} = G, k_{0} = 1,
∩_{i∈N}M_{i} = 1,

and for each i the following four conditions hold,

(1) M_{i} is a normal finite-index subgroup of G;

(2) M_{i} is the direct product of k_{i} copies of the subgroup L_{i}, i.e. there are subgroups L_{i}^{(1)},

…, L_{i}^{(k_{i})} of G such that

M_{i} = L_{i}^{(1)} × … × L_{i}^{(k_{i})},  (1.1)

and each of the factors is isomorphic to L_{i};

(3) k_{i} properly divides k_{i+1}, i.e. m_{i+1} = \frac{k_{i+1}}{k_{i}} ≥ 2, and the product decomposition (1.1) of M_{i+1} refines the product decomposition (1.1) of M_{i}, meaning that each factor L_{i}^{(j)} of M_{i} contains m_{i+1} factors of M_{i+1}, namely the factors L_{i+1}^{(l)} for l = (j − 1)m_{i+1} + 1, …, jm_{i+1};

(4) conjugations by elements of G transitively permute the factors in the product decomposition (1.1).

Notice that the set of subgroups L_{i}^{(j)} ordered by inclusion forms a locally finite rooted spherically homogeneous tree T. The group G acts by automorphisms on T and is transitive on each layer of T (comprised by all subgroups L_{i}^{(j)} for a fixed i).

In general, this action may not be faithful. However, in case of just infinite branch groups, which are what we are primarily interested in, it is faithful. Otherwise, as is easily seen, the kernel of the action would be a non-trivial normal subgroup of infinite index. Thus, a just infinite branch group is a group of automorphisms of a rooted spherically homogeneous tree.

Let us specify what conditions a group of tree automorphisms must meet in order to be branch. Notice that if A = (A_{1}, A_{2}, …) is a sequence of finite sets of cardinality 2 or greater, then the set of all finite sequences

a_{1}a_{2} … a_{n}, a_{i} ∈ A_{i}, n ∈ N,

forms a rooted spherically homogeneous tree, where the sequences correspond to vertices, the root is the empty sequence, and two vertices are joined by an edge if they have the form a_{1}a_{2} … a_{n} and a_{1}a_{2} … a_{n}a_{n+1} for some n and some a_{i} ∈ A_{i}. For example, the tree corresponding to a branch group G can be constructed over a sequence of arbitrary sets A_{i} such that the cardinality of A_{i} is equal to m_{i}.

The number n is called the length of a vertex u = a_{1}a_{2} … a_{n} and is denoted by |u|. The set of all vertices of length n is called the nth level of T.

Of special interest is the case when all sets A_{i} are equal to one set X. In this case the tree is called regular, and the set of its vertices is precisely the set of all words in the alphabet X.

Just as we constructed a tree T over an infinite sequence of sets A = (A_{1}, A_{2}, …), we can construct a tree over a finite sequence (A_{1}, A_{2}, …, A_{n}). This tree will be denoted by T_{[n]}; geometrically, it is obtained by removing everything that lies below the nth level.

Suppose that u = â_{1}â_{2} … â_{n} is a vertex. The set of all vertices â_{1}â_{2} … â_{n}a_{n+1}a_{n+2} … a_{n+m}, where m ∈ N and a_{n+i} range over the set A_{n+i}, forms a subtree of T. We will denote that subtree by T_{u}. It is easy to see that T_{u} is naturally isomorphic to the tree T_{(n)} constructed over the sequence (A_{n+1}, A_{n+2}, …), via the map

â_{1}â_{2} … â_{n}a_{n+1}a_{n+2} … a_{n+m} ↦ a_{n+1}a_{n+2} … a_{n+m},

This map also allows to identify subtrees T_{v} for all vertices v of length n, with one fixed tree T_{(n)}.

Of course, if T is regular then for any vertex u T_{u} is isomorphic to T.

Consider now an arbitrary subgroup X in Aut T and a vertex v of T. The rigid stabilizer of v in X is the subgroup

rist_{X}(v) = {x ∈ X| for any u ∈ T \ T_{v} u^{x} = u}.

We also denote by rist_{X}(n) the subgroup ∏_{|v|=n} rist_{X}(n). This is, of course, a normal subgroup in X.

We are now ready to give an equivalent definition of a just-infinite branch group.

Definition 1.2. Let G be a group of automorphisms of a spherically homogenous tree T. G is called a branch group, if it is transitive on each level, and for every integer n ≥ 1 rist_{G}(n) has finite index in G.

There are several notions that are particularly useful in studying branch groups. Let G be a branch group and let v be a vertex of T. The stabilizer of v in G is the subgroup

st_{G}(v) = {x ∈ G|v^{x} = v}.

Denote also by st_{G}(n) the subgroup ∩_{|v|=n} st_{G}(v), which keeps all vertices of level n fixed.

Subgroups st_{G}(n) are called principal congruence subgroups in G. A subgroup of G which contains a principal congruence subgroup is called a congruence subgroup.

Definition 1.3. Group G is said to possess the congruence property if any its finite index subgroup is a congruence subgroup.

Notice that to every x ∈ st_{Aut T} (v) we can assign a unique automorphism x_{v} ∈ Aut T_{v} which is obtained by taking the restriction of x onto the subtree T_{v}. Moreover, for each vertex v of length n x_{v} can be considered as an element of Aut T_{(n)}. Thus, for each vertex v there is a fixed homomorphism φ_{v} : st_{Aut T} (v) → Aut T_{(|v|)}. We will denote φ_{v}(st_{G}(v)) by G_{v} and call it an upper companion group of G (at vertex v).

In general, for different vertices u, v of the same length G_{u} and G_{v} are conjugate in Aut T_{(|u|)} but may not coincide. Let us fix for each pair of vertices u, v of the same length a conjugating element ψ_{u,v} such that G_{u}^{ψ_{u,v}} = G_{v}, and ψ_{u,u} = 1 for any u. Nonetheless, in the examples we will be considering G_{u} and G_{v} do coincide for any two vertices of the same length, and all elements ψ_{u,v} will be equal to the trivial element.

Definition 1.4. A branch group G acting on a regular tree T is fractal, if for each vertex u of T G_{u} = G.

As will be evidenced many times throughout this paper, the property of a group to be fractal is very helpful in inductive proofs.

It is clear that the direct product of φ_{v} taken over all vertices v of length n defines a monomorphism of st_{Aut T} (n) into the direct product of |A_{1}||A_{2}| … |A_{n}| copies of Aut T_{(n)}. This product map will be denoted by Φ_{n}.

Of course, collections of maps φ_{u} and Φ_{n} vary with and are determined by the tree T. Since at each given moment a tree in question will be fixed, we will keep the same notation for maps on different trees.

The above maps give a way to describe some, but not all, automorphisms of T. A more general way is given by an observation that for any vertex u of length n x induces a permutation x(u) of A_{n+1}, which acts according to the rule a ↦ ((ua)^{x})_{n+1}. Such permutations are called accompanying permutations of x. It is easy to see that they fully determine x.

Let us now describe our major examples.

1.1 Grigorchuk’s groups

The periodic Grigorchuk’s groups G_{ω} are known as the first examples of groups of intermediate growth [6]. They are just-infinite branch groups ([6]) acting on the infinite binary tree. That tree can be constructed as a tree over a sequence of sets (X, X, …) for X = {0, 1}.

Groups G_{ω} are parameterized by infinite sequences ω = (ω_{1}, ω_{2}, …) such that for each i ω_{i} ∈ {0, 1, 2}, and each of the numbers 0, 1, 2 occurs in ω infinitely many times. The set of all such ω is denoted by Ω_{0}. The expression σω denotes the sequence (ω_{2}, ω_{3}, …). Clearly, σω is also in Ω_{0}.

Each Grigorchuk’s group G_{ω}, ω ∈ Ω_{0}, is generated by four automorphisms a, b_{ω}, c_{ω}, d_{ω}. (Although just three of these automorphisms are actually enough). The only nontrivial accompanying permutation of a is at the root vertex and is the cyclic permutation, denoted by π, of two elements. The other three automorphisms have infinitely many nontrivial accompanying permutations, but all of them are located in vertices u_{n} = \underbrace{11 … 1}_{n} 0, n =

0, 1, 2, . . .. Permutation b_{ω}(u_{n}) is trivial if ω_{n+1} = 2 and is equal to π otherwise. Likewise, permutation c_{ω}(u_{n}) is trivial if ω_{n+1} = 1 and is equal to π otherwise, and permutation d_{ω}(u_{n}) is trivial if ω_{n+1} = 0 and is equal to π otherwise.

Thus, for any vertex u of the form u = \underbrace{11 … 1}_{n} or u = \underbrace{11 … 1}_{n} 0 = u_{n} generators b_{ω}, c_{ω},

d_{ω} are in st_{G_{ω}}(u), and we have

φ_{u}(b_{ω}) = \{ \begin{matrix}a, & if u = u_{n} for some n and ω_{n+1} ≠ 2 \\ b_{σ_{n}ω}, & if u = \underbrace{11 … 1}_{n} \end{matrix}

φ_{u}(c_{ω}) = \{ \begin{matrix}a, & if u = u_{n} for some n and ω_{n+1} ≠ 1 \\ c_{σ_{n}ω}, & if u = \underbrace{11 … 1}_{n} \end{matrix}

φ_{u}(d_{ω}) = \{ \begin{matrix}a, & if u = u_{n} for some n and ω_{n+1} ≠ 0 \\ d_{σ_{n}ω}, & if u = \underbrace{11 … 1}_{n} \end{matrix}

In particular, for any vertex u φ_{u}(G_{ω}) = G_{σ^{n}ω} for n = |u|.

It also follows from the above definitions that exactly one of generators b_{ω}, c_{ω}, d_{ω} has trivial accompanying permutation at u_{0} = (0). That generator will be called the d-generator of G_{ω}. Now, let m(ω) be the maximal number n such that the finite sequence ω_{1}ω_{2} … ω_{n} is constant. Suppose that y_{ω} is the generator of G_{ω} such that y_{σ^{m(ω)}ω} is the d-generator of G_{σ^{m(ω)}ω}. Then y_{ω} is called the c-generator of G_{ω}. That of generators b_{ω}, c_{ω}, d_{ω} which is neither the d-generator nor the c-generator is called b-generator.

Directly applying Proposition 1 from [12], we get the following fact.

Lemma 1.1. Let G_{ω} be a periodic Grigorchuk’s group. Then

G_{ω}/[G_{ω}, G_{ω}] ≅< a > × < b_{ω}, c_{ω} >≅ Z_{2} × Z_{2} × Z_{2}.

Corollary 1.1. Let x_{ω} be the d-generator of G_{ω} and let y_{ω} be its c-generator. Then the factorgroup of G_{ω} by the normal closure N of x_{ω} is isomorphic to the group generated by a and y_{ω}.

Proof. It is sufficient to show that the intersection < a, y_{ω} > ∩N is trivial. Notice that N is contained in st_{G_{ω}}(m(ω)+1). On the other hand, the intersection < a, y_{ω} > ∩ st_{G_{ω}}(m(ω)+1) is generated by element (ay_{ω})^{2^{m(ω)+1}}. Since the order of this element is equal to 2, it is sufficient to prove that the element does not belong to N.

The image of (ay_{ω})^{2^{m(ω)+1}} under Φ_{m(ω)+1} equals (\underbrace{y_{σ^{m(ω)+1}ω}, y_{σ^{m(ω)+1}ω}, …, y_{σ^{m(ω)+1}ω}}_{2^{m(ω)+1}}). How-

ever, since x_{ω} is the d-generator of G_{ω}, for any vertex u of length m(ω) φ_{u}(N) is in the normal closure of x_{σ^{m(ω)}ω}. And since x_{σ^{m(ω)}ω} is not the d-generator of G_{σ^{m(ω)}ω}, for any vertex u of length m(ω)+1 φ_{u}(N) is in the normal closure of the set {a, x_{σ^{m(ω)+1}ω}}. Hence the image of any φ_{u}(N) in the factor-group of G_{σ^{m(ω)+1}ω} by its derived subgroup is in the subgroup

(< a, x_{σ^{m(ω)+1}ω} > [G_{σ^{m(ω)+1}ω}, G_{σ^{m(ω)+1}ω}])/[G_{σ^{m(ω)+1}ω}, G_{σ^{m(ω)+1}ω}].

By Lemma 1.1 we get that the image of y_{σ^{m(ω)+1}ω} is not in this subgroup, hence (ay_{ω})^{2^{m(ω)+1}} is not in N.

Let us introduce length function ∂_{ω}() on each G_{ω}. Each element of G_{ω} can be represented by a word of the form,

a^{ϵ}x_{1}a … ax_{m}a^{μ},

where x_{1}, …, x_{m} ∈ {b_{ω}, c_{ω}, d_{ω}}, and ϵ, μ are in GF(2). The length of such a word, by definition, is equal to m. The length ∂_{ω}(x) is defined as the minimum of lengths of words representing x.

1.2 Periodic GGS-groups

The term GGS-group was introduced by Gilbert Baumslag [3] and refers to Rostislav Grigorchuk, Narain Gupta and Said Sidki. Those groups act on regular p-trees, where p ≥ 3. We will always assume that p is prime.

Each GGS-group is fully defined by a vector

ᾱ = (α_{1}, α_{2}, …, α_{p−1}) ∈ \underbrace{GF(p) ⊕ GF(p) ⊕ … ⊕ GF(p)}_{p−1}  (1.2)

in the following way.

Construct the regular p-tree T_{p} as the tree over the sequence (X, X, …), X = {0, 1, 2, …, p − 1}. Each GGS-group Gᾱ acting on T_{p} is generated by two automorphisms a and b. The only accompanying permutation of a is at the root and is the cyclic permutation (01 … p − 1), i.e. a cyclically permutes edges coming out of the root. The second generator is in the stabilizer of level 1, and

Φ_{1}(b) = (a^{α_{1}}, a^{α_{2}}, …, a^{α_{p−1}}, b).

The vector (1.2) is called the accompanying vector of G.

By [17], all GGS-groups are just infinite branch; they are periodic if and only if ∑_{i=1}^{p−1} α_{i} = 0. Notice that for any vertex u φ_{u}(Gᾱ) = Gᾱ. Hence all Gᾱ are fractal as well.

By Proposition 1 from [12], we have

Lemma 1.2. Let Gᾱ be a GGS-group. Then

Gᾱ/[Gᾱ, Gᾱ] ≅< a > × < b >≅ Z/pZ × Z/pZ.

Later on we will need the following length function on Gᾱ. Each element x ∈ Gᾱ can be represented (not uniquely) by a word of the form,

a^{δ_{1}}b^{β_{1}} … a^{δ_{m}}b^{β_{m}}a^{δ_{m+1}},

where δ_{2}, …, δ_{m}, β_{1}, …, β_{m} ∈ GF(p) \ {0} and δ_{1}, δ_{m+1} ∈ GF(p). The number m will be called the length of such a word. The length ∂_{Gᾱ}(x) is defined as the minimal length of all words representing x.

Also, denote by ∂_{a}(x) the sum of all δ_{i} and by ∂_{b}(x) the sum of all β_{i} in some representation of x. By Lemma 1.2, the sums are independent of the choice of a particular representation. Indeed, pair (∂_{a}(x), ∂_{b}(x)) is the image of x under the composition of the natural projection Gᾱ → Gᾱ/[Gᾱ, Gᾱ] and the isomorphisms Gᾱ/[Gᾱ, Gᾱ] ≅< a > × < b >≅ Z/pZ × Z/pZ of Lemma 1.2. Thus, ∂_{a}(x) and ∂_{b}(x) are well-defined.

1.3 EGS-groups

The term EGS-group stands for “extended Gupta-Sidki group”, but the word “extended” should not be understood in the usual algebraic sense. While each EGS-group contains some specific GGS-group as a subgroup, it is not an extension of it. However, in a certain sense a periodic GGS-group, or, more precisely, its accompanying vector, does define a unique EGS-group.

Let Gᾱ be a periodic GGS-group with generators a and b and accompanying vector (α_{1}, α_{2}, …, α_{p−1}). The corresponding EGS-group Γᾱ is generated by automorphisms a, b, and the automorphism c such that

Φ_{1}(c) = (c, a^{α_{1}}, a^{α_{2}}, …, a^{α_{p−1}}).

The group Gᾱ is then called the associated GGS-group of Γᾱ.

Let us establish a few needed properties of EGS-groups. First of all, by Theorem 2 of [14], all EGS-groups are periodic. Just as with GGS-groups, for any vertex u φ_{u}(Γᾱ) = Γᾱ.

Lemma 1.3. All EGS-groups are just infinite branch groups.

Proof. Consider a EGS-group Γᾱ with accompanying vector ᾱ = (α_{1}, …, α_{p−1}). Let us prove first that for any vertex u rist_{Γᾱ}(u)_{u} contains the third term γ_{3}(Γᾱ) of the lower central series of Γᾱ. To this end, we will need to check that ᾱ is not ε-diagonal, i.e. it is not proportional to a vector of form (ε, ε^{2}, …, ε^{p−1}), where ε would be a unity root of pth degree in the field GF(p^{n}) for a natural n.

By definition of a EGS-group, α_{1} + … + α_{p−1} = 0. But the sum of coordinates of a vector (ε, ε^{2}, …, ε^{p−1}) is equal to

ε + ε^{2} + … + ε^{p−1} = \{ \begin{matrix}ε\frac{1−ε^{p−1}}{1−ε} , & if ε ≠ 1 \\ p − 1, & if ε = 1. \end{matrix}

Thus, if ε = 1, this sum is not equal to 0. If ε ≠ 1, then clearly 1 − ε^{p−1} ≠ 0, and the sum is not equal to 0, either.

We obtain that ᾱ is not ε-diagonal. This, together with Theorem 2.7.1 of [15], implies that Φ_{1}(γ_{3}(Gᾱ)) contains the direct product \underbrace{γ_{3}(Gᾱ) × … × γ_{3}(Gᾱ)}_{p} (recall that Gᾱ is

the associated GGS-group of Γᾱ). By the same theorem, we obtain that if Fᾱ is the subgroup of Γᾱ generated by elements a and c, then Φ_{1}(γ_{3}(Fᾱ)) contains the direct product

\underbrace{γ_{3}(Fᾱ) × … × γ_{3}(Fᾱ)}_{p}.

Notice that γ_{3}(Γᾱ) is the normal closure in Γᾱ of the following set, [[a, b], a], [[a, b], b], [[a, b], c], [[a, c], a], [[a, c], b], [[a, c], c], [[b, c], a], [[b, c], b], [[b, c], c]. It follows from the above that Φ_{1}(γ_{3}(Gᾱ)) contains elements ([[a, b], a], \underbrace{1, …, 1}_{p−1}), ([[a, b], b], \underbrace{1, …, 1}_{p−1}),

([[a, c], a], \underbrace{1, …, 1}_{p−1}), ([[a, c], c], \underbrace{1, …, 1}_{p−1}). Further, Φ_{1}([b^{a}, c]) = ([b, c], \underbrace{1, …, 1}_{p−1}). Hence

Φ_{1}(γ_{3}(Γᾱ)) contains elements ([[b, c], a], \underbrace{1, …, 1}_{p−1}), ([[b, c], b], \underbrace{1, …, 1}_{p−1}), ([[b, c], c], \underbrace{1, …, 1}_{p−1}) as

well.

Finally, assume that n is a number such that α_{n} ≠ 0. Then

Φ_{1}([[b^{a}, b^{a^{p−n+1}}], c]) = ([[b, a^{α_{n}}], c], \underbrace{1, …, 1}_{p−2}, [[a^{α_{n}}, b], a^{α_{n}}]),

Φ_{1}([[c, b^{a^{p−n+1}}], b^{a}]) = ([[c, a^{α_{n}}], b], \underbrace{1, …, 1}_{p−2}, [[a^{α_{n}}, b], a^{α_{n}}]).

Thus, we can conclude that for one, and hence for any, vertex u of length 1 rist_{Γᾱ}(u)_{u} contains γ_{3}(Γᾱ). Moreover, Φ_{1}(γ_{3}(Γᾱ)) contains the direct product \underbrace{γ_{3}(Γᾱ) × … × γ_{3}(Γᾱ)}_{p}.

Hence, we have the desired inclusion for all vertices.

Since Γᾱ is periodic and finitely generated, γ_{3}(Γᾱ) has finite index. Therefore for every n Φ_{n}^{−1}(\underbrace{γ_{3}(Γᾱ) × … × γ_{3}(Γᾱ)}_{p^{n}}) has finite index in Γᾱ. Indeed, it has finite index in st_{Γᾱ}(n),

because the image Φ_{n}(st_{Γᾱ}(n)) is contained in the direct product of p^{n} copies of Γᾱ. But st_{Γᾱ}(n) obviously has finite index in Γᾱ, so we get that Φ_{n}^{−1}(\underbrace{γ_{3}(Γᾱ) × … × γ_{3}(Γᾱ)}_{p^{n}}) has finite

index in Γᾱ.

Thus, we have proven that Γᾱ is a branch group. However, Γᾱ is periodic and finitely generated, hence γ_{3}(Γᾱ) is also periodic and finitely generated. Therefore, index |γ_{3}(Γᾱ) : [γ_{3}(Γᾱ), γ_{3}(Γᾱ)]| is finite. Then by Theorem 4 of [7] Γᾱ is just infinite.

In particular, we have that all Γᾱ are fractal.

Lemma 1.4. Let Γᾱ be a EGS-group. Then

Γᾱ/[Γᾱ, Γᾱ] ≅< a > × < b > × < c >≅ Z/pZ × Z/pZ × Z/pZ.

Proof. Consider a group H =< a, b, c|a^{p}, b^{p}, c^{p} >, the free product of three copies of Z/pZ. Each element x of H is uniquely represented in the form

a^{s_{1}}b^{β_{1}}c^{δ_{1}} … a^{s_{m}}b^{β_{m}}c^{δ_{m}},  (1.3)

where s_{i}, β_{i}, δ_{i} ∈ GF(p), and for each i = 1, 2, …, m − 1 either β_{i} or δ_{i} is not equal to 0, for each i = 1, 2, …, m − 1 either δ_{i} or s_{i+1} is not equal to 0, and for each i = 2, 3, …, m either s_{i} or β_{i} is not equal to 0. The amount of non-zero numbers among β_{1}, β_{2}, …, β_{m}, δ_{1}, δ_{2}, …, δ_{m} is called the length of x and is denoted by ∂(x). Denote the sum of all β_{i} by ∂_{b}(x) and the sum of all δ_{i} by ∂_{c}(x).

Consider the subgroup L(H) that consists of all elements x such that s_{1}+s_{2}+…+s_{m} = 0. This is clearly a normal subgroup generated by elements b, b^{a}, …, b^{a^{p−1}}, c, c^{a}, …, c^{a^{p−1}}. Moreover, each element of L(H) is uniquely represented in the form

b^{β_{1}a^{s_{1}}}c^{δ_{1}a^{t_{1}}} … b^{β_{m}a^{sm}}c^{δ_{m}a^{tm}},  (1.4)

where s_{i}, t_{i} β_{1}, δ_{m} ∈ GF(p), and β_{2}, …, β_{m}, δ_{1}, …, δ_{m−1} are in GF(p) \ {0}. Let us denote by ∂_{b,i}(x) the sum of the β_{j} such that s_{j} = i, and by ∂_{c,i}(x) the sum of the δ_{j} such that t_{j} = i.

Consider a map Φ̃ : L(H) → \underbrace{H × … × H}_{p} acting via the rule:

Φ̃(b) = (a^{α_{1}}, …, a^{α_{p−1}}, b),
Φ̃(c) = (c, a^{α_{1}}, …, a^{α_{p−1}}),

and Φ̃(b^{a^{i}}) and Φ̃(c^{a^{i}}) are obtained from, respectively, Φ̃(b^{a^{i−1}}) and Φ̃(c^{a^{i−1}}) by one cyclic shift to the right.

Let us show first that

Φ̃^{−1}(\underbrace{[H, H] × … × [H, H]}_{p}) ≤ [H, H].

Consider an arbitrary element x ∈ L(H). Suppose that Φ̃(x) = (x_{1}, x_{2}, …, x_{p−1}, x_{0}). Then ∂_{b}(x_{i}) = ∂_{b,i}(x) and ∂_{c}(x_{i}) = ∂_{c,i}(x). Hence

∑_{i∈GF(p)} ∂_{b}(x_{i}) = ∂_{b}(x)

and

∑_{i∈GF(p)} ∂_{c}(x_{i}) = ∂_{c}(x).

Suppose that x is not in [H, H]. Then either ∂_{b}(x) or ∂_{c}(x) is not equal to zero. By the above equalities, not all of numbers ∂_{b}(x_{i}), ∂_{c}(x_{i}) are equal to zero. Hence Φ̃(x) is not in ([H, H]×…×[H, H]), which is equivalent to saying that Φ̃^{−1}([H, H]×…×[H, H]) ≤ [H, H].

Consider now the following sequence of subgroups. K_{0} = 1; K_{n} = Φ̃^{−1}(K_{n−1}×…×K_{n−1}), n ≥ 1. Take K = ∪_{n=0}^{∞}K_{n}. It is easy to show that K is a normal subgroup of H. It follows, for instance, from Proposition 1 of [14] that Γᾱ is isomorphic to the factor-group H/K. Hence Γᾱ/[Γᾱ, Γᾱ] ≅ H/[H, H], and we have the desired statement.

Later on we will need a length function ∂_{Γᾱ}() on each Γᾱ. Since Γᾱ is a homomorphic image of H from the above lemma, each its element x can be represented (not uniquely) by a word of form (1.3). Define the length ∂_{Γᾱ}(x) of x to be the minimal length of all such words (as defined in the proof of Lemma 1.4) representing x.

Notice that the restriction of length function ∂_{Γᾱ}() to the associated GGS-group gives length function ∂_{Gᾱ}() introduced in Section 1.2.

Also, for some word (1.3) representing x, denote the sum of all s_{i} by ∂_{a}(x), the sum of all β_{i} by ∂_{b}(x), and the sum of all δ_{i} by ∂_{c}(x). By Lemma 1.4, these numbers do not depend on the choice of a particular word representing x. Indeed, the triple (∂_{a}(x), ∂_{b}(x), ∂_{c}(x)) is the image of x under the composition of the natural projection Γᾱ → Γᾱ/[Γᾱ, Γᾱ] and the isomorphisms Γᾱ/[Γᾱ, Γᾱ] ≅< a > × < b > × < c >≅ Z/pZ × Z/pZ × Z/pZ. Hence, the numbers are well-defined.

1.4 Scheme of the proof

The key ingredient in studying maximal subgroups of infinite index is a simple observation that their presence in a group is equivalent to that group having proper subgroups everywhere dense in the profinite topology (Lemma 4.1 and Proposition 5.1). It is such subgroups that we will in fact attempt to locate throughout the paper (ultimately failing at that).

The main idea of the proof that our examples 1.1, 1.2, and 1.3 do not contain proper everywhere dense subgroups is to study possible upper companion groups of such subgroups. We will eventually establish that their companion groups are also everywhere dense (in the corresponding upper companion groups of their respective groups, see Proposition 4.1), and that for an everywhere dense subgroup to be proper, all of its upper companion groups have to be proper as well (Lemmas 4.2, 4.4, and 4.7).

Two lengthy sections, 2 and 3, contain technical results (Corollaries 2.1 and 3.2) necessary for proving the reverse, i.e. that any everywhere dense subgroup has an upper companion group coinciding with the corresponding upper companion group of the whole group (Propositions 4.2, 4.3, and 4.4). This allows us to establish Theorems 4.1, 4.2, and 4.3, which easily yield the main result of the paper, Theorem 5.1.

2 Upper companion groups of subgroups in Grigorchuk’s groups: one special case

We start with Grigorchuk’s 2-groups G_{ω}. Let H_{ω} denote the following group:

< a, b_{ω}, c_{ω}, d_{ω}|a^{2}, b_{ω}^{2}, c_{ω}^{2}, d_{ω}^{2}, b_{ω}c_{ω}d_{ω} > .

Up to isomorphisms, these are just an infinite collection of exemplars of the free product Z_{2} ∗ (Z_{2} × Z_{2}).

A generator of a group H_{ω} is called its d-generator, c-generator, or b-generator, respectively, if it is the d-generator, c-generator, or b-generator, of the corresponding group G_{ω}.

Each element z of H_{ω} is uniquely represented in the form

a^{i}x_{1}ax_{2}a … ax_{m}a^{j},  (2.1)

where x_{1}, x_{2}, …, x_{m} ∈ {b_{ω}, c_{ω}, d_{ω}}, i, j ∈ GF(2). Denote by ∂_{ω}(z) the number m (it will be called the length of z), by ∂_{b_{ω}}(z) the number of occurrences of b_{ω} in z, by ∂_{c_{ω}}(z) the number of occurrences of c_{ω} in z, by ∂_{d_{ω}}(z) the number of occurrences of d_{ω} in z, and by ∂_{a}(z) the number of occurrences of a in z.

We say that z has an inner occurrence of the d-, c-, or b-generator, if among x_{2}, …, x_{m−1} of (2.1) there is the d-, c-, or b-generator, respectively. z has a boundary occurrence of the d-, c-, or b-generator, if among x_{1}, x_{m} there is the d-, c-, or b-generator, respectively.

Denote by L(H_{ω}) the set of all z such that ∂_{a}(z) is even (clearly, it is a subgroup of index 2). It is easy to see that L(H_{ω}) is generated by elements b_{ω}, c_{ω}, d_{ω}, b_{ω}^{a}, c_{ω}^{a}, d_{ω}^{a}; in fact, it is the free product of the subgroups generated by these elements.

Define a pair of maps φ̃_{1}^{ω}, φ̃_{2}^{ω} : L(H_{ω}) → H_{σω}, in the following way:

\{ \begin{matrix}φ̃_{1}^{ω}(b_{ω}^{a}) & = & b_{σω} \\ φ̃_{1}^{ω}(c_{ω}^{a}) & = & c_{σω} \\ φ̃_{1}^{ω}(d_{ω}^{a}) & = & d_{σω} \end{matrix} , \{ \begin{matrix}φ̃_{2}^{ω}(b_{ω}) & = & b_{σω} \\ φ̃_{2}^{ω}(c_{ω}) & = & c_{σω} \\ φ̃_{2}^{ω}(d_{ω}) & = & d_{σω} \end{matrix}

Besides, the image of the d-generator under φ̃_{1}^{ω} is equal to 1, and the images of the other two generators are equal to a. Likewise, the image of the d-generator conjugated by a under φ̃_{2}^{ω} is equal to 1, and the images of the a-conjugates of the other two-generators are equal to a. For instance, if d_{ω} is the d-generator, c_{ω} is the c-generator, and b_{ω} is the b-generator, then we have

\{ \begin{matrix}φ̃_{1}^{ω}(b_{ω}) & = & a \\ φ̃_{1}^{ω}(c_{ω}) & = & a \\ φ̃_{1}^{ω}(d_{ω}) & = & 1 \\ φ̃_{1}^{ω}(b_{ω}^{a}) & = & b_{σω} \\ φ̃_{1}^{ω}(c_{ω}^{a}) & = & c_{σω} \\ φ̃_{1}^{ω}(d_{ω}^{a}) & = & d_{σω} \end{matrix} , \{ \begin{matrix}φ̃_{2}^{ω}(b_{ω}) & = & b_{σω} \\ φ̃_{2}^{ω}(c_{ω}) & = & c_{σω} \\ φ̃_{2}^{ω}(d_{ω}) & = & d_{σω} \\ φ̃_{2}^{ω}(b_{ω}^{a}) & = & a \\ φ̃_{2}^{ω}(c_{ω}^{a}) & = & a \\ φ̃_{2}^{ω}(d_{ω}^{a}) & = & 1 \end{matrix}

Obviously, the extensions by homomorphism of these maps to the whole L(H_{ω}) are welldefined.

Let us introduce a map

Θ_{ω} : [H_{ω}, H_{ω}] → [H_{σω}, H_{σω}],

which will play a key part in the sequel. It acts via the rule

Θ_{ω}(z) = aφ̃_{1}^{ω}(z)aφ̃_{2}^{ω}(z).

We need, of course, to verify that the map is well-defined, i.e. the image of a derived subgroup [H_{ω}, H_{ω}] under it is contained in the derived subgroup [H_{σω}, H_{σω}]. Indeed, it is easy to see that {1, a, b_{ω}, c_{ω}, d_{ω}, ab_{ω}, ac_{ω}, ad_{ω}} is a Schreier system of representatives for [H_{ω}, H_{ω}]. Hence [H_{ω}, H_{ω}] is generated by elements [a, b_{ω}] = (ab_{ω})^{2}, [a, c_{ω}] = (ac_{ω})^{2}, [a, d_{ω}] = (ad_{ω})^{2}.

Each of these elements has the property that its images under φ̃_{1}^{ω} and φ̃_{2}^{ω} belong to the same coset of [H_{σω}, H_{σω}]. Since the factor-group H_{σω}/[H_{σω}, H_{σω}] is of exponent 2, the product of those images belongs to [H_{σω}, H_{σω}]. Certainly, the same is true for any other element of [H_{ω}, H_{ω}].

Let us consider how maps Θ_{ω} interact with the corresponding length functions. First we show that in some specific cases applying the maps causes length to decrease.

Lemma 2.1. Let z belong to [H_{ω}, H_{ω}]. Then the following inequality holds

∂_{σω}(Θ_{ω}(z)) ≤ ∂_{ω}(z).

In addition,

1) if z has at least one inner occurrence of the d-generator (this always happens when ∂_{d_{ω}}(z) ≥ 3), then

∂_{σω}(Θ_{ω}(z)) ≤ ∂_{ω}(z) − 1;

2) if ∂_{σω}(Θ_{ω}(z)) = ∂_{ω}(z), then ∂_{b_{σω}}(Θ_{ω}(z)) = ∂_{b_{ω}}(z), ∂_{c_{σω}}(Θ_{ω}(z)) = ∂_{c_{ω}}(z), ∂_{d_{σω}}(Θ_{ω}(z)) = ∂_{d_{ω}}(z);

3) if at least one of the numbers ∂_{b_{ω}}(z), ∂_{c_{ω}}(z), ∂_{d_{ω}}(z) is greater than or equal to 3, then there exists n such that ∂_{σ^{n+1}ω}(Θ_{σ^{n}ω} ∘ … ∘ Θ_{ω}(z)) < ∂_{ω}(z).

Proof. Inequality ∂_{σω}(Θ_{ω}(z)) ≤ ∂_{ω}(z) and 2) follow from the fact that to each occurrence of b_{ω}, c_{ω}, d_{ω} in z corresponds exactly one occurrence of, respectively, b_{σω}, c_{σω}, d_{σω} in either φ̃_{1}^{ω}(z) or φ̃_{2}^{ω}(z). Statement 1) is verified directly. Let us prove 3).

Without losing any generality, we can assume that d_{ω} is the d-generator of H_{ω}, andc_{ω} is the c-generator. Since at least one of the numbers ∂_{d_{ω}}(z), ∂_{c_{ω}}(z), ∂_{b_{ω}}(z) is greater than or equal to 3, we have three cases to consider.

If ∂_{d_{ω}}(z) ≥ 3, then by 1) we have that ∂_{σω}(Θ_{ω}(z)) < ∂_{ω}(z). Suppose that ∂_{d_{ω}}(z) ≤ 2, but ∂_{c_{ω}}(z) ≥ 3. Apply to z subsequently Θ_{ω}, Θ_{σω}, …, Θ_{σ^{m(ω)−1}ω} (recall that m(ω) is the maximal length of a constant initial segment of ω). Then either upon application of these maps we receive an element of smaller length, or by 2) we get

∂_{c_{σ^{m(ω)}ω}}(Θ_{σ^{m(ω)−1}ω} ∘ … ∘ Θ_{ω}(z)) = ∂_{c_{ω}}(z) ≥ 3.

Since c_{σ^{m(ω)}ω} is the d-generator of H_{σ^{m(ω)}ω}, it follows from 1) that

∂_{σ^{m(ω)+1}ω}(Θ_{σ^{m(ω)}ω} ∘ … ∘ Θ_{ω}(z)) < ∂_{ω}(z).

If both ∂_{d_{ω}}(z) ≤ 2 and ∂_{c_{ω}}(z) ≤ 2, we have that ∂_{b_{ω}}(z) ≥ 3. Let n(ω) be the minimal number such that the segment (ω_{1}, …, ω_{n(ω)}) of ω contains all three numbers 0,1,2. Notice that in this case b_{σ^{n(ω)−1}ω} is the d-generator of H_{σ^{n(ω)−1}ω}. Apply to z the maps Θ_{ω}, Θ_{σω}, …, Θ_{σ^{n(ω)−2}ω}. Then either upon application of these maps we receive an element of smaller length, or by 2) we get

∂_{b_{σ^{n(ω)−1}ω}}(Θ_{σ^{n(ω)−2}ω} ∘ … ∘ Θ_{ω}(z)) = ∂_{b_{ω}}(z) ≥ 3.

Since b_{σ^{n(ω)−1}ω} is the d-generator of H_{σ^{n(ω)−1}ω}, it follows from 1) that

∂_{σ^{n(ω)}ω}(Θ_{σ^{n(ω)−1}ω} ∘ … ∘ Θ_{ω}(z)) < ∂_{ω}(z).

This concludes the proof of the lemma.

Let us consider a general case now.

Proposition 2.1. Let Θ_{ω} : [H_{ω}, H_{ω}] → [H_{σω}, H_{σω}] be the maps introduced above. Then for any z ∈ [H_{ω}, H_{ω}] there exists n such that

Θ_{σ^{n}ω} ∘ … ∘ Θ_{ω}(z) ∈< (ay_{σ^{n+1}ω})^{2} >,

where y_{σ^{n+1}ω} is the d-generator of H_{σ^{n+1}ω}.

Proof. The proof is by induction on length ∂_{ω}(z). As above, assume that d_{ω} is the d-generator and c_{ω} is the c-generator. The base of induction: ∂_{ω}(z) ≤ 2. If ∂_{ω}(z) = 0 and thus z = 1, there’s nothing to prove. Obviously, no element of [H_{ω}, H_{ω}] can have length 1. For ∂_{ω}(z) = 2 we have 6 cases: (ad_{ω})^{2}, (d_{ω}a)^{2}, (ac_{ω})^{2}, (c_{ω}a)^{2}, (ab_{ω})^{2}, (b_{ω}a)^{2}. Now, since d_{ω} is the d-generator, (ad_{ω})^{4} = 1 and hence (d_{ω}a)^{2} = (ad_{ω})^{2}.

Consider the remaining cases:

Θ_{ω} : (ac_{ω})^{2} ↦ (ac_{σω})^{2}, (c_{ω}a)^{2} ↦ (c_{σω}a)^{2},
(ab_{ω})^{2} ↦ (ab_{σω})^{2}, (b_{ω}a)^{2} ↦ (b_{σω}a)^{2}.

Thus,

Θ_{σ^{m(ω)−1}ω} ∘ … ∘ Θ_{σω} ∘ Θ_{ω} : (ac_{ω})^{2} ↦ (ac_{σ^{m(ω)}ω})^{2}, (c_{ω}a)^{2} ↦ (c_{σ^{m(ω)}ω}a)^{2},
Θ_{σ^{n(ω)−2}ω} ∘ … ∘ Θ_{σω} ∘ Θ_{ω} : (ab_{ω})^{2} ↦ (ab_{σ^{n(ω)}ω})^{2}, (b_{ω}a)^{2} ↦ (b_{σ^{n(ω)−1}ω}a)^{2}.

Since by definitions of m(ω), n(ω) c_{σ^{m(ω)}ω} and b_{σ^{n(ω)−1}ω} are the d-generators of H_{σ^{m(ω)}ω} and H_{σ^{n(ω)−1}ω}, respectively, we get the base of induction. The inductive step. In carrying out the inductive step, the following remark will be important to us. An element z ∈ H_{ω} belongs to [H_{ω}, H_{ω}] if and only if ∂_{b_{ω}}(z) ≡ ∂_{c_{ω}}(z) ≡ ∂_{d_{ω}}(z)(mod 2) (this is easily derived from the definition of H_{ω}).

We have several possibilities depending on the length of z. Case 1. Assume that ∂_{ω}(z) = 3. Then by the preceding remark there are twelve possibilities, c_{ω}ad_{ω}ab_{ω}, b_{ω}ad_{ω}ac_{ω}, d_{ω}ab_{ω}ac_{ω}, d_{ω}ac_{ω}ab_{ω}, c_{ω}ab_{ω}ad_{ω}, b_{ω}ac_{ω}ad_{ω}, ac_{ω}ad_{ω}ab_{ω}a, ab_{ω}ad_{ω}ac_{ω}a, ac_{ω}ab_{ω}ad_{ω}a, ab_{ω}ac_{ω}ad_{ω}a, ad_{ω}ab_{ω}ac_{ω}a, ad_{ω}ac_{ω}ab_{ω}a. Let us consider them all.

We have

Θ_{ω}(c_{ω}ad_{ω}ab_{ω}) = 1,
Θ_{ω}(b_{ω}ad_{ω}ac_{ω}) = 1,
Θ_{ω}(d_{ω}ab_{ω}ac_{ω}) = (ac_{σω})^{2},
Θ_{ω}(d_{ω}ac_{ω}ab_{ω}) = (ab_{σω})^{2}.

The result of applying our maps to the next two elements depends on m(ω). Indeed, Θ_{ω}(c_{ω}ab_{ω}ad_{ω}) = b_{σω}ac_{σω}ad_{σω}. If m(ω) = 1, the resulting element has an inner occurrence of the d-generator. If m(ω) > 1, then Θ_{σω}(b_{σω}ac_{σω}ad_{σω}) = c_{σ^{2}ω}ab_{σ^{2}ω}ad_{σ^{2}ω}. Thus

Θ_{σ^{m(ω)−1}ω} ∘ … ∘ Θ_{ω}(c_{ω}ab_{ω}ad_{ω}) = \{ \begin{matrix}b_{σ^{m(ω)}ω}ac_{σ^{m(ω)}ω}ad_{σ^{m(ω)}ω}, & if m(ω) = 2k + 1 \\ c_{σ^{m(ω)}ω}ab_{σ^{m(ω)}ω}ad_{σ^{m(ω)}ω}, & if m(ω) = 2k \end{matrix}

Since c_{σ^{m(ω)}ω} is the d-generator, in the first case after applying Θ_{σ^{m(ω)}ω} the length will decrease by Lemma 2.1. In the second case we have

Θ_{σ^{m(ω)}ω}(c_{σ^{m(ω)}ω}ab_{σ^{m(ω)}ω}ad_{σ^{m(ω)}ω}) = (ad_{σ^{m(ω)+1}ω})^{2}.

In both cases we can use the inductive assumption.

Similarly, Θ_{ω}(b_{ω}ac_{ω}ad_{ω}) = c_{σω}ab_{σω}ad_{σω}. If m(ω) = 1, we have Θ_{σω}(c_{σω}ab_{σω}ad_{σω}) = (ad_{σ^{2}ω})^{2}. If m(ω) > 1, then Θ_{σω}(c_{σω}ab_{σω}ad_{σω}) = b_{σ^{2}ω}ac_{σ^{2}ω}ad_{σ^{2}ω}. Thus

Θ_{σ^{m(ω)−1}ω} ∘ … ∘ Θ_{ω}(b_{ω}ac_{ω}ad_{ω}) = \{ \begin{matrix}c_{σ^{m(ω)}ω}ab_{σ^{m(ω)}ω}ad_{σ^{m(ω)}ω}, & if m(ω) = 2k + 1 \\ b_{σ^{m(ω)}ω}ac_{σ^{m(ω)}ω}ad_{σ^{m(ω)}ω}, & if m(ω) = 2k \end{matrix}

Since c_{σ^{m(ω)}ω} is the d-generator, in the latter case after applying Θ_{σ^{m(ω)}ω} we get an element of smaller length by Lemma 2.1. In the first case we have Θ_{σ^{m(ω)}ω}(c_{σ^{m(ω)}ω}ab_{σ^{m(ω)}ω}ad_{σ^{m(ω)}ω}) = (ad_{σ^{m(ω)+1}ω})^{2}. Hence in both cases we can use the inductive assumption.

Further, we have

Θ_{ω}(ac_{ω}ad_{ω}ab_{ω}a) = 1,
Θ_{ω}(ab_{ω}ad_{ω}ac_{ω}a) = 1,
Θ_{ω}(ac_{ω}ab_{ω}ad_{ω}a) = (ac_{σω})^{2},
Θ_{ω}(ab_{ω}ac_{ω}ad_{ω}a) = (ab_{σω})^{2}.

Now let us consider ad_{ω}ab_{ω}ac_{ω}a. We have

Θ_{ω}(ad_{ω}ab_{ω}ac_{ω}a) = ad_{σω}ac_{σω}ab_{σω}a.

Thus, if m(ω) = 1, the latter element has an inner occurrence of the d-generator, therefore upon applying to it Θ_{σ^{2}ω} we get an element of smaller length. And if m(ω) > 1, then Θ_{σ^{2}ω}(ad_{σω}ac_{σω}ab_{σω}a) = ad_{σ^{2}ω}ab_{σ^{2}ω}ac_{σ}^{2}ωa. Therefore,

Θ_{σ^{m(ω)−1}ω} ∘ … ∘ Θ_{ω}(ad_{ω}ab_{ω}ac_{ω}a) = \{ \begin{matrix}ad_{σ^{m(ω)}ω}ac_{σ^{m(ω)}ω}ab_{σ^{m(ω)}ω}a, & if m(ω) = 2k + 1 \\ ad_{σ^{m(ω)}ω}ab_{σ^{m(ω)}ω}ac_{σ^{m(ω)}ω}a, & if m(ω) = 2k. \end{matrix}

Since c_{σ^{m(ω)}ω} is the d-generator, in the former case by Lemma 2.1 we get a decrease in the length right after applying Θ_{σ^{m(ω)}ω}. In the latter case we have Θ_{σ^{m(ω)}ω}(ad_{σ^{m(ω)}ω}ab_{σ^{m(ω)}ω}ac_{σ^{m(ω)}ω}a) = (ad_{σ^{m(ω)+1}ω})^{2}, and we can apply the inductive assumption.

Finally, Θ_{ω}(ad_{ω}ac_{ω}ab_{ω}a) = ad_{σω}ab_{σω}ac_{σω}a. Again, if m(ω) = 1, then Θ_{σω}(ad_{σω}ab_{σω}ac_{σω}a) = (ad_{σ^{2}ω})^{2}. If m(ω) > 1, then Θ_{σω}(ad_{σω}ab_{σω}ac_{σω}a) = ad_{σ^{2}ω}ac_{σ^{2}ω}ab_{σ^{2}ω}a. Therefore,

Θ_{σ^{m(ω)−1}ω} ∘ … ∘ Θ_{ω}(ad_{ω}ac_{ω}ab_{ω}a) = \{ \begin{matrix}ad_{σ^{m(ω)}ω}ab_{σ^{m(ω)}ω}ac_{σ^{m(ω)}ω}a, & if m(ω) = 2k + 1 \\ ad_{σ^{m(ω)}ω}ac_{σ^{m(ω)}ω}ab_{σ^{m(ω)}ω}a, & if m(ω) = 2k \end{matrix}

Since c_{σ^{m(ω)}ω} is the d-generator, in the latter case by Lemma 2.1 we get a decrease in the length right after applying Θ_{σ^{m(ω)}ω}. In the former case we have Θ_{σ^{m(ω)}ω}(ad_{σ^{m(ω)}ω}ab_{σ^{m(ω)}ω}ac_{σ^{m(ω)}ω}a) = (ad_{σ^{m(ω)+1}ω})^{2}. In both cases, we can use the inductive assumption. Thus, we have considered all possibilities for ∂_{ω}(z) = 3. Case 2. Assume that ∂_{ω}(z) = 4. If at least one of the numbers ∂_{b_{ω}}(z), ∂_{c_{ω}}(z), ∂_{d_{ω}}(z) is greater than 2, then by Lemma 2.1 there exists n such that ∂_{σ^{n+1}ω}(Θ_{σ^{n}ω} ∘ … ∘ Θ_{ω}(z)) < ∂_{ω}(z). If each of these numbers does not exceed 2, then one of them must be zero and the other two equal to 2. If ∂_{d_{ω}}(z) = 0, then either ∂_{σ^{m(ω)}}(Θ_{σ^{m(ω)−1}ω} ∘ … ∘ Θ_{ω}(z)) < ∂_{ω}(z), or ∂_{c_{σ^{m(ω)}ω}}(Θ_{σ^{m(ω)−1}ω} ∘ … ∘ Θ_{ω}(z)) = ∂_{c_{ω}}(z) = 2. Therefore it suffices to consider only the cases when ∂_{ω}(z) = 4 and ∂_{d_{ω}}(z) = 2.

By Lemma 2.1, if z has an inner occurrence of d_{ω}, then ∂_{σω}(Θ_{ω}(z)) < ∂_{ω}(z). Hence we should only look at those instances when d_{ω} has only boundary occurrences. There are just four such elements: d_{ω}c_{ω}^{a}c_{ω}d_{ω}^{a}, d_{ω}^{a}c_{ω}c_{ω}^{a}d_{ω}, d_{ω}b_{ω}^{a}b_{ω}d_{ω}^{a}, d_{ω}^{a}b_{ω}b_{ω}^{a}d_{ω}. Let us consider them in order. Case 2.1: z = d_{ω}c_{ω}^{a}c_{ω}d_{ω}^{a}. We have that Θ_{ω}(d_{ω}c_{ω}^{a}c_{ω}d_{ω}^{a}) = c_{σω}^{a}d_{σω}d_{σω}^{a}c_{σω}. If m(ω) > 1, the latter element has an inner occurrence of the d-generator, so upon applying to it Θ_{σ^{2}ω} we get an element of smaller length.

Let m(ω) = 1. Then c_{σω} is the d-generator, and

Θ_{σω}(c_{σω}^{a}d_{σω}d_{σω}^{a}c_{σω}) = c_{σ^{2}ω}^{a}d_{σ^{2}ω}d_{σ^{2}ω}^{a}c_{σ^{2}ω}.

Now, if ω_{3} = 0, i.e. if d_{σ^{2}ω} is the d-generator again, the obtained element has an inner occurrence of the d-generator. If ω_{3} is equal to 1 or 2, then

Θ_{σ^{2}ω}(c_{σ^{2}ω}^{a}d_{σ^{2}ω}d_{σ^{2}ω}^{a}c_{σ^{2}ω}) = c_{σ^{3}ω}^{a}d_{σ^{3}ω}d_{σ^{3}ω}^{a}c_{σ^{3}ω}.

Let n be the minimal number such that ω_{n} = 0 and n > 1. Then by the above remarks

Θ_{σ^{n−2}ω} ∘ … ∘ Θ_{ω}(d_{ω}c_{ω}^{a}c_{ω}d_{ω}^{a}) = c_{σ^{n−1}ω}^{a}d_{σ^{n−1}ω}d_{σ^{n−1}ω}^{a}c_{σ^{n−1}ω},

and the resulting element has an inner occurrence of the d-generator. Case 2.2: z = d_{ω}^{a}c_{ω}c_{ω}^{a}d_{ω}. We have

Θ_{ω}(d_{ω}^{a}c_{ω}c_{ω}^{a}d_{ω}) = d_{σω}^{a}c_{σω}c_{σω}^{a}d_{σω}.

Obviously,

Θ_{σ^{m(ω)−1}ω} ∘ … ∘ Θ_{ω}(d_{ω}^{a}c_{ω}c_{ω}^{a}d_{ω}) = d_{σ^{m(ω)}ω}^{a}c_{σ^{m(ω)}ω}c_{σ^{m(ω)}ω}^{a}d_{σ^{m(ω)}ω},

and the resulting element has an inner occurrence of the d-generator.

Case 2.3: z = d_{ω}b_{ω}^{a}b_{ω}d_{ω}^{a}. Then we have

Θ_{ω}(d_{ω}b_{ω}^{a}b_{ω}d_{ω}^{a}) = b_{σω}^{a}d_{σω}d_{σω}^{a}b_{σω}.

If m(ω) > 1, the latter element has an inner occurrence of the d-generator, hence by applying Θ_{σ^{2}ω} to it we get an element of smaller length.

Suppose that m(ω) = 1. Then c_{σω} is the d-generator, so

Θ_{σω}(b_{σω}^{a}d_{σω}d_{σω}^{a}b_{σω}) = b_{σ^{2}ω}^{a}d_{σ^{2}ω}d_{σ^{2}ω}^{a}b_{σ^{2}ω}.

If ω_{3} = 0, i.e. if d_{σ^{2}ω} is the d-generator, the latter element has an inner occurrence of the d-generator. If ω_{3} equals 1 or 2, then

Θ_{σ^{2}ω}(b_{σ^{2}ω}^{a}d_{σ^{2}ω}d_{σ^{2}ω}^{a}b_{σ^{2}ω}) = b_{σ^{3}ω}^{a}d_{σ^{3}ω}d_{σ^{3}ω}^{a}b_{σ^{3}ω}.

Let n be the minimal number such that ω_{n} = 0 and n > 1. Then by the above

Θ_{σ^{n−2}ω} ∘ … ∘ Θ_{ω}(d_{ω}b_{ω}^{a}b_{ω}d_{ω}^{a}) = b_{σ^{n−1}ω}^{a}d_{σ^{n−1}ω}d_{σ^{n−1}ω}^{a}b_{σ^{n−1}ω},

and the resulting element has an inner occurrence of the d-generator. Case 2.4: z = d_{ω}^{a}b_{ω}b_{ω}^{a}d_{ω}. Here we have

Θ_{ω}(d_{ω}^{a}b_{ω}b_{ω}^{a}d_{ω}) = d_{σω}^{a}b_{σω}b_{σω}^{a}d_{σω}.

Since ω_{2} equals either 0 or 1,

Θ_{σω}(d_{σω}^{a}b_{σω}b_{σω}^{a}d_{σω}) = d_{σ^{2}ω}^{a}b_{σ^{2}ω}b_{σ^{2}ω}^{a}d_{σ^{ω}}).

Thus,

Θ_{σ^{n(ω)−2}ω} ∘ … ∘ Θ_{ω}(d_{ω}^{a}b_{ω}b_{ω}^{a}d_{ω}) = d_{σ^{n(ω)−1}ω}^{a}b_{σ^{n(ω)−1}ω}b_{σ^{n(ω)−1}ω}^{a}d_{σ^{n(ω)−1}ω},

and the resulting element has an inner occurrence of the d-generator. Thus, all four possibilities have been considered. Case 3. We proceed to assume that ∂_{ω}(z) = 5. The latter condition, however, implies that at least one of the following numbers ∂_{b_{ω}}(z), ∂_{c_{ω}}(z), ∂_{d_{ω}}(z) is greater than 2. Indeed, since z is in the derived subgroup, all three numbers have to be equal modulo 2. As their sum equals 5, they all have to be odd, and therefore one of them has to be greater than 2. Then by Lemma 2.1 there is an n such that ∂_{σ^{n+1}ω}(Θ_{σ^{n}ω} ∘ … ∘ Θ_{ω}(z)) < ∂_{ω}(z), and we can apply the inductive assumption. Case 4. Suppose that ∂_{ω}(z) = 6. Then either one of the numbers ∂_{b_{ω}}(z), ∂_{c_{ω}}(z), ∂_{d_{ω}}(z) is greater than 2, and by Lemma 2.1 there is n such that ∂_{σ^{n+1}ω}(Θ_{σ^{n}ω} ∘ … ∘ Θ_{ω}(z)) < ∂_{ω}(z) or ∂_{d_{ω}}(z) = ∂_{c_{ω}}(z) = ∂_{b_{ω}}(z) = 2. In the latter case, suppose z has inner occurrences of d_{ω}. Then we have ∂_{σω}(Θ_{ω}(z)) < ∂_{ω}(z). Hence we should only consider the cases when z has only boundary occurrences of d_{ω}. Then there are twelve possibilities.

Consider the following table.

\begin{matrix}N & z & Θ_{ω}(z) & N & z & Θ_{ω}(z) \\ 4.1 & d_{ω}b_{ω}^{a}b_{ω}c_{ω}^{a}c_{ω}d_{ω}^{a} & b_{σω}^{a}c_{σω}d_{σω}^{a}d_{σω}b_{σω}^{a}c_{σω} & 4.7 & d_{ω}^{a}b_{ω}b_{ω}^{a}c_{ω}c_{ω}^{a}d_{ω} & d_{σω}^{a}b_{σω}c_{σω}^{a}b_{σω}c_{σω}^{a}d_{σω} \\ 4.2 & d_{ω}b_{ω}^{a}c_{ω}b_{ω}^{a}c_{ω}d_{ω}^{a} & b_{σω}^{a}b_{σω}d_{σω}^{a}d_{σω}c_{σω}^{a}c_{σω} & 4.8 & d_{ω}^{a}b_{ω}c_{ω}^{a}b_{ω}c_{ω}^{a}d_{ω} & d_{σω}^{a}c_{σω}c_{σω}^{a}b_{σω}b_{σω}^{a}d_{σω} \\ 4.3 & d_{ω}b_{ω}^{a}c_{ω}c_{ω}^{a}b_{ω}d_{ω}^{a} & b_{σω}^{a}c_{σω}d_{σω}^{a}d_{σω}c_{σω}^{a}b_{σω} & 4.9 & d_{ω}^{a}c_{ω}c_{ω}^{a}b_{ω}b_{ω}^{a}d_{ω} & d_{σω}^{a}c_{σω}b_{σω}^{a}c_{σω}b_{σω}^{a}d_{σω} \\ 4.4 & d_{ω}c_{ω}^{a}c_{ω}b_{ω}^{a}b_{ω}d_{ω}^{a} & c_{σω}^{a}b_{σω}d_{σω}^{a}d_{σω}c_{σω}^{a}b_{σω} & 4.10 & d_{ω}^{a}c_{ω}b_{ω}^{a}c_{ω}b_{ω}^{a}d_{ω} & d_{σω}^{a}b_{σω}b_{σω}^{a}c_{σω}c_{σω}^{a}d_{σω} \\ 4.5 & d_{ω}c_{ω}^{a}b_{ω}c_{ω}^{a}b_{ω}d_{ω}^{a} & c_{σω}^{a}c_{σω}d_{σω}^{a}d_{σω}b_{σω}^{a}b_{σω} & 4.11 & d_{ω}^{a}b_{ω}c_{ω}^{a}c_{ω}b_{ω}^{a}d_{ω} & d_{σω}^{a}c_{σω}b_{σω}^{a}b_{σω}c_{σω}^{a}d_{σω} \\ 4.6 & d_{ω}c_{ω}^{a}b_{ω}b_{ω}^{a}c_{ω}d_{ω}^{a} & c_{σω}^{a}b_{σω}d_{σω}^{a}d_{σω}b_{σω}^{a}c_{σω} & 4.12 & d_{ω}^{a}c_{ω}b_{ω}^{a}b_{ω}c_{ω}^{a}d_{ω} & d_{σω}^{a}b_{σω}c_{σω}^{a}c_{σω}b_{σω}^{a}d_{σω} \end{matrix} .

In cases 4.1-4.5 we have inner occurrences of c_{σω} and d_{σω}. Since c_{ω} is the c-generator, in those cases we always (irrespective of ω) have an inner occurrence of a d-generator.

Now consider case 4.6. If ω_{2} ≠ 1, we have an inner occurrence of the d-generator. Suppose that ω_{2} = 1 and let n > 2 be the smallest integer exceeding 2 and such that ω_{n} ≠ 1. It is easy to see that Θ_{σ^{n−2}ω} ∘ … ∘ Θ_{ω}(z) has the form

c_{σ^{n−1}ω}^{a} ∗ ∗ ∗ ∗c_{σ^{n−1}ω},

where exactly two stars are from the set {d_{σ^{n−1}ω}, d_{σ^{n−1}ω}^{a}}, and the other two are from {b_{σ^{n−1}ω}, b_{σ^{n−1}ω}^{a}}. Hence the resulting element has an inner occurrence of a d-generator.

If m(ω) = 1, then in cases 4.7-4.12 Θ_{ω}(z) has an inner occurrence of the d-generator. Otherwise the equalities

Θ_{σω} ∘ Θ_{ω}(d_{ω}^{a}b_{ω}b_{ω}^{a}c_{ω}c_{ω}^{a}d_{ω}) = d_{σ^{2}ω}^{a}c_{σ^{2}ω}c_{σ^{2}ω}^{a}b_{σ^{2}ω}b_{σ^{2}ω}^{a}d_{σ^{2}ω},
Θ_{σ^{2}ω} ∘ Θ_{σω} ∘ Θ_{ω}(d_{ω}^{a}b_{ω}b_{ω}^{a}c_{ω}c_{ω}^{a}d_{ω}) = d_{σ^{3}ω}^{a}c_{σ^{3}ω}b_{σ^{3}ω}^{a}c_{σ^{3}ω}b_{σ^{3}ω}^{a}d_{σ^{3}ω},
Θ_{σ^{3}ω} ∘ Θ_{σ^{2}ω} ∘ Θ_{σω} ∘ Θ_{ω}(d_{ω}^{a}b_{ω}b_{ω}^{a}c_{ω}c_{ω}^{a}d_{ω}) = d_{σ^{4}ω}^{a}b_{σ^{4}ω}b_{σ^{4}ω}^{a}c_{σ^{4}ω}c_{σ^{4}ω}^{a}d_{σ^{4}ω}

and

Θ_{σω} ∘ Θ_{ω}(d_{ω}^{a}b_{ω}c_{ω}^{a}c_{ω}b_{ω}^{a}d_{ω}) = d_{σ^{2}ω}^{a}b_{σ^{2}ω}c_{σ^{2}ω}^{a}c_{σ^{2}ω}b_{σ^{2}ω}^{a}d_{σ^{2}ω}

imply that for any element of the form d_{σ^{m(ω)}ω}^{a} ∗∗∗∗d_{σ^{m(ω)}ω}, where the stars are in one-to-one correspondence with the set {b_{σ^{m(ω)}ω}, b_{σ^{m(ω)}ω}^{a}, c_{σ^{m(ω)}ω}, c_{σ^{m(ω)}ω}^{a}}, Θ_{σ^{m(ω)−1}ω} ∘ … ∘ Θ_{ω}(z) has a similar form and therefore has an inner occurrence of the d-generator. Case 5. Finally, suppose that ∂_{ω}(z) > 6. Then at least one of the numbers ∂_{b_{ω}}(z), ∂_{c_{ω}}(z), ∂_{d_{ω}}(z) is greater than 2, so by Lemma 2.1 there is n such that ∂_{σ^{n+1}ω}(Θ_{σ^{n}ω} ∘ … ∘ Θ_{ω}(z)) < ∂_{ω}(z) This (finally) proves the proposition.

Proposition 2.1 allows to introduce a “norm” || ||_{ω} on [H_{ω}, H_{ω}].

Definition 2.1. We put ||z||_{ω} to be the smallest n such that Θ_{σ^{n−1}ω} ∘ … ∘ Θ_{ω}(z) belongs to the subgroup generated by (ay_{σ^{n}ω})^{2}, where y_{σ^{n}ω} is the d-generator of H_{σ^{n}ω}.

Theorem 2.1. Let x_{ω}, y_{ω} be two different elements of the set {b_{ω}, c_{ω}, d_{ω}}, let z be an arbitrary element of [H_{ω}, H_{ω}], and let X =< x_{ω}, y_{ω}az >. Then there exists n such that

φ̃_{2}^{σ^{n−1}ω}(L(φ̃_{2}^{σ^{n−2}ω}(L(… (φ̃_{2}^{ω}(L(X))) …)))) = H_{σ^{n}ω}.

Proof. We will assume that d_{ω} is the d-generator and c_{ω} is the c-generator. There are three possibilities, < b_{ω}, d_{ω}az >=< b_{ω}, c_{ω}az >, < c_{ω}, d_{ω}az >=< c_{ω}, b_{ω}az >, < d_{ω}, c_{ω}az >=< d_{ω}, b_{ω}az >.

The proof is by induction on ||z||_{ω} simultaneously for all H_{ω}. The base of induction is ||z||_{ω} = 0, i.e. z = 1 or z = (ad_{ω})^{2}.

Let z = 1. Consider the subgroup < b_{ω}, d_{ω}az >. Since φ̃_{2}^{ω}((d_{ω}a)^{2}) = d_{σω}, φ̃_{2}^{ω}(b_{ω}^{d_{ω}a}) = a, φ̃_{2}^{ω}(b_{ω}) = b_{σω}, we have φ̃_{2}^{ω}(L(X)) ∋ d_{ω}, a, b_{ω}, i.e. φ̃_{2}^{ω}(L(X)) ≥ H_{σω}.

Now consider the subgroup < c_{ω}, d_{ω}az >. Since φ̃_{2}^{ω}((d_{ω}a)^{2}) = d_{σω}, φ̃_{2}^{ω}(c_{ω}^{d_{ω}a}) = a, φ̃_{2}^{ω}(c_{ω}) = c_{σω}, we have φ̃_{2}^{ω}(L(X)) ∋ d_{ω}, a, c_{ω}, i.e. φ̃_{2}^{ω}(L(X)) ≥ H_{σω}.

Finally, consider < d_{ω}, c_{ω}az >. Notice that the order of c_{ω}a is equal to 2^{m(ω)+2}, and

φ̃_{2}^{σ^{m(ω)}ω} … φ̃_{2}^{σω}φ̃_{2}^{ω}((c_{ω}a)^{2^{m(ω)+1}}) = c_{σ^{m(ω)+1}ω}.

Further,

φ̃_{2}^{σ^{m(ω)−1}ω} … φ̃_{2}^{σω}φ̃_{2}^{ω}(d_{ω}^{(c_{ω}a)^{2^{m(ω)}}}) = d_{σ^{m(ω)}ω}^{a}.
Hence φ̃_{2}^{σ^{m(ω)}ω}(L(… φ̃_{2}^{σω}(L(φ̃_{2}^{ω}(L(X)))) …))  ∋  c_{σ^{m(ω)+1}ω}, a, d_{σ^{m(ω)+1}ω},  i.e.

φ̃_{2}^{σ^{m(ω)}ω}(L(… φ̃_{2}^{σω}(L(φ̃_{2}^{ω}(L(X)))) …)) ≥ H_{σω}.

Let us now consider the case when z = (ad_{ω})^{2}. We have that < b_{ω}, d_{ω}a(ad_{ω})^{2} >=< b_{ω}, ad_{ω} >, this case has already been considered. Next, < c_{ω}, d_{ω}a(ad_{ω})^{2} >=< c_{ω}, ad_{ω} >. This case also has been already considered.

Finally, < d_{ω}, c_{ω}a(ad_{ω})^{2} >=< d_{ω}, b_{ω}ad_{ω} >=< d_{ω}, b_{ω}a >. Again this case has been considered, and the base of induction is completed.

Let us proceed to the inductive step. Consider the subgroup < b_{ω}, d_{ω}az >=< b_{ω}, c_{ω}az >. Since φ̃_{2}^{ω}(b_{ω}) = b_{σω}, φ̃_{2}^{ω}((c_{ω}az)^{2}) = c_{σω}aΘ_{ω}(z), we get that φ̃_{2}^{ω}(L(X)) ≥< b_{σω}, c_{σω}aΘ_{ω}(z) >. Since, obviously, ||Θ_{ω}(z)||_{σω} < ||z||_{ω}, we can apply the inductive assumption.

Next consider the subgroup < c_{ω}, d_{ω}az >=< c_{ω}, b_{ω}az >. Since φ̃_{2}^{ω}(c_{ω}) = c_{σω}, φ̃_{2}^{ω}((b_{ω}az)^{2}) = b_{σω}aΘ_{ω}(z), we have that φ̃_{2}^{ω}(L(X)) ≥< c_{σω}, b_{σω}aΘ_{ω}(z) >, and we can apply the inductive assumption.

Finally, consider the subgroup < d_{ω}, b_{ω}az >. Since φ̃_{2}^{ω}(d_{ω}) = d_{σω}, φ̃_{2}^{ω}((b_{ω}az)^{2}) = b_{σω}aΘ_{ω}(z), we have that φ̃_{2}^{ω}(L(X)) ≥< d_{σω}, b_{σω}aΘ_{ω}(z) >, and we can apply the inductive assumption.

Returning to groups G_{ω}, we have th following important fact.

Corollary 2.1. Let x_{ω}, y_{ω} be two different generators in {b_{ω}, c_{ω}, d_{ω}} in the generating set of G_{ω}, z ∈ [G_{ω}, G_{ω}]. Then there exists n such that X_{u} = G_{σ^{n}ω}, where X =< x_{ω}, y_{ω}az >,

u = \underbrace{11 … 1}_{n}.

Proof. Notice that every G_{ω} is a factor-group of H_{ω}. Let ε_{ω} : H_{ω} → G_{ω} be the natural projection. Then by definition of homomorphisms φ̃_{1}^{ω}, φ̃_{2}^{ω} the following diagrams are commutative:

\begin{aligned}& L(H_{ω}) & →^{ε_{ω}} & st_{G_{ω}}(1) \\ φ̃_{1}^{ω} & ↓ & & ↓ & φ_{0} \\ & H_{ω} & →^{ε_{ω}} & G_{ω} \end{aligned}  \begin{aligned}& L(H_{ω}) & →^{ε_{ω}} & st_{G_{ω}}(1) \\ φ̃_{2}^{ω} & ↓ & & ↓ & φ_{1} \\ & H_{ω} & →^{ε_{ω}} & G_{ω} \end{aligned}

Hence the required statement follows from Theorem 2.1.

Remark 2.1. Applying the same reasoning as in the above proof, we obtain from Proposition 2.1 that if z be an arbitrary element of [G_{ω}, G_{ω}], then there exists n such that Θ_{σ^{n}ω} ∘ … ∘ Θ_{ω}(z) ∈< (ay_{σ^{n+1}ω})^{2} >, where y_{σ^{n+1}ω} is the d-generator of G_{σ^{n+1}ω}. Hence a “norm” similar to the one of Definition 2.1 can be introduced on [G_{ω}, G_{ω}] as well.

3 Upper companion groups of subgroups in GGSgroups: one special case

Let Gᾱ be a periodic GGS-group with accompanying vector

ᾱ = (α_{1}, …, α_{p−1}) ∈ \underbrace{GF(p) × … × GF(p)}_{p−1} .

Since Gᾱ is transitive on each layer of vertices, there exists (in general, not unique) n such that α_{n} ≠ 0. Fix some such number for each vector ᾱ and denote it by n(ᾱ). Then, up to replacement of b with its power, we can assume that α_{n(ᾱ)} = n(ᾱ) for any GGS-group.

Let us fix a vector ᾱ defining a periodic GGS-group for the duration of this Section. For simplicity, we will write just n for the number n(ᾱ).

Let us consider a group

H =< a, b|a^{p}, b^{p} >,

i.e., the free product Z/pZ ∗ Z/pZ. Any element z of H is uniquely represented in the form

a^{s_{1}}b^{β_{1}} … a^{s_{m}}b^{β_{m}}a^{s_{m+1}},  (3.1)

where s_{1}, s_{m+1} ∈ GF(p), and s_{2}, …, s_{m}, β_{1}, …, β_{m} ∈ GF(p) \ {0}. We define the length of z to be m and denote it by ∂(z).

The set of all z such that s_{1} + s_{2} + … + s_{m+1} = 0 is a subgroup (of finite index p), which will be denoted by L(H). If X ≤ H, then L(X) is defined to be the intersection X ∩ L(H).

Notice that any z ∈ L(H) admits a unique representation in the form

b^{β_{1}a^{i_{1}}} … b^{β_{m}a^{im}},  (3.2)

where β_{1}, …, β_{m} ∈ GF(p) \ {0}, i_{1}, …, i_{m} ∈ GF(p), and i_{j} ≠ i_{j+1} for j = 1, 2, …, m − 1. Here m is equal to ∂(z) (as defined above). Finally, let ∂_{i}(z) denote the number of i_{j} such that i_{j} = i.

Define two homomorphisms, φ̃_{n} and φ̃_{p}, acting from L(H) to H as follows:

φ̃_{n}(b^{a^{i}}) = \{ \begin{matrix}b, & i = n \\ a^{α_{n−i}}, & otherwise \end{matrix}

φ̃_{p}(b^{a^{i}}) = \{ \begin{matrix}b, & i = 0 \\ a^{α_{p−i}}, & otherwise \end{matrix}

The indices above are taken modulo p. Now, since L(H) is the free product of the cyclic groups generated by b^{a^{i}}, i ∈ GF(p), the above maps can be extended to the whole L(H).

Theorem 3.1. Let z ∈ [H, H] and let K be the subgroup of H generated by b and a^{n}z. Then there exists N (which depends on z) such that

φ̃_{p}(L(φ̃_{p}(L(… (φ̃_{p}(L(K))) …)))) = H

(φ̃_{p} is applied N times.)

Proof. Note first that

φ̃_{p}(b^{z^{−1}a^{−n}}) = φ̃_{n}(z)a^{n}φ̃_{n}(z)^{−1}.  (3.3)

Therefore, if we introduce a function Θ_{n(ᾱ)} : [H, H] → [H, H]:

Θ_{n(ᾱ)}(z) = a^{−n}φ̃_{n}(z)a^{n}φ̃_{n}(z)^{−1},

then by (3.3) we have that

φ̃_{p}(L(< b, a^{n}z >)) ≥< b, a^{n}Θ_{n(ᾱ)}(z) > .  (3.4)

Moreover, it is obvious that

∂(Θ_{n(ᾱ)}(z)) ≤ 2∂_{n}(z),  (3.5)

which we can use to proceed by induction on ∂(z). The base of induction: ∂(z) ≤ 2. Since z is in [H, H], either ∂(z) = 2, or ∂(z) = 0 and z = 1. In the latter case there is nothing to prove, and we assume that ∂(z) = 2. Then z has the form,

z = b^{αa^{i}}b^{−αa^{j}},

where α ∈ GF(p) \ {0}, i, j ∈ GF(p) and i ≠ j. Thus, there are three possibilities.

(a) If neither i nor j are equal to n, then φ̃_{n}(z) = a^{β} for some β ∈ GF(p). In this case Θ_{n(ᾱ)}(z) = 1, and φ̃_{p}(L(K)) ≥< b, a^{n} >= H.

(b) Suppose that i = n, i.e.

z = b^{αa^{n}}b^{−αa^{j}}.

Then j ≠ n, and

φ̃_{n}(z) = b^{α}a^{−αα_{n−j}}.

Although Θ_{n(ᾱ)} = b^{αa^{n}}b^{−α}, we have that φ̃_{p}(b^{z^{−1}a^{−n}}) = b^{α}a^{n}b^{−α}, and φ̃_{p}(L(K)) ≥< b, b^{α}a^{n}b^{−α} >=< b, a^{n} >= H.

(c) Suppose that j = n, i.e.

z = b^{αa^{i}}b^{−αa^{n}}.

Then i ≠ n, and

φ̃_{n}(z) = a^{αα_{n−i}}b^{−α},
Θ_{n(ᾱ)}(z) = b^{−αa^{n−αα_{n−i}}}b^{αa^{−αα_{n−i}}}.

If neither n − αα_{n−i} nor −αα_{n−i} are equal to n, then we have (a). If n − αα_{n−i} = n, i.e. α_{n−i} = 0, then we have (b).

Finally, assume that −αα_{n−i} = n, i.e. α_{n−i} is invertible and α = −\frac{n}{α_{n−i}}. Then

Θ_{n(ᾱ)}(z) = b^{−αa^{2n}}b^{αa^{n}}.  (3.6)

Again,

φ̃_{n}(Θ_{n(ᾱ)}(z)) = a^{−αα_{p−n}}b^{α},

which implies

Θ_{n(ᾱ)}^{2}(z) = b^{αa^{n+αα_{p−n}}}b^{−αa^{αα_{p−n}}}.

If, as above, neither of the numbers n + αα_{p−n}, −αa^{αα_{p−n}} is equal n, then we have (a). If n + αα_{p−n} equals n, we have (b).

Assume that αα_{p−n} = n, i.e. α_{p−n} is invertible and α = \frac{n}{α_{p−n}}. Then we should only consider z = b^{\frac{n}{α_{p−n}} a^{2n}}b^{− \frac{n}{α_{p−n}} a^{n}}.

In this case we see that

φ̃_{p}(z) = a^{n}b^{− \frac{n}{α_{p−n}}} ,

and thus

Θ_{n(ᾱ)}(z) = b^{− \frac{n}{α_{p−n}}} a^{n}b^{\frac{n}{α_{p−n}}} a^{−n} = b^{− \frac{n}{α_{p−n}}} b^{\frac{n}{α_{p−n}} a^{−n}}.

Since n is invertible, −n is not equal to n, and we get case (a). This completes the base of induction. The inductive step. Here again we have three main cases. To describe them, we need the following definition.

Definition 3.1. We say that z is of type one, if ∂_{n}(z) ≠ \frac{∂(z)}{2}, is of type two, if ∂_{n}(z) = \frac{∂(z)}{2} and β_{1} = n (see (3.2)), and is of type three, if ∂_{n}(z) = \frac{∂(z)}{2} and β_{∂(z)} = n.

Case 1: z is of type one. Then there are two possibilities: ∂_{n}(z) < \frac{∂(z)}{2} and ∂_{n}(z) > \frac{∂(z)}{2}. If ∂_{n}(z) < \frac{∂(z)}{2}, then by (3.5)

∂(Θ_{n(ᾱ)}(z)) ≤ 2∂_{n}(z) < ∂(z),

and we can apply the inductive assumption.

The second possibility is that ∂_{n}(z) > \frac{∂(z)}{2}. It implies that ∂(z) is odd and ∂_{n}(z) = \frac{∂(z)+1}{2}. If this is the case, we apply the observation that

φ̃_{p}((a^{n}z)^{p}) = φ̃_{n}(z)φ̃_{n}(z^{a^{−n}}) … φ̃_{n}(z^{a^{−(p−1)n}}),

which follows from equality (true for any x ∈ L(H))

φ̃_{p}(x^{a^{−n}}) = φ̃_{n}(x).

Thus, we get

φ̃_{p}(b^{(a^{n}z)^{p−1}}) = φ̃_{p}(b^{z^{−1}a^{−n}(a^{n}z)^{p}}) = (φ̃_{n}(z)a^{n}φ̃_{n}(z)^{−1})^{φ̃_{n}(z)φ̃_{n}(z^{a^{−n}})...φ̃_{n}(z^{a^{−(p−1)n}})}

= φ̃_{n}(z^{a^{−(p−1)n}})^{−1} … φ̃_{n}(z^{a^{−n}})^{−1}a^{n}φ̃_{n}(z^{a^{−n}}) … φ̃_{n}(z^{a^{−(p−1)n}}).

Since n is invertible, we have ∂(φ̃_{n}(z^{a^{−n}}) … φ̃_{n}(z^{a^{−(p−1)n}})) ≤ ∂_{1}(z) + … + ∂_{n−1}(z) + ∂_{n+1}(z) + … + ∂_{p}(z) = ∂(z) − ∂_{n}(z). Therefore,

∂(a^{−n}φ̃_{p}(b^{(a^{n}z)^{p−1}})) ≤ 2(∂(z) − ∂_{n}(z)) = 2∂(z) − (∂(z) + 1) < ∂(z).

Since

φ̃_{p}(L(K)) ≥< b, a^{n}(a^{−n}φ̃_{p}(b^{(a^{n}z)^{p−1}})) >,

and a^{−n}φ̃_{p}(b^{(a^{n}z)^{p−1}}) is in [H, H], we can apply the inductive assumption. Case 2: z is of type two. Then z has the form,

z = b^{β_{1}a^{n}}b^{γ_{1}a^{i_{1}}}b^{β_{2}a^{n}}b^{γ_{2}a^{i_{2}}} … b^{β_{m}a^{n}}b^{γ_{m}a^{im}}.

Therefore

φ̃_{n}(z) = b^{β_{1}}a^{δ_{1}} … b^{β_{m}}a^{δ_{m}}

for some δ_{1}, …, δ_{m} ∈ GF(p). Now,

φ̃_{p}(b^{z^{−1}a^{−n}b^{β_{1}}}) = b^{−β_{1}} φ̃_{n}(z)a^{n}φ̃_{n}(z)^{−1}b^{β_{1}} = a^{δ_{1}}b^{β_{2}} … a^{δ_{m−1}}b^{β_{m}}a^{n}b^{−β_{m}}a^{−δ_{m−1}} … b^{−β_{2}}a^{−δ_{1}}. Thus,

∂(a^{−n}φ̃_{p}(b^{z^{−1}a^{−n}b^{β_{1}}})) ≤ 2m − 2 = ∂(z) − 2,

and we can use the inductive assumption as before.

Case 3: z is of type three. Then z has the form,

z = b^{s_{1}a^{γ_{1}}}b^{β_{1}a^{n}}b^{s_{2}a^{γ_{2}}}b^{β_{2}a^{n}} … b^{s_{m}a^{γm}}b^{β_{m}a^{n}}.

Let us write

φ̃_{n}(z) = a^{δ_{1}}b^{β_{1}}a^{δ_{2}}b^{β_{2}} … a^{δ_{m}}b^{β_{m}}.

Lemma 3.1. Let z be of type three, and φ̃_{n}(z) = a^{δ_{1}}b^{β_{1}}a^{δ_{2}}b^{β_{2}} … a^{δ_{m}}b^{β_{m}}. Then Θ_{n(ᾱ)}(z) is of type three if and only if δ_{i} = (−1)^{i}n for i = 1, 2, …, m.

Proof. If z is of type three, then

z = b^{s_{1}a^{γ_{1}}}b^{β_{1}a^{n}}b^{s_{2}a^{γ_{2}}}b^{β_{2}a^{n}} … b^{s_{m}a^{γm}}b^{β_{m}a^{n}} = a^{−γ_{1}}b^{s_{1}}a^{γ_{1}−n}b^{β_{1}}a^{n−γ_{2}} … a^{n−γ_{m}}b^{s_{m}}a^{γ_{m}−n}b^{β_{m}}a^{n}. Of course, vector

(−γ_{1}, γ_{1} − n, n − γ_{2}, γ_{2} − n, …, n − γ_{m}, γ_{m} − n, n) ∈ \underbrace{GF(p) × … × GF(p)}_{2m+1}  (3.7)

is uniquely determined by z. Vice versa, any arbitrary element w ∈ H,

w = a^{μ_{1}}b^{t_{1}} … a^{μ_{l}}b^{t_{l}}a^{μ_{l+1}},

is determined by a pair of vectors

(μ_{1}, …, μ_{l+1}) ∈ (GF(p) \ \{0\}) × \underbrace{GF(p) × … × GF(p)}_{l−1} ×(GF(p) \ \{0\}),

(t_{1}, …, t_{l}) ∈ \underbrace{(GF(p) \ \{0\}) × … × (GF(p) \ \{0\})}_{l−1} .

As follows from (3.7), w is of type three if and only if the following conditions hold: (i) l is even; (ii) μ_{l+1} = n; (iii) μ_{2i−1} + μ_{2i} = 0, i = 2, 3, …, \frac{l}{2}; (iv) μ_{1} + μ_{2} = −n.

Let us now calculate Θ_{n(ᾱ)}(z) for z of type three:

Θ_{n(ᾱ)}(z) = a^{δ_{1}−n}b^{β_{1}}a^{δ_{2}} … a^{δ_{m}}b^{β_{m}}a^{n}b^{−β_{m}}a^{−δ_{m}} … a^{−δ_{2}}b^{−β_{1}}a^{−δ_{1}}.

We will consider two cases.

1) m is odd. Then conditions (i)-(iv) are equivalent to the following system of equations:

\{ \begin{matrix}−δ_{1} & = & n \\ −δ_{2} − δ_{3} & = & 0 \\ & ⋯ \\ −δ_{m−1} − δ_{m} & = & 0 \\ n + δ_{m} & = & 0 \\ & ⋯ \\ δ_{3} + δ_{4} & = & 0 \\ δ_{1} + δ_{2} − n & = & −n \end{matrix} ⇔ \{ \begin{matrix}δ_{1} & = & −n \\ δ_{1} + δ_{2} & = & 0 \\ δ_{2} + δ_{3} & = & 0 \\ δ_{3} + δ_{4} & = & 0 \\ & ⋯ \\ δ_{m−2} + δ_{m−1} & = & 0 \\ δ_{m−1} + δ_{m} & = & 0 \\ δ_{m} + n & = & 0 \end{matrix}

The latter system is obtained from the former by multiplying some equations by −1 and then changing the order of equations; the second equation of the latter system is obtained from the last equation of the former system by eliminating −n.

It is easy to see now that the latter system is equivalent to δ_{i} = (−1)^{i}n for all i = 1, 2, …, m. We are done with the first case.

2) The second case is when m is even. Then conditions (i)-(iv) are equivalent to the following system of equations:

\{ \begin{matrix}−δ_{1} & = & n \\ −δ_{2} − δ_{3} & = & 0 \\ & ⋯ \\ −δ_{m−2} − δ_{m−1} & = & 0 \\ −δ_{m} + n & = & 0 \\ δ_{m−1} + δ_{m} & = & 0 \\ & ⋯ \\ δ_{3} + δ_{4} & = & 0 \\ δ_{1} + δ_{2} − n & = & −n \end{matrix} ⇔ \{ \begin{matrix}δ_{1} & = & −n \\ δ_{1} + δ_{2} & = & 0 \\ δ_{2} + δ_{3} & = & 0 \\ δ_{3} + δ_{4} & = & 0 \\ & ⋯ \\ δ_{m−2} + δ_{m−1} & = & 0 \\ δ_{m−1} + δ_{m} & = & 0 \\ δ_{m} − n & = & 0 \end{matrix}

Just as before, this is equivalent to δ_{i} = (−1)^{i}n for all i = 1, 2, …, m. Since we have considered all possible cases, the lemma is proven.

Corollary 3.1. Let Θ_{n(ᾱ)}(z) = b^{s_{1}a^{γ_{1}}}b^{β_{1}a^{n}}b^{s_{2}a^{γ_{2}}}b^{β_{2}a^{n}} … b^{s_{m}a^{γm}}b^{β_{m}a^{n}} be of type three. Then:

1) if m is even, then

γ_{i} = \{ \begin{matrix}2n, & i = 1, …, \frac{m}{2} \\ 0, & i = \frac{m}{2} + 1, …, m \end{matrix}

2) if m is odd, then

γ_{i} = \{ \begin{matrix}2n, & i = 1, …, \frac{m+1}{2} \\ 0, & i = \frac{m+3}{2}, …, m \end{matrix}

The corollary is verified directly.

We will consider the following three cases separately: m = 2, m = 3, and m ≥ 4. Case 3.1: m = 2. In this case z has the form

z = b^{s_{1}a^{γ_{1}}}b^{β_{1}a^{n}}b^{s_{2}a^{γ_{2}}}b^{β_{2}a^{n}},

and therefore

φ̃_{n}(z) = a^{δ_{1}}b^{β_{1}}a^{δ_{2}}b^{β_{2}}.

If Θ_{n(ᾱ)}(z) is of type one or two, we go to Cases 1 or, respectively, 2 above. Assume that Θ_{n(ᾱ)}(z) is of type three. Then by Lemma 3.1

δ_{1} = −n, δ_{2} = n,

and by Corollary 3.1

Θ_{n(ᾱ)}(z) = b^{β_{1}a^{2n}}b^{β_{2}a^{n}}b^{−β_{2}}b^{−β_{1}a^{n}}.

Let us again consider φ̃_{n}(Θ_{n(ᾱ)}(z)):

φ̃_{n}(Θ_{n(ᾱ)}(z)) = a^{β_{1}α_{p−n}}b^{β_{2}}a^{−β_{2}n}b^{−β_{1}}.

If Θ_{n(ᾱ)}^{2}(z) is of type one or two, we go to Cases 1 or 2 again. Suppose that Θ_{n(ᾱ)}^{2}(z) is of type three. Then by Lemma 3.1

β_{1}α_{p−n} = −n, −β_{2}n = n.

Clearly, when α_{p−n} is not invertible, this is impossible, so Θ_{n(ᾱ)}^{2}(z) will necessarily be of either type one or type two. Assume that α_{p−n} is invertible, then Θ_{n(ᾱ)}^{2}(z) is of type three if and only if

β_{1} = −\frac{n}{α_{p−n}} , β_{2} = −1.

In this case

Θ_{n(ᾱ)}^{2}(z) = b^{−a^{2n}}b^{\frac{n}{α_{p−n}} a^{n}}b^{− \frac{n}{α_{p−n}}} b^{a^{n}},

and therefore

φ̃_{n}(Θ_{n(ᾱ)}^{2}(z)) = a^{−α_{p−n}}b^{\frac{n}{α_{p−n}}} a^{− \frac{n^{2}}{α_{p−n}}} b.

By Lemma 3.1, for Θ_{n(ᾱ)}^{3}(z) to be of type three, it is necessary that

−α_{p−n} = −n, −\frac{n^{2}}{α_{p−n}} = n.

Since n is invertible, −n is not equal to n. Therefore, the equalities contradict each other. Hence, Θ_{n(ᾱ)}^{3}(z) is always of either type one or type two, and we can proceed to either Case 1 or Case 2. Case 3.2: m = 3. Then z has form

z = b^{s_{1}a^{γ_{1}}}b^{β_{1}a^{n}}b^{s_{2}a^{γ_{2}}}b^{β_{2}a^{n}}b^{s_{3}a^{γ_{3}}}b^{β_{3}a^{n}},

and therefore

φ̃_{n}(z) = a^{δ_{1}}b^{β_{1}}a^{δ_{2}}b^{β_{2}}a^{δ_{3}}b^{β_{3}}.

By Lemma 3.1 Θ_{n(ᾱ)}(z) is of type three if and only if δ_{1} = −n, δ_{2} = n, and δ_{3} = −n. If this is the case, then

Θ_{n(ᾱ)}(z) = b^{β_{1}a^{2n}}b^{β_{2}a^{n}}b^{β_{3}a^{2n}}b^{−β_{3}a^{n}}b^{−β_{2}}b^{−β_{1}a^{n}},

and

φ̃_{n}(Θ_{n(ᾱ)}(z)) = a^{β_{1}α_{p−n}}b^{β_{2}}a^{β_{3}α_{p−n}}b^{−β_{3}}a^{−β_{2}n}b^{−β_{1}}.

Again, Θ_{n(ᾱ)}^{2}(z) is of type three if and only if

α_{p−n}β_{1} = −n, β_{3}α_{p−n} = n, −β_{2}n = −n.

If α_{p−n} is not invertible, this is impossible; assume that it is. Then for

β_{1} = −\frac{n}{α_{p−n}} , β_{2} = 1, β_{3} = \frac{n}{α_{p−n}}

Θ_{n(ᾱ)}^{2}(z) is of type three, and

Θ_{n(ᾱ)}^{2}(z) = b^{a^{2n}}b^{− \frac{n}{α_{p−n}} a^{n}}b^{\frac{n}{α_{p−n}} a^{2n}}b^{− \frac{n}{α_{p−n}} a^{n}}b^{\frac{n}{α_{p−n}}} b^{−a^{n}}.

Furthermore,

φ̃_{n}(Θ_{n(ᾱ)}^{2}(z)) = a^{α_{p−n}}b^{− \frac{n}{α_{p−n}}} a^{n}b^{− \frac{n}{α_{p−n}}} a^{\frac{n^{2}}{α_{p−n}}} b^{−1},

and we see that Θ_{n(ᾱ)}^{3}(z) is of type three if and only if α_{p−n} = −n. If this is so, then

φ̃_{n}(Θ_{n(ᾱ)}^{2}(z)) = a^{−n}ba^{n}ba^{−n}b^{−1},

and we have

Θ_{n(ᾱ)}^{3}(z) = b^{a^{2n}}b^{a^{n}}b^{−a^{2n}}b^{a^{n}}b^{−1}b^{−a^{n}}.

Notice that

φ̃_{n}(Θ_{n(ᾱ)}^{3}(z)) = a^{−n}ba^{n}ba^{−n}b^{−1},

and therefore Θ_{n(ᾱ)}^{4}(z) = Θ_{n(ᾱ)}^{3}(z).

The end conclusion is that if z is of type 3 and has length 6 (i.e. m = 3) then either one of the elements Θ_{n(ᾱ)}(z), Θ_{n(ᾱ)}^{2}(z), Θ_{n(ᾱ)}^{3}(z) is not of type three, or Θ_{n(ᾱ)}^{3}(z) and all subsequent Θ_{n(ᾱ)}^{N}(z) are equal to b^{a^{2n}}b^{a^{n}}b^{−a^{2n}}b^{a^{n}}b^{−1}b^{−a^{n}}. Therefore we should check the statement of the theorem specifically for z = b^{a^{2n}}b^{a^{n}}b^{−a^{2n}}b^{a^{n}}b^{−1}b^{−a^{n}}.

If z = b^{a^{2n}}b^{a^{n}}b^{−a^{2n}}b^{a^{n}}b^{−1}b^{−a^{n}}, then

φ̃_{p}(z) = a^{2α_{p−n}}b^{−1}a^{−α_{p−n}}.

Because we assume that α_{p−n} = −n (i.e., α_{p−n} = −α_{n}), we have

φ̃_{p}(b^{−a^{n}z}) = a^{n}(b^{a^{2n}}b^{−a^{n}}).

And since

φ̃_{p}(L(K)) ≥< b, φ̃_{p}(b^{−a^{n}z}) >,

we can apply the inductive assumption. Case 3.3: m ≥ 4.

Lemma 3.2. Let z be of type three, and let ∂(z) ≥ 8. Then at least one of the elements Θ_{n(ᾱ)}(z), Θ_{n(ᾱ)}^{2}(z), Θ_{n(ᾱ)}^{3}(z) is not of type three.

Proof. As before, there are two cases: when m is even and when m is odd. Case 3.3.1: m is even. Let m be 2k, k ≥ 2,

φ̃_{n}(z) = a^{δ_{1}}b^{β_{1}} … a^{δ_{2k}}b^{β_{2k}}.

By Lemma 3.1, Θ_{n(ᾱ)}(z) is of type three if and only if δ_{i} = (−1)^{i}n. If these equalities hold, then

Θ_{n(ᾱ)}(z) = b^{β_{1}a^{2n}}b^{β_{2}a^{n}} … b^{β_{2k−1}a^{2n}}b^{β_{2k}a^{n}}b^{−β_{2k}}b^{−β_{2k−1}a^{n}} … b^{−β_{2}}b^{−β_{1}a^{n}},

and

φ̃_{n}(Θ_{n(ᾱ)}(z)) = a^{β_{1}α_{p−n}}b^{β_{2}} … a^{β_{2k−1}α_{p−n}}b^{β_{2k}}a^{−β_{2k}n}b^{−β_{2k−1}} … a^{−β_{2}n}b^{−β_{1}}

(remember that α_{n} = n).

Suppose Θ_{n(ᾱ)}^{2}(z) is of type three. Then, in particular, −β_{4}n = −n, i.e.

β_{4} = 1,  (3.8)

and

Θ_{n(ᾱ)}^{2}(z) = b^{β_{2}a^{2n}}b^{β_{4}a^{n}} … b^{−β_{3}a^{2n}}b^{−β_{1}a^{n}}b^{β_{1}}b^{β_{3}a^{n}} … b^{−β_{4}}b^{−β_{2}a^{n}}.

Since

φ̃_{n}(Θ_{n(ᾱ)}^{2}(z)) = a^{β_{2}α_{p−n}}b^{β_{4}} … a^{−β_{3}α_{p−n}}b^{−β_{1}}a^{β_{1}n}b^{β_{3}} … a^{−β_{4}n}b^{−β_{2}}

and m is even, then Θ_{n(ᾱ)}^{3}(z) is of type three only if −β_{4}n = n, i.e. only if

β_{4} = −1.

But this contradicts to (3.8), therefore Θ_{n(ᾱ)}^{3}(z) cannot have type three. Case 3.3.2: m is odd. Let m be 2k + 1, k ≥ 2,

φ̃_{n}(z) = a^{δ_{1}}b^{β_{1}} … a^{δ_{2k+1}}b^{β_{2k+1}}.

By Lemma 3.1, Θ_{n(ᾱ)}(z) is of type three if and only if δ_{i} = (−1)^{i}n. If these equalities take place, then

Θ_{n(ᾱ)}(z) = b^{β_{1}a^{2n}}b^{β_{2}a^{n}} … b^{β_{2k}a^{n}}b^{β_{2k+1}a^{2n}}b^{−β_{2k+1}a^{n}}b^{−β_{2k}} … b^{−β_{2}}b^{−β_{1}a^{n}},

and

φ̃_{n}(Θ_{n(ᾱ)}(z)) = a^{β_{1}α_{p−n}}b^{β_{2}} … b^{β_{2k}}a^{β_{2k+1}α_{p−n}}b^{−β_{2k+1}}a^{−β_{2k}n} … a^{−β_{2}n}b^{−β_{1}}

(remember that α_{n} = n).

Suppose Θ_{n(ᾱ)}^{2}(z) is of type three. Then, in particular, −β_{4}n = n, i.e.

β_{4} = −1,  (3.9)

and

Θ_{n(ᾱ)}^{2}(z) = b^{β_{2}a^{2n}}b^{β_{4}a^{n}} … b^{−β_{3}a^{n}}b^{−β_{1}a^{2n}}b^{β_{1}a^{n}}b^{β_{3}} … b^{−β_{4}}b^{−β_{2}a^{n}}.

Further, since

φ̃_{n}(Θ_{n(ᾱ)}^{2}(z)) = a^{β_{2}α_{p−n}}b^{β_{4}} … b^{−β_{3}}a^{−β_{1}α_{p−n}}b^{β_{1}}a^{β_{3}n} … a^{−β_{4}n}b^{−β_{2}}

and m is odd, then Θ_{n(ᾱ)}^{3}(z) is of type three only if −β_{4}n = −n, i.e. only if

β_{4} = 1.

But this contradicts to (3.9), therefore Θ_{n(ᾱ)}^{3}(z) cannot have type three.

Since we have considered all possibilities, the lemma is proven.

This completes the inductive step, so the theorem is proven.

Remark 3.1. Notice that the proof works as well in case of a non-prime p and a GGS-group Gᾱ for which there exists an invertible number n such that α_{n} is also invertible.

Let us return to the GGS-group Gᾱ we’d started with.

Corollary 3.2. Let z be an arbitrary element of [Gᾱ, Gᾱ], and let K be the subgroup of Gᾱ generated by b and a^{n}z. Then there exists a vertex u such that K_{u} = Gᾱ.

Proof. Notice that Gᾱ is a factor-group of H. Let εᾱ : H → Gᾱ be the natural projection. Then by definition of homomorphisms φ̃_{n}, φ̃_{p} the following diagrams are commutative:

\begin{aligned}& L(H) & →^{εᾱ} & st_{Gᾱ}(1) \\ φ̃_{n} & ↓ & & ↓ & φ_{n−1} \\ & H & →^{εᾱ} & Gᾱ \end{aligned}  \begin{aligned}& L(H) & →^{εᾱ} & st_{Gᾱ}(1) \\ φ̃_{p} & ↓ & & ↓ & φ_{p−1} \\ & H & →^{εᾱ} & Gᾱ \end{aligned}

Hence the required statement follows from Theorem 3.1.

4 Dense subgroups

As noted in Section 1.4, we intend to study maximal subgroups of our groups in question via investigating their subgroups everywhere dense in the profinite topology. One such subgroup clearly exists in any group; it is the group itself. This section is devoted to the question of existence of proper everywhere dense subgroups.

Let us give a criterion for a subgroup to be everywhere dense. Let G be a just infinite group and let M be an arbitrary subgroup. For every normal subgroup N in G denote by ε_{N} the natural projection G → G/N. Then M is everywhere dense in the profinite topology if and only if for any normal subgroup N of G

ε_{N}(M) = ε_{N}(G).  (4.1)

Thus, in order to find a proper everywhere dense subgroup, we need to find a proper subgroup satisfying condition (4.1). This condition tells us how it acts on all finite trees T_{[n]}. The next natural question to ask is probably how it acts on subtrees, i.e. how large its upper companion groups are. These things probably do not determine the subgroup entirely, but they are good for the beginning.

In the previous two sections we have already established some results about upper companion groups of subgroups of some special form. Below we develop those results. First, we need the following lemma.

Lemma 4.1. Let G ≤ Aut T be a group acting transitively on each layer of T. Then any everywhere dense proper subgroup M of G has infinite index.

Proof. Suppose an everywhere dense subgroup M has finite index. Then it contains a normal subgroup N which is also of finite index. Since by definition ε_{N}(M) = ε_{N}(G), we have M = G.

Proposition 4.1. Let T be a rooted spherically periodic tree, let G ≤ Aut T be a just infinite group acting transitively on each layer of T, and let M be an everywhere dense subgroup of G. Then we have (a) M is transitive on each layer of vertices;

(b) for any vertex u of T M_{u} is everywhere dense in G_{u}.

Proof. Let us prove (a). Since for any natural number n G is transitive on the set of vertices of level n, the factor-group G/ st_{G}(n) acts transitively on the level n vertices of the finite tree T_{[n]}. Then by (4.1)

ε_{st_{G}(n)}(M) = ε_{st_{G}(n)}(G) = G/ st_{G}(n).

Hence ε_{st_{G}(n)}(M) is also transitive on the level n vertices of the finite tree T_{[n]}. Since ε_{st_{G}(n)}(M) = M/ st_{M}(n), M is transitive on the set of vertices of level n.

Let us prove (b) now. Take a normal subgroup N ◁ G_{u} and consider the subgroup

N̄ = G ∩ Φ_{|u|}^{−1}( ∏_{v: |v|=|u|} N^{ψ_{u,v}}).

This is a normal subgroup, since N is normal in G_{u}. Then by (4.1)

εN̄(M) = εN̄(G).

Since N̄ ≤ st_{G}(u), we also have that

εN̄(st_{M}(u)) = εN̄(st_{G}(u)).  (4.2)

Denote ε_{N}^{u} the natural projection G_{u} → G_{u}/N. Since N̄ is in the kernel of the composition map ε_{N}^{u} ∘ φ_{u} : st_{G}(u) → ε_{N}^{u}(G_{u}), this map can be factorized to a map φ̂_{u} : εN̄(st_{G}(u)) → ε_{N}^{u}(G_{u}) such that

φ̂_{u} ∘ εN̄ = ε_{N}^{u} ∘ φ_{u}.

Hence, by (4.2) we have

ε_{N}^{u}(M_{u}) = φ̂_{u}(εN̄(st_{M}(u))) = φ̂_{u}(εN̄(st_{G}(u))) = ε_{N}^{u}(G_{u}),

which is what we need.

Now we proceed to separate treatment of our three examples. However, we more or less follow the same general scheme. We first prove that any upper companion group of a proper everywhere dense subgroup is a proper subgroup (and therefore is also a proper everywhere dense subgroup, by the above proposition), see Lemmas 4.2, 4.4, and 4.7. In cases of Grigorchuk’s groups and GGS-groups we then prove that each everywhere dense subgroup has an upper companion group containing a subgroup from Corollary 2.1 and Corollary 3.2, respectively, and thus has an upper companion group which is not proper. In order to establish the same fact for EGS-groups, we use the fact that every EGS-group contains a GGS-group.

4.1 Grigorchuk’s groups

Lemma 4.2. Let M an everywhere dense proper subgroup of G_{ω}. Then for any vertex u of T st_{M}(u)_{u} is a proper subgroup of G_{σ^{n}ω}, n = |u|.

Proof. Assume the contrary, i.e. that there exists a vertex u such that st_{M}(u)_{u} = G_{σ^{n}ω}, n = |u|. By Proposition 4.1 it is sufficient to assume that |u| = 1.

Since M is everywhere dense, M is transitive on the set of vertices of length 1. Therefore st_{M}(v)_{v} also equals G_{σω}, where v is the other vertex of length 1.

Let M_{1} = rist_{M}(u)_{u}. Since rist_{M}(u) ◁ st_{M}(u), we have M_{1} ◁ st_{M}(u)_{u} = G_{σω}. As any nontrivial normal subgroup of G_{ω} has finite index, either M_{1} has finite index in G_{σω} or M_{1} = 1.

If M_{1} is a finite index subgroup of G_{σω}, the following subgroup

< rist_{M}(u), rist_{M}(v) >,

has finite index in the direct product Φ_{1}^{−1}(G_{σω} × G_{σω}), hence has finite index in st_{G_{ω}}(1). This implies that index |G_{ω} : M| is finite, in contradiction with Lemma 4.1.

Suppose M_{1} = 1. In this case rist_{M}(1) = 1. This means that st_{M}(1) can be represented in the following way,

st_{M}(1) ≅ {(x, x^{ϕ})|x ∈ G_{σω}},

where ϕ : G_{σω} → G_{σω} is an automorphism.

By definition of an everywhere dense subgroup there exists y ∈ G ∩ Φ_{ω}^{−1}([G_{σω}, G_{σω}] × [G_{σω}, G_{σω}]) such that d_{ω}y ∈ M, where d_{ω} is the d-generator of G_{ω}. Hence (φ_{0}(y))^{ϕ} = d_{σω}φ_{1}(y), and φ_{0}(y), φ_{1}(y) ∈ [G_{σω}, G_{σω}]. This means that the automorphism of G_{σω}/[G_{σω}, G_{σω}] induced by ϕ takes the trivial coset to coset d_{σω}[G_{σω}, G_{σω}]. This does contradict Lemma 1.1. Thus, the lemma is proven.

Lemma 4.3. Let M be an everywhere dense subgroup of G_{ω}. Then there exist an element x ∈ M and a vertex u ∈ T such that X_{u} contains one of the elements b_{σ^{n}ω}, c_{σ^{n}ω}, d_{σ^{n}ω}, where n = |u| and X is the cyclic subgroup generated by x.

Proof. Let x belong to one of the following cosets b_{ω}[G_{ω}, G_{ω}], c_{ω}[G_{ω}, G_{ω}], d_{ω}[G_{ω}, G_{ω}]. Let us prove by induction on ∂_{ω}(x) that there exists a vertex u such that X_{u} contains one of the elements b_{σ^{n}ω}, c_{σ^{n}ω}, d_{σ^{n}ω}, n = |u|.

Without losing any generality, we can assume that d_{ω} is the d-generator and c_{ω} is the c-generator. The base of induction is when ∂_{ω}(x) = 1, in which case the statement is obvious.

Let ∂_{ω}(x) = m > 1. By choice, x = y_{ω}z, where y_{ω} ∈ {b_{ω}, c_{ω}, d_{ω}} and z ∈ [G_{ω}, G_{ω}]. Consider elements φ_{0}(x), φ_{1}(x), φ_{0}(z) and φ_{1}(z). It follows from Lemma 1.1 that the latter two elements belong to the same coset of [G_{σω}, G_{σω}], namely, to one of the cosets [G_{σω}, G_{σω}], d_{σω}[G_{σω}, G_{σω}], ab_{σω}[G_{σω}, G_{σω}], ac_{σω}[G_{σω}, G_{σω}]. Thus, depending on y_{ω}, we get twelve possibilities, listed in the following table (each row corresponds to a fixed choice of y_{ω} and each column to a fixed choice of the coset of φ_{0}(z); the cells contain the cosets of pair (φ_{0}(x), φ_{1}(x))).

\begin{aligned}& 1 & d_{σω} & ac_{σω} & ab_{σω} \\ b_{ω} & (a, b_{σω}) & (ad_{σω}, c_{σω}) & (c_{σω}, ad_{σω}) & (b_{σω}, a) \\ c_{ω} & (a, c_{σω}) & (ad_{σω}, b_{σω}) & (c_{σω}, a) & (b_{σω}, ad_{σω}) \\ d_{ω} & (1, d_{σω}) & (d_{σω}, 1) & (ac_{σω}, ab_{σω}) & (ab_{σω}, ac_{σω}) \end{aligned} .

Thus, if x ∈ b_{ω}[G_{ω}, G_{ω}] ∪ c_{ω}[G_{ω}, G_{ω}], one of the elements φ_{0}(x), φ_{1}(x) also belongs to b_{σω}[G_{ω}, G_{ω}] ∪ c_{σω}[G_{ω}, G_{ω}]. Since both φ_{0}(x), φ_{1}(x) a strictly smaller length, we can apply the inductive assumption.

Further, if x ∈ d_{ω}[G_{ω}, G_{ω}], then either one of φ_{0}(x), φ_{1}(x) is in d_{σω}[G_{ω}, G_{ω}], and we use the inductive assumption, or one of the elements φ_{0}(x), φ_{1}(x) is in ac_{σω}[G_{ω}, G_{ω}], and the other in ab_{σω}[G_{ω}, G_{ω}]. In this case we consider x^{2^{m(ω)}}.

Clearly, x^{2^{m(ω)}} ∈ st_{G_{ω}}(m(ω) + 1). By definition of m(ω) c_{σ^{m(ω)}ω} is the d-generator of G_{σ^{m(ω)}ω}, therefore there is at least one vertex u of length (m(ω) + 1) such that φ_{u}(x^{2^{m(ω)}}) belongs to the coset c_{σ^{m(ω)+1}ω}[G_{σ^{m(ω)+1}ω}, G_{σ^{m(ω)+1}ω}].

Let us prove by induction on n = 1, 2, …, m(ω) that for any vertex u of length n + 1 ∂_{σ^{n+1}ω}(φ_{u}(x^{2^{n}})) is strictly less than ∂_{ω}(x). Notice that for any ω, for any t ∈ G_{ω} \ st_{G_{ω}}(1) and for any j = 0, 1

∂_{σω}(φ_{j}(t^{2})) ≤ ∂_{ω}(t).  (4.3)

Let us apply inequality (4.3) to σω and an element t = φ_{i}(x), where i = 0 or 1. We have

∂_{σ^{2}ω}(φ_{j}(φ_{i}(x)^{2})) ≤ ∂_{σω}(φ_{i}(x)) < ∂_{ω}(x),  (4.4)

where the latter inequality follows from the fact that for any x ∈ st_{G_{ω}}(1) ∂_{σω}(φ_{i}(x)) ≤ \frac{∂_{ω}(x)+1}{2}, and \frac{∂_{ω}(x)+1}{2} is strictly less than ∂_{ω}(x) as soon as ∂_{ω}(x) > 1.

Inequality (4.4) means that for any vertex u of length 2

∂_{σ^{2}ω}(φ_{u}(x^{2})) < ∂_{ω}(x).

Thus, the desired statement is true for n = 1. Suppose now that m(ω) > 1. Let v be a vertex of length m(ω) and u be either v0 or v1. Again, applying (4.3) to σ^{m(ω)}ω and t = φ_{v}(x^{2^{m(ω)−1}}) we get

∂_{σ^{m(ω)+1}ω}(φ_{u}(x^{2^{m(ω)}})) ≤ ∂_{σ^{m(ω)}ω}(φ_{v}(x^{2^{m(ω)−1}})) < ∂_{ω}(x).

Indeed, for any n < m(ω) x^{2^{n}} is not in st_{G_{ω}}(n + 2) and therefore φ_{v}(x^{2^{m(ω)−1}}) is not in st_{G_{σ^{m(ω)}ω}}(1). Thus, we can apply the inductive assumption to obtain that the length of φ_{u}(x^{2^{m(ω)}}) is strictly less than the length of x. Now we can apply the (initial) inductive assumption to prove the lemma.

Proposition 4.2. Let M be an everywhere dense subgroup of a Grigorchuk’s 2-group G_{ω}. Then there exists a vertex u such that M_{u} = G_{σ^{|u|}ω}.

Proof. By Lemma 4.3, there exists a vertex v such that M_{v} contains an element x_{σ^{m}ω} from the set {b_{σ^{m}ω}, c_{σ^{m}ω}, d_{σ^{m}ω}}, m = |v|. Since M_{v} is everywhere dense, it contains an element of the form y_{σ^{m}ω}az, where y_{σ^{m}ω} ≠ x_{σ^{m}ω} is from the set {b_{σ^{m}ω}, c_{σ^{m}ω}, d_{σ^{m}ω}}, z ∈ [G_{σ^{m}ω}, G_{σ^{m}ω}]. Then by Corollary 2.1 there is a vertex w such that (M_{v})_{w} = G_{σ^{n}ω}, where n = |w| + m. Hence, there is a vertex u = vw such that M_{u} = G_{σ^{n}ω}.

Theorem 4.1. No G_{ω} contains everywhere dense proper subgroups.

Proof. Follows from contradiction between Proposition 4.2 and Lemma 4.2.

4.2 GGS-groups

Lemma 4.4. Let Gᾱ be a GGS-group and let M be its everywhere dense proper subgroup. Then for each u ∈ T M_{u} is a proper subgroup of Gᾱ.

Proof. By Proposition 4.1 it is sufficient to consider |u| = 1.

Let u be an arbitrary vertex of length 1. Suppose M_{u} = Gᾱ. Then, since by Proposition 4.1 M is transitive on the set of vertices of length 1, we have that for every other vertex v of length 1 M_{v} also equals Gᾱ.

Let us prove by induction on n = 1, 2, …, p − 1 that M has a trivial intersection with any of the following subgroups of Aut T:

∏_{u∈U} rist_{Aut T} (u),

where U is some set of vertices of length 1 and the cardinality of U is equal to n.

The base of induction is n = 1. Consider M_{u} ∩ rist_{Aut T} (u)_{u} =: L. Clearly, L ◁ Gᾱ and M ≥ Φ_{1}^{−1}(\underbrace{L × … × L}_{p}). If L ≠ 1, then, since Gᾱ is just infinite, L has finite index

in Gᾱ, which implies Φ_{1}^{−1}(\underbrace{L × … × L}_{p}) has finite index in Gᾱ. This, however, contradicts

Lemma 4.1. Thus, L = 1 and

M ∩ rist_{Aut T} (u) = 1.

Denote

M(n, U) := M ∩ ∏_{u∈U} rist_{Aut T} (u),

where U is some set of vertices of length 1 and the cardinality of U is equal to n. Assume now that for k = 1, …, n and for each set U M(k, U) = 1. Let k = n + 1 and choose U of cardinality n + 1.

Suppose that M(n + 1, U) is not trivial. Choose an i ∈ {0, 1, …, p − 1} such that φ_{i}(M(n + 1, U)) ≠ 1 and a j such that φ_{i−j}(M(n + 1, U)) = 1 and φ_{i+j}(M(n + 1, U)) ≠ 1 (such a j exists because n+1 ≥ 2 and p is prime; indeed, starting with a couple of nontrivial subgroups φ_{k}(M(n+1, U)) and φ_{k+l}(M(n+1, U)), we consider subgroups φ_{k−l}(M(n+1, U)), φ_{k−2l}(M(n+1, U)), …, φ_{k−(p−1)l}(M(n+1, U)). Since numbers k, k−l, k−2l, …, k−(p−1)l are all the numbers 0, 1, 2, …, p − 1 but written in a different order, sooner or later we will meet a nontrivial subgroup). Denote φ_{i}(M(n + 1, U)) by L_{1} and φ_{i+j}(M(n + 1, U)) by L_{2}.

Notice that since every element of Gᾱ acts either trivially or transitively on the set of all vertices of length 1, for any vertex u of length 1 each set M(n + 1, U) is normal in st_{M}(u). Thus, since for every such u M_{u} = Gᾱ, L_{1} and L_{2} are normal in Gᾱ. Since M is everywhere dense, it contains an element of the form a^{j}x, x ∈ st_{Gᾱ}(1). Since L_{1} is normal in Gᾱ, we have

φ_{i+j}(M(n + 1, U)^{a^{j}x}) = L_{1}^{φ_{i+j}(x)} = L_{1}.

It follows from the similar equality

φ_{i}(M(n + 1, U)^{a^{j}x}) = φ_{i−j}(M(n + 1, U))^{φ_{i}(x)} = 1

that subgroups M(n + 1, U) and M(n + 1, U)^{a^{j}x} commute. Indeed,

[M(n + 1, U), M(n + 1, U)^{ax}] ≤ M(n, U′),

for some set U′ of cardinality n (U′ is contained in U \ {i}). But M(n, U′) is trivial by inductive assumption, so L_{1} and L_{2} commute as well.

Since L_{1} and L_{2} are normal and by assumption nontrivial, they both have finite index in Gᾱ. Therefore their intersection also has finite index in Gᾱ. However, it is a finite index subgroup in a finitely generated group and therefore is finitely generated. But it is also Abelian and a subgroup of a p-group, hence, it is finite. This implies that Gᾱ is finite, which is a contradiction. Thus, M(n + 1, U) is trivial. The induction is finished.

Thus, the intersection of M with any ∏_{u∈U} rist_{Aut T} (u), where the cardinality of U is p − 1, is trivial. This means that M consists of elements of the form:

(g, g^{ψ_{1}}, g^{ψ_{2}}, …, g^{ψ_{p−1}}),

where g ∈ Gᾱ, ψ_{1}, ψ_{2}, …, ψ_{p−1} are some automorphisms of Gᾱ. Let us consider those automorphisms.

Being everywhere dense, M contains elements of the form bx_{1}, b^{a}x_{2}, …, b^{a^{p−1}}x_{p}, where

x_{1}, x_{2}, …, x_{p} are in Gᾱ ∩ Φ_{1}^{−1}(\underbrace{[Gᾱ, Gᾱ] × … × [Gᾱ, Gᾱ]}_{p}). Thus,

Φ_{1}(bx_{1}) = (a^{α_{1}}x_{11}, a^{α_{2}}x_{12}, …, bx_{1p}),
Φ_{1}(b^{a}x_{2}) = (bx_{21}, a^{α_{1}}x_{22}, …, a^{α_{p−1}}x_{2p}),

⋯

Φ_{1}(b^{a^{p−1}}x_{p}) = (a^{α_{2}}x_{p1}, a^{α_{3}}x_{p2}, …, a^{α_{1}}x_{pp}),

where all x_{ij} ∈ [Gᾱ, Gᾱ]. Let ψ̄_{1} be the induced automorphism Gᾱ/[Gᾱ, Gᾱ] → Gᾱ/[Gᾱ, Gᾱ]. Then we have

a^{α_{1}}[Gᾱ, Gᾱ] ↦ a^{α_{2}}[Gᾱ, Gᾱ]
a^{α_{2}}[Gᾱ, Gᾱ] ↦ a^{α_{3}}[Gᾱ, Gᾱ]
⋯
a^{α_{p−2}}[Gᾱ, Gᾱ] ↦ a^{α_{p−1}}[Gᾱ, Gᾱ]
a^{α_{p−1}}[Gᾱ, Gᾱ] ↦ b[Gᾱ, Gᾱ]
b[Gᾱ, Gᾱ] ↦ a^{α_{1}}[Gᾱ, Gᾱ]

If α_{1} = 0, we obtain a contradiction right away. Suppose α_{1} ≠ 0. Then ψ̄_{1} takes a[Gᾱ, Gᾱ] to a^{\frac{α_{2}}{α_{1}}} [Gᾱ, Gᾱ]. Hence it takes α_{p−1}[Gᾱ, Gᾱ] to a^{\frac{α_{2}α_{p−1}}{α_{1}}} [Gᾱ, Gᾱ], not to b[Gᾱ, Gᾱ]. So we get a contradiction anyway, and this proves the lemma.

Lemma 4.5. Let M be an everywhere dense subgroup of a GGS-group Gᾱ. Then there exist an element x ∈ M and a vertex u ∈ T such that X_{u} ∋ b, where X is the cyclic subgroup generated by x.

Proof. By definition of an everywhere dense subgroup M contains an element x of the form b^{i}y, y ∈ [Gᾱ, Gᾱ], i ≠ 0. Let us prove by induction on n = ∂_{Gᾱ}(x) that for every such x there exists a vertex u such that X_{u} ∋ b, where X is as above.

The base of induction when ∂_{Gᾱ}(x) = 1 is obvious. Suppose ∂_{Gᾱ}(x) = n > 1. Notice that ∂_{b}(φ_{0}(x)) + ∂_{b}(φ_{1}(x)) + … + ∂_{b}(φ_{p−1}(x)) = ∂_{b}(x) ≠ 0. Therefore there exists j such that ∂_{b}(φ_{j}(x)) ≠ 0.

Observe that

∂_{Gᾱ}(φ_{j}(x)) ≤ \frac{∂_{Gᾱ}(x) + 1}{2} < ∂_{Gᾱ}(x).

Thus, if φ_{j}(x) is in st_{Gᾱ}(1), then φ_{j}(x) has the desired form b^{k}y′, k ≠ 0, y′ ∈ [Gᾱ, Gᾱ], and we can apply the inductive assumption.

Suppose that φ_{j}(x) is not in st_{Gᾱ}(1). However,

∂_{b}(φ_{l}(φ_{j}(x)^{p})) = ∂_{b}(φ_{j}(x)) ≠ 0

for all l = 0, 1, …, p − 1 and

∂_{a}(φ_{l}(φ_{j}(x)^{p})) = (α_{1} + … + α_{p−1})∂_{b}(φ_{j}(x)) = 0,

since Gᾱ is periodic. Thus, φ_{l}(φ_{j}(x)^{p}) has the desired form. And since

∂_{Gᾱ}(φ_{l}(φ_{j}(x)^{p})) ≤ ∂_{Gᾱ}(φ_{j}(x)) < ∂_{Gᾱ}(x),

we can apply the inductive assumption.

Proposition 4.3. Let M be an everywhere dense subgroup of a GGS-group Gᾱ. Then there exists a vertex u such that M_{u} = Gᾱ.

Proof. By Proposition 4.5 there exists a vertex v such that M_{v} ∋ b. By Proposition 4.1 M_{v} is everywhere dense as well, hence it contains an element of the form a^{n(ᾱ)}z, z ∈ [Gᾱ, Gᾱ]. Thus, M_{v} contains a subgroup generated by elements b and a^{n(ᾱ)}z. By Corollary 3.2 there exists a vertex w such that (M_{v})_{w} = Gᾱ. In other words, if u = vw, then M_{u} = Gᾱ.

This leads us to the central conclusion about GGS-groups.

Theorem 4.2. No GGS-group contains everywhere dense proper subgroups.

Proof. Indeed, there is a contradiction between Proposition 4.3 and Lemma 4.4, which proves the theorem.

4.3 EGS-groups

Lemma 4.6. Let M be an everywhere dense subgroup of Γᾱ. Then there exists a proper everywhere dense subgroup M′ of Γᾱ such that M′ ∩ Gᾱ either is everywhere dense in the profinite topology of Gᾱ or is a subgroup of finite index in Gᾱ.

Proof. If M ∩ Gᾱ coincides with Gᾱ, then this intersection is everywhere dense, and there is nothing to prove. Suppose that M ∩Gᾱ is a proper subgroup of Gᾱ. Denote by M′ a maximal subgroup in Γᾱ containing M. This is, of course, a proper everywhere dense subgroup, and M′ ∩ Gᾱ is either a maximal subgroup of Gᾱ or Gᾱ itself.

Suppose that M′ ∩ Gᾱ is a proper subgroup of infinite index. Then by Proposition 5.1 it is everywhere dense in Gᾱ.

Lemma 4.7. Let Γᾱ be a EGS-group and let M be its everywhere dense proper subgroup. Then for each u ∈ T M_{u} is a proper subgroup of Γᾱ.

Proof. The proof repeats word-for-word that of Lemma 4.4.

Lemma 4.8. Let M be an everywhere dense subgroup of a EGS-group Γᾱ. Then there exist an element x ∈ M and a vertex u ∈ T such that X_{u} ∋ c, where X is the cyclic subgroup generated by x.

Proof. By definition of an everywhere dense subgroup, M contains an element x of the form c^{α}y, where y belongs to the subgroup generated by b and the derived subgroup [Γᾱ, Γᾱ] and α ≠ 0. Let us prove by induction on n = ∂_{Γᾱ}(x) that for every such x there exists a vertex u such that X_{u} ∋ c.

The base of induction is ∂_{Γᾱ}(x) ≤ 2. If ∂_{Γᾱ}(x) = 1, the statement is obvious. Suppose ∂_{Γᾱ}(x) = 2. Then either x = c^{αa^{i}}b^{βa^{j}} or x = b^{βa^{j}}c^{αa^{i}}.

Let us consider two separate cases: when j = i + 1 and when j ≠ i + 1. Assume that j = i + 1. Then φ_{i}(x) = c^{α}b^{β} or φ_{i}(x) = b^{β}c^{α}, respectively, and φ_{0}(φ_{i}(x)) = c^{α}a^{α_{1}β} or φ_{0}(φ_{i}(x)) = a^{α_{1}β}c^{α}, respectively. Thus, φ_{0} ∘ φ_{0} ∘ φ_{i}(x^{p}) = c^{α}, which is what we need.

Assume that j ≠ i + 1. Then φ_{i}(x) = c^{α}a^{βα_{j−i−1}} or φ_{i}(x) = a^{βα_{j−i−1}}c^{α}, respectively (index j − i − 1 is taken modulo p). If α_{j−i−1} = 0, we already have the desired element. If α_{j−i−1} ≠ 0, then φ_{0}(φ_{i}(x^{p})) = c^{α}.

Suppose ∂_{Γᾱ}(x) = n > 1. Observe that either

∂_{Γᾱ}(φ_{i}(x)) ≤ ∂_{Γᾱ}(x) − 1,

or else x has the form

x = c^{γ_{1}a^{i}}b^{β_{1}a^{i+1}} … c^{γ_{m}a^{i}}b^{β_{m}a^{i+1}},  (4.5)

where γ_{1}, β_{m}, i ∈ GF(p), and γ_{2}, …, γ_{m}, β_{1}, …, β_{m−1} ∈ GF(p). (Notice that γ_{1} +…+γ_{m} = α.)

In the latter case we have that

φ_{i}(x) = c^{γ_{1}}b^{β_{1}} … c^{γ_{m}}b^{β_{m}},

and therefore

∂_{Γᾱ}(φ_{0}(φ_{i}(x))) ≤ m ≤ \frac{∂_{Γᾱ}(x) + 1}{2} < ∂_{Γᾱ}(x),

while ∂_{c}(φ_{0}(φ_{i}(x))) = γ_{1} + … + γ_{m} = α and ∂_{a}(φ_{0}(φ_{i}(x))) = α_{1}(β_{1} + … + β_{m}). If α_{1}(β_{1} + … + β_{m}) = 0, we can apply the inductive assumption.

Suppose that α_{1}(β_{1}+…+β_{m}) ≠ 0. Then ∂_{c}(φ_{0}∘φ_{0}∘φ_{i}(x^{p})) = α and ∂_{a}(φ_{0}∘φ_{0}∘φ_{i}(x^{p})) = 0. Notice also that

∂_{Γᾱ}(φ_{0} ∘ φ_{0} ∘ φ_{i}(x^{p})) ≤ ∂_{Γᾱ}(φ_{0}(φ_{i}(x))) < ∂_{Γᾱ}(x).

Hence, we can apply the inductive assumption.

Suppose that x does not have the form (4.5). This means, in particular, that for any i

∂_{Γᾱ}(φ_{i}(x)) ≤ \frac{∂_{Γᾱ}(x) + 1}{2} < ∂_{Γᾱ}(x).

Notice that ∂_{c}(φ_{0}(x)) + ∂_{c}(φ_{1}(x)) + … + ∂_{c}(φ_{p−1}(x)) = ∂_{c}(x) ≠ 0. Therefore there exists j such that ∂_{c}(φ_{j}(x)) ≠ 0.

If φ_{j}(x) is in st_{Γᾱ}(1), φ_{j}(x) has the desired form c^{β}z, where β ≠ 0 and z is in the subgroup generated by b and [Γᾱ, Γᾱ]. Therefore we can apply the inductive assumption.

Now assume that φ_{j}(x) is not in st_{Γᾱ}(1). However,

∂_{c}(φ_{k}(φ_{j}(x)^{p})) = ∂_{c}(φ_{j}(x)) ≠ 0

for all k = 0, 1, …, p − 1 and

∂_{a}(φ_{k}(φ_{j}(x)^{p})) = (α_{1} + … + α_{p−1})∂_{a}(φ_{j}(x)) = 0,

since Γᾱ is periodic.

Thus, φ_{k}(φ_{j}(x^{p})) has the desired form. And since

∂_{Γᾱ}(φ_{k}(φ_{j}(x^{p}))) ≤ ∂_{Γᾱ}(φ_{j}(x)) < ∂_{Γᾱ}(x),

we can apply the inductive assumption.

Proposition 4.4. Let M be an everywhere dense subgroup of a EGS-group Γᾱ. Then there exists a vertex u such that M_{u} = Γᾱ.

Proof. By Lemma 4.6, we can assume that M ∩ Gᾱ either has a finite index in Gᾱ or is everywhere dense in it. Suppose that M ∩ Gᾱ has finite index. Then it contains a normal subgroup N. By Proposition 4.1 of [13], there is a vertex u such that N_{u}, and hence M_{u}, contains Gᾱ. Now, if M ∩ Gᾱ is everywhere dense in Gᾱ, then by Theorem 4.2, it has to coincide with Gᾱ. In particular, we get that for every vertex u M_{u} contains Gᾱ.

Let us choose a vertex u such that M_{u} contains Gᾱ. By Proposition 4.1, M_{u} is everywhere dense in Γᾱ. By Proposition 4.8 there exists a vertex v such that (M_{u})_{v} ∋ c. Thus, M_{w} contains a subgroup generated by c and Gᾱ, where w = uv. This means that M_{w} = Γᾱ.

Theorem 4.3. No EGS-group contains everywhere dense proper subgroups.

Proof. Indeed, there is a contradiction between Proposition 4.4 and Lemma 4.7, which proves the theorem.

5 Maximal subgroups

It is important to notice in studying maximal subgroups of our examples that all of them are just infinite p-groups for a prime p. Hence it is very easy to describe which of their maximal subgroups are normal. Those are exactly all subgroups of index p (for the suitable p).

Indeed, a normal maximal subgroup of either of those groups cannot be of infinite index, because the groups are just infinite, and so any their normal subgroup has finite index. On the other hand, suppose G is a p-group from one of the series 1.1, 1.2, or 1.3. Any maximal subgroup M of finite index in G contains a normal subgroup N of finite index. The factorgroup G/N by that subgroup is a finite p-group, and the image of M/N in that factor-group is again a maximal subgroup. But the maximal subgroups of finite p-groups are exactly the subgroups of index p. Hence M has index p in the whole group. Reversing this argument shows that every subgroup of index p is maximal in G.

Moreover, being a finite p-group, G/N is nilpotent. Since every subgroup of a nilpotent group is subnormal, M/N, being of prime index, is normal in G/N. Thus, M is normal in G.

It follows that the existence of non-normal maximal subgroups in the groups under study is equivalent to the existence of maximal subgroups of infinite index. In Section 1.4 we mentioned that it is, in its turn, equivalent to the existence of proper everywhere dense subgroups. The following easy proposition provides a precise proof to this claim.

Proposition 5.1. Any maximal subgroup M of infinite index in a group G is everywhere dense in the topology generated by all subgroups of finite index in G.

Proof. Let N be a normal subgroup of finite index. M is maximal, therefore the difference M \ N is non-empty, and MN = G. Hence,

ε_{N}(G) = G/N = MN/N ≅ M/(M ∩ N) = ε_{N}(M).

Since ε_{N}(M) ≤ ε_{N}(G), we have ε_{N}(M) = ε_{N}(G).

Corollary 5.1. Neither Grigorchuk’s groups nor GGS-groups nor EGS-groups contain maximal subgroups of infinite index.

Combining this corollary with the above discussion, we obtain our main result.

Theorem 5.1. All maximal subgroups of Grigorchuk’s 2-groups, GGS-groups, and EGSgroups are normal of prime index.

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