# A Branch Group in a Class of Non-Contracting Weakly Regular Branch Groups

Sagar Saha  Department of Mathematics, Indian Institute of Technology
Guwahati, Guwahati, India<EMAIL_ADDRESS>and  K. V. Krishna  Department
of Mathematics, Indian Institute of Technology Guwahati, Guwahati, India
<EMAIL_ADDRESS>
###### Abstract.

We provide a class of non-contracting groups containing an infinite family of
fractal and weakly regular branch groups, and study certain properties
including abelianization, just infiniteness, and word problem. We present an
example of a branch group in this class and show that it is of exponential
growth. It seems this is the first example of a non-contracting branch group
constructed explicitly.

###### Key words and phrases:

Groups acting on trees, Self-similar group, Branch group, Non-contracting
group

###### 2010 Mathematics Subject Classification:

20E08

##  1\. Introduction

The exploration of groups acting on regular rooted trees gained significant
importance since the pioneering work of Grigorchuk in 1980’s. Notably, a
simple counterexample to the General Burnside Problem was provided through the
Grigorchuk group, which is an automorphism group of a 2-regular rooted tree
[15]. The Grigorchuk group also unveiled interesting properties, e.g., it is
amenable but not elementary amenable, just infinite, and importantly, it is
the first example of a group of intermediate word growth [13]. Subsequently,
many generalizations and different examples of groups acting on regular rooted
trees were studied. For instance, in [16], Gupta and Sidki introduced a family
of $p$-groups for each odd prime $p$, and in [5], Bartholdi and Sunic studied
a family of groups generalizing the Grigorchuk group.

Following the emergence of similar examples that share common features, there
was an attempt to categorize these instances into a cohesive framework which
enables a more organized approach to their studies. This endeavor resulted in
the classification of these instances into self-similar groups, and (weakly)
branch groups. The groups falling in the intersection of these classes
received special attention in the literature as they have many tools enabling
to study properties like just infiniteness, fractalness, maximal subgroups,
congruence subgroup problems, amenability, L-presentations (cf. [2, 6, 9, 10,
20, 23]). The aforesaid examples are in the intersection of these classes. In
addition, many other examples were extensively studied in the literature
(e.g., [4, 14]). Almost all the examples have a specific property, which
defines contracting groups.

In contrast to contracting groups, the literature has very limited examples of
non-contracting weakly branch groups. The groups constructed explicitly in [7]
by Dahmani and in [17] by Mamaghani were only two examples available until
recently. While the Hanoi tower group $\Gamma_{3}$ is a contracting regular
branch group, its generalization $\Gamma_{d}$ (defined based on the game on d
pegs, for $d\geq 4$) is a family of non-contracting weakly branch groups (cf.
[12, 22]). For any $d\geq 4$, it is not known whether or not $\Gamma_{d}$ is a
branch group. In [19], Noce constructed an infinity family of non-contracting
weakly branch groups acting on a $d$-regular rooted tree, for $d\geq 2$. Due
to their non-contracting nature, not much on any of the aforesaid groups is
available in the literature.

In this work, we provide yet another class of non-contracting weakly regular
branch groups aiming to establish a branch group in the class. In Section 3,
we present a class of non-contracting self-similar groups acting on
$d$-regular rooted trees, for $d\geq 3$, and study certain properties of their
elements. In Section 4, we consider the class of groups when $d$ is odd, and
prove that each group in the class is fractal, weakly branch over its
commutator subgroup. Further, we establish that the abelianization of each
group in this class is isomorphic to $\mathds{Z}^{d}$, and hence the groups
are not just infinite. We also provide an efficient algorithm for the word
problem of this class. For $d=3$, in Section 5, we show that the semigroup
generated by $A$ (the defining generating set of the group) is free and
accordingly observe that the group is of exponential growth. Moreover, we
study the structure of its rigid stabilizer and show that the group is a
branch group.

##  2\. Preliminaries

We now recall the fundamental notions and fix the notations used in this work.
We also present a few required properties from the literature. For more
details, one may refer to [3, 11, 18].

For an integer $d\geq 2$, consider the set $X=\\{1,\ldots,d\\}$ of first $d$
positive integers. Let $X^{*}$ be the free monoid of the words over $X$ with
respect to concatenation and the empty word be denoted by $\varepsilon$. The
$d$-regular rooted tree over $X$, denoted by $T_{X}$, is a graph with the
vertex set $X^{*}$, rooted at $\varepsilon$, and two vertices $u,v\in X^{*}$
are adjacent if and only if $u=vx$ for some $x\in X$. In what follows, $T_{X}$
is simply denoted by $T$. The length of a word $w$ over any set is denoted by
$|w|$, which is the number of symbols from the set appearing in $w$ including
their repetitions. Note that the set $X^{k}$ of all words of length $k$ are
the vertices at the level $k$ of $T$. For $u\in X^{*}$, we write $T_{u}$ to
denote the subtree of $T$ rooted at $u$ consisting of all vertices $v$ such
that $u$ is a prefix of $v$, i.e., $v=uu^{\prime}$ for some $u^{\prime}\in
X^{*}$. Consider the set $\text{Aut}(T)$ of all graph automorphisms of $T$
preserving the root. Note that $\text{Aut}(T)$ is a group with respect to
composition of maps, $gh(u)=h(g(u))$ for $g,h\in\text{Aut}(T)$ and $u\in
X^{*}$. We shall denote the identity element of $\text{Aut}(T)$ by $e$. For
$g\in\text{Aut}(T)$, note that $g$ is length preserving, and $g(X^{k})=X^{k}$
for all $k$.

Let $u\in X^{*}$ and $g\in\text{Aut}(T)$. Since $g$ is length preserving
automorphism, for each $v\in X^{*}$, there exists unique $v^{\prime}\in X^{*}$
such that $g(uv)=g(u)v^{\prime}$. Define $g|_{u}:T\to T$ by
$g|_{u}(v)=v^{\prime}$, called the section of $g$ at $u$. Note that
$g|_{u}\in\text{Aut}(T)$. The sections satisfy the following properties: For
any $g,h\in\text{Aut}(T)$ and $u,v\in X^{*}$, we have $g|_{uv}=g|_{u}|_{v}$,
$gh|_{u}=g|_{u}h|_{g(u)}$ and $g^{-1}|_{u}=(g|_{g^{-1}(u)})^{-1}$.

We define the map $\psi$ from $\text{Aut}(T)$ to the wreath product
$\text{Aut}(T)\wr S_{d}$ by

$\psi(g)=(g|_{1},g|_{2},\ldots,g|_{d})\lambda_{g},$

where $S_{d}$ is the permutation group on $X$, and $\lambda_{g}$ is the
induced action of $g$ on the set $X$. For $g\in\text{Aut}(T)$, the expression
$\psi(g)$ is called the wreath recursion of $g$. Clearly, $\psi$ is an
isomorphism. Accordingly, we identify $\text{Aut}(T)$ with $\text{Aut}(T)\wr
S_{d}$ and we will often write $g$ in place of $\psi(g)$ so that
$g=(g|_{1},g|_{2},\ldots,g|_{d})\lambda_{g}$.

Note that any subgroup $G$ of $\text{Aut}(T)$, denoted $G\leq\text{Aut}(T)$,
acts on the tree $T$. A subgroup $G\leq\text{Aut}(T)$ is said to be
spherically transitive if it is transitive on each level of $T$, i.e., for
each $k$ and for all $u,v\in X^{k}$, there exists an element $g\in G$ such
that $g(u)=v$.

For each level $k$, the $k$-th level stabilizer, denoted by
$\text{St}_{G}(\widehat{k})$, is the stabilizer of the set $X^{k}$, i.e.,
$\bigcap_{u\in X^{k}}\text{St}_{G}(u)$, the intersection of the stabilizers of
vertices in the $k$-th level. The normal subgroup $\text{St}_{G}(\widehat{k})$
is precisely the kernel of the induced action of $G$ on the finite set
$X^{k}$, and hence it has finite index in $G$.

The rigid stabilizer of a vertex $u$, denoted by $\text{Rist}_{G}(u)$, is the
subgroup of $G$ containing all the elements that act trivially on the
complement of the subtree $T_{u}$. The $k$-th level rigid stabilizer, denoted
by $\text{Rist}_{G}(\widehat{k})$, is the subgroup
$\left\langle\bigcup_{|u|=k}\text{Rist}_{G}(u)\right\rangle$.

######  Definition 2.1.

Let $G$ be a spherically transitive subgroup of $\text{Aut}(T)$. The group $G$
is said to be a branch group if the index of $\text{Rist}_{G}(\widehat{k})$ in
$G$ is finite for every level $k$. The group $G$ is called a weakly branch
group if, for all $k$, $\text{Rist}_{G}(\widehat{k})$ is nontrivial, and hence
$|\text{Rist}_{G}(\widehat{k})|=\infty$.

######  Definition 2.2.

A subgroup $G\leq\text{Aut}(T)$ is said to be self-similar if $g|_{u}\in G$,
for every $g\in G$ and $u\in X^{*}$.

Notice that for a self-similar group $G\leq\text{Aut}(T)$ the restriction of
the map $\psi$ to $G$

$\psi:G\to G\wr S_{d}$

embeds $G$ into the wreath product $G\wr S_{d}$.

Let $G$ be a self-similar group and $g\in\text{St}_{G}(\widehat{1})$. The
wreath recursion of $g$ is given by $\psi(g)=(g|_{1},g|_{2},\ldots,g|_{d})$
and the induced homomorphism

$\psi_{1}:\text{St}_{G}(\widehat{1})\to\stackrel{{\scriptstyle
d}}{{G\times\cdots\times G}}$

is an embedding. Due to self-similarity of $G$, the homomorphism $\psi_{1}$
extends to all $k\in\mathds{N}$, the set of positive integers, in a natural
way such that

$\psi_{k}:\text{St}_{G}(\widehat{k})\to\stackrel{{\scriptstyle
d^{k}}}{{G\times\cdots\times G}}$

is injective for all $k\in\mathds{N}$. For convenience, we often identify an
element $g\in\text{St}_{G}(\widehat{k})$ with its image $\psi_{k}(g)$. For any
vertex $u\in X^{*}$ we can define the projection $\pi_{u}:\text{St}_{G}(u)\to
G$ by $\pi_{u}(g)=g|_{u}$, where $g\in\text{St}_{G}(u)$. Clearly, for every
$u\in X^{*}$, $\pi_{u}$ is a homomorphism.

######  Definition 2.3.

A self-similar group $G\leq\text{Aut}(T)$ is said to be fractal if
$\pi_{u}(\text{St}_{G}(u))=G$, for all vertices $u$ in $T$.

To prove that a group is fractal, it suffices to show that the above condition
is satisfied by the vertices of the first level of the tree, as per the
following proposition.

######  Proposition 2.4 ([1]).

If a self-similar group $G\leq\textup{Aut}(T)$ is transitive on the first
level and $\pi_{x}(\textup{St}_{G}(x))=G$ for some $x\in X$, then $G$ is
fractal and spherically transitive.

######  Definition 2.5.

A self-similar group $G\leq\text{Aut}(T)$ is said to be weakly regular branch
over a nontrivial subgroup $K\leq G$ if $G$ is spherically transitive and

$\stackrel{{\scriptstyle d}}{{K\times\cdots\times
K}}\leq\psi(K\cap\text{St}_{G}(\widehat{1}))$

If, additionally, $K$ is of finite index in $G$, then $G$ is said to be
regular branch over $K$.

######  Definition 2.6.

A self-similar group $G\leq\text{Aut}(T)$ is said to be contracting if there
exists a finite subset $S\subseteq G$ satisfying the following: for every
$g\in G$ there is a $k\in\mathds{N}$ such that $g|_{u}$ belongs to $S$, for
all vertices $u$ in $T$ with $|u|\geq k$; otherwise, it is said to be non-
contracting.

######  Proposition 2.7 ([8]).

Let $G\leq\text{Aut}(T)$ be a self-similar group. Suppose that there exist
$g\in G$ and $u\in X^{*}$ such that, $g|_{u}=g$, $g(u)=u$, and $g$ has
infinite order, then $G$ is non-contracting.

We use the following notations in the rest of the paper. For elements $g,h$ of
a group, we write $g^{h}:=h^{-1}gh$ to denote the conjugate of $g$ by $h$, and
$[g,h]:=g^{-1}h^{-1}gh$ to denote the commutator of $g$ and $h$.

For a set $A$, let $\tilde{A}=A\cup A^{-1}$, where $A^{-1}=\\{a^{-1}\ :\ a\in
A\\}$, the set of formal inverses of elements of $A$. Let $w$ be a word over
$\tilde{A}$. For $p\in A$, we write $|w|_{p}$ to denote the exponent sum of
$p$ in $w$. Further, $|w|_{A}=\sum_{a\in A}|w|_{a}$. For any $j\in\mathds{N}$,
we write $\overline{j}\in\\{1,\ldots,d\\}$ such that $\overline{j}\equiv j\
(\text{mod}\;d)$.

##  3\. A Class of Non-Contracting Groups

For $d\geq 3$, consider the group $G_{d}\leq\text{Aut}(T)$ generated by the
elements of $A=\\{a_{1},a_{2},\dots,a_{d}\\}$ which are defined recursively as
follows:

$\displaystyle a_{1}$ $\displaystyle=$
$\displaystyle(a_{1},a_{2},e,\dots,e)(1\ 2)$ $\displaystyle a_{2}$
$\displaystyle=$ $\displaystyle(e,a_{2},a_{3},e,\dots,e)(2\ 3)$
$\displaystyle\vdots$ $\displaystyle a_{d-1}$ $\displaystyle=$
$\displaystyle(e,\dots,e,a_{d-1},a_{d})(d-1\ d)$ $\displaystyle a_{d}$
$\displaystyle=$ $\displaystyle(a_{1},e,\dots,e,a_{d})(d\ 1)$

Clearly, $G_{d}$ is a self-similar group. In this section, we prove that
$G_{d}$ is a non-contracting group and, further, study certain properties of
their elements. We begin with the following remarks regarding the elements of
$G_{d}$, which are useful in the sequel.

######  Remark 3.1.

For a word $w$ over $A$, if the wreath recursion of $w$ is
$(w_{1},w_{2},\ldots,w_{d})\lambda_{w}$, then $|w_{1}w_{2}\cdots w_{d}|=2|w|$.

######  Remark 3.2.

For a word $w$ over $\tilde{A}$, let the wreath recursion of $w$ be
$(w_{1},w_{2},\ldots,w_{d})\lambda_{w}$. Then we have the following:

  1. (i)

$|w_{1}w_{2}\cdots w_{d}|_{A}=2|w|_{A}$.

  2. (ii)

For $1\leq i\leq d$, if $|w|_{a_{i}}=s_{i}$, then $|w_{1}w_{2}\cdots
w_{d}|_{a_{i}}=s_{i}+s_{\overline{i-1}}$.

  3. (iii)

If $d$ is odd and $|w_{1}w_{2}\cdots w_{d}|_{a_{i}}=t_{i}$, for $1\leq i\leq
d$, then

$|w|_{a_{i}}=\frac{1}{2}\sum_{j=1}^{d}t_{j}-\sum_{j=1}^{\frac{d-1}{2}}t_{\overline{i+2j}}$

for all $i$.

  4. (iv)

In particular, when $d$ is odd, if $|w_{1}w_{2}\cdots w_{d}|_{a_{i}}=0$ for
all $i$, then $|w|_{a_{i}}=0$, for all $i$.

######  Remark 3.3.

If $d$ even, Remark 3.2(iv) does not hold, in general. For instance, consider
the word $w=a_{2}a_{1}a_{3}^{-1}a_{2}a_{1}^{-1}a_{4}a_{2}^{-1}a_{1}^{-1}$ in
$G_{4}$. Note that $w|_{1}=abb^{-1}a^{-1}$, $w|_{2}=bd^{-1}db^{-1}$,
$w|_{3}=cbb^{-1}c^{-1}$ and $w_{4}=c^{-1}ca^{-1}a$. Although
$|w|_{i}|_{a_{j}}=0$, for all $1\leq i,j\leq 4$, we have $|w|_{a_{i}}\neq 0$
for any $i$.

The following lemma is useful to prove $G_{d}$ is non-contracting.

######  Lemma 3.4.

The order of the product $a_{1}a_{2}\cdots a_{d}$ is infinite in $G_{d}$.

###### Proof.

Let $g=a_{1}a_{2}\cdots a_{d}$ and $h=a_{2}a_{1}a_{d}a_{d-1}\cdots
a_{4}a_{3}$. Note that the section
$g|_{2}=a_{1}|_{2}a_{2}|_{1}a_{3}|_{1}\cdots a_{d}|_{1}=a_{2}a_{1}$ and, for
each $i\geq 3$, the section

$g|_{i}=a_{1}|_{i}a_{2}|_{i}a_{3}|_{i}\cdots a_{i-1}|_{i}a_{i}|_{i-1}\cdots
a_{d}|_{i-1}=a_{i}.$

The action of $g$ on the first level of $T$ is given by the permutation $(1\
2)(2\ 3)\cdots(d\ 1)$, i.e., $(d\ d-1\ldots 3\ 2)$. Therefore,
$g^{d-1}\in\text{St}_{G_{d}}(\widehat{1})$ and
$g^{d-1}|_{2}=g|_{2}g|_{d}g|_{d-1}\cdots g|_{3}=a_{2}a_{1}a_{d}a_{d-1}\cdots
a_{3}=h$.

Also, $h|_{1}=a_{2}|_{1}a_{1}|_{1}a_{d}|_{2}a_{d-1}|_{2}\cdots
a_{3}|_{2}=a_{1}$, $h|_{2}=a_{2}|_{2}a_{1}|_{3}a_{d}|_{3}a_{d-1}|_{3}\cdots
a_{3}|_{3}=a_{2}a_{3}$, for $4\leq i\leq d-1$,
$h|_{i}=a_{2}|_{i}a_{1}|_{i}a_{d}|_{i}\cdots a_{i}|_{i}a_{i-1}|_{i+1}\cdots
a_{3}|_{i+1}=a_{i}$, and
$h|_{d}=a_{2}|_{d}a_{1}|_{d}a_{d}|_{d}a_{d-1}|_{1}\cdots a_{3}|_{1}=a_{d}.$
The action of $h$ on the first level of $T$ is given by the permutation $(2\
3)(1\ 2)(d\ 1)(d-1\ d)\cdots(3\ 4)$, i.e., $(1\ 2\ 4\ 5\ldots d)$. Therefore,
$h^{d-1}\ \in\ \text{St}_{G_{d}}(\widehat{1})$ and
$h^{d-1}|_{1}=h|_{1}h|_{2}h|_{4}\cdots h|_{d}=a_{1}a_{2}a_{3}a_{4}\cdots
a_{d}=g.$

Accordingly, for $j\geq 2$, note that

$\displaystyle g^{(d-1)^{j}}|_{2}$
$\displaystyle=(g^{d-1}|_{2})^{(d-1)^{j-1}}=h^{(d-1)^{j-1}},\ \text{and}$
$\displaystyle h^{(d-1)^{j}}|_{1}$
$\displaystyle=(h^{d-1}|_{1})^{(d-1)^{j-1}}=g^{(d-1)^{j-1}}.$

If possible, suppose the order of $g$ is $r<\infty$. Then $d-1$ divides $r$;
because if $d-1\notdivides r$ then
$g^{r}\notin\text{St}_{G_{d}}(\widehat{1})$. Let $r=m(d-1)^{n}$ for some
$m,n\in\mathds{N}$ such that $d-1\notdivides m$. Then, if $n$ is even,
$g^{m(d-1)^{n}}|_{\underset{\text{length}\ n}{2121\dots
21}}=g^{m}\notin\text{St}_{G_{d}}(\widehat{1})$, and, if $n$ is odd,
$g^{m(d-1)^{n}}|_{\underset{\text{length}\ n}{2121\dots
2}}=h^{m}\notin\text{St}_{G_{d}}(\widehat{1})$. In both the cases we arrive at
a contradiction to the fact that any section of $g^{r}$ must be in
$\text{St}_{G_{d}}(\widehat{1})$. Hence, the order of the product
$a_{1}a_{2}\cdots a_{d}$ is infinite. ∎

######  Theorem 3.5.

The group $G_{d}$ is non-contracting.

###### Proof.

Note that the section $(a_{1}a_{2}\cdots a_{d})|_{1}=a_{1}a_{2}\cdots a_{d}$
and $a_{1}a_{2}\cdots a_{d}(1)=1$. Also, by Lemma 3.4, since the order of
$a_{1}a_{2}\cdots a_{d}$ is infinite, the group $G_{d}$ is non-contracting by
Proposition 2.7. ∎

We now establish the orders of certain elements of $G_{d}$ and show that any
nonempty word over $A$ does not represent the identity element of $G_{d}$ (see
Theorem 3.9).

######  Lemma 3.6.

For $1\leq i\leq d$, the order of the product $a_{1}a_{2}\cdots
a_{i-1}a_{i}a_{i-1}\cdots a_{3}a_{2}$ is infinite in $G_{d}$.

###### Proof.

For $1\leq i\leq d$, let $h_{i}=a_{1}a_{2}\cdots a_{i-1}a_{i}a_{i-1}\cdots
a_{3}a_{2}$. First we show that the order of $h_{d}$ is infinite. If possible,
suppose the order of $h_{d}$ is finite, say $r$. Note that the action of
$h_{d}$ on the first level of $T$ is given by the permutation $(1\ 2)(2\
3)\cdots(d\ 1)(d\ d-1)\cdots(3\ 2)$, i.e., $(d\ d-1\ldots 3\ 2)(2\ 3\ \ldots
d)$, which is identity permutation. Therefore,
$h_{d}\in\text{St}_{G_{d}}(\widehat{1})$. Further, note that
$h_{d}|_{1}=a_{1}a_{2}\cdots a_{d}$. Accordingly,
$e=h_{d}^{r}|_{1}=(h_{d}|_{1})^{r}=(a_{1}a_{2}\cdots a_{d})^{r}$, which
contradicts Lemma 3.4 that the order of $a_{1}a_{2}\cdots a_{d}$ is infinite.
Hence, the order of $h_{d}$ is infinite.

We now prove the result through the following statement: for $1\leq i\leq
d-1$, the order of $h_{i+1}$ is infinite implies that the order of $h_{i}$ is
infinite. Suppose the order of $h_{i+1}$ is infinite.

Note that the sections $h_{i}|_{1}=a_{1}|_{1}a_{2}|_{2}\cdots
a_{i}|_{i}a_{i-1}|_{1}\cdots a_{3}|_{1}a_{2}|_{1}=a_{1}a_{2}\cdots a_{i}$,
$h_{i}|_{2}=a_{1}|_{2}a_{2}|_{1}\cdots a_{i}|_{1}a_{i-1}|_{1}\cdots
a_{2}|_{1}=a_{2}$, and $h_{i}|_{i+1}=a_{1}|_{i+1}\cdots
a_{i}|_{i+1}a_{i-1}|_{i}\cdots a_{2}|_{3}$ $=a_{i+1}a_{i}\cdots a_{3}$. Also,
the action of $h_{i}$ on the first level of $T$ is given by the permutation
$(1\ 2)\cdots(i\ i+1)(i-1\ i)\cdots(2\ 3)$, i.e., $(i+1\ 2\ 1)$. Therefore,
$h_{i}^{3}\in\text{St}_{G_{d}}(\widehat{1})$. Observe that
$h_{i}^{3}|_{1}=h_{i}|_{1}h_{i}|_{i+1}h_{i}|_{2}=a_{1}a_{2}\cdots
a_{i}a_{i+1}a_{i}a_{i-1}\cdots a_{3}a_{2}=h_{i+1}$.

If possible, suppose the order of $h_{i}$ is $s<\infty$. Then $3$ divides $s$;
because if $3\notdivides s$, then
$h_{i}^{s}\notin\text{St}_{G_{d}}(\widehat{1})$. Let $s=3t$ for some
$t\in\mathds{N}$. Thus, we have
$e=h_{i}^{s}|_{1}=(h_{i}^{3}|_{1})^{t}=h_{i+1}^{t}$, which contradicts our
assumption that the order of $h_{i+1}$ is infinite. Hence, the order of
$h_{i}$ is infinite. ∎

######  Lemma 3.7.

For $1\leq i<d$, the order of the product $a_{1}a_{2}\cdots a_{i}$ is infinite
in $G_{d}$.

###### Proof.

For $1\leq i<d$, let $g_{i}=a_{1}a_{2}\cdots a_{i}$. Note that the section
$g_{i}|_{1}=a_{1}|_{1}a_{2}|_{2}\cdots a_{i}|_{i}$ $=a_{1}a_{2}\cdots a_{i}$,
and, for $2\leq j\leq i+1$, the section $g_{i}|_{j}=a_{1}|_{j}\cdots
a_{j-1}|_{j}a_{j}|_{j-1}\cdots a_{i}|_{j-1}$ $=a_{j}$. Also, the action of
$g_{i}$ on the first level of $T$ is given by the permutation $(1\ 2)(2\
3)\cdots(i\ i+1)$, i.e., $(i+1\ i\ldots 2\ 1)$. Thus,
$g_{i}^{i+1}\in\text{St}_{G_{d}}(\widehat{1})$. Note that
$g_{i}^{i+1}|_{1}=g_{i}|_{1}g_{i}|_{i+1}g_{i}|_{i}\cdots
g_{i}|_{2}=a_{1}a_{2}\cdots a_{i}a_{i+1}a_{i}a_{i-1}\cdots a_{3}a_{2}=h\
(\textit{say})$.

If possible, suppose the order of $g_{i}$ is finite, say $r$. Then $i+1$
divides $r$; because if $i+1\notdivides r$ then
$g_{i}^{r}\notin\text{St}_{G_{d}}(\widehat{1})$. Let $r=(i+1)s$, for some
$s\in\mathds{N}$. Thus, we have
$e=g_{i}^{r}|_{1}=(g_{i}^{i+1}|_{1})^{s}=h^{s}$, which contradicts Lemma 3.6
that the order of $h$ is infinite. Hence, the order of $g_{i}$ is infinite. ∎

######  Corollary 3.8.

The order of $a_{1}$ is infinite in $G_{d}$. Hence, by symmetry, the order of
$a_{i}$ is infinite in $G_{d}$, for $2\leq i\leq d$.

######  Theorem 3.9.

If $w$ is a nonempty word over $A$, then $w$ does not represent the identity
element of $G_{d}$.

###### Proof.

We prove the statement by induction on the length $|w|$. For $1\leq i\leq d$,
since $a_{i}$ is non-identity element of $G_{d}$, the statement is true for
$|w|=1$. Let $w$ be a word over $A$ of length $n+1$. If $w$ is of the form
$a_{i}^{n+1}$, for some $i\in\\{1,\ldots,d\\}$, then the statement follows
from Corollary 3.8. Otherwise, $w$ contains a subword of the form
$a_{i}a_{j}$, for $i\neq j$. In the following, we observe that a section of
$a_{i}a_{j}$ has length one by considering $j$ in the cases
$\overline{i+1},\overline{i-1}$, or otherwise. Note that, for $1\leq i\leq d$,

$\begin{array}[]{rcl}a_{i}a_{\overline{i+1}}|_{l}&=&\left\\{\begin{array}[]{ll}a_{i}a_{\overline{i+1}},&\text{if
}\ l=i;\\\ a_{l},&\text{if}\ l\in\\{\overline{i+1},\overline{i+2}\\};\\\
e,&\text{otherwise}.\end{array}\right.\\\ &&\\\
a_{i}a_{\overline{i-1}}|_{l}&=&\left\\{\begin{array}[]{ll}a_{l},&\text{if}\
l\in\\{i,\overline{i-1}\\};\\\ a_{\overline{i+1}}a_{i},&\text{if }\
l=\overline{i+1};\\\ e,&\text{otherwise}.\end{array}\right.\end{array}$

For $d>3$, if $1\leq i,j\leq d$ with
$j\notin\\{\overline{i-1},i,\overline{i+1}\\}$, we have

$\begin{array}[]{rcl}a_{i}a_{j}|_{l}&=&\left\\{\begin{array}[]{ll}a_{l},&\text{if
}\ l\in\\{i,j,\overline{i+1},\overline{j+1}\\};\\\
e,&\text{otherwise}.\end{array}\right.\end{array}$

Thus, there exists at least one $l$ such that $|a_{i}a_{j}|_{l}|=1$.
Consequently, there exists $k$ such that $1\leq|w|_{k}|<|w|$. Therefore, by
inductive hypothesis $w|_{k}$ is non-identity in $G_{d}$. Hence, $w$ is non-
identity in $G_{d}$. ∎

##  4\. A Class of Fractal and Weakly Regular Branch Groups

We now restrict the values of $d$ to odd integers and consider the
corresponding class of non-contracting groups defined in Section 3. We
establish some important properties of the groups in this class, including
just infiniteness, and word problem. Further, we show that a group in the
class is fractal, spherically transitive, and weakly regular branch group over
its commutator subgroup.

######  Theorem 4.1.

Let $d$ be odd. Suppose $w$ is a word over $\tilde{A}$ representing the
identity element of $G_{d}$. Then $|w|_{p}=0$, for all $p\in A$.

###### Proof.

We prove the statement by induction on the length $|w|$. In view of Theorem
3.9, if $w$ is a word over $A$ or over $A^{-1}$, then $w$ is a non-identity
element of $G_{d}$. It can be easily verified that every reduced word $w$ such
that $1\leq|w|\leq 2$ represents a non-identity element of $G_{d}$. Suppose
the statement is true for all words of length $\leq n$. Let us consider a word
$w$ of length $n+1$ that represents the identity element in $G_{d}$. Then
clearly $w$ contains at least one symbol from $A$ and one symbol from
$A^{-1}$. Note that we can get a circular permutation on the symbols of $w$
obtained by applying conjugation on $w$. Let $w^{\prime}$ be such a conjugate
of $w$ such that $|w|=|w^{\prime}|$ and $w^{\prime}=w_{0}a_{i}^{-1}a_{j}$, for
some $i,j$.

If $i=j$, then $w^{\prime}=w_{0}$. Since $|w_{0}|<|w|$, by inductive
hypothesis, $|w_{0}|_{p}=0$ for all $p\in A$. Thus, $|w^{\prime}|_{p}=0$ for
all $p\in A$. Since $w^{\prime}$ is a conjugate of $w$, we have $|w|_{p}=0$
for all $p\in A$.

Suppose $i\neq j$. In the following, we observe that for each
$k\in\\{1,\ldots,d\\}$ section $a_{i}^{-1}a_{j}|_{k}$ has length at most one
by considering $j$ in the cases $\overline{i+1},\overline{i-1}$, or otherwise.
Note that, for $1\leq i\leq d$,

$\begin{array}[]{rcl}a_{i}^{-1}a_{\overline{i+1}}|_{l}&=&\left\\{\begin{array}[]{ll}a_{i}^{-1},&\text{if
}\ l=\overline{i+1};\\\ a_{\overline{i+2}},&\text{if}\ l=\overline{i+2};\\\
e,&\text{otherwise}.\end{array}\right.\\\ &&\\\
a_{i}^{-1}a_{\overline{i-1}}|_{l}&=&\left\\{\begin{array}[]{ll}a_{l},&\text{if}\
l=\overline{i-1};\\\ a_{\overline{i+1}}^{-1},&\text{if }\ l=i;\\\
e,&\text{otherwise}.\end{array}\right.\end{array}$

For $d>3$, if $1\leq i,j\leq d$ with
$j\notin\\{\overline{i-1},i,\overline{i+1}\\}$, we have

$\begin{array}[]{rcl}a_{i}^{-1}a_{j}|_{l}&=&\left\\{\begin{array}[]{ll}a_{\overline{i+1}}^{-1},&\text{if
}\ l=i;\\\ a_{i}^{-1},&\text{if }\ l=\overline{i+1};\\\ a_{l},&\text{if }\
l\in\\{j,\overline{j+1}\\};\\\
e,&\text{otherwise}.\end{array}\right.\end{array}$

Thus, for all $l\in\\{1,\ldots,d\\}$,

(1) $|a_{i}^{-1}a_{j}|_{l}|\leq 1$

For $l\in\\{1,\ldots,d\\}$, note that

$|w^{\prime}|_{l}|=|w_{0}a_{i}^{-1}a_{j}|_{l}|\leq|w_{0}|_{l}|+|a_{i}^{-1}a_{j}|_{w_{0}(l)}|\leq|w_{0}|+1<|w_{0}|+2=|w^{\prime}|.$

Since $w^{\prime}$ represents the identity element in $G_{d}$,
$w^{\prime}|_{l}=e$ for all $l$. By inductive hypothesis, the exponent sum
$|w^{\prime}|_{l}|_{a_{k}}$ is zero, for all $k,l$. Hence, for all $p\in A$,
$|w^{\prime}|_{1}w^{\prime}|_{2}\cdots w^{\prime}|_{d}|_{p}=0$. Since $d$ is
odd, by Remark 3.2(iii), we have $|w^{\prime}|_{p}=0$ for all $p\in A$, and
hence, $|w|_{p}=0$ for all $p\in A$. ∎

######  Remark 4.2.

Theorem 4.1 is not true when $d$ is even. In fact, the word $w$ given in
Remark 3.3 is a relator in $G_{4}$, whereas $|w|_{p}\neq 0$ for any $p\in A$.

In the following theorem we obtain the abelianization of $G_{d}$ and
accordingly show that $G_{d}$ is not just infinite.

######  Theorem 4.3.

For odd values of $d$, the quotient group $G_{d}/G_{d}^{\prime}$ is isomorphic
to $\mathds{Z}\times\stackrel{{\scriptstyle d}}{{\cdots}}\times\mathds{Z}$.

###### Proof.

We prove that for any $(i_{1},i_{2},\ldots,i_{d})\neq(0,0,\ldots,0)$ with
$i_{j}\in\mathds{Z}$, we have $a_{1}^{i_{1}}a_{2}^{i_{2}}\cdots
a_{d}^{i_{d}}\notin G_{d}^{\prime}$. On the contrary, suppose
$w=a_{1}^{i_{1}}a_{2}^{i_{2}}\cdots a_{d}^{i_{d}}\in G_{d}^{\prime}$ for some
$i_{j}\in\mathds{Z}$ ($1\leq j\leq d$) with at least one $i_{j}\neq 0$. Since
every element of $G_{d}^{\prime}$ is generated by commutators and their
conjugates, $w$ equals a word $w^{\prime}$ in $G_{d}$ such that the exponent
sum $|w^{\prime}|_{p}=0$ for all $p\in A$. Thus, $w^{\prime}w^{-1}$ is a
relator in $G_{d}$ with $|w^{\prime}w^{-1}|_{p}\neq 0$ for some $p\in A$,
which contradicts Theorem 4.1. Hence, $\\{a_{1}^{i_{1}}a_{2}^{i_{2}}\cdots
a_{d}^{i_{d}}:i_{1},i_{2},\ldots i_{d}\in\mathds{Z}\\}$ is set of
representatives for the cosets of $G_{d}/G_{d}^{\prime}$ so that
$G_{d}/G_{d}^{\prime}$ is isomorphic to
$\mathds{Z}\times\stackrel{{\scriptstyle d}}{{\cdots}}\times\mathds{Z}$. ∎

######  Corollary 4.4.

For odd values of $d$, the group $G_{d}$ is not just infinite.

Recall that a self-similar group $G$ acting on $X^{*}$ is said to be finite-
state if for any $g\in G$ the set $\\{g|_{v}:v\in X^{*}\\}$ is finite. The
word problem is solvable for finitely generated finite-state self-similar
groups, but the algorithm runs in exponential time (for instance, see [24]).
For contracting groups there is a polynomial-time algorithm [21]. Although
$G_{d}$ is non-contracting, we now present an efficient algorithm for the
group $G_{d}$, when $d$ is odd, which is inspired by the algorithm given in
[14] for a contracting group.

######  Theorem 4.5.

For odd values of $d$, the word problem for $G_{d}$ is solvable via an
efficient algorithm.

###### Proof.

Consider the following procedure Id-Br($u$) for which a word $u$ over
$\tilde{A}$ is an input.

Id-Br($u$) =

  1. (i)

If $u\in\text{St}_{G_{d}}(\widehat{1})$, then go to step (ii). Otherwise,
$u\neq e$.

  2. (ii)

If $|u|_{p}=0$, for all $p\in A$, then go to step (iii). Otherwise, $u\neq e$.

  3. (iii)

If $u$ has a subword of the form $p^{-1}q$, for some $p,q\in A$, then go to
step (v). Otherwise, go to step (iv).

  4. (iv)

Consider a conjugate $u^{\prime}$ of $u$ through a cyclic shift such that
$u^{\prime}$ has a subword of the form $p^{-1}q$, for some $p,q\in A$. Replace
$u$ by $u^{\prime}$ and go to step (v).

  5. (v)

Compute the wreath recursion of $u$, say $(u_{1},u_{2},\ldots,u_{d})$.

Let $w(\neq\varepsilon)$ be a word over $\tilde{A}$. If the wreath recursion
of $w$ is $(w_{1},w_{2},\ldots,w_{d})\lambda_{w}$, then note that $w=e$ if and
only if $w_{i}=e$, for all $i\in\\{1,2,\ldots,d\\}$ and $\lambda_{w}$ is the
identity permutation. Give input $w$ to the procedure Id-Br( ) and iteratively
call the procedure on each branch (obtained in step 5) if it is not empty. In
any iteration, if the procedure returns that the input is not $e$, then the
given word $w\neq e$. Otherwise, after a finite number of iterations the
process results that $w=e$. The process converges because we have the length
reduction of each branch. Indeed, for $p,q\in A$, we have
$|p^{-1}q|_{i}|<|p^{-1}q|$, for all $i\in\\{1,2,\ldots,d\\}$ (cf. Equation 1).
∎

######  Theorem 4.6.

For odd values of $d$, the group $G_{d}$ is fractal and spherically
transitive.

###### Proof.

Consider the element $g=a_{1}a_{2}\cdots a_{d-1}\in G_{d}$ and note that the
action of $g$ on the first level of $T$ is given by the permutation
$\lambda_{g}=(1\ 2)(2\ 3)\cdots(d-1\ d)=(d\ d-1\dots 1)$. Therefore, for all
$x,y\in X$, $\lambda_{g}^{i}(x)=y$ for some $i$. Hence, $G_{d}$ is transitive
on the first level of $T$. In view of Proposition 2.4, we show that
$\pi_{1}(\textup{St}_{G_{d}}(1))=G_{d}$ so that $G_{d}$ is fractal and
spherically transitive. For this, in the following, we explicitly construct
elements within the subset $\text{St}_{G_{d}}(\widehat{1})$ whose images under
$\pi_{1}$ give the generators $a_{i}$’s of $G_{d}$.

For $2\leq i\leq d-1$, let $s_{i}=(a_{i}^{2})^{g_{i-1}}$, where
$g_{i-1}=a_{i-1}\cdots a_{1}$. For all $i$, clearly
$s_{i}\in\text{St}_{G_{d}}(\widehat{1})$. Note that the section
$g_{i-1}|_{i}=a_{i-1}|_{i}a_{i-2}|_{i-1}a_{i-3}|_{i-2}\cdots
a_{1}|_{2}=a_{i}\cdots a_{2}$ and, since $g_{i-1}^{-1}(1)=i$, we have
$g_{i-1}^{-1}|_{1}=(g_{i-1}|_{i})^{-1}={a_{2}}^{-1}\cdots{a_{i}}^{-1}$.
Further, note that $\pi_{1}(s_{2})=s_{2}|_{1}=a_{3}a_{2}$ and, for $3\leq
i\leq d-1$,
$\pi_{1}(s_{i})=s_{i}|_{1}=g_{i-1}^{-1}|_{1}a_{i}^{2}|_{i}g_{i-1}|_{i}=({a_{2}}^{-1}\cdots{a_{i}}^{-1})(a_{i}a_{i+1})(a_{i}\cdots
a_{2})=a^{-1}_{2}\cdots a^{-1}_{i-1}a_{i+1}\ \cdots a_{2}.$

Let $h=a_{2}a_{3}\cdots a_{d}a_{1}$. Note that $\lambda_{h}=(2\ 3)(3\
4)\cdots(d\ 1)(1\ 2)=(d\ d-1\ \dots\break\dots\ 3\ 1)$ and hence
$h^{d-1}\in\text{St}_{G_{d}}(\widehat{1})$. To calculate $\pi_{1}(h^{d-1})$,
first observe that $h|_{1}=a_{2}|_{1}a_{3}|_{1}\cdots
a_{d}|_{1}a_{1}|_{d}=a_{1}$, $h|_{3}=a_{2}|_{3}a_{3}|_{2}\cdots
a_{d}|_{2}a_{1}|_{2}=a_{3}a_{2}$, and, for $4\leq j\leq d$,
$h|_{j}=a_{2}|_{j}\cdots a_{j-1}|_{j}a_{j}|_{j-1}\cdots
a_{d}|_{j-1}a_{1}|_{j-1}=a_{j}$. Thus,
$\pi_{1}(h^{d-1})=h|_{1}h|_{d}h|_{d-1}\cdots
h|_{3}\break=a_{1}a_{d}a_{d-1}\cdots a_{3}a_{2}$.

Now note that for $1\leq i\leq\frac{d-1}{2}$, $\pi_{1}(s_{2}s_{4}\cdots
s_{2i})=a_{2i+1}a_{2i}\cdots a_{2}$. Accordingly,

$\pi_{1}(h^{d-1}(s_{2}s_{4}\cdots
s_{d-1})^{-1})=a_{1}\in\pi_{1}(\text{St}_{G_{d}}(\widehat{1})).$

Let $s_{1}=(s_{2}s_{4}\cdots s_{d-1})h^{-(d-1)}a_{1}^{2}$. Then
$\pi_{1}(s_{1})=a_{2}$ as $\pi_{1}(a_{1}^{2})=a_{1}a_{2}$. Further, for $1\leq
i\leq\frac{d-1}{2}$,

$\pi_{1}(s_{1}s_{3}\cdots s_{2i-1})=a_{2i}a_{2i-1}\cdots a_{2}.$

A direct computation gives

$\displaystyle a_{2i}$ $\displaystyle=$
$\displaystyle\pi_{1}\left((s_{1}s_{3}\cdots s_{2i-1})(s_{2}\cdots
s_{2i-2})^{-1}\right),\ 2\leq i\leq\frac{d-1}{2};$ $\displaystyle a_{2i+1}$
$\displaystyle=$ $\displaystyle\pi_{1}\left((s_{2}s_{4}\cdots
s_{2i})(s_{1}s_{3}\cdots s_{2i-1})^{-1}\right),\ 1\leq i\leq\frac{d-1}{2}.$

Hence, for all $i\in\\{1,\dots,d\\}$,
$a_{i}\in\pi_{1}(\text{St}_{G_{d}}(\widehat{1}))$. ∎

We use the following notation in the proof of Theorem 4.8 for presenting it in
an elegant manner.

######  Notation 4.7.

Let the wreath recursion of an element $g=(g_{1},\ldots,g_{d})\lambda_{g}$. If
$g_{i}=e$ for most of the components, then $g$ is simply denoted by
$\\{\\!\\!\\{\underset{i_{1}}{g_{i_{1}}},\ldots,\underset{i_{n}}{g_{i_{n}}}\\}\\!\\!\\}\lambda_{g}$,
where $g_{i_{l}}$ is non-identity element appearing at $i_{l}$-th component.
For example, if $g=(a,e,e,b,e,e,c)\lambda_{g}$, then $g$ can be denoted by
$\\{\\!\\!\\{\underset{7}{c},\underset{1}{a},\underset{4}{b}\\}\\!\\!\\}\lambda_{g}$,
irrespective of the order in which the non-identity elements appear in wreath
recursion of $g$, as their positions are given.

######  Theorem 4.8.

For odd values of $d$, the group $G_{d}$ is weakly regular branch over its
commutator subgroup $G_{d}^{{}^{\prime}}$.

###### Proof.

In view of Theorem 4.6, it is sufficient to prove
$G_{d}^{{}^{\prime}}\times\dots\times
G_{d}^{{}^{\prime}}\leq\psi_{1}(\text{St}_{G_{d}}(\widehat{1})\ \cap\
G_{d}^{{}^{\prime}})$. For this, we construct elements in
$\text{St}_{G_{d}}(\widehat{1})\ \cap\ G_{d}^{{}^{\prime}}$ whose images under
$\psi_{1}$ give all the generators of $G_{d}^{{}^{\prime}}\times\dots\times
G_{d}^{{}^{\prime}}$.

For $i,j\in\\{1,2,\dots,d\\}$, note that the commutators $[a_{i},a_{j}]=e$,
when$|i-j|\neq 1$ and $|i-j|\neq d-1$. Hence, the commutator group
$G_{d}^{{}^{\prime}}$ is generated by
$[a_{1},a_{2}],[a_{2},a_{3}],\ldots,[a_{d-1},a_{d}],[a_{d},a_{1}]$ and their
conjugates, written $[a_{i},a_{j}]^{g}$, by any element $g\in G_{d}$.

For $1\leq i\leq d$, let $\beta_{i}=[a_{i},a_{\overline{i+1}}]$. We compute
the wreath recursion of the following elements in the succession:

$\displaystyle\beta_{i}$
$\displaystyle=\\{\\!\\!\\{\underset{i}{a_{\overline{i+1}}^{-1}},\underset{\overline{i+1}}{a_{\overline{i+1}}}\\}\\!\\!\\}(i\
\overline{i+1}\ \overline{i+2})$
$\displaystyle\beta_{i}\beta_{\overline{i+1}}$
$\displaystyle=\\{\\!\\!\\{\underset{i}{a_{\overline{i+1}}^{-1}a_{\overline{i+2}}^{-1}},\underset{\overline{i+1}}{a_{\overline{i+1}}a_{\overline{i+2}}}\\}\\!\\!\\}(i\
\overline{i+2})(\overline{i+1}\ \overline{i+3})$
$\displaystyle[a_{i}^{2},a_{\overline{i+1}}]$
$\displaystyle=\\{\\!\\!\\{\underset{\overline{i+1}}{a_{i}^{-1}a_{\overline{i+1}}^{-1}},\underset{\overline{i+2}}{a_{i}a_{\overline{i+1}}}\\}\\!\\!\\}$
$\displaystyle[a_{i}^{2},a_{\overline{i+1}}]^{a_{i}}$
$\displaystyle=\\{\\!\\!\\{\underset{i}{a_{\overline{i+1}}^{-1}a_{i}^{-1}},\underset{\overline{i+2}}{a_{i}a_{\overline{i+1}}}\\}\\!\\!\\}$

For $1\leq i\leq d$, consider
$\xi_{i}=[a_{\overline{i+1}}^{2},a_{\overline{i+2}}](\beta_{i}\beta_{\overline{i+1}})^{-1}\in
G_{d}$. Note that the wreath recursion of $\xi_{i}$ is given by

(2) $\xi_{i}=(e,\dots,e)(i\ \overline{i+2})(\overline{i+1}\ \overline{i+3}).$

In case of the group $G_{3}$, let $g_{i}=\xi_{i}$ (for $1\leq i\leq 3$). In
case of $G_{d}$ (for $d\geq 5$), let
$g_{i}=(\xi_{\overline{i+1}}\xi_{i})(\xi_{\overline{i+3}}\xi_{\overline{i+2}})\cdots(\xi_{\overline{i+(d-4)}}\xi_{\overline{i+(d-5)}})\xi_{\overline{i+(d-3)}}$
(for $1\leq i\leq d$). For $1\leq j\leq d$, since
$\xi_{\overline{j+1}}\xi_{j}$ sends $j$ to $\overline{j+2}$ and fixes
$\overline{j+1}$, we have $g_{i}(i)=\overline{i-1}$ and
$g_{i}(\overline{i+1})=\overline{i+1}$, for any $1\leq i\leq d$.

Note that, for $1\leq i\leq d$,

$\displaystyle([a_{i}^{2},a_{\overline{i+1}}]^{a_{i}})^{\xi_{i}^{-1}}$
$\displaystyle=\\{\\!\\!\\{\underset{i}{a_{i}a_{\overline{i+1}}},\underset{\overline{i+2}}{a_{\overline{i+1}}^{-1}a_{i}^{-1}}\\}\\!\\!\\}$
$\displaystyle[a_{i}^{2},a_{\overline{i+1}}]^{g_{\overline{i+1}}}$
$\displaystyle=\\{\\!\\!\\{\underset{i}{a_{i}^{-1}a_{\overline{i+1}}^{-1}},\underset{\overline{i+2}}{a_{i}a_{\overline{i+1}}}\\}\\!\\!\\}$

Hence,
$[a_{i}^{2},a_{\overline{i+1}}]^{g_{\overline{i+1}}}([a_{i}^{2},a_{\overline{i+1}}]^{a_{i}})^{\xi_{i}^{-1}}=\\{\\!\\!\\{\underset{i}{[a_{i},a_{\overline{i+1}}]}\\}\\!\\!\\},\
\textup{for}\ 1\leq i\leq d.$ Further, in the wreath recursion of
$[a_{i}^{2},a_{\overline{i+1}}]^{g_{\overline{i+1}}}([a_{i}^{2},a_{\overline{i+1}}]^{a_{i}})^{\xi_{i}^{-1}}$,
we can move the generators $[a_{i},a_{i+1}]$ to any $j$-th position by taking
appropriate conjugates by a product of $\xi_{i}$’s. Clearly, these elements
are in $\text{St}_{G_{d}}(\widehat{1})\ \cap\ G_{d}^{\prime}$.

Thus, the images of the above-produced elements under $\psi_{1}$ are the
generators of $G_{d}^{\prime}\times\cdots\times G_{d}^{\prime}$ of the form
$(e,\ldots,e,\underset{j}{[a_{i},a_{\overline{i+1}}]},e,\ldots,e)$ having
commutators at any $j$-th component. Since $G_{d}$ is fractal, we also get the
generators of the form
$(e,\ldots,e,\underset{j}{[a_{i},a_{\overline{i+1}}]^{g}},e,\ldots,e)$ having
conjugates of the commutators at any $j$-th component. Hence, for any
$(\alpha_{1},\alpha_{2},\dots,\alpha_{d})\in G_{d}^{\prime}\times\dots\times
G_{d}^{\prime}$, there exists $\alpha\in\text{St}_{G_{d}}(\widehat{1})\cap
G_{d}^{\prime}$ such that
$\psi_{1}(\alpha)=(\alpha_{1},\alpha_{2},\dots,\alpha_{d})$. ∎

In view of Theorem 4.3, we have the following corollary of Theorem 4.8.

######  Corollary 4.9.

For odd values of $d$, the group $G_{d}$ is not a regular branch group over
$G_{d}^{\prime}$.

Note that by Equation 2, in the group $G_{3}$, the order of $\xi_{i}$ (for
$1\leq i\leq 3$) is $3$ and in the group $G_{d}$ ($d\geq 4$ is odd), the order
of $\xi_{i}$ (for $1\leq i\leq d$) is 2. Further, since $G_{d}$ has infinite
order elements (see Section 3), we have the following remark.

######  Remark 4.10.

For odd values of $d$, the group $G_{d}$ is neither torsion nor torsion-free.

##  5\. A Branch Group

In this section, we consider $d=3$ and show that the group $G_{d}$ is a branch
group. This provides an explicit example in the class of non-contracting
weakly regular branch groups established in Section 4.

Let $A=\\{a,b,c\\}$ and consider the group $G_{3}$ generated by $A$. Recall
that the wreath recursions of $a,b$ and $c$ are as per the following:

$\displaystyle a$ $\displaystyle=$ $\displaystyle(a,b,e)(1\ 2)$ $\displaystyle
b$ $\displaystyle=$ $\displaystyle(e,b,c)(2\ 3)$ $\displaystyle c$
$\displaystyle=$ $\displaystyle(a,e,c)(3\ 1)$

We rephrase Remark 3.2(iii) for $G_{3}$ and present in the following remark.

######  Remark 5.1.

For a word $w$ over $\tilde{A}$, let the wreath recursion of $w$ be
$(w_{1},w_{2},w_{3})\lambda_{w}$. If $|w_{1}w_{2}w_{3}|_{a}=t_{1}$,
$|w_{1}w_{2}w_{3}|_{b}=t_{2}$ and $|w_{1}w_{2}w_{3}|_{c}=t_{3}$, then
$|w|_{a}=\frac{t_{1}+t_{2}-t_{3}}{2},\ |w|_{b}=\frac{t_{2}+t_{3}-t_{1}}{2}$
and $|w|_{c}=\frac{t_{3}+t_{1}-t_{2}}{2}$.

######  Remark 5.2.

In the group $G_{3}$, the elements $\xi_{i}$, for $1\leq i\leq 3$, given in
Equation 2 coincide and are equal to $(e,e,e)(1\ 3\ 2)$. We denote the element
by $\xi$.

######  Remark 5.3.

For $w\in G_{3}$ with $|w|=3$, note that $\lambda_{w}=\lambda_{p}$ for some
$p\in A$. Accordingly, if $|w|$ is odd, then $\lambda_{w}$ is not identity so
that $w\not\in\text{St}_{G_{3}}(\widehat{1})$. Thus, if
$w\in\text{St}_{G_{3}}(\widehat{1})$, then $|w|$ is even and hence $|w|_{A}$
is even.

######  Lemma 5.4.

Let $p_{1},p_{2},q_{1},q_{2}\in A$ with $p_{2}\neq q_{2}$. For each
permutation $\sigma$ on $\\{1,2,3\\}$, there exists $j\in\\{1,2,3\\}$ such
that the non-identity (i.e., $\neq e$) sections $p_{1}p_{2}|_{j}$ and
$q_{1}q_{2}|_{\sigma(j)}$ are not identical and they are not of the form $pq$
and $q$, for some $p,q\in A$.

###### Proof.

By considering the possibilities of two length words, we prove the result
using brute force method. Consider the wreath recursions of all two length
words over $A$ as per the following, where $\lambda_{1}=(1\ 3\
2),\lambda_{2}=(1\ 2\ 3)$.

$\begin{array}[]{ccccc}ab=(ab,b,c)\lambda_{1}&&ba=(a,b,cb)\lambda_{2}&&a^{2}=(ab,ba,e)\\\
bc=(a,bc,c)\lambda_{1}&&cb=(ac,b,c)\lambda_{2}&&b^{2}=(e,bc,cb)\\\
ca=(a,b,ca)\lambda_{1}&&ac=(a,ba,c)\lambda_{2}&&c^{2}=(ac,e,ca)\end{array}$

We discuss the result in the following three cases.

Case-I: $p_{1}\neq p_{2}$ and $q_{1}\neq q_{2}$. It is evident from the list
that two among the sections $p_{1}p_{2}|_{i}$ ($1\leq i\leq 3$) end with one
letter and the third section ends with a different letter. Accordingly, there
exists $j$ such that the section $p_{1}p_{2}|_{j}$ is not identical with the
section $q_{1}q_{2}|_{\sigma(j)}$.

Case-II: $p_{1}=p_{2}$ and $q_{1}=q_{2}$. Let us consider $a^{2}$ and $b^{2}$.
Note that $a^{2}|_{1}=ab$ is not identical with the sections $bc$ and $cb$ of
$b^{2}$. In case $\sigma(1)=1$, then clearly $a^{2}|_{2}=ba$ is not identical
with the sections $bc$ and $cb$ of $b^{2}$. Similarly, for all $p,q\in A$, it
can be observed that there exists $j$ such that the section $p^{2}|_{j}$ is
not identical with the section $q^{2}|_{\sigma(j)}$.

Case-III: $p_{1}=p_{2}=p$ and $q_{1}\neq q_{2}$. Note that two among the
sections $p^{2}|_{i}$ ($1\leq i\leq 3$) are of the form $pq$ and $qp$, for
$q\in A$. If, for some $i$, $q_{1}q_{2}|_{i}$ ends with $p$ (or $q$), then
other two sections of $q_{1}q_{2}$ never end with $q$ (or $p$, respectively).
Hence, there exists $j$ such that the section $p^{2}|_{j}$ is not identical
with the section $q_{1}q_{2}|_{\sigma(j)}$.

In all three cases, it is evident that the resultant sections are not of the
form $p^{\prime}q^{\prime}$ and $q^{\prime}$, for some
$p^{\prime},q^{\prime}\in A$. ∎

######  Theorem 5.5.

The semigroup $S$ generated by $A$ is free of rank $3$.

###### Proof.

By Theorem 3.9, if $w$ is a word over $A$, then $w$ cannot be a relator in
$G_{3}$ and hence in $S$. Also, by Theorem 4.1, if $u,v$ are words over $A$
such that $|u|_{p}\neq|v|_{p}$ for some $p\in A$, then $uv^{-1}\neq e$ in
$G_{3}$ and hence $u=v$ is not a relation in $S$.

Accordingly, it is sufficient to show that the group $G_{3}$ does not have any
relation of the form $u=v$ for nonempty and non-identical words $u$ and $v$
over $A$ such that $|u|_{p}=|v|_{p}$, for all $p\in A$. On the contrary,
suppose such relations exist in $G_{3}$ and $u^{\prime}=v^{\prime}$ is such a
relation with $|u^{\prime}|+|v^{\prime}|$ is minimum.

Observe that $|u^{\prime}|=|v^{\prime}|\geq 3$ and
$u^{\prime}|_{i}=v^{\prime}|_{i}$, for all $i\in\\{1,2,3\\}$. Assume that,
$u^{\prime}$ is of the form $p_{0}u_{0}p_{1}p_{2}$ and $v^{\prime}$ is of the
form $q_{0}v_{0}q_{1}q_{2}$, where $p_{0},p_{1},p_{2},q_{1},q_{2},q_{3}\in A$
and $u_{0},v_{0}$ are words over $A$. Note that $p_{0}\neq q_{0}$; otherwise,
$u_{0}p_{1}p_{2}=v_{0}q_{1}q_{2}$ with
$|u_{0}p_{1}p_{2}|+|v_{0}q_{1}q_{2}|<|u^{\prime}|+|v^{\prime}|$, which
contradicts our assumption. Similarly, $p_{2}\neq q_{2}$. Recall the wreath
recursions of $a,b$ and $c$ and note that the sections
$p_{0}|_{i_{1}}=q_{0}|_{i_{1}}=p$, $p_{0}|_{i_{2}}=e$, $q_{0}|_{i_{2}}=q$,
$p_{0}|_{i_{3}}=r$ and $q_{0}|_{i_{3}}=e$, for distinct
$i_{1},i_{2},i_{3}\in\\{1,2,3\\}$ and $p,q,r\in A$. Consequently,
$u^{\prime}|_{i_{1}}=pu_{1}$ and $v^{\prime}|_{i_{1}}=pv_{1}$, for some words
$u_{1},v_{1}$ over $A$,
$|u^{\prime}|_{i_{2}}|+|v^{\prime}|_{i_{2}}|<|u^{\prime}|+|v^{\prime}|$, and
$|u^{\prime}|_{i_{3}}|+|v^{\prime}|_{i_{3}}|<|u^{\prime}|+|v^{\prime}|$.

Note that $u^{\prime}|_{i}=(p_{0}u_{0})|_{i}(p_{1}p_{2})|_{\sigma_{1}(i)}$,
and $v^{\prime}|_{i}=(q_{0}v_{0})|_{i}(q_{1}q_{2})|_{\sigma_{2}(i)}$, where
$\sigma_{1}=\lambda_{p_{0}u_{0}}$ and $\sigma_{2}=\lambda_{q_{0}v_{0}}$. Since
$p_{2}\neq q_{2}$, by Lemma 5.4, there exists $j\in\\{1,2,3\\}$ such that the
sections $p_{1}p_{2}|_{\sigma_{1}(j)}$ and $q_{1}q_{2}|_{\sigma_{2}(j)}$ are
not identical and they are not of the form $p^{\prime}q^{\prime}$ and
$q^{\prime}$, for some $p^{\prime},q^{\prime}\in A$. Hence, $u^{\prime}|_{j}$
and $v^{\prime}|_{j}$ are not identical.

In case $j=i_{1}$, since $u^{\prime}|_{j}=v^{\prime}|_{j}$, we have
$u_{1}=v_{1}$ and they are not identical with
$|u_{1}|+|v_{1}|<|u^{\prime}|_{j}|+|v^{\prime}|_{j}|\leq|u^{\prime}|+|v^{\prime}|$.
If for some $p^{\prime}\in A$, $|u_{1}|_{p^{\prime}}\neq|v_{1}|_{p^{\prime}}$,
then we get a contradiction by Theorem 4.1; else, it contradicts our
assumption that $|u^{\prime}|+|v^{\prime}|$ is minimum.

In case $j=i_{2}$ or $i_{3}$, we clearly have
$u^{\prime}|_{j}=v^{\prime}|_{j}$ and they are not identical with
$|u^{\prime}|_{j}|+|v^{\prime}|_{j}|<|u^{\prime}|+|v^{\prime}|$. If for some
$p^{\prime}\in A$, $|u|_{j}|_{p^{\prime}}\neq|v|_{j}|_{p^{\prime}}$, then we
get a contradiction by Theorem 4.1; else, it contradicts our assumption that
$|u^{\prime}|+|v^{\prime}|$ is minimum.

Hence, the semigroup $S$ generated by $A$ is free. ∎

Since the subsemigroup $S$ generated by $A$ in $G_{3}$ is free, we have the
following corollary.

######  Corollary 5.6.

The group $G_{3}$ is of exponential growth.

In $G_{3}$, for $k\geq 1$, consider the set $H_{k}=\\{w\in
G_{3}\;:\;|w|_{A}\equiv 0\ (\text{mod}\ 2^{k+1})\\}$. Clearly, $H_{k}$ is a
normal subgroup of $G_{3}$.

######  Lemma 5.7.

For every $g\in G_{3}$, there exists $h\in H_{k}$ and $i\in\mathds{N}$ with
$0\leq i<2^{k+1}$ such that $g=a^{i}h$.

###### Proof.

For $w\in G_{3}^{\prime}$, since $|w|_{p}=0$ for all $p\in A$, we have
$G_{3}^{\prime}\leq H_{k}$. Accordingly, for any $g\in G_{3}$, we have

$\displaystyle g$ $\displaystyle=$ $\displaystyle
a^{n_{1}}b^{n_{2}}c^{n_{3}}h_{1},\ \text{for some}\ h_{1}\in G_{3}^{\prime}$
$\displaystyle=$ $\displaystyle
a^{n_{1}}b^{n_{2}}c^{n_{3}}(c^{-n_{3}}b^{-n_{2}}a^{n_{2}+n_{3}})h_{2},\
\text{where}\ h_{2}=(c^{-n_{3}}b^{-n_{2}}a^{n_{2}+n_{3}})^{-1}h_{1}\in H_{k}$
$\displaystyle=$ $\displaystyle a^{(n_{1}+n_{2}+n_{3})}h_{2}$ $\displaystyle=$
$\displaystyle a^{i}h,\ \text{for some}\ h\in H_{k}\ \text{and}\ 0\leq
i<2^{k+1}\ \text{with}\ i\equiv n_{1}+n_{2}+n_{3}\ (\text{mod}\ 2^{k+1}).$

∎

######  Theorem 5.8.

For all $k\in\mathds{N}$,
$\psi(\textup{Rist}_{G_{3}}(\widehat{k}))=\stackrel{{\scriptstyle
3^{k}}}{{H_{k}\times\dots\times H_{k}}}$.

###### Proof.

We prove the statement by induction on $k$. For $k=1$, it is sufficient to
show that $\psi(\text{Rist}_{G_{3}}(1))=H_{1}\times\\{e\\}\times\\{e\\}$ so
that $\psi(\text{Rist}_{G_{3}}(\widehat{1}))=H_{1}\times H_{1}\times H_{1}$.

Let $\psi(\text{Rist}_{G_{3}}(1))=K\times\\{e\\}\times\\{e\\}$. Note that $K$
is a subgroup of $G_{3}$. We will show that $K=H_{1}$. Since $G_{3}$ is weakly
regular branch group over $G_{3}^{\prime}$, we have $G_{3}^{\prime}\leq K$.
Also, note that $c^{-1}a,(ab)^{2}\in K$ because
$c^{-1}b^{-1}ac\xi=(c^{-1}a,e,e)$ and $(a^{b}\xi^{2})^{2}=((ab)^{2},e,e)$.
Since $G_{3}^{\prime}\leq K$, clearly $a^{2}b^{2}\in K$. By symmetry, we can
observe that $a^{-1}b,b^{-1}c,b^{2}c^{2},a^{2}c^{2}\in K$ and hence
$a^{4},b^{4},c^{4}\in K$.

Let $w\in H_{1}$ and $|w|_{a}=n_{1},|w|_{b}=n_{2},|w|_{c}=n_{3}$. Thus,
$n_{1}+n_{2}+n_{3}=0\;(\text{mod}\ 4)$. Observe that

$\displaystyle w$ $\displaystyle=$ $\displaystyle
a^{n_{1}}b^{n_{2}}c^{n_{3}}h_{1},\ \text{for some}\ h_{1}\in G_{3}^{\prime}$
$\displaystyle=$ $\displaystyle a^{n_{1}}b^{n_{2}}c^{-n_{1}-n_{2}+4t}h_{1},\
\text{for some}\;t\in\mathds{Z}$ $\displaystyle=$ $\displaystyle
c^{-n_{1}}a^{n_{1}}c^{-n_{2}}b^{n_{2}}h_{2},\ \text{for some}\ h_{2}\in
K,\text{as}\;G_{3}^{\prime}\leq K\;\text{and}\;c^{4}\in K$ $\displaystyle=$
$\displaystyle(c^{-1}a)^{n_{1}}(c^{-1}b)^{n_{2}}h_{3},\ \text{for some}\
h_{3}\in K.$

Hence, since $c^{-1}a$ and $b^{-1}c$ are in $K$, we have $w\in K$ so that
$H_{1}\subseteq K$.

For reverse inequality, first we observe that $a,a^{2},a^{3}\notin K$. If
$a,a^{3}\in K$, then $(a,e,e),(a^{3},e,e)\in G_{3}$, which are not possible by
Remark 5.1. Further, if $a^{2}\in K$, then
$h=(a^{2},e,e)\in\text{St}_{G_{3}}(\widehat{1})$. However, by Remark 5.1, we
get $|h|_{A}=1$, which is a contradiction to Remark 5.3.

Now, let $w\notin H_{1}$ and $|w|_{a}=m_{1},|w|_{b}=m_{2},|w|_{c}=m_{3}$.
Thus, $m_{1}+m_{2}+m_{3}=s+4t$, for some $t\in\mathds{Z}$ and $0<s\leq 3$. If
possible, suppose $w\in K$. Then

$\displaystyle w$ $\displaystyle=$ $\displaystyle
a^{m_{1}}b^{m_{2}}c^{m_{3}}g_{1},\ \text{for some}\ g_{1}\in G_{3}^{\prime}$
$\displaystyle=$ $\displaystyle a^{m_{1}}b^{m_{2}+m_{3}}g_{2},\ \text{where }\
g_{2}=(c^{-m_{3}}b^{m_{3}})^{-1}g_{1}\in K$ $\displaystyle=$ $\displaystyle
a^{m_{1}+m_{2}+m_{3}}g_{3},\ \text{where }\
g_{3}=(b^{-(m_{2}+m_{3})}a^{m_{2}+m_{3}})^{-1}g_{2}\in K$ $\displaystyle=$
$\displaystyle a^{s}g_{4},\ \text{where}\ g_{4}=a^{4t}g_{3}\in K.$

Hence, $a^{s}\in K$, which is a contradiction. Therefore, $K\subseteq H_{1}$
so that $\text{Rist}_{G_{3}}(1)=H_{1}\times\\{e\\}\times\\{e\\}$.

For $k\geq 1$, suppose
$\textup{Rist}_{G_{3}}(\widehat{k})=\stackrel{{\scriptstyle
3^{k}}}{{H_{k}\times\dots\times H_{k}}}$. Accordingly,
$\textup{Rist}_{G_{3}}(\underset{\text{length }k}{11\ldots
1})=H_{k}\times\\{e\\}\stackrel{{\scriptstyle
3^{k}-1}}{{\times\dots\times}}\\{e\\}$. We show that
$\textup{Rist}_{G_{3}}(\underset{k+1}{11\ldots
1})=H_{k+1}\times\\{e\\}\stackrel{{\scriptstyle
3^{k+1}-1}}{{\times\dots\times}}\\{e\\}$.

Let $g\in\textup{Rist}_{G_{3}}(\underset{\text{length }k+1}{11\ldots 1})$.
Clearly, $g\in\textup{Rist}_{G_{3}}(\underset{\text{length }k}{11\ldots 1})$.
Then there exist $g^{\prime},g^{\prime\prime}\in G_{3}$ such that
$\psi_{k+1}(g)=(g^{\prime},e,\ldots,e)$ and
$\psi_{k}(g)=(g^{\prime\prime},e,\ldots,e)$ with
$\psi_{1}(g^{\prime\prime})=(g^{\prime},e,e)$. Hence, by inductive hypothesis,
we have $g^{\prime\prime}\in H_{k}$ so that $|g^{\prime\prime}|_{A}\equiv 0\
(\textup{mod}\ 2^{k+1})$. Since $\psi(g^{\prime\prime})=(g^{\prime},e,e)$, by
Remark 3.2(i), we have $|g^{\prime}|_{A}=2|g^{\prime\prime}|_{A}$ so that
$|g^{\prime}|_{A}\equiv 0\ (\textup{mod}\ 2^{k+2})$. Hence, $g^{\prime}\in
H_{k+1}$ so that $\textup{Rist}_{G_{3}}(\underset{\text{length }k+1}{11\ldots
1})\subseteq H_{k+1}\times\\{e\\}\stackrel{{\scriptstyle
3^{k+1}-1}}{{\times\dots\times}}\\{e\\}$.

Conversely, suppose $h\in H_{k+1}$. Clearly, $h\in H_{1}$ so that there exists
$h^{\prime}\in\textup{Rist}_{G_{3}}(1)$ such that $\psi(h^{\prime})=(h,e,e)$.
Since $|h|_{A}\equiv 0\ (\textup{mod}\ 2^{k+2})$, again by Remark 3.2(i),
$|h^{\prime}|_{A}\equiv 0\ (\textup{mod}\ 2^{k+1})$, i.e., $h^{\prime}\in
H_{k}$. Thus, by inductive hypothesis, there exists
$g\in\textup{Rist}_{G_{3}}(\underset{\text{length }k}{11\ldots 1})$ such that
$\psi_{k}(g)=(h^{\prime},e,\ldots,e)$ and hence
$\psi_{k+1}(g)=(h,e,\ldots,e)$. Hence,
$\textup{Rist}_{G_{3}}(\underset{\text{length }k+1}{11\ldots
1})=H_{k+1}\times\stackrel{{\scriptstyle
3^{k+1}-1}}{{\\{e\\}\times\dots\times\\{e\\}}}$. Consequently, we have
$\textup{Rist}_{G_{3}}(\widehat{k+1})=\stackrel{{\scriptstyle
3^{k+1}}}{{H_{k+1}\times\dots\times H_{k+1}}}$. ∎

######  Theorem 5.9.

The group $G_{3}$ is a branch group.

###### Proof.

We have to show that the subgroup $\textup{Rist}_{G_{3}}(\widehat{k})$ is of
finite index in $G_{3}$, for all $k\in\mathds{N}$. Since
$\textup{St}_{G_{3}}(\widehat{k})$ is of finite index in $G_{3}$, we show that
$\textup{Rist}_{G_{3}}(\widehat{k})$ is of finite index in
$\textup{St}_{G_{3}}(\widehat{k})$, for all $k\in\mathds{N}$.

For $g\in\textup{St}_{G_{3}}(\widehat{k})$, let
$\psi_{k}(g)=(g_{1},\ldots,g_{3^{k}})$. By Lemma 5.7, for every $g_{i}$ there
exist $h_{i}\in H_{k}$ and $j_{i}\in\mathds{N}$ with $0\leq j_{i}<2^{k+1}$
such that $g_{i}=a^{j_{i}}h_{i}$. Note that
$(h_{1},\ldots,h_{3^{k}})\in\textup{Rist}_{G_{3}}(\widehat{k})$. Since
$(a^{j_{1}}h_{1},\ldots,a^{j_{3^{k}}}h_{3^{k}})$ and
$(h_{1}^{-1},\ldots,h_{3^{k}}^{-1})$ are in
$\textup{St}_{G_{3}}(\widehat{k})$, we have $(a^{j_{1}},\ldots
a^{j_{3^{k}}})\in\textup{St}_{G_{3}}(\widehat{k})$. Hence, for every
$g\in\textup{St}_{G_{3}}(\widehat{k})$, $\psi_{k}(g)=(a^{j_{1}},\ldots
a^{j_{3^{k}}})(h_{1},\ldots,h_{3^{k}})$ so that the coset representatives for
the quotient group
$\textup{St}_{G_{3}}(\widehat{k})/\textup{Rist}_{G_{3}}(\widehat{k})$ are from

$\\{(a^{i_{1}},\ldots,a^{i_{3^{k}}})\in\textup{St}_{G_{3}}(\widehat{k}):i_{1},\ldots,i_{3^{k}}\in\\{0,\ldots,2^{k+1}-1\\}\\}.$

Hence, for every $k\in\mathds{N}$, the index of
$\textup{Rist}_{G_{3}}(\widehat{k})$ in $\textup{St}_{G_{3}}(\widehat{k})$ is
bounded by the number ${(2^{k+1})}^{3^{k}}$. ∎

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