2. Preliminaries

By $\mathbb{N}$ we denote the set of positive integers, and by $\mathbb{N}_0$ the set of non-negative integers.

Let $m\in \mathbb{N}_{\ge 2}$ and let $T = T_m$ be the $m$ -adic tree, that is, a rooted tree where all vertices have $m$ children. Using the alphabet $X = \{0,1,\ldots,m-1\}$ , the vertices $u_\omega$ of $T$ are labelled bijectively by the elements $\omega$ of the free monoid $X^*$ in the following natural way: the root of $T$ is labelled by the empty word and is denoted by $\epsilon$ , and for each word $\omega \in X^*$ and letter $x \in X$ , there is an edge connecting $u_\omega$ to $u_{\omega x}$ . More generally, we say that $u_\omega$ precedes $u_\lambda$ whenever $\omega$ is a prefix of $\lambda$ .

There is a natural length function on $X^*$ , which is defined as follows: the words $\omega$ of length $\lvert \omega \rvert = n$ , representing vertices $u_\omega$ that are at distance $n$ from the root, are the $n$ th level vertices and constitute the $n$ th layer of the tree.

We denote by $T_u$ the full rooted subtree of $T$ that has its root at a vertex $u$ and includes all vertices succeeding $u$ . For any two vertices $u = u_\omega$ and $v = u_\lambda$ , the map $u_{\omega \tau } \mapsto u_{\lambda \tau }$ , induced by replacing the prefix $\omega$ by $\lambda$ , yields an isomorphism between the subtrees $T_u$ and $T_v$ .

Now each $f\in \textrm{Aut}\, T$ fixes the root, and the orbits of $\textrm{Aut}\, T$ on the vertices of the tree $T$ are the layers of the tree $T$ . The image of a vertex $u$ under $f$ will be denoted by $uf$ . The automorphism $f$ induces a faithful action on $X^*$ given by $(u_\omega )f = u_{\omega f}$ . For $\omega \in X^*$ and $x \in X$ , we have $(\omega x)f = (\omega f) x^{\prime}$ , for $x^{\prime} \in X$ uniquely determined by $\omega$ and $f$ . This induces a permutation $f^\omega$ of $X$ which satisfies

\begin{equation*} (\omega x) f= (\omega f)xf^\omega, \quad \text {and consequently} \quad (u_{\omega x})f = u_{(\omega f) xf^\omega }. \end{equation*}

More generally, for an automorphism $f$ of $T$ , since the layers are invariant under $f$ , for $u\in X^*$ , the equation

\begin{equation*} (uv)f=(uf)vf_u \qquad {\text {for every }v\in X^*,} \end{equation*}

defines a unique automorphism $f_u$ of $T$ called the section of $f$ at $u$ . This automorphism can be viewed as the automorphism of $T$ induced by $f$ upon identifying the rooted subtrees of $T$ at the vertices $u$ and $uf$ with the tree $T$ . As seen here, we often do not differentiate between $X^*$ and vertices of $T$ .

2.1. Subgroups of $\textrm{Aut}\, T$

Let $G$ be a subgroup of $\textrm{Aut}\, T$ acting level-transitively, that is, transitively on every layer of $T$ . The vertex stabiliser $\textrm{st}_G(u)$ is the subgroup consisting of elements in $G$ that fix the vertex $u$ . For $n \in \mathbb{N}$ , the $n$ th level stabiliser $\textrm{St}_G(n)= \bigcap _{\lvert \omega \rvert =n} \textrm{st}_G(u_\omega )$ is the subgroup consisting of automorphisms that fix all vertices at level $n$ .

Each $g\in \textrm{St}_{\textrm{Aut}\, T} (n)$ can be completely determined in terms of its restrictions to the subtrees rooted at vertices at level $n$ . There is a natural isomorphism

\begin{equation*} \psi _n \colon \textrm {St}_{\textrm {Aut} T}(n) \longrightarrow \prod \nolimits _{\lvert \omega \rvert = n} \textrm {Aut} T_{u_\omega } \cong \textrm {Aut} T \times \overset {m^n}{\cdots } \times \textrm {Aut} T \end{equation*}

defined by sending $g\in \textrm{St}_{\textrm{Aut}\, T} (n)$ to its tuple of $m^n$ sections. For conciseness, we will omit the use of $\psi _1$ , and simply write $g=(g_1,\ldots,g_m)$ for $g\in \text{St}_{\textrm{Aut}\, T}(1)$ .

Let $\omega \in X^n$ be of length $n$ . We further define

\begin{equation*} \varphi _\omega :\textrm {st}_{\textrm {Aut} T}(u_\omega ) \longrightarrow \textrm {Aut} T_{u_\omega } \cong \textrm {Aut} T \end{equation*}

to be the map sending $f\in \textrm{st}_{\textrm{Aut}\, T}(u_\omega )$ to the section $f_{u_\omega }$ .

A group $G \leq \textrm{Aut}\, T$ is said to be self-similar if for all $f\in G$ and all $\omega \in X^*$ the section $f_{u_{\omega }}$ belongs to $G$ . We will denote $G_{\omega }$ to be the subgroup $\varphi _{\omega }(\textrm{st}_G(u_{\omega }))$ .

Recall that for a group $G$ generated by a finite symmetric subset $S$ (i.e. a set for which $S = S^{-1}$ ), for every $g \in G$ , the length of $g$ with respect to $S$ is

\begin{equation*} |g| = \min \{n \geq 0 \mid g = s_{1} \cdots s_{n}, \text { for }s_{1}, \dots, s_{n} \in S\}. \end{equation*}

Now assume that $G$ is a self-similar subgroup of $\textrm{Aut}\, T$ . Then for every $g\in G$ and every $n\in \mathbb{N}_0\cup \{0\}$ , let

\begin{equation*} \ell _n(g) = \max \{ |g_u| \mid |u|=n \}. \end{equation*}

If there exist $\lambda \lt 1$ and $C,L\in \mathbb{N}$ such that

\begin{equation*} \ell _n(g) \le \lambda |g|+C, \quad \text {for every $n\gt L$ and every $g\in G$,} \end{equation*}

then we say that the group $G$ is contracting with respect to $S$ .

Let $G$ be a subgroup of $\textrm{Aut}\, T$ acting level-transitively. Here, the vertex stabilisers at every level are conjugate under $G$ . We say that the group $G$ is fractal if $G_{\omega }\,{:\!=}\,\varphi _\omega (\textrm{st}_G(u_\omega ))=G$ for every $\omega \in X^*$ , after the natural identification of subtrees. Furthermore, we say that the group $G$ is super strongly fractal if, for each $n\in \mathbb{N}$ , we have $\varphi _\omega (\text{St}_G(n))=G$ for every word $\omega \in X^n$ of length $n$ .

The rigid vertex stabiliser of $u$ in $G$ is the subgroup $\textrm{rist}_G(u)$ consisting of all automorphisms in $G$ that fix all vertices of $T$ not succeeding $u$ . The rigid $n$ th level stabiliser is the direct product of the rigid vertex stabilisers of the vertices at level $n$ :

\begin{equation*} \textrm {Rist}_G(n) = \prod \nolimits _{\lvert \omega \rvert = n} \textrm {rist}_G(u_\omega ) \trianglelefteq G. \end{equation*}

We recall that a level-transitive group $G$ is a branch group if $\textrm{Rist}_G(n)$ has finite index in $G$ for every $n \in \mathbb{N}$ ; and $G$ is weakly branch if $\textrm{Rist}_G(n)$ is non-trivial for every $n \in \mathbb{N}$ . If, in addition, the group $G$ is self-similar and there exists a subgroup $1 \ne K \leq G$ with $K\times \overset{m}\cdots \times K \subseteq{\psi _1}(\textrm{St}_K(1))$ and $\lvert G \,{:}\, K \rvert \lt \infty$ , then $G$ is said to be regular branch over $K$ . If in the previous definition the condition $\lvert G\,{:}\, K \rvert \lt \infty$ is omitted, then $G$ is said to be weakly regular branch over $K$ .

2.2. A basic result

Here we record a general result that will be useful in the sequel. For $g\in \textrm{Aut}\, T$ , recall that $g^\epsilon$ denotes the action induced by $g$ at the root of $T$ .

Lemma 2.1. For a self-similar group $G\le \textrm{Aut}\, T$ , let $z = (z_0,\dots,z_{m-1})z^\epsilon \in G^{\prime}$ . Then $z_0\cdots z_{m-1}\in G^{\prime}$ .

Proof. It suffices to prove the result for a basic commutator $[g,h]$ , where $g,h\in G$ . Write $g=(g_0,\ldots,g_{m-1})g^\epsilon$ and $h=(h_0,\ldots,h_{m-1})h^\epsilon$ . For notational convenience, let us write $\tau =(g^\epsilon )^{-1}$ and $\kappa =(h^\epsilon )^{-1}$ , and for $\alpha \in \text{Sym}(X)$ and $x\in X$ we write $x^\alpha$ for the image $\alpha (x)$ of $x$ under $\alpha$ . As

\begin{align*} &[g,h]\\ &= \tau (g_0^{-1},\dots,g_{m-1}^{-1}) \kappa (h_0^{-1},\dots,h_{m-1}^{-1}) (g_0,\dots,g_{m-1})g^\epsilon (h_0,\dots,h_{m-1})h^\epsilon \\ & = (g_{0^\tau }^{-1},\dots,g_{(m-1)^\tau }^{-1}) (h_{0^{\tau \kappa }}^{-1},\dots,h_{(m-1)^{\tau \kappa }}^{-1}) (g_{0^{\tau \kappa }},\dots,g_{(m-1)^{\tau \kappa }})(h_{0^{\tau \kappa{g^\epsilon }}},\dots,h_{(m-1)^{\tau \kappa{g^\epsilon }}})\tau \kappa{g^\epsilon h^\epsilon }, \end{align*}

the result follows.

3. Properties of $\mathfrak{K(v)}$

For any two integers $i,j,$ let $[i, j]$ denote the set $\{i,i+1,\dots,j-1,j\}$ . In the following sections, we fix $m,s\in \mathbb{N}_{\geq 2}$ . Let $X = \{0,1,\dots,m-1\}$ and let ${\mathfrak{v}}= x_0\cdots x_{s-2}$ be a word in $X^*$ . Recall that the group $\mathfrak{K(v)}$ is generated by the elements $a_0,\dots,a_{s-1}$ , where

\begin{align*} a_0 = (1,\dots,1,a_{s-1}) \sigma,{\qquad \text{and}\qquad a_{i+1} = (1,\overset{x_i-1}{\ldots },1, a_{i}, 1,\ldots,1),} \end{align*}

for $i \in{[0, s-2]}.$

To prove Theorem 1.1(i), we use the following characterisation of iterated monodromy groups by Nekrashevych [Reference Nekrashevych21, Theorem 6.10.8] according to the notation given in [Reference Jones18, Theorem 5.8]:

We refer the reader to [Reference Nekrashevych21] or [Reference Jones18] for any unexplained terminology.

Proof of Theorem 1.1(i). We will show that the above four conditions hold for $\mathfrak{K(v)}$ . Conditions (1) and (2) of Theorem 3.1 are clear from the definition of the generators $a_i$ . For condition (3), notice that $a_0$ is the only generator of $\mathfrak{K(v)}$ that acts non-trivially on $X$ . Therefore, according to [Reference Jones18, Section 5.4], the multi-set of permutations of $X$ is given by $\{\sigma \}$ . Furthermore, the cycle diagram associated with the multi-set $\{\sigma \}$ is an oriented 2-dimensional CW complex, whose set of 0-cells is $X$ , and with one 2-cell obtained by connecting the elements of $X$ by the action of $\sigma$ . Hence, the cycle diagram is contractible. Thus, by [Reference Jones18, Definition 5.7] the multi-set of permutations defined by the set of states of $\mathfrak{K(v)}$ acting on $X$ is tree-like.

For condition (4), the restriction $(a_i)_{v_i}=a_i$ implies that the $v_i$ have to be at level $ns$ for some positive integer $n$ . However, for any $a_i$ , after $i$ levels, the section is $a_0$ , which does not then fix any vertex below it. So we can never find vertices $v_i$ such that $a_i(v_i)=v_i$ . So (4) vacuously holds.

Proof. Observe that $a_0^m=(a_{s-1},\dots,a_{s-1})$ and, for any $i \in{[1, s-1]}$ and $j \in{[0, m-1]}$ ,

\begin{align*}{\varphi _{j}\bigg (}a_i^{a_0^{j-x_{i-1}}}{\bigg )}=a_{i-1}. \end{align*}

It follows that ${\varphi _x(\textrm{St}}_{\mathfrak{K(v)}}(1))=\mathfrak{K(v)}$ for all $x \in X$ . For $n\in \mathbb{N}$ , we write $n=\ell s+r$ , where $\ell \in \mathbb{N}_0$ and $r\in [0,s-1]$ . Using similar arguments with the elements

\begin{equation*} a_0^{m^{\ell +1}}, a_1^{m^{\ell +1}}, \ldots, a_{r-1}^{m^{\ell +1}}, a_{r}^{m^{\ell }},\ldots, a_{s-1}^{m^{\ell }}\in \textrm {St}_{\mathfrak {K(v)}}(n), \end{equation*}

and their conjugates, we deduce that ${\varphi _w(\textrm{St}}_{\mathfrak{K(v)}}(n))=\mathfrak{K(v)}$ for all $w \in X^n$ .

The second statement is immediate from the fact that ${\varphi _x(\textrm{st}}_{\mathfrak{K(v)}}(x))=\mathfrak{K(v)}$ for some $x \in X$ , and since $a_0$ , and hence, $\mathfrak{K(v)}$ , acts transitively on the first layer.

Next, we record an elementary but useful result. Recall that ${\mathfrak{v}} = x_0\cdots x_{s-2}$ is a word in $X^*$ .

Proof. The first statement is a straightforward computation. For the next statement, note that for $i\in [1,s-1]$ ,

\begin{equation*} [a_i,a_0]=\begin {cases} (a_{s-1}^{-1}a_{i-1}a_{s-1},1,\ldots,1,a_{i-1}^{-1})&\text {if }x_{i-1}= m-1,\\ (1,\overset {x_{i-1}-1}\ldots,1,a_{i-1}^{-1},a_{i-1},1,\ldots,1)&\text {if }x_{i-1}\ne m-1, \end {cases} \end{equation*}

so $a_0$ does not commute with any other $a_i$ . The result then follows, using the fact that

\begin{equation*} [a_i,a_\ell ]=(1,\ldots,1,[a_{i-1},a_{\ell -1}],1,\ldots,1). \end{equation*}

Proof. We use that $\mathfrak{K(v)}^{\prime} = \langle [a_i,a_j] \mid i,j \in{[0, s-1]}\rangle ^{\mathfrak{K(v)}}$ . We have

\begin{align*} \Big [a_i^{a_0^{m-1-x_{i-1}}}, a_j^{a_0^{m-1-x_{j-1}}}\Big ] &= (1,\dots,1,[a_{i-1},a_{j-1}]),\\ [a_i, a_0^m]^{a_0^{m-1-x_{i-1}}} &= (1,\dots,1,[a_{i-1}, a_{s-1}]), \end{align*}

for $i,j\in{[1, s-1]}.$ Therefore, by transitivity and fractalness, we have $\mathfrak{K(v)}^{\prime} \times \overset{m}\cdots \times \mathfrak{K(v)}^{\prime} \leq \psi (\mathfrak{K(v)}^{\prime})$ .

For $i \in{[0, s-1]}$ , let $A_i = \langle a_j\mid{j\in [0,s-1]\backslash \{i\}} \rangle ^{\mathfrak{K(v)}}$ .

Proof. We prove simultaneously for all $i$ that $\mathfrak{K(v)}/A_i \cong \mathbb{Z}.$ Assume for a contradiction that, for some $n \in \mathbb{N}$ that $a_i^n \in A_i$ , for some $i \in{[0, s-1]}.$ Choose $n \in \mathbb{N}$ minimal with respect to the property that $a_i^n \in A_i$ for some $i \in{[0, s-1]}$ . If $i =0$ , then $a_{0}^n \in \textrm{St}_{\mathfrak{K(v)}}(1)$ and, in particular $n \equiv 0{\pmod m}$ . We have

\begin{align*} a_0^n = (a_{s-1}^{{n}/{m}}, \dots, a_{s-1}^{{n}/{m}}), \end{align*}

and hence, $a_{s-1}^{{n}/{m}} \in A_{s-1}.$ This is a contradiction to the minimality of $n$ . Now suppose that $i \neq 0$ . Then by considering appropriate sections of $a_i^n$ , we see that $a_0^n \in A_0$ , which cannot happen as shown above. Therefore, we conclude that ${\mathfrak{K(v)}}/A_i \cong \mathbb{Z}$ for all $i \in{[0, s-1]}$ .

Proof. Analogous to the proof of [Reference Di Domenico, Fernández-Alcober, Noce and Thillaisundaram8, Theorem 3.5(i)], we clearly have $A_0\le \textrm{Rist}_{\mathfrak{K(v)}}(1)\le \textrm{St}_{\mathfrak{K(v)}}(1)=A_0\langle a_0^m \rangle$ , where the last equality follows from the fact that $a_i\in \textrm{St}_{\mathfrak{K(v)}}(1)$ for all $i\ne 0$ , and that $a_0^n\in \textrm{St}_{ \mathfrak{K(v)}}(1)$ if and only if $n\equiv 0\pmod m$ . So $\textrm{Rist}_{\mathfrak{K(v)}}(1)=A_0\langle a_0^{mn} \rangle$ for some $n$ . Since

\begin{equation*} \psi (A_0\langle a_0^{mn} \rangle )=(A_{s-1}\times \cdots \times A_{s-1})\langle (a_{s-1}^n,\ldots,a_{s-1}^n) \rangle \end{equation*}

and $a_{s-1}$ has infinite order modulo $A_{s-1}$ by Lemma 3.5, it follows from the definition of the rigid stabiliser that $n=0$ . Hence, $\textrm{Rist}_{\mathfrak{K(v)}}(1)=A_0$ , which has infinite index in $\mathfrak{K(v)}$ , so $\mathfrak{K(v)}$ is not branch.

We shall adopt the convention that the subscripts of the $a_i$ ’s are taken modulo $s$ . Set $S = \{a_i^{\pm 1} \mid i \in{[0, s-1]}\}$ and then $\mathfrak{K(v)}= \langle S \rangle$ . For each word $w \in S^*$ , the length $|w|$ is the usual word length of $w$ over the alphabet $S$ .

Proof. We proceed by induction on the length of $w$ . Let $w$ be a non-trivial word in $S$ representing the identity in $\mathfrak{K(v)}$ . It is immediate from Lemma 3.5 that $|w| \geq 2$ and $w$ must contain non-trivial powers of at least two distinct $a_i,a_j \in S$ . For the base case $|w|=2$ , we have that $w=a_i^{\epsilon _i}a_j^{\epsilon _j}$ for distinct $i,j\in [0,s-1]$ and $\epsilon _i,\epsilon _j\in \{\pm 1\}$ . From considering the image of $w$ in $\mathfrak{K(v)}/A_i$ , we obtain a contradiction to $\epsilon _i$ being non-zero.

Assume that the result holds for all words of length $n \geq 2$ , and let $|w|=n+1$ . By realising $w$ in $\mathfrak{K(v)}$ , we can see that the exponent sum of $a_0$ in $w$ must be zero modulo $m$ . By abuse of notation, we write

\begin{equation*}\psi (w) = (w_0,\dots,w_{m-1}),\end{equation*}

where $w_i$ are reduced words determined by the appropriate sections of the letters of $w$ . Since $w$ represents the identity in $\mathfrak{K(v)}$ , each $w_k$ also represents the identity in $\mathfrak{K(v)}$ . It follows from the definition of the generators $a_i$ that $\sum \limits _{k=0}^{m-1}|w_{k}| \leq |w|$ . In particular, if $|w_{k}| = |w|$ , for some $k \in{[0, m-1]}$ , then the $w_j$ are trivial for all $j \in{[0, m-1]}\backslash \{k\}$ , otherwise $|w_{k}| \lt |w|$ for all ${k} \in{[0, m-1]}$ .

Assume that $|w_{k}| \lt |w|$ for all $k \in{[0, m-1]}$ . By the induction hypothesis, for all $i \in{[0, s-1]}$ the exponent sum of $a_i$ is zero in $w_{k}$ for all ${k} \in{[0, m-1]}$ . Since the $a_i$ in $w_{k}$ are obtained from $a_{i+1}$ in $w$ , we conclude that the exponent sum of $a_{i+1}$ in $w$ is zero for all $i \in{[0, s-1]}$ .

Now, assume that there exists $k \in{[0, m-1]}$ such that $|w_k| = |w|$ . By replacing $w$ with $w_k$ , repeatedly if necessary, we may assume that $w$ does involve the letters $a_0^{\pm 1}$ . Then, if $w$ has $a_0^{\pm 1}$ as a letter, then the total exponent of $a_0$ is congruent to 0 modulo $m$ . Suppose first that the total exponent of $a_0$ is non-zero. Since the total exponent of $a_0$ is congruent to $0$ modulo $m$ , we can find a subword $a_0^{\pm 1}w^{\prime} a_0^{\pm 1}$ such that the exponent sum of $a_0$ in $w^{\prime}$ is zero. By realising this subword in $\mathfrak{K(v)}$ , we immediately see a length reduction among the sections, and it follows by induction that every section $w_k$ has exponent sum of $a_i$ equal to zero for all $i\in [0,s-1]$ . So in particular, the case that total exponent of $a_0$ in $w$ is non-zero cannot occur. So assume that the total exponent of $a_0$ is zero. Upon considering each $w_k$ , we may assume that the total exponent of $a_0$ in $w_k$ is zero, else we are done by the above argument. Hence, the total exponent of $a_1$ in $w$ is zero. Recursively, we deduce that the total exponent in $w$ of any $a_i$ is zero. The result then follows.

We proceed to prove parts (iv) and (v) of Theorem 1.1.

Proof. For every $g \in \mathfrak{K(v)}$ , there exist integers $n_0,\dots,n_{s-1}$ such that $g \equiv a_0^{n_0}\cdots a_{s-1}^{n_{s-1}}$ modulo $\mathfrak{K(v)}^{\prime}$ . It is enough to show that, if $a_0^{n_0}\cdots a_{s-1}^{n_{s-1}} \in \mathfrak{K(v)}^{\prime}$ , for some $n_0,\dots,n_{s-1} \in \mathbb{Z}$ , then $n_0=\cdots =n_{s-1} = 0$ .

Assume that $a_0^{n_0}\cdots a_{s-1}^{n_{s-1}} \in \mathfrak{K(v)}^{\prime}$ for some $n_0,\dots,n_{s-1} \in \mathbb{Z}$ . It is clear that every element in $\mathfrak{K(v)}^{\prime}$ can be written as a word in $S$ and the exponent sum of each $a_i$ is zero. It follows from Lemma 3.7 that the exponent sums of the $a_i$ ’s in all words representing the same element in $\mathfrak{K(v)}$ are the same. This, in particular, applies to $a_0^{n_0}\cdots a_{s-1}^{n_{s-1}}$ , and hence, $n_0 = \cdots = n_{s-1} = 0$ .

Proof. It follows immediately from [Reference Di Domenico, Fernández-Alcober, Noce and Thillaisundaram8, Theorem 3.6] that $\mathfrak{K(v)}$ is torsion-free as the quotient group $\mathfrak{K(v)}/\mathfrak{K(v)}^{\prime}$ is torsion-free and the subgroup $\mathfrak{K(v)}^{\prime}$ is contained in $\textrm{St}_{\mathfrak{K(v)}}(1)$ .

Next, we finish the proof of Theorem 1.1. Recall that for $g\in \mathfrak{K(v)}$ , the $n$ th level sections $g_0,\ldots,g_{m^n-1}$ of $g$ are determined from expressing $g$ as the product

\begin{equation*} g=\psi _n^{-1}((g_0,\ldots,g_{m^n-1}))\tau _g, \end{equation*}

where $\psi _n^{-1}((g_0,\ldots,g_{m^n-1}))\in \textrm{St}_{\mathfrak{K(v)}}(n)$ and $\tau _g\in \textrm{Sym}(X^n)$ .

Proof. By [Reference Di Domenico, Fernández-Alcober, Noce and Thillaisundaram8, Lemma 3.7], it suffices to prove that

(3.1) \begin{equation} \ell _s(g)\le \frac{2}{3}|g|+1, \quad \text{for every $g\in \mathfrak{K(v)}$.} \end{equation}

We can further reduce the problem to showing that $\ell _s(h)\le 2$ for every $h\in \mathfrak{K(v)}$ of length $3$ . Indeed, writing $|g|=3\alpha +\beta$ with $\beta \in \{0,1,2\}$ and $g=h_1\cdots h_\alpha f$ with $|h_1|=\cdots =|h_\alpha |=3$ and $|f|=\beta$ , we see that (3.1) immediately follows from the subadditivity of $\ell _s$ ; cf. [Reference Di Domenico, Fernández-Alcober, Noce and Thillaisundaram8, Equation (3.3)].

Let us then consider an arbitrary element $h\in \mathfrak{K(v)}$ of length $3$ and prove that $\ell _s(h)\le 2$ . Observe that $h$ is of the form

1. (i) $a_i^{\epsilon _i}a_j^{\epsilon _j}a_k^{\epsilon _k}$ for $i\ne j$ and $j\ne k$ ; or

2. (ii) $a_i^{2\epsilon _i}a_j^{\epsilon _j}$ or $a_i^{\epsilon _i}a_j^{2\epsilon _j}$ for $i\ne j$ ; or

3. (iii) $a_i^{3\epsilon _i}$ ,

where $\epsilon _i, \epsilon _j, \epsilon _k\in \{\pm 1\}$ . For case (iii), it is clear that $\ell _{i+1}(h)\le 2$ , since the only non-trivial $i$ th level section is $a_0^{3\epsilon _i}$ . Similarly for case (ii).

For case (i), suppose for a contradiction that $\ell _n(h)=3$ for all $n\in \mathbb{N}$ . Then note that for every $n$ , the element $h$ has exactly one non-trivial $n$ th level section, which is $a_{i-n}^{\epsilon _i}a_{j-n}^{\epsilon _j}a_{k-n}^{\epsilon _k}$ . As usual, the indices of the generators are viewed modulo $s$ , so we effectively only consider the first $(s-1)$ st level sections.

Consider now the non-trivial $i$ th level section, which is $a_0^{\epsilon _i}a_{j-i}^{\epsilon _j}a_{k-i}^{\epsilon _k}$ . Suppose $k=i$ . If $\epsilon _i=\epsilon _k$ , then clearly the $(i+1)$ st sections decrease in length. So suppose $\epsilon _i\ne \epsilon _k$ . Upon considering the only non-trivial $j$ th level section $a_{i-j}^{\epsilon _i}a_{0}^{\epsilon _j}a_{i-j}^{\epsilon _k}$ , we see that $\ell _{j+1}(h)\le 2$ . So we suppose that $k\ne i$ . Consider first the case $\epsilon _i=1$ . Recall that the non-trivial $i$ th level section of $h$ is $a_0a_{j-i}^{\epsilon _j}a_{k-i}^{\epsilon _k}$ . In order for $\ell _{i+1}(h)$ to still be 3, we need

\begin{equation*} a_{j-i}=(a_{j-i-1},1,\ldots,1)\qquad \text {and}\qquad a_{k-i}=(a_{k-i-1},1,\ldots,1). \end{equation*}

If the defining word $\mathfrak{v}$ of the group $\mathfrak{K(v)}$ does not correspond to the above generators, then this implies that our assumption for this case (i) does not hold, and hence, the section lengths eventually decrease, giving us $\ell _s(h)\le 2$ . So we proceed by assuming that $\mathfrak{v}$ tallies with the above generators and with the other generators singled out below.

Next, we similarly see that in order for $\ell _{s+2i-j}(h)$ to still be 3, we need $a_{j-i}a_{2j-2i}^{\epsilon _j}a_{j+k-2i}^{\epsilon _k}$ to have exactly one non-trivial section. Thus,

\begin{equation*} a_{2j-2i}=(a_{2j-2i-1},1,\ldots,1)\qquad \text {and}\qquad a_{j+k-2i}=(a_{j+k-2i-1},1,\ldots,1). \end{equation*}

In particular, proceeding in this manner, we may assume that

\begin{equation*} a_{\theta (j-i)}=(a_{\theta (j-i)-1},1,\ldots,1)\end{equation*}

for all $\theta \in \mathbb{N}$ . Now let $\theta =s-1$ , and so $a_{i-j}=(a_{i-j-1},1,\ldots,1)$ . Observe that the only non-trivial $j$ th level section of $h$ is $a_{i-j}a_0^{\epsilon _j}a_{k-j}^{\epsilon _k}$ . To avoid a decrease in length at the next level, we must have $\epsilon _j=-1$ and $a_{k-j}=(1,\ldots,1,a_{k-j-1})$ . Similarly, for $\ell _{s+2j-k}(h)$ to still be 3, we need $a_{k-2j+i}a_{k-j}^{-1}a_{2(k-j)}^{\epsilon _k}$ to have exactly one non-trivial section. Thus,

\begin{equation*} a_{k-2j+i}=(1,\ldots,1,a_{k-2j+i-1})\qquad \text {and}\qquad a_{2(k-j)}=(1,\ldots,1,a_{2(k-j)-1}). \end{equation*}

As deduced above, since $a_{k-i}=(a_{k-i-1},1,\ldots,1)$ we also then have that

\begin{equation*} a_{\theta (k-i)}=(a_{\theta (k-i)-1},1,\ldots,1)\qquad \text {and}\qquad a_{\theta (k-j)}=(1,\ldots,1,a_{\theta (k-j)-1}) \end{equation*}

for all $\theta \in \mathbb{N}$ . As before, we let $\theta =s-1$ , and consider the only non-trivial $k$ th level section of $h$ , which is $a_{i-k}a_{j-k}^{-1}a_{0}^{\epsilon _k}$ . Since

\begin{equation*} a_{(s-1)(k-i)}=a_{i-k}=(a_{i-k-1},1,\ldots,1)\qquad \text {and}\qquad a_{(s-1)(k-j)}=a_{j-k}=(1,\ldots,1,a_{j-k-1}) \end{equation*}

Then, we see that $\ell _{k+1}(h)\le 2$ , and we are done in this case $\epsilon _i=1$ .

The remaining case $\epsilon _i=-1$ follows similarly. Hence, the proof is complete.

Proof. We proceed as in the proof of [Reference Di Domenico, Fernández-Alcober, Noce and Thillaisundaram8, Theorem 6.1], which is based on [Reference Grigorchuk and Żuk15, Lemma 4]. Let $u$ and $w$ be two different words representing the same element in the semigroup generated by $a_0,\dots,a_{s-1}$ , and with $\rho =\max (|u|,|w|)$ minimal. Clearly, $\rho \ge 2$ . Now for any word $z$ in the semigroup, let $|z|_{a_0}$ denote the $a_0$ -length in $z$ , that is, the number of occurrences of $a_0$ in $z$ . Note that we have $|u|_{a_0}\equiv |w|_{a_0} \pmod m$ . In the rest of the proof, we may consider the sections of $u$ and $w$ as defined in the proof of Lemma 3.7.

Suppose first that $u$ contains no $a_0$ s. We may assume that $w$ contains at least one $a_0$ . Indeed, if both $u$ and $w$ have no $a_0$ ’s, by replacing both words with their corresponding non-trivial sections respectively, the condition of both words having no $a_0$ ’s cannot hold indefinitely, as then both words $u$ and $w$ would be the trivial word. Thus $w$ contains at least one $a_0$ . Since $|w|_{a_0}$ must then be a non-zero multiple of $m$ , one deduces that $w_j$ , for $j\in [0,m-1]$ , is a non-empty word, where $\psi (w)=(w_0,\ldots,w_{m-1})$ . Indeed, every component of $\psi (w)$ contains an $a_{s-1}$ . Certainly $|w_j|\lt \rho$ . Therefore, $u$ must have only one section with a non-empty word, as otherwise, it will contradict the minimality of $\rho$ . Suppose this section is in the $i$ th component, for some $i\in [0,m-1]$ . From considering the $j$ th section of $u$ and $w$ , for $j\ne i$ , we obtain a contradiction to the minimality of $\rho$ . So the number of occurrences of $a_0$ in $u$ is at least one.

If the number of occurrences of $a_0$ in $u$ , respectively $w$ , is at least 2, then there are at least two sections of $u$ , respectively $w$ , that contain $a_{s-1}$ . Hence, all sections of $u$ , respectively $w$ , have length strictly less than $|u|$ , respectively $|w|$ , which contradicts the minimality of $\rho$ . So suppose that both $u$ and $w$ have exactly one occurrence of $a_0$ . Repeating this argument for the sections of $u$ and $w$ , we may assume that every generator appears at most once in $u$ and $w$ and that the same generators appear in both $u$ and $w$ . In other words, the word $w$ is a reordering of the letters in $u$ . Furthermore, as seen above, both $u$ and $w$ have only one non-empty section, which is therefore the rightmost section. Indeed, if this is not the case, then either $u$ or $w$ have more than one non-trivial section, then we contradict the minimality of $\rho$ . Without loss of generality, we may suppose that $a_0$ is the $i$ th letter of the word $u$ , but the $j$ th letter of the word $w$ where $j\lt i$ . The restrictions on the sections of $u$ imply that the first $i$ letters of $u$ have non-empty section in the rightmost component, and all other letters have non-empty section in the leftmost component. At least one of these first $i-1$ letters of $u$ appears to the right of $a_0$ in the word $w$ . Thus, there is a non-empty section in the second last component, which yields a contradiction to the minimality of $\rho$ . Hence, we are done.

Lastly, we end this section with a result that will be useful for the final section.

Proof. Notice first that, for $i \in{[1, s-1]}$ with $x_{i-1} \neq m-1$ , we have

\begin{align*} [a_i,a_0]^{a_0^{m-2-x_{i-1}}} = (1,\dots,1,a_{i-1}^{-1},a_{i-1}). \end{align*}

When $x_{i-1} = m-1$ , we get

\begin{align*} [a_0^{-1}, a_i] = (1,\dots,1,a_{i-1}^{-1},a_{i-1}). \end{align*}

Together with Lemma 3.4, we see that the elements of the form

\begin{align*} (1,\dots,1,{[a_i,a_j,a_k]}) \end{align*}

are contained in $\psi (\mathfrak{K(v)}^{\prime\prime})$ , for $i,j \in{[0, s-1]}$ and $k \in{[0, s-2]}$ .

We have to find the elements of the form $(1,\dots,1,{[a_i,a_j,a_{s-1}]} )$ . Observe from Lemma 3.4 that

\begin{equation*} \mathfrak {K(v)}^{\prime\prime}\times \overset {m}{\cdots }\times \mathfrak {K(v)}^{\prime\prime}\le \psi (\mathfrak {K(v)}^{\prime\prime}). \end{equation*}

Then from the Hall-Witt identity, we deduce that

\begin{align*} 1&=[a_j,a_i^{-1},a_{s-1}]^{a_i} [a_i,a_{s-1}^{-1},a_j]^{a_{s-1}}[a_{s-1},a_j^{-1},a_i]^{a_j}\\ &=[a_i,a_j,a_{s-1}^{a_i}] [a_{s-1},a_i,a_j^{a_{s-1}}][a_j,a_{s-1},a_i^{a_j}]\\ &\equiv [a_i,a_j,a_{s-1}] [a_{s-1},a_i,a_j][a_j,a_{s-1},a_i] \pmod{\mathfrak{K(v)}^{\prime\prime}}. \end{align*}

Hence, if $i,j\ne s-1$ , from considering the product of

\begin{equation*}(1,\dots,1,{[a_{s-1},a_i,a_j] } )\qquad \text {and}\qquad (1,\dots,1,{[a_j,a_{s-1},a_i] } )\end{equation*}

together with an appropriate element from $\mathfrak{K(v)}^{\prime\prime}\times \overset{m}{\cdots }\times \mathfrak{K(v)}^{\prime\prime}$ , we obtain

\begin{equation*} (1,\dots,1,{[a_i,a_j,a_{s-1}] } )\in \psi {(\mathfrak {K(v)}^{\prime\prime})}. \end{equation*}

If $i=s-1$ but $j\ne s-1$ , we will show that ${[a_{s-1},a_j},a_{s-1}]=1$ . To this end, in light of Lemma 3.3 it suffices to consider ${[a_{s-1-j},a_0},a_{s-1-j}]$ . Recall from the proof of Lemma 3.3 that

\begin{equation*} [a_{s-1-j},a_0]=\begin {cases} (a_{s-1}^{-1}a_{s-j-2}a_{s-1},1,\ldots,1,a_{s-j-2}^{-1})&\text {if }x_{s-j-2}= m-1,\\ (1,\overset {x_{s-j-2}-1}\ldots,1,a_{s-j-2}^{-1},a_{s-j-2},1,\ldots,1)&\text {if }x_{s-j-2}\ne m-1. \end {cases} \end{equation*}

Hence, it is now clear that ${[a_{s-1-j},a_0},a_{s-1-j}]=1$ .

If $j=s-1$ but $i\ne s-1$ , from the Hall-Witt identity, we have the equivalence

\begin{equation*} 1\equiv [a_{i},a_{s-1},a_{s-1}] [a_{s-1},a_i,a_{s-1}] \pmod {\mathfrak {K(v)}^{\prime\prime}}. \end{equation*}

Hence, we are done from the previous case.

The result now follows from the level-transitivity and the fractalness of $\mathfrak{K(v)}$ .

4. Length-reducing properties

As before, we have ${\mathfrak{v}} = x_0 \cdots x_{s-2}$ is a word in the alphabet $X=\{0,1,\dots,m-1\}$ and $\mathfrak{K(v)}$ is the group associated with $\mathfrak{v}$ . In this section, we establish some length-reducing properties for some of the groups $\mathfrak{K(v)}$ . Notice that for every $g \in \mathfrak{K(v)}$ , the local action $g^{\epsilon }$ of $g$ at the root is an element of $\langle \sigma \rangle$ . Hence, for conciseness, we denote $g^{\epsilon }$ by $\sigma _g$ . If $g \in \mathfrak{K(v)}$ then $|g|$ denotes the minimal length of all words in the alphabet $S$ representing $g$ . A word $w \in S^*$ is called a geodesic word if $\vert w \vert = \vert g \vert$ , where $g$ is the image of the word $w$ in $\mathfrak{K(v)}$ .

Proof. The proof proceeds by induction on the length of $g$ . Clearly, the result is true if $\vert g \vert =0$ and $\vert g \vert =1.$ Assume that $\vert g \vert \gt 1.$ Let $w \in S^*$ be a geodesic word representing $g$ . The word $w$ can be written as $w=bw^{\prime}$ for some $b \in S$ and $w^{\prime}\in S^*$ such that $w^{\prime}$ is reduced. Then $\vert w^{\prime} \vert \lt \vert w \vert$ and $w^{\prime}$ does not represent $g$ in $\mathfrak{K(v)}$ . Denote by $g^{\prime}$ the corresponding element in $\mathfrak{K(v)}$ . Then $\vert g^{\prime} \vert \leq \vert w^{\prime} \vert \lt \vert w \vert = \vert g \vert$ . We obtain

\begin{align*} (g_0,\dots,g_{m-1})\sigma _{g}= g & = b g^{\prime} = (b_0,\dots,b_{m-1})\sigma _{b}(g^{\prime}_0,\dots,g^{\prime}_{m-1})\sigma _{g^{\prime}} \\ & = ({b}_0g^{\prime}_{0^{\sigma _{b}}},\dots,{b}_{m-1}g^{\prime}_{(m-1)^{\sigma _{b}}})\sigma _{b}\sigma _{g^{\prime}}, \end{align*}

which implies $g_k ={b}_kg^{\prime}_{k^{\sigma _{b}}}$ for all $k \in{[0, m-1]}$ . It follows by induction that

\begin{equation*} \sum \limits _{k =0}^{m-1} \vert g_k \vert = \sum \limits _{k =0}^{m-1} \vert {b}_kg^{\prime}_{k^{\sigma _{b}}} \vert \leq \sum \limits _{k =0}^{m-1} \vert {b}_k \vert + \sum \limits _{k =0}^{m-1} \vert g^{\prime}_{k^{\sigma _{b}}} \vert \leq \vert {b} \vert + \vert g^{\prime} \vert \leq \vert {b} \vert + \vert w^{\prime} \vert = \vert w \vert = \vert g \vert. \end{equation*}

Proof. Observe that

\begin{align*} g^m =\big (g_{0}g_{0^{\sigma ^i}}g_{0^{\sigma ^{2i}}}\cdots g_{0^{\sigma ^{(m-1)i}}}\,,\,\dots \,,\,g_{m-1}g_{(m-1)^{\sigma ^i}}g_{(m-1)^{\sigma ^{2i}}}\cdots g_{(m-1)^{\sigma ^{(m-1)i}}}\big ). \end{align*}

By setting $\alpha _k = g_{k}g_{k^{\sigma ^i}}\cdots g_{k^{\sigma ^{(m-1)i}}}$ , for each $k \in{[0, m-1]}$ , we obtain

\begin{equation*} \vert \alpha _k \vert = \vert g_{k}g_{k^{\sigma ^i}}\cdots g_{k^{\sigma ^{(m-1)i}}} \vert \leq \sum \limits _{\ell =0}^{m-1} \vert g_{\ell } \vert \leq \vert g \vert, \end{equation*}

where the last inequality follows from Lemma 4.1.

Proof. By assumption, the word ${b}_1\cdots{b}_\ell$ contains a subword of the form $a_{0}wa_{0}^{-1}$ , where $w$ is a non-trivial reduced word in the alphabet $S$ . We assume, without loss of generality, that $w$ is a reduced word in the alphabet $S\setminus \{a_0^{\,\pm 1}\}$ . Let $w$ represent an element $h$ in $\mathfrak{K(v)}$ . Since ${b}_1 \cdots{b}_\ell \in S^*$ is a geodesic word, the word $w$ is also geodesic and so $|h| = |w|$ . Notice that $\vert a_{0}w a_{0}^{-1} \vert = \vert w \vert + 2$ . Realising the word $a_{0}wa_{0}^{-1}$ in $\mathfrak{K(v)}$ gives

\begin{align*} a_{0}wa_{0}^{-1}=(\varphi _{1}(h),\dots,\varphi _{m-2}(h),\varphi _{m-1}(h),1); \end{align*}

indeed, recall that by assumption we have

\begin{equation*} h = (1, \varphi _{1}(h),\dots,\varphi _{m-2}(h),\varphi _{m-1}(h)). \end{equation*}

By Lemma 4.1, we get $\sum _{i=1}^{m-1}|\varphi _{i}(h)| \le |h| = |w|$ . Therefore, we conclude that

\begin{equation*} \sum \limits _{k =0}^{m-1} \vert g_k \vert \leq \vert g \vert -2 \lt \vert g \vert. \end{equation*}

Proof. Note that we may assume that there are no $1\le i\lt j\le \ell$ such that $b_i=a_0^{-1}$ and $b_j=a_0$ , as then the desired length reduction is clear. It then follows that

\begin{equation*} w\,{:\!=}\,{b}_1 \cdots {b}_\ell =*a_0(*a_0)\overset {\lambda }\cdots (*a_0)(*a_0^{-1})\overset {\lambda }\cdots (*a_0^{-1})*\qquad \text {if }\epsilon =1 \end{equation*}

and

\begin{equation*} w\,{:\!=}\,{b}_1 \cdots {b}_\ell =(*a_0)\overset {\lambda }\cdots (*a_0)(*a_0^{-1})\overset {\lambda }\cdots (*a_0^{-1})*a_0^{-1}*\qquad \text {if }\epsilon =-1, \end{equation*}

for some $\lambda \in \mathbb{N}_0$ and the $*\in \langle a_1,\ldots,a_{s-1}\rangle$ are arbitrary.

Let

\begin{equation*} w^{\prime}=(*a_0)\overset {\lambda }\cdots (*a_0)(*a_0^{-1})\overset {\lambda }\cdots (*a_0^{-1})*. \end{equation*}

Write $w^{\prime}=(w_0^{\prime},\ldots,w_{m-1}^{\prime})$ and $\lambda =\mu m+\chi$ for some $\mu \in \mathbb{N}_0$ and $\chi \in [0,m-1]$ . If $\mu \ne 0$ , the first-level sections of $w^{\prime}$ have the following form

\begin{align*} w_0^{\prime} &=\varphi _0(*)(a_{s-1}\varphi _0(*))\overset{\mu -1}{\cdots }(a_{s-1}\varphi _0(*))a_{s-1}\varphi _0(*)a_{s-1}^{-1} (\varphi _0(*)a_{s-1}^{-1})\overset{\mu -1}{\cdots }(\varphi _0(*)a_{s-1}^{-1}) \varphi _0(*)\\ w_1^{\prime}&=(a_{s-1}\varphi _0(*))\overset{\mu -1}{\cdots }(a_{s-1}\varphi _0(*))a_{s-1}\varphi _0(*)a_{s-1}^{-1} (\varphi _0(*)a_{s-1}^{-1})\overset{\mu -1}{\cdots }(\varphi _0(*)a_{s-1}^{-1})\\ &\,\,\,\vdots \\ w_{m-1-\chi }^{\prime}&=(a_{s-1}\varphi _0(*))\overset{\mu -1}{\cdots }(a_{s-1}\varphi _0(*))a_{s-1}\varphi _0(*)a_{s-1}^{-1} (\varphi _0(*)a_{s-1}^{-1})\overset{\mu -1}{\cdots }(\varphi _0(*)a_{s-1}^{-1})\\ w_{m-\chi }^{\prime}&=(a_{s-1}\varphi _0(*))\overset{\mu }{\cdots }(a_{s-1}\varphi _0(*))a_{s-1}\varphi _0(*)a_{s-1}^{-1} (\varphi _0(*)a_{s-1}^{-1})\overset{\mu }{\cdots }(\varphi _0(*)a_{s-1}^{-1})\\ &\,\,\,\vdots \\ w_{m-1}^{\prime}&=(a_{s-1}\varphi _0(*))\overset{\mu }{\cdots }(a_{s-1}\varphi _0(*))a_{s-1}\varphi _0(*)a_{s-1}^{-1} (\varphi _0(*)a_{s-1}^{-1})\overset{\mu }{\cdots }(\varphi _0(*)a_{s-1}^{-1}). \end{align*}

When $\mu = 0$ and hence $\chi \ne 0$ , we get

\begin{align*} w_0^{\prime} &= \varphi _0(*)\\ w_1^{\prime} &= 1\\ &\,\,\,\vdots \\ w_{m-1-\chi }^{\prime} &= 1 \\ w_{m-\chi }^{\prime} &= a_{s-1}\varphi _0(*)a_{s-1}^{-1}\\ &\,\,\,\vdots \\ w_{m-1}^{\prime} &= a_{s-1}\varphi _0(*)a_{s-1}^{-1}. \end{align*}

Hence, a straightforward computation shows that $|g_0g_1\cdots g_{m-1}|\lt |g|$ if $\epsilon =1$ , and that $\vert g_0g_{m-1}\cdots g_{1} \vert \lt \vert g \vert$ if $\epsilon =-1$ , as required.

5. Maximal subgroups

It follows from Proposition 5.1 below together with [Reference Francoeur10, Proposition 2.21] that the group $\mathfrak{K(v)}$ admits maximal subgroups of infinite index if and only if it admits a proper subgroup $H \lt \mathfrak{K(v)}$ such that $HN = \mathfrak{K(v)}$ for every non-trivial normal subgroup $N \trianglelefteq \mathfrak{K(v)}$ . A subgroup $H \leq \mathfrak{K(v)}$ satisfying the above condition is called a prodense subgroup. As seen below, we prove that $\mathfrak{K(v)}$ , for a constant word $\mathfrak{v}$ , does not admit any proper prodense subgroup, which proves Theorem 1.2.

Proof. As $\mathfrak{K(v)}$ has exponential word growth from Lemma 3.11, it follows from Bass [Reference Bass5] and Guivarc’h [Reference Guivarc’h16] that $\mathfrak{K(v)}$ is not virtually nilpotent. To see that every proper quotient of $\mathfrak{K(v)}$ is virtually nilpotent, by [Reference Francoeur10, Theorem 4.10], it suffices to prove that $\mathfrak{K(v)}/\mathfrak{K(v)}^{\prime\prime}$ is virtually nilpotent. Set $N = \psi ^{-1}(\gamma _3(\mathfrak{K(v)}) \times \cdots \times \gamma _3(\mathfrak{K(v)}))$ . From Lemma 3.12, we have $N \leq \mathfrak{K(v)}^{\prime\prime} \lt \textrm{St}_{\mathfrak{K(v)}}(1) \lt{\mathfrak{K(v)}}$ . Therefore, $\psi$ induces a homomorphism

\begin{equation*} {\widetilde {\psi }}\,{:}\, \textrm {St}_{\mathfrak {K(v)}}(1)/N \longrightarrow \mathfrak {K(v)}/\gamma _3(\mathfrak {K(v)}) \times \overset {m}\cdots \times \mathfrak {K(v)}/\gamma _3(\mathfrak {K(v)}). \end{equation*}

Since $\widetilde{\psi }$ is injective and $\widetilde{\psi }(\textrm{St}_{\mathfrak{K(v)}}(1)/N)$ is nilpotent (being a subgroup of a nilpotent group), we obtain that $\textrm{St}_{\mathfrak{K(v)}}(1)/N$ is nilpotent. This implies that $\textrm{St}_{\mathfrak{K(v)}}(1)/{\mathfrak{K(v)}}^{\prime\prime}$ is nilpotent as it is a quotient of $\textrm{St}_{\mathfrak{K(v)}}(1)/N$ . As the subgroup $\textrm{St}_{\mathfrak{K(v)}}(1)$ has finite index in $\mathfrak{K(v)}$ , the group $\textrm{St}_{\mathfrak{K(v)}}(1)/\mathfrak{K(v)}^{\prime\prime}$ has finite index in $\mathfrak{K(v)}/\mathfrak{K(v)}^{\prime\prime}$ , and hence, $\mathfrak{K(v)}/\mathfrak{K(v)}^{\prime\prime}$ is virtually nilpotent. The last part of the result follows from [Reference Francoeur9, Corollary 5.1.3].

Hereafter, for $g,h \in \mathfrak{K(v)}$ , the equivalence $g \equiv h \mod \mathfrak{K(v)}^{\prime}$ will simply be denoted by $g \equiv h$ . Notice that for every $z \in \mathfrak{K(v)}^{\prime}$ , we have $\sigma _z =1$ and $\mathfrak{K(v)}^{\prime} \leq \textrm{St}_{\mathfrak{K(v)}}(1)$ . Recall also from Subsection 2.1 the map $\varphi _u$ for $u\in X^*$ . For convenience, we introduce the following notation. For $i \in [0,s-2]$ and $j \in [0,m-1]$ , we define

\begin{align*} \delta _{j}(a_i) = \begin{cases} a_i &\text{ if } x_i = j,\\ 1 &\text{otherwise}, \end{cases} \end{align*}

and extend $\delta _j$ to a homomorphism on $S^*$ .

Proof. Since $g \equiv a_{s-1}^{\epsilon _{s-1}} \cdots a_{0}^{\epsilon _{0}}$ , there exists an element $(z_0,\dots,z_{m-1})= z \in \mathfrak{K(v)}^{\prime}$ such that $g = a_{s-1}^{\epsilon _{s-1}} \cdots a_{0}^{\epsilon _{0}} z$ . Suppose that $\epsilon _0 = 1$ . We have that the section of $g$ at the vertex $j$ is

\begin{align*} (a_{s-1}^{\epsilon _{s-1}} \cdots a_{0}^{\epsilon _{0}} z)_j= \begin{cases} \delta _j(a_{s-2}^{\epsilon _{s-1}}\cdots a_0^{\epsilon _1} )z_{j+1} & \text{ if } j \in [0,m-2],\\ \delta _{m-1}(a_{s-2}^{\epsilon _{s-1}}\cdots a_0^{\epsilon _1} ) a_{s-1}z_0 & \text{ if } j = m-1. \end{cases} \end{align*}

Therefore, for $j \in [0,m-1]$ , the element $\varphi _j(g^m)$ is equal to the product

\begin{align*} &\big (\delta _j(a_{s-2}^{\epsilon _{s-1}} \cdots a_0^{\epsilon _1} )z_{j+1}\big )\cdot \,\cdots \,\cdot \big ( \delta _{m-2}( a_{s-2}^{\epsilon _{s-1}} \cdots a_0^{\epsilon _1})z_{m-1} \big )\cdot \\ &\qquad \qquad \big (\delta _{m-1}(a_{s-2}^{\epsilon _{s-1}}\cdots a_0^{\epsilon _1}) a_{s-1}z_0\big )\cdot \\ &\big (\delta _0(a_{s-2}^{\epsilon _{s-1}}\cdots a_0^{\epsilon _1})z_{1}\big )\cdot \,\cdots \,\cdot \big ( \delta _{j-1}(a_{s-2}^{\epsilon _{s-1}}\cdots a_0^{\epsilon _1})z_{j}\big ). \end{align*}

Since $\delta _k(a_i^{\epsilon _{i+1}})$ is non-trivial for only one $k \in [0,m-1]$ , we get that $\varphi _j(g^m) \equiv a_{s-2}^{\epsilon _{s-1}} \cdots a_{0}^{\epsilon _{1}}a_{s-1}^{\epsilon _{0}}$ for all $j \in{[0, m-1]}$ , since $z_0 \cdots z_{m-1} \in \mathfrak{K(v)}^{\prime}$ by Lemma 2.1. The case $\epsilon _0 = -1$ follows in a similar way. The result then follows upon repeating the above process.

In the following, we denote by $\textrm{Sym}(s)$ the symmetric group on $[0,s-1]$ .

Proof. Let $i \in{[0, s-1]}$ be such that $0 = i ^{\pi }$ . Then

\begin{align*} g&= a_{(s-1)^{\pi }}^{\epsilon _{(s-1)^{\pi }}}\cdots a_{(i+1)^{\pi }}^{\epsilon _{(i+1)^{\pi }}}a_0^{\epsilon _{0}}a_{(i-1)^{\pi }}^{\epsilon _{(i-1)^{\pi }}}\cdots a_{0^{\pi }}^{\epsilon _{0^{\pi }}}\\ &={\begin{cases}(1,\overset{t-1}\dots,1, a_{(i-1)^{\pi }-1}^{\epsilon _{(i-1)^{\pi }}}\cdots a_{0^{\pi }-1}^{\epsilon _{0^{\pi }}},a_{(s-1)^{\pi }-1}^{\epsilon _{(s-1)^{\pi }}}\cdots a_{(i+1)^{\pi }-1}^{\epsilon _{(i+1)^{\pi }}}, 1,\ldots,1,a_{s-1})\sigma & \text{ if } \epsilon _0 =1\text{ and }t\gt 0,\\ (a_{(s-1)^{\pi }-1}^{\epsilon _{(s-1)^{\pi }}}\cdots a_{(i+1)^{\pi }-1}^{\epsilon _{(i+1)^{\pi }}}, 1,\ldots,1,a_{s-1}a_{(i-1)^{\pi }-1}^{\epsilon _{(i-1)^{\pi }}}\cdots a_{0^{\pi }-1}^{\epsilon _{0^{\pi }}})\sigma & \text{ if } \epsilon _0 =1\text{ and }t=0,\\ (a_{s-1}^{-1},1,\overset{t-1}\dots,1,a_{(s-1)^{\pi }-1}^{\epsilon _{(s-1)^{\pi }}}\cdots a_{(i+1)^{\pi }-1}^{\epsilon _{(i+1)^{\pi }}}, a_{(i-1)^{\pi }-1}^{\epsilon _{(i-1)^{\pi }}}\cdots a_{0^{\pi }-1}^{\epsilon _{0^{\pi }}},1,\ldots,1)\sigma ^{-1} & \text{ if } \epsilon _0 =-1\text{ and }t\gt 0,\\ (a_{(s-1)^{\pi }-1}^{\epsilon _{(s-1)^{\pi }}}\cdots a_{(i+1)^{\pi }-1}^{\epsilon _{(i+1)^{\pi }}}a_{s-1}^{-1}, a_{(i-1)^{\pi }-1}^{\epsilon _{(i-1)^{\pi }}}\cdots a_{0^{\pi }-1}^{\epsilon _{0^{\pi }}},1,\ldots,1)\sigma ^{-1} & \text{ if } \epsilon _0 =-1\text{ and }t=0. \end{cases} } \end{align*}

By taking the $m$ th power of the element $g$ we get

\begin{align*} g^m= {\begin{cases}(*,\overset{t}\dots,*,a_{(s-1)^{\pi }-1}^{\epsilon _{(s-1)^{\pi }}}\cdots a_{(i+1)^{\pi }-1}^{\epsilon _{(i+1)^{\pi }}} a_{s-1}a_{(i-1)^{\pi }-1}^{\epsilon _{(i-1)^{\pi }}}\cdots a_{0^{\pi }-1}^{\epsilon _{0^{\pi }}},\#,\ldots,\#) & \text{ if } \epsilon _0 =1\text{ and }t\gt 0,\\ (a_{(s-1)^{\pi }-1}^{\epsilon _{(s-1)^{\pi }}}\cdots a_{(i+1)^{\pi }-1}^{\epsilon _{(i+1)^{\pi }}} a_{s-1}a_{(i-1)^{\pi }-1}^{\epsilon _{(i-1)^{\pi }}}\cdots a_{0^{\pi }-1}^{\epsilon _{0^{\pi }}},\#,\ldots,\#) & \text{ if } \epsilon _0 =1\text{ and }t=0,\\ (*,\overset{t}\dots,*,a_{(s-1)^{\pi }-1}^{\epsilon _{(s-1)^{\pi }}}\cdots a_{(i+1)^{\pi }-1}^{\epsilon _{(i+1)^{\pi }}}a_{s-1}^{-1} a_{(i-1)^{\pi }-1}^{\epsilon _{(i-1)^{\pi }}}\cdots a_{0^{\pi }-1}^{\epsilon _{0^{\pi }}},\#,\ldots,\#) & \text{ if } \epsilon _0 =-1\text{ and }t\gt 0,\\ (a_{(s-1)^{\pi }-1}^{\epsilon _{(s-1)^{\pi }}}\cdots a_{(i+1)^{\pi }-1}^{\epsilon _{(i+1)^{\pi }}}a_{s-1}^{-1} a_{(i-1)^{\pi }-1}^{\epsilon _{(i-1)^{\pi }}}\cdots a_{0^{\pi }-1}^{\epsilon _{0^{\pi }}},\#,\ldots,\#)\sigma ^{-1} & \text{ if } \epsilon _0 =-1\text{ and }t=0. \end{cases}} \end{align*}

where

\begin{align*} * = \begin{cases}a_{(i-1)^{\pi }-1}^{\epsilon _{(i-1)^{\pi }}}\cdots a_{0^{\pi }-1}^{\epsilon _{0^{\pi }}}a_{(s-1)^{\pi }-1}^{\epsilon _{(s-1)^{\pi }}}\cdots a_{(i+1)^{\pi }-1}^{\epsilon _{(i+1)^{\pi }}}a_{s-1} & \text{ if } \epsilon _0 =1, \\ \qquad &\\ a_{s-1}^{-1}a_{(i-1)^{\pi }-1}^{\epsilon _{(i-1)^{\pi }}}\cdots a_{0^{\pi }-1}^{\epsilon _{0^{\pi }}}a_{(s-1)^{\pi }-1}^{\epsilon _{(s-1)^{\pi }}}\cdots a_{(i+1)^{\pi }-1}^{\epsilon _{(i+1)^{\pi }}} & \text{ if } \epsilon _0 = -1, \end{cases} \end{align*}

and

\begin{align*} \# = \begin{cases}a_{s-1}a_{(i-1)^{\pi }-1}^{\epsilon _{(i-1)^{\pi }}}\cdots a_{0^{\pi }-1}^{\epsilon _{0^{\pi }}}a_{(s-1)^{\pi }-1}^{\epsilon _{(s-1)^{\pi }}}\cdots a_{(i+1)^{\pi }-1}^{\epsilon _{(i+1)^{\pi }}} & \text{ if } \epsilon _0 =1, \\ &\\ a_{(i-1)^{\pi }-1}^{\epsilon _{(i-1)^{\pi }}}\cdots a_{0^{\pi }-1}^{\epsilon _{0^{\pi }}}a_{(s-1)^{\pi }-1}^{\epsilon _{(s-1)^{\pi }}}\cdots a_{(i+1)^{\pi }-1}^{\epsilon _{(i+1)^{\pi }}}a_{s-1}^{-1} & \text{ if } \epsilon _0 = -1. \end{cases} \end{align*}

In particular, we have $\varphi _{t}(g^{m})=a_{(s-1)^{\pi }-1}^{\epsilon _{(s-1)^{\pi }}}\cdots a_{0^{\pi }-1}^{\epsilon _{0^{\pi }}}$ , and the result follows recursively.

We recall that $H_u$ denotes the subgroup $\varphi _u(\textrm{st}_H(u))$ for a vertex $u \in X^*$ . By Lemma 3.2, the group $\mathfrak{K(v)}$ is fractal, so $ \mathfrak{K(v)}_u= \mathfrak{K(v)}$ for all $u \in X^*$ .

Proof. (i) Let $g = a_{(s-1)^{\pi }}^{\epsilon _{(s-1)^{\pi }}}\cdots a_{0^{\pi }}^{\epsilon _{0^{\pi }}} \in H$ . Then, $0 = i ^{\pi }$ for some $i \in{[0, s-1]}$ . Observe from the proof of Lemma 5.3 that

\begin{equation*}\varphi _{t}(g^m)=a_{(s-1)^{\pi }-1}^{\epsilon _{(s-1)^{\pi }}}\cdots a_{(i+1)^{\pi }-1}^{\epsilon _{(i+1)^{\pi }}}a_{s-1}^{\epsilon _{0}}a_{(i-1)^{\pi }-1}^{\epsilon _{(i-1)^{\pi }}}\cdots a_{0^{\pi }-1}^{\epsilon _{0^{\pi }}}\end{equation*}

and that $\varphi _j(g^m)$ is a cyclic permutation of $\varphi _{t}(g^m)$ for every $j \in{[0, m-1]}$ . By repeating the process of taking powers we get that ${\psi }_s(g^{m^s})=(g_0,\dots,g_{m^s-1})$ with $g_{\tau }= g$ , for the component $\tau$ corresponding to the vertex $t^s$ , and $g_k$ is a cyclic permutation of the word $a_{(s-1)^{\pi }}^{\epsilon _{(s-1)^{\pi }}}\cdots a_{0^{\pi }}^{\epsilon _{0^{\pi }}}$ for $k\in [ 0,m^s-1]$ .

(ii) We will only consider the case $t\gt 0$ , as the case $t=0$ is analogous. In particular, if $\epsilon _0 =1$ , we note from the proof of Lemma 5.3 that

\begin{equation*} \varphi _{t-1}(g^m)=a_{(i-1)^{\pi }-1}^{\epsilon _{(i-1)^{\pi }}}\cdots a_{0^{\pi }-1}^{\epsilon _{0^{\pi }}}a_{(s-1)^{\pi }-1}^{\epsilon _{(s-1)^{\pi }}}\cdots a_{(i+1)^{\pi }-1}^{\epsilon _{(i+1)^{\pi }}}a_{s-1}, \end{equation*}

and by Lemma 5.3 we see that

\begin{align*} \varphi _{(t-1)t^{s-1}}(g^m) = a_{(i-1)^{\pi }}^{\epsilon _{(i-1)^{\pi }}}\cdots a_{0^{\pi }}^{\epsilon _{0^{\pi }}}a_{(s-1)^{\pi }}^{\epsilon _{(s-1)^{\pi }}}\cdots a_{(i+1)^{\pi }}^{\epsilon _{(i+1)^{\pi }}}a_{0}. \end{align*}

More generally, suppose that $\epsilon _j =1$ for $j\in [0,s]$ and let $\ell \in [0,s]$ be such that $\ell ^\pi =j$ . Writing $v_j= t\overset{j}\cdots t$ and $w_j=(t-1)t\,\overset{s-j-1}\cdots \, t$ , we recall from Lemma 5.3 that

\begin{equation*} \varphi _{v_j}(g^{m^j})=a_{(s-1)^{\pi }-j}^{\epsilon _{(s-1)^{\pi }}}\cdots a_{(\ell +1)^{\pi }-j}^{\epsilon _{(\ell +1)^{\pi }}}a_{0}a_{(\ell -1)^{\pi }-j}^{\epsilon _{(\ell -1)^{\pi }}}\cdots a_{0^{\pi }-j}^{\epsilon _{0^{\pi }}}. \end{equation*}

Then similar to the above, we see that

\begin{align*} \varphi _{v_jw_j}(g^m) = a_{(\ell -1)^{\pi }}^{\epsilon _{(\ell -1)^{\pi }}}\cdots a_{0^{\pi }}^{\epsilon _{0^{\pi }}}a_{(s-1)^{\pi }}^{\epsilon _{(s-1)^{\pi }}}\cdots a_{(\ell +1)^{\pi }}^{\epsilon _{(\ell +1)^{\pi }}}a_{j}, \end{align*}

and as $u_j\,{:\!=}\,v_jw_j$ is a vertex of level $s$ , we have that $H_{u_j}$ contains a cyclic permutation of $a_{(s-1)^{\pi }}^{\epsilon _{(s-1)^{\pi }}}\cdots a_{0^{\pi }}^{\epsilon _{0^{\pi }}}$ that ends with $a_j$ on the right.

Now, by using Lemma 5.3 repeatedly, one can see that the result holds for level $ns$ of $T$ , for $n\gt 1$ .

Proof. The proof proceeds by induction on the length of $g$ . Recall from Lemma 3.8 that $\mathfrak{K(v)}/\mathfrak{K(v)}^{\prime}=\langle a_0\mathfrak{K(v)}^{\prime},\ldots,a_{s-1}\mathfrak{K(v)}^{\prime}\rangle \cong \mathbb{Z}^s$ . Hence, if $g$ is equivalent to $a_{s-1}^{\epsilon _{s-1}}\cdots a_{0}^{\epsilon _{0}}$ then $|g| \gt s-1$ , since any word containing each of the distinct generators of $\mathfrak{K(v)}$ has length at least $s$ . Assume that $\vert g \vert = s$ . Then

\begin{equation*} g \in \big \{a_{(s-1)^{\pi }}^{\epsilon _{(s-1)^{\pi }}}\cdots a_{0^{\pi }}^{\epsilon _{0^{\pi }}} \mid \pi \in \textrm {Sym}(s)\big \} \end{equation*}

and the result follows trivially by choosing $u$ as the root vertex. Now, assume that $\vert g \vert \gt s$ . Since the exponent sum of $a_0$ in any word representing $g$ is $\epsilon _0$ , we can write $g = (g_0,\dots, g_{m-1})\sigma ^{\epsilon _0}$ with $g_0, \dots,g_{m-1}\in \mathfrak{K(v)}$ . We get

\begin{equation*} g^m = \begin {cases}(g_0g_1\cdots g_{m-1},\,g_1\cdots g_{m-1}g_0\,,\,\dots \,,\, g_{m-1}g_0g_1\cdots g_{m-2}) & \text { if } \epsilon _0 =1,\\ (g_0g_{m-1}g_{m-2}\cdots g_1,\,g_1g_0g_{m-1}\cdots g_2\,,\,\dots \,,\, g_{m-1}g_{m-2}\cdots g_{0}) & \text { if } \epsilon _0 = -1. \end {cases} \end{equation*}

For every $k \in{[0, m-1]}$ , we set $\alpha _k = \varphi _{k}(g^m)$ . It follows from Lemma 5.2 that

\begin{equation*} \alpha _k \equiv a_{s-2}^{\epsilon _{s-1}}\cdots a_0^{\epsilon _1} a_{s-1}^{\epsilon _0} \end{equation*}

for all $k \in{[0, m-1]}$ . Furthermore, $\vert \alpha _k \vert \leq \vert g \vert$ for all $k \in{[0, m-1]}$ by Lemma 4.2. If there exists $k \in{[0, m-1]}$ such that $\vert \alpha _k \vert \lt \vert g \vert$ , then it follows by induction that there exists a vertex $u$ of level $ns$ in $T$ , for some $n \in \mathbb{N}_0$ , and $g^{\prime} \in \textrm{st}_{\langle \alpha _k \rangle }(u)$ such that

\begin{equation*} \varphi _u(g^{\prime})=a_{(s-1)^{\pi }-1}^{\epsilon _{(s-1)^{\pi }}}a_{(s-2)^{\pi }-1}^{\epsilon _{(s-2)^{\pi }}} \cdots a_{0^{\pi }-1}^{\epsilon _{0^{\pi }}} \end{equation*}

for some $\pi \in \textrm{Sym}(s)$ . Using Lemma 5.3, we get that

\begin{equation*}\varphi _{t^{s-1} }((g^{\prime})^{m^{s-1}}) = a_{(s-1)^{\pi }}^{\epsilon _{(s-1)^{\pi }}}a_{(s-2)^{\pi }}^{\epsilon _{(s-2)^{\pi }}} \cdots a_{0^{\pi }}^{\epsilon _{0^{\pi }}},\end{equation*}

and hence, the result follows.

Assume that $\vert \alpha _k \vert = \vert g \vert$ for all $k \in{[0, m-1]}.$ Since $\vert \alpha _k \vert \leq \sum \limits _{\ell =0}^{m-1} \vert g_{\ell } \vert$ , in particular, we get

\begin{equation*} \sum \limits _{\ell =0}^{m-1} \vert g_{\ell } \vert = \vert g \vert. \end{equation*}

Let $w_g \in S^*$ be a geodesic word representing $g$ . Since for each $i\in [0,s-1]$ the element $a_i^{\epsilon _i}$ contributes $a_{i-1}^{\epsilon _i}$ in exactly one component, we can obtain words $w_{g_k}$ representing $g_k$ by substituting $a_i^{\epsilon _i}$ in $w_g$ with $a_{i-1}^{\epsilon _i}$ in the appropriate component. Notice that $\vert g_k\vert \leq \vert w_{g_k} \vert$ for every $k \in{[0, m-1]}$ . Moreover, the words $w_{g_k}$ are geodesic. Indeed,

\begin{equation*} \sum \limits _{k = 0}^{m-1}\vert g_k\vert \leq \sum \limits _{k = 0}^{m-1}\vert w_{g_k}\vert \leq \vert w_g \vert = \vert g \vert = \sum \limits _{\ell =0}^{m-1} \vert g_{\ell } \vert, \end{equation*}

which forces that $\vert w_{g_k} \vert = \vert g_k \vert .$ Now, set

\begin{equation*} w_{\alpha _k} = \begin {cases} w_{g_k}w_{g_{k+1}}\cdots w_{g_{k+m-1}} & \text { if } \epsilon _0 = 1,\\ w_{g_k}w_{g_{k-1}}\cdots w_{g_{k-(m-1)}} & \text { if } \epsilon _0 = -1. \end {cases} \end{equation*}

Clearly, $w_{\alpha _k}$ represents $\alpha _k$ . Therefore, $\vert \alpha _k \vert \leq \vert w_{\alpha _k} \vert$ . Furthermore,

\begin{equation*} \vert w_{\alpha _k} \vert \leq \sum \limits _{\ell =0}^{m-1}\vert w_{g_{\ell }} \vert = \sum \limits _{\ell =0}^{m-1} \vert g_{\ell } \vert = \vert g \vert = \vert \alpha _k \vert. \end{equation*}

Thus, $\vert \alpha _k \vert = \vert w_{\alpha _k} \vert$ and $w_{\alpha _k}$ is a geodesic word.

Now, we claim that in order to prove the result, it suffices to consider the situation in which for every $i \in{[0, s-1]}$ there exists a unique $k \in{[0, m-1]}$ such that $w_{g_k}$ contains a non-trivial power of $a_i$ . First, we consider the case when $i=0$ . Assume to the contrary that there exist distinct $k_1,k_2 \in{[0, m-1]}$ such that $w_{g_{k_1}}$ and $w_{g_{k_2}}$ contain non-trivial powers of $a_0$ . We can reduce to the following two cases.

Case 1: Suppose that there exist distinct $k_1,k_2 \in{[0, m-1]}$ such that $w_{g_{k_1}}$ and $w_{g_{k_2}}$ contain $a_0$ and $a_0^{-1}$ , respectively. Then for some $k\in{[0, m-1]}$ , the word $w_{\alpha _k}$ contains a subword of the form $a_0 w a_0^{-1}$ with $w \in S^*$ . If $\alpha _k^m = (\beta _0,\dots,\beta _{m-1})$ then, by Lemma 4.2, Lemma 4.3 and Lemma 4.4, we obtain that $\vert \beta _\ell \vert \lt \vert \alpha \vert$ for $\ell \in{[0, m-1]}$ . Again, the result follows by induction.

Case 2: Suppose there exist distinct $k_1,k_2 \in{[0, m-1]}$ such that $w_{g_{k_1}}$ and $w_{g_{k_2}}$ contain $a_0$ . Recall that $\mathfrak{K(v)}/\mathfrak{K(v)}^{\prime}=\langle a_0\mathfrak{K(v)}^{\prime},\ldots,a_{s-1}\mathfrak{K(v)}^{\prime}\rangle \cong \mathbb{Z}^s$ . Hence, as the exponent sum of $a_{1}$ in any word representing $g$ is $\epsilon _1$ , the exponent sum of $a_0$ in $w_{\alpha _k}$ is equal to $\epsilon _1$ for all $k \in{[0, m-1]}$ . This implies that there exists $k_3 \in{[0, m-1]}$ such that $w_{g_{k_3}}$ contains $a_{0}^{-1}$ , and we are in the previous case. Analogously, the same argument works if both $w_{g_{k_1}}$ and $w_{g_{k_2}}$ contain $a_0^{-1}$ .

We reduce to the case where there exists a unique $k \in{[0, m-1]}$ such that $w_{g_k}$ contains a non-trivial power of $a_0$ . By inducting on $i \in{[0, s-1]}$ , assume that there exists a unique $k \in{[0, m-1]}$ such that $w_{g_k}$ contains a non-trivial power of $a_{i-1}$ . Then suppose that there exist distinct $k_1,k_2 \in{[0, m-1]}$ such that $w_{g_{k_1}}$ and $w_{g_{k_2}}$ contain non-trivial powers of $a_{i}$ . We can find $k_3 \in{[0, m-1]}$ such that $w_{\alpha _{k_3}}$ contains a subword of the form $a_{i} ^{\ell _1}wa_{i} ^{\ell _2}$ , where $\ell _1,\ell _2 \in \mathbb{Z}\backslash \{0\}$ and $w \in S^*$ with exponent sum of $a_0$ in $w$ is not equal to $0 \bmod m$ . Thanks to Lemma 5.2, we may replace $g$ with $\alpha _{k_3}$ . Then we find more than one $w_{g_k}$ containing non-trivial powers of $a_{i-1}$ , contradicting the induction hypothesis and hence proving the claim.

Thus, we reduce to the situation in which for every $i \in{[0, s-1]}$ there exists a unique $k \in{[0, m-1]}$ such that $w_{g_k}$ contains a non-trivial power of $a_i$ . An easy computation yields that $w_{g}$ does not contain a subword of the form $a_{i}^{\ell _1}wa_{i}^{\ell _2}$ , for some $i \in{[0, s-1]}$ where $\ell _1,\ell _2 \in \mathbb{Z}\backslash \{0\}$ and $w \in S^*$ with the exponent sum of $a_0$ in $w$ being not equal to $0 \bmod m$ . Hence, we conclude that $w_g$ must be of the form

\begin{equation*} {w_1}(a_{i_1},\dots, a_{i_r})a_0^{\epsilon _0} {w_2}(a_{i_{r+1}},\dots, a_{i_{s-1}}), \end{equation*}

where $w_1$ and $w_2$ are words in the given elements, and $\{i_1,\dots,i_r,i_{r+1},\dots,i_{s-1}\}={[1, s-1]}$ such that the intersection $\{i_1,\dots,{i_r} \} \cap \{i_{r+1},\dots,i_{s-1}\}$ is empty. Consider the element $\alpha _{t}$ obtained from the element $g$ above. Then, the corresponding $w_{\alpha _t}$ has the form

\begin{equation*} w_1(a_{i_{1}-1},\dots, a_{i_{r}-1})a_{s-1}^{\epsilon _0} w_2(a_{i_{r+1}-1},\dots, a_{i_{s-1}-1}), \end{equation*}

and continuing the above procedure with this word yields the element $a_{(s-1)^{\pi }}^{\epsilon _{(s-1)^{\pi }}}\cdots a_{0^{\pi }}^{\epsilon _{0^{\pi }}} \in H_u$ for some $u$ of level $ns$ in $T$ , for some $n \in \mathbb{N}_0$ .

Proof. Note that $H\mathfrak{K(v)}^{\prime} =\mathfrak{K(v)}$ as $H$ is a prodense subgroup. Therefore, there exists an element $z \in \mathfrak{K(v)}^{\prime}$ such that $ a_{s-1}\cdots a_0z\in H$ . By an application of Proposition 5.5, we can find $u\in T$ such that $H_{u}$ contains $a_{(s-1)^{\pi }}\cdots a_{0^{\pi }}$ for some $\pi \in \textrm{Sym}(s)$ . We set $g=a_{(s-1)^{\pi }}\cdots a_{0^{\pi }}$ . Thanks to [Reference Francoeur10, Lemma 3.1], the subgroup $H_u$ is again a prodense subgroup of $\mathfrak{K(v)}$ . Without loss of generality, we replace $H$ with $H_u$ .

Again, as $H$ is prodense, for some $\widetilde{z}\in \mathfrak{K(v)}^{\prime}$ we similarly have $a_{s-1}\cdots a_1a_0^{-1}\widetilde{z}\in H$ . By Proposition 5.5, there exists a vertex $u$ at level $ns$ , for some $n \in \mathbb{N}$ , such that $H_u$ contains an element $h_0$ of the form

\begin{equation*} h_0 = a_{(s-1)^{\tau _0}}^{\,\epsilon _{(s-1)^{\tau _{0}}}}\cdots a_{0^{\tau _0}}^{\,\epsilon _{0^{\tau _0}}}, \end{equation*}

where $\tau _0 \in \textrm{Sym}(s)$ , with $\epsilon _{i^{\tau _0}}=-1$ if $i^{\tau _0}=0$ and $\epsilon _{i^{\tau _0}}=1$ otherwise. Now, by Lemma 5.4(i), the subgroup $H_u$ also contains some cyclic permutation of the element $g$ . By abuse of notation, we replace $g$ with this cyclic permutation of $g$ . We again replace $H$ with $H_u$ . Now $H$ contains the elements $g$ and $h_0$ . Repeating this argument $s-1$ times, we may assume that $H$ contains the elements $g,h_0,\dots,h_{s-1}$ , where

\begin{equation*} h_j = a_{(s-1)^{\tau _j}}^{\,\epsilon _{(s-1)^{\tau _{j}}}}\cdots a_{0^{\tau _j}}^{\,\epsilon _{0^{\tau _j}}}, \end{equation*}

where $\tau _j \in \textrm{Sym}(s)$ with $\epsilon _{i^{\tau _j}}=-1$ if $i^{\tau _j}=j$ and $\epsilon _{i^{\tau _j}}=1$ otherwise.

For the remainder of the proof, we will assume that $t\gt 0$ ; the case $t=0$ is analogous, as accounted for in Remark 5.7. Appealing to Lemma 5.4(ii), we now choose a vertex $v$ , with $v$ of level $\widetilde{n}s$ for some $\widetilde{n} \in \mathbb{N}$ , such that the cyclic permutation of $g$ that is contained in $H_v$ ends with $a_0$ on the right. We rename this element $g$ . So we have $g\in H_v$ and by Lemma 5.4(i) we have a cyclic permutation of each of the elements $h_0,\ldots,h_{s-1}$ in $H_v$ . By abuse of notation, we rename these cyclic permutations $h_0,\ldots,h_{s-1}$ respectively. As before we replace $H$ with $H_v$ . Now $H$ contains the elements $g,h_0,\dots,h_{s-1}$ , where $g$ ends with $a_0$ on the right.

For each $n \in \mathbb{N}_0$ , let $v_n=t \,\overset{n}\dots \, t$ . It follows from Lemma 5.3 that for $d\in \mathbb{N}$ , $\varphi _{v_{ds}}(g^{m^{ds}})=g$ and $\varphi _{v_{ds}}(h_i^{\,m^{ds}})=h_i$ where $i\in [0,s-1]$ . Furthermore, for any element $f \in \mathfrak{K(v)}$ of the form

\begin{equation*} f=a_{\iota _1}^{\epsilon _1} \cdots a_{\iota _b}^{\epsilon _b}, \end{equation*}

for pairwise distinct $\iota _1,\ldots,\iota _{b}\in [0,s-1]$ with ${b}\in [1,s]$ and $\epsilon _i\in \{\pm 1\}$ , we can consider its contribution to $H_{v_n}$ . Specifically, if $f \in \textrm{St}_{\mathfrak{K(v)}}(1)$ , we simply consider its image under $\varphi _{t}$ . If $f \notin \textrm{St}_{\mathfrak{K(v)}}(1)$ , then we consider $\varphi _{t}(f^m)$ . We refer to this general process as projecting along the path $t_{\infty }\,{:\!=}\,tt\cdots$ . By projecting along this path, we observe that if $f \in H_{v_j}$ , for $j \in \mathbb{N}$ , then $f \in H_{v_{j+s}}$ ; compare the proof of Lemma 5.3. This observation will be used repeatedly throughout the proof without special mention.

The strategy of the proof is now to consider the contributions from

\begin{equation*} \langle g\rangle \,,\,\langle h_0\rangle \,,\,\ldots \,,\,\langle h_{s-1}\rangle \end{equation*}

to $H_{v_n}$ and to multiply them appropriately to separate the generators $a_0,\ldots,a_{s-1}$ . More specifically, if for some $n\in \mathbb{N}$ , suppose we have non-trivial distinct elements $\alpha,\beta \in H_{v_n}$ of the form

\begin{equation*} \alpha =a_{i_1}^{\epsilon _1}\cdots a_{i_q}^{ \epsilon _q},\qquad \beta =a_{j_1}^{\delta _1}\cdots a_{j_r}^{\delta _r} \end{equation*}

where $\epsilon _i,\delta _j\in \{\pm 1\}$ and $2\le{q},r\le s$ , with $i_1,\ldots,i_{q}\in [0,s-1]$ pairwise distinct, and also $j_1,\ldots,j_r\in [0,s-1]$ pairwise distinct. We consider two situations below, where we assume always that $\widetilde{\alpha },\widetilde{\beta }$ are non-trivial.

1. (i) If $\alpha = \widetilde{\alpha }a_0$ and $\beta = \widetilde{\beta }a_0^{-1}\widehat{\beta }$ , then

   \begin{equation*} \beta \alpha =\widetilde {\beta }a_0^{-1}\widehat {\beta }\widetilde {\alpha }a_0 \end{equation*}
   yields
   \begin{equation*} \varphi _{t}(\widetilde {\beta })\in H_{v_{n+1}}, \end{equation*}
   and hence
   \begin{equation*} \widetilde {\beta }\in H_{v_{n+s}}\qquad \text {and}\qquad a_0^{-1}\widehat {\beta }\in H_{v_{n+s}}. \end{equation*}

2. (ii) If $ \alpha = a_0\widetilde{\alpha }$ and $\beta = \widehat{\beta }a_0^{-1}\widetilde{\beta }$ , from

   \begin{equation*} \alpha \beta =a_0\widetilde {\alpha }\widehat {\beta }a_0^{-1}\widetilde {\beta }, \end{equation*}
   we obtain
   \begin{equation*} \varphi _{t}(\widetilde {\beta })\in H_{v_{n+1}}, \end{equation*}
   and similarly,
   \begin{equation*} \widetilde {\beta }\in H_{v_{n+s}}\qquad \text {and}\qquad \widehat {\beta }a_0^{-1}\in H_{v_{n+s}}. \end{equation*}

In other words, upon replacing $H_{v_n}$ with $H_{v_{n+s}}$ we have split $\beta \in H_{v_{n+s}}$ into two non-trivial parts. The plan is to repeatedly perform such operations as in (i) and (ii) above to keep splitting the products of generators. Eventually we will end up with $a_0,\ldots,a_{s-1}\in H_u$ for some $u$ , which gives $H_u=\mathfrak{K(v)}$ and equivalently that $H=\mathfrak{K(v)}$ , as required.

We begin by first considering the contributions from $\langle g\rangle$ and $\langle h_0\rangle$ along the path $t_{\infty }$ of the tree. For convenience, write

\begin{equation*} g=a_{i_1}\cdots a_{i_{s-1}}a_0\qquad \text {and}\qquad h_0=a_{j_1}\cdots a_{j_{d-1}}a_0^{-1}a_{j_{d+1}}\cdots a_{j_s}, \end{equation*}

for some $d\in [1,s]$ , where $\{i_1,\ldots,i_{s-1}\}=\{j_1,\ldots,j_{d-1},j_{d+1},\ldots,j_s\}=[1,s-1]$ .

Case 1: Suppose $1\lt d\lt s$ . Then we are in situation (i) from above, and it follows that

\begin{equation*} a_{j_1-1}\cdots a_{j_{d-1}-1}\in H_{v_1}\quad \text {and}\quad a_{s-1}^{-1}a_{j_{d+1}-1}\cdots a_{j_s-1}\in H_{v_1}. \end{equation*}

We will now use (the projections of) these two parts of $h_0$ to split $g$ into two non-trivial parts.

Let $j\,{:\!=}\,j_{d-1}$ . We consider the contribution of $a_{j_1}\cdots a_{j_{d-1}}$ to $H_{v_j}$ . In other words, we project along the path $t_{\infty }$ down to level $j$ , which gives

\begin{equation*} a_{j_1-j}\cdots a_{j_{d-2}-j }a_{0}\in H_{v_j}. \end{equation*}

Recalling that $g=a_{i_1}\cdots a_{i_{s-1}}a_0$ , we have that $i_r=j$ for some $r\in [1,s-1]$ . Then setting

\begin{equation*} \beta ^{-1}\,{:\!=}\, \varphi _{v_j}(g^{m^j})=a_{i_1-j}\cdots a_{i_{r-1}-j}a_0a_{i_{r+1}-j}\cdots a_{i_{s-1}-j}a_{s-j}\in H_{v_j} \end{equation*}

and

\begin{equation*} \alpha \,{:\!=}\, a_{j_1-j}\cdots a_{j_{d-2}-j }a_{0}\in H_{v_j}, \end{equation*}

it follows from situation (i) that

\begin{equation*} a_{i_{r+1}-j-1}\cdots a_{i_{s-1}-j-1}a_{s-j-1}\in H_{v_{j+1}}\quad \text {and}\quad a_{i_1-j-1}\cdots a_{i_{r-1}-j-1}a_{s-1}\in H_{v_{j+1}}, \end{equation*}

so we have split $g$ into two non-trivial parts.

We now use the two parts of $g$ to split the parts of $h_0$ further. For clarity, let us first project to $v_s$ . Here in $H_{v_s}$ we have the elements

\begin{equation*} a_{j_1}\cdots a_{j_{d-2}}a_j,\qquad a_0^{-1}a_{j_{d+1}}\cdots a_{j_s},\qquad a_{i_1}\cdots a_{i_{r-1}}a_j,\qquad a_{i_{r+1}}\cdots a_{i_{s-1}}a_0. \end{equation*}

The left two elements are the two parts of $h_0$ , and the right two are those of $g$ . Without loss of generality, we replace $H$ with $H_{v_s}$ .

Subcase (a): Suppose $1\lt r\lt s-1$ . Let $k\,{:\!=}\,i_1$ . Then either $k=j_{q}$ for ${q}\in [1,d-2]$ or $k=j_{q}$ for ${q}\in [d+1,s]$ . Suppose the former; a similar argument works for the latter. If ${q}\gt 1$ , we let $\beta ^{-1}$ be the $k$ th level projection of $a_{j_1}\cdots a_{j_{d-2}}a_j$ (as usual along the path $t_{\infty }$ ) and $\alpha$ be that of $a_ka_{i_2}\cdots a_{i_{r-1}}a_j$ , which by (ii) gives, upon replacing $H$ with $H_{v_s}$ , the following elements in $H$ :

\begin{equation*} a_{j_1}\cdots a_{j_{q-1}},\quad a_ka_{j_{q+1}}\cdots a_{j_{d-2}}a_j,\quad a_0^{-1}a_{j_{d+1}}\cdots a_{j_s},\quad a_ka_{i_2}\cdots a_{i_{r-1}}a_j,\quad a_{i_{r+1}}\cdots a_{i_{s-1}}a_0. \end{equation*}

If ${q}=1$ , before splitting, we have instead the following elements in $H$ :

\begin{equation*} a_ka_{j_2}\cdots a_{j_{d-2}}a_j,\qquad a_0^{-1}a_{j_{d+1}}\cdots a_{j_s},\qquad a_ka_{i_2}\cdots a_{i_{r-1}}a_j,\qquad a_{i_{r+1}}\cdots a_{i_{s-1}}a_0. \end{equation*}

Hence, we let ${\ell }\,{:\!=}\,i_{r+1}$ and let $c\in [2,d-2]\cup [d+1,s]$ be such that $j_c={\ell }$ . We consider the $\ell$ th projection of $a_{\ell }a_{i_{r+2}}\cdots a_{i_{s-1}}a_0$ multiplied accordingly with that of $a_j^{-1}a_{j_{d-2}}^{-1} \cdots a_{j_2}^{-1} a_k^{-1}$ or $a_{j_s}^{-1} \cdots a_{j_{d+1}}^{-1} a_0$ . This is situation (ii).

Subcase (b): Suppose $r=1$ . Then we have the following elements in $H$ :

\begin{equation*} a_{j_1}\cdots a_{j_{d-2}},\qquad a_0^{-1}a_{j_{d+1}}\cdots a_{j_s},\qquad a_j,\qquad a_{i_{2}}\cdots a_{i_{s-1}}a_0. \end{equation*}

If $i_2\ne j_1$ , let $k\,{:\!=}\,{j_1}$ , and we proceed according to (ii), with $\alpha$ being the $k$ th projection of $a_ka_{j_2}\cdots a_{j_{d-2}}$ and $\beta$ that of $( a_{i_{2}}\cdots a_{i_{s-1}}a_0)^{-1}$ . If $i_2= j_1$ , we let instead $k\,{:\!=}\,j_s$ and consider the $k$ th projection of $(a_{i_{2}}\cdots a_{i_{s-1}}a_0)^{-1}$ multiplied with that of $a_0^{-1}a_{j_{d+1}}\cdots a_{j_{s-1}}a_k$ ; that is, situation (i).

Subcase (c): Suppose $r=s-1$ . Here we have the following elements in $H$ :

\begin{equation*} a_{j_1}\cdots a_{j_{d-2}}a_j,\qquad a_{j_{d+1}}\cdots a_{j_s},\qquad a_{i_1}\cdots a_{i_{s-2}}a_j,\qquad a_0. \end{equation*}

If $i_1\ne j_1$ , we let $k\,{:\!=}\,{j_1}$ , and proceed as in (ii), taking $\alpha$ to be the $k$ th projection of $a_ka_{j_2}\cdots a_{j_{d-2}}a_j$ and $\beta$ that of $( a_{i_{1}}\cdots a_{i_{s-2}}a_j)^{-1}$ . If $i_1= j_1$ , we instead let $k\,{:\!=}\,j_{d+1}$ and likewise following (ii) we consider the $k$ th level projection of $ a_ka_{j_{d+2}}\cdots a_{j_s}$ multiplied with that of $\big ( a_{i_1}\cdots a_{i_{s-2}}a_j \big )^{-1}$ .