2 Preliminaries

In the present section, we recall the notion of branch groups and establish prerequisites for the rest of the paper. For more information, see [Reference Bartsholdi, Grigorchuk and Šunik2, Reference Grigorchuk11].

2.1 The p-regular rooted tree and its automorphisms

Let T be the p-regular rooted tree, for an odd prime p. Let $X=\{1,2,\ldots ,p\}$ be an alphabet on p letters. The set of vertices of T can be identified with the free monoid $X^*$ , and we will freely use this identification without special mention. The root of T corresponds to the empty word $\varnothing $ , and for each word $v\in X^*$ and letter x, an edge connects v to $vx$ . There is a natural length function on $X^*$ , and the words w of length $|w| = n$ , representing vertices that are at distance n from the root, form the nth layer of the tree. The boundary $\partial T$ consisting of all infinite simple rooted paths is in one-to-one correspondence with the p-adic integers.

For u a vertex, we write $T_u$ for the full rooted subtree of T that has its root at u and includes all vertices v with u a prefix of v. For any two vertices u and v, the subtrees $T_u$ and $T_v$ are isomorphic under the map that deletes the prefix u and replaces it by the prefix v. We refer to this identification as the natural identification of subtrees and write $T_n$ to denote the subtree rooted at a generic vertex of level n.

We observe that every automorphism of T fixes the root and that the orbits of $\text {Aut }T$ on the vertices of the tree T are precisely its layers. Let $f \in \text {Aut } T$ be an automorphism of T. The image of a vertex u under f is denoted by $u^f$ . For a vertex u, considered as a word over X, and a letter $x \in X$ , we have $(ux)^f=u^fx^{\prime }$ where $x^{\prime } \in X$ is uniquely determined by u and f. This gives a permutation $f(u)$ of X, so that

$$ \begin{align*} (ux)^f = u^f x^{f(u)}. \end{align*} $$

The permutation $f(u)$ is called the label of f at u. The automorphism f is called rooted if $f(u)=1$ for $u\ne \varnothing $ . The automorphism f is called directed, with directed path $\ell $ for some $\ell \in \partial T$ , if the support $\{u \mid f(u)\ne 1 \}$ of its labelling is infinite and contains only vertices at distance $1$ from $\ell $ .

The section of f at a vertex u is the unique automorphism $f_u$ of $T \cong T_{|u|}$ given by the condition $(uv)^f = u^f v^{f_u}$ for $v \in X^*$ .

2.2 Subgroups of $\text {Aut } T$

Let $G\le \text {Aut } T$ . The vertex stabilizer $\text {st}_G(u)$ is the subgroup consisting of elements in G that fix the vertex u. For $n \in \mathbb {N}$ , the nth level stabilizer $\text {St}_G(n)= \cap _{|v|=n} \text {st}_G(v)$ is the subgroup of automorphisms that fix all vertices at level n. We emphasize the difference in the notation of vertex stabilizers and level stabilizers, which aims to avoid confusion from the fact that the first-level vertices of T are often identified with the elements of X. Note that elements in $\text {St}_G(n)$ fix all vertices up to level n and that $\text {St}_G(n)$ has finite index in G.

The full automorphism group $\text {Aut }T$ is a profinite group:

$$ \begin{align*} \text{Aut } T= \varprojlim_{n\to\infty} \text{Aut } T_{[n]}, \end{align*} $$

where $T_{[n]}$ denotes the subtree of T on the finitely many vertices up to level n. The topology of $\text {Aut } T$ is defined by the open subgroups $\text {St}_{\text{Aut}}(n)$ , for $n \in \mathbb {N}$ . For $G\le \text {Aut } T$ , we say that the subgroup G has the congruence subgroup property if for every subgroup H of finite index in G, there exists some $n\in \mathbb {N}$ such that $\text {St}_G(n)\leq H$ .

For $n\in \mathbb {N}$ , every $g\in \text {St}_{\text {Aut } T} (n)$ can be identified with a collection $g_1,\ldots ,g_{p^n}$ of elements of $\text {Aut } T_n$ , where $p^n$ is the number of vertices at level n. Denoting the vertices of T at level n by $u_1, \ldots , u_{p^n}$ , there is a natural isomorphism

$$ \begin{align*} \text{St}_{\text{Aut } T}(n) \cong \prod\nolimits_{i=1}^{p^n} \text{Aut }T_{u_i} \cong \text{Aut }T_n \times \overset{p^n}{\cdots} \times \text{Aut } T_n. \end{align*} $$

Recall that $\text {Aut } T_n$ is isomorphic to $\text {Aut } T$ via the natural identification of subtrees. Therefore, the decomposition $(g_1,\ldots ,g_{p^n})$ of g defines an embedding

$$ \begin{align*} \psi_n \colon \text{St}_{\text{Aut } T}(n) \rightarrow \prod\nolimits_{i=1}^{p^n} \text{Aut } T_{u_i} \cong \text{Aut } T \times \overset{p^n}{\cdots} \times \text{Aut } T. \end{align*} $$

For convenience, we will write $\psi =\psi _1$ .

For $\omega \in X^*$ , we further define

$$ \begin{align*} \varphi_\omega :\text{st}_{\text{Aut } T}(\omega) \rightarrow \text{Aut } T_{\omega} \cong \text{Aut } T \end{align*} $$

to be the natural restriction of $f\in \text {st}_{\text {Aut } T}(\omega )$ to its section $f_\omega $ .

We write $G_u=\varphi _u(\text {st}_G(u))$ for the restriction of the vertex stabilizer $\text {st}_G(u)$ to the subtree rooted at a vertex u. We say that G is self-similar if $G_u$ is contained in G for every vertex u, and we say that G is fractal if $G_u$ equals G for every vertex u, after the natural identification of subtrees.

The subgroup $\text {rist}_G(u)$ , consisting of all automorphisms in G that fix all vertices v of T not having u as a prefix, is called the rigid vertex stabilizer of u in G. The rigid nth level stabilizer is the product

$$ \begin{align*} \text{Rist}_G(n)=\prod\nolimits_{i=1}^{p^n} \text{rist}_G(u_i) \trianglelefteq G, \end{align*} $$

of the rigid vertex stabilizers of the vertices $u_1, \ldots , u_{p^n}$ at level n.

Let G be a subgroup of $\text {Aut } T$ acting spherically transitively, i.e., transitively on every layer of T. Then, G is a weakly branch group if $\mathrm {Rist}_G(n)$ is nontrivial for every $n \in \mathbb {N}$ , and G is a branch group if $\mathrm {Rist}_G(n)$ has finite index in G for every $n \in \mathbb {N}$ . If, in addition, the group G is self-similar and $1 \not = K \leq G$ with $K\times \cdots \times K \subseteq \psi (K\cap \text {St}_G(1))$ and $\lvert G : K \rvert < \infty $ , then G is said to be regular branch over K. If the condition $\lvert G : K \rvert < \infty $ in the previous definition is omitted, then G is said to be weakly regular branch over K.

3 The GGS-groups

By a, we denote the rooted automorphism, corresponding to the p-cycle $(1\, 2\, \cdots\, p)\in \text {Sym}(p)$ , that cyclically permutes the vertices $u_1, \ldots , u_p$ at the first level. Given a nonzero vector $ \mathbf {e} =(e_{1}, e_{2},\ldots , e_{p-1})\in (\mathbb {F}_p)^{p-1}$ , we recursively define a directed automorphism $b \in \text {St}_{\text {Aut } T}(1)$ via

$$ \begin{align*} \psi(b)=(a^{e_{1}}, a^{e_{2}},\ldots,a^{e_{p-1}},b). \end{align*} $$

We call the subgroup $G=G_{\mathbf {e}}=\langle a, b \rangle $ of $\text {Aut } T$ the GGS-group associated with the defining vector $\mathbf {e}$ . We observe that $\langle a \rangle \cong \langle b \rangle \cong C_p$ are cyclic groups of order p. Hence, whenever we refer to an exponent of a or of b, it goes without saying that it is an element of $\mathbb {F}_p$ .

A GGS-group G acts spherically transitively on the tree T, and every section of every element of G is contained in G. Moreover, a GGS-group G is fractal.

If $G_{\mathbf {e}}=\langle a, b\rangle $ is a GGS-group corresponding to the defining vector $\mathbf {e}$ , then G is an infinite p-group if and only if $ \sum \nolimits _{j=1}^{p-1} e_{j}= 0$ in $\mathbb {F}_p$ (cf. [Reference Grigorchuk11, Reference Vovkivsky16]).

We also call $G=\langle a,b\rangle $ a generalized Fabrykowski–Gupta group if

$$ \begin{align*} \psi(b)=(a,1,\ldots,1,b) \end{align*} $$

(see [Reference Francoeur6]). By [Reference Fernández-Alcober and Zugadi-Reizabal5, Theorem 2.16], it follows that a GGS-group whose defining vector contains only one nontrivial element is conjugate in $\text {Aut } T$ to a generalized Fabrykowski–Gupta group.

We write $\mathcal {G}= \langle a,b \rangle $ with $\psi (b) = (a,\ldots ,a,b)$ , for the GGS-group arising from the constant defining vector $(1,\ldots ,1)$ . It is known that $\mathcal {G}$ is weakly regular branch [Reference Fernández-Alcober and Zugadi-Reizabal5, Lemma 4.2] but not branch [Reference Fernández-Alcober, Garrido and Uria-Albizuri4, Theorem 3.7].

For all other GGS-groups $G \ne \mathcal {G}$ , we have from [Reference Fernández-Alcober and Zugadi-Reizabal5] that G is regular branch over $\gamma _3(G)$ . Furthermore, from [Reference Fernández-Alcober, Garrido and Uria-Albizuri4], a GGS-group G has the congruence subgroup property and is just infinite if and only if $G \ne \mathcal {G}$ ; we recall that an infinite group G is just infinite if all its proper quotients are finite.

The following result is useful.

As evident in the proof of the above result, all actions of group elements on the tree T will be taken on the right.

3.1 Length function

Here, we recall the following items from [Reference Pervova15]: the abelianization $G/G^{\prime }$ of a GGS-group G and a natural length function on elements of G.

Let $G =\langle a,b\rangle $ be a GGS-group acting on the p-regular rooted tree T. We consider

$$ \begin{align*} H = \langle \hat{a}, \hat{b}\mid \hat{a}^p=\hat{b}^p=1 \rangle, \end{align*} $$

the free product $\langle \hat {a}\rangle * \langle \hat {b} \rangle $ of cyclic groups $\langle \hat {a}\rangle \cong C_p$ and $\langle \hat {b} \rangle \cong C_p$ . There is a unique epimorphism $\pi \colon H \rightarrow G$ such that $\hat {a} \mapsto a$ and $\hat {b} \mapsto b$ , which induces an isomorphism from $H/H^{\prime }\cong \langle \hat a \rangle \times \langle \hat {b} \rangle \cong C_p^{2}$ to $G/G^{\prime }$ (see [Reference Pervova15]).

Let $h \in H$ . Recall that h can be uniquely represented in the form

$$ \begin{align*} h = \hat{a}^{\alpha_1} \cdot \hat{b}^{\beta_{1}} \cdot \hat{a}^{\alpha_2} \cdot \hat{b}^{\beta_{2}}\cdots \hat{a}^{\alpha_m} \cdot \hat{b}^{\beta_{m}} \cdot \hat{a}^{\alpha_{m+1}}, \end{align*} $$

where $m\in \mathbb {N}\cup \{0\}$ , $\alpha _1, \ldots , \alpha _{m+1}\in \mathbb {F}_p$ with $ \alpha _i \ne 0$ for $i\in \{2,\ldots ,m\}$ , and $\beta _{1}, \ldots , \beta _{m} \in \mathbb {F}_p^*$ .

We denote by $|h| = m$ the length of h, with respect to the factor $\langle \hat {b} \rangle $ . Clearly, for $h_1,h_2 \in H$ , we have

(3.1) $$ \begin{align} |h_1 h_2| \leq |h_1| + |h_2|. \end{align} $$

Then, for $G=\langle a,b \rangle $ a GGS-group, the length of $g \in G$ is

$$ \begin{align*} |g| = \min \{ |h| \mid h \in \pi^{-1}(g) \}. \end{align*} $$

Based on (3.1), we deduce that for $g_1,g_2 \in G$ ,

(3.2) $$ \begin{align} |g_1 g_2| \leq |g_1| + |g_2|. \end{align} $$

We end this section with the following result from [Reference Alexoudas, Klopsch and Thillaisundaram1], specialized to the setting of GGS-groups:

Lemma 3.2 [Reference Alexoudas, Klopsch and Thillaisundaram1, Lemma 4.4]

Let G be a GGS-group, and $g \in \text {St}_G(1)$ with $\psi (g)=(g_1,\ldots ,g_{p})$ . Then, $\sum _{j=1}^{p} |g_j| \le |g|$ , and $|g_j| \le \frac {|g|+1}{2}$ for each $j\in \{1,\ldots ,p\}$ .

In particular, if $|g|>1$ , then $|g_j| < |g|$ for every $j \in \{1,\ldots ,p\}$ .

3.2 The GGS-group defined by the constant vector

Due to the group $\mathcal {G}$ not being branch, some further properties of $\mathcal {G}$ need to be established for the next section.

By [Reference Fernández-Alcober and Zugadi-Reizabal5, Lemma 4.2], the group $\mathcal {G}$ is weakly regular branch over $K^{\prime }$ , where $K=\langle (ba^{-1})^{a^i}\mid i\in \{0,\ldots ,p-1\}\rangle ^{\mathcal {G}}$ .

Lemma 3.3 Let $\mathcal {G}$ and K be as above. Then, $K^{\prime }$ is a subdirect product of $K\times \overset {p}\cdots \times K$ and $\psi (K^{\prime \prime })\ge \gamma _3(K)\times \overset {p}\cdots \times \gamma _3(K)$ .

Proof This is similar to [Reference Garrido and Uria-Albizuri9, Proof of Proposition 17]. We first show that $K^{\prime }$ is a subdirect product of $K\times \overset {p}\cdots \times K$ . Indeed, writing $y_i=(ba^{-1})^{a^i}$ for $0\le i\le p-1$ , we have, for $0\le i< j\le p-1$ , that

$$ \begin{align*} \psi([y_i,y_j])&=\psi((b^{-1})^{a^{i-1}} (b^{-1})^{a^{j-2}} b^{a^{i-2}}b^{a^{j-1}})\\[2pt] &=\begin{cases} (1,\overset{i-3}\ldots,1,a^{-2}ba ,b^{-1}a,1,\overset{j-i-2}\ldots,1 ,a^{-1}b^{-1}a^2,a^{-1}b,1,\ldots,1) &\text{if }j>i+1,\\[2pt] (1,\overset{i-3}\ldots,1,a^{-2}ba ,b^{-2}a^2,a^{-1}b,1,\ldots,1) & \text{if }j=i+1, \end{cases}\\[2pt] &=\begin{cases} (1,\overset{i-3}\ldots,1,y_2 ,y_1^{-1},1,\overset{j-i-2}\ldots,1 ,y_2^{-1},y_1,1,\ldots,1) & \hspace{5pt}\quad\qquad\qquad\text{if }j>i+1,\\[2pt] (1,\overset{i-3}\ldots,1,y_2 ,(y_0^{-1}y_1^{-1})^a,y_1,1,\ldots,1) & \hspace{5pt}\quad\qquad\qquad\text{if }j=i+1, \end{cases} \end{align*} $$

hence the result.

Thus, for every $k_1\in K$ , there is some $g_1\in K^{\prime }$ with $\psi (g_1)=(k_1,*,\ldots ,*)$ . As $\mathcal {G}$ is weakly regular branch over $K^{\prime }$ , for every $k_2\in K^{\prime }$ , there is some $g_2\in K^{\prime }$ with $\psi (g_2)=(k_2,1,\ldots ,1)$ . Therefore, $\psi ([g_1,g_2])=([k_1,k_2],1,\ldots ,1)$ , and the second part of the lemma follows.▪

Proposition 3.4 Let $\mathcal {G}$ be the GGS-group defined by the constant vector. Then, every proper quotient of $\mathcal {G}$ is virtually nilpotent and has maximal subgroups only of finite index.

Proof For the first statement, using [Reference Francoeur7, Theorem 4.10], it suffices to show that $\mathcal {G}/K^{\prime \prime }$ is virtually nilpotent. As $\mathcal {G}^{\prime }/K^{\prime \prime }$ is of finite index in $\mathcal {G}/K^{\prime \prime }$ , it suffices to show that $\mathcal {G}^{\prime }/K^{\prime \prime }$ is nilpotent. Because $\psi (\mathcal {G}^{\prime })\le K\times \overset {p}\cdots \times K$ by [Reference Fernández-Alcober and Zugadi-Reizabal5, Lemma 4.2(iii)], the result follows from the second part of the previous lemma.

The second statement is immediate, because every virtually nilpotent group has maximal subgroups only of finite index: indeed, a nilpotent group G has $\Phi (G)$ containing $G^{\prime }$ (see [Reference Zassenhaus17]). Then, every maximal subgroup of a nilpotent group G is normal and hence of finite index. As the property of having only maximal subgroups of finite index passes to finite extensions [Reference Francoeur6, Corollary 5.1.3], we have that virtually nilpotent groups have maximal subgroups only of finite index.▪

4 Prodense subgroups

In the present section, we lay out the strategy for proving Theorem 1.1, where it suffices to restrict to nontorsion GGS-groups in light of [Reference Pervova15].

We recall that a subgroup H of G is prodense if $G = NH$ for every nontrivial normal subgroup N of G. Let G be a finitely generated group such that every proper quotient of G has maximal subgroups only of finite index. Then, by [Reference Francoeur7, Proposition 2.21], every maximal subgroup of infinite index in G is prodense, and every proper prodense subgroup is contained in a maximal subgroup of infinite index.

We now recall a key result concerning prodense subgroups of weakly branch groups.

Note that every proper quotient of a branch GGS-group has maximal subgroups only of finite index, because proper quotients of branch groups are virtually abelian (cf. [Reference Francoeur7, Proposition 2.22]). This, and Proposition 3.4, allows us to use the above proposition to show that nontorsion GGS-groups G do not possess proper prodense subgroups: for a prodense subgroup M of G, we will show that there exists a vertex u of T such that $M_u=G$ . This then shows that M is not proper.

For G a nontorsion GGS-group and M a prodense subgroup of G, it remains to show that there exists a vertex u of T such that $M_u=G$ . The following result was proved in a more general setting in [Reference Alexoudas, Klopsch and Thillaisundaram1], but it was stated for dense instead of prodense subgroups M. However, the proof, and hence the result, still holds for prodense subgroups.

It follows from the above result that it suffices to show that $b\in M_u$ for M a prodense subgroup of G and u a vertex of T. This will be done in the next section.

5 Obtaining b

For $G=\langle a,b\rangle $ a nontorsion GGS-group with defining vector $\mathbf {e}=(e_1,\ldots ,e_{p-1})$ , we write $\lambda :=\sum _{i=1}^{p-1} e_i\ne 0$ . Furthermore, we write $g\equiv _{G^{\prime }} h$ when $gh^{-1}\in G^{\prime }$ , for elements $g,h\in G$ .

Proof As $g\equiv _{{G}^{\prime }} b^{t}$ , it follows that

$$ \begin{align*} g=(b^{j_1})^{a^{l_1}}(b^{j_2})^{a^{l_2}}\cdots (b^{j_n})^{a^{l_n}}, \end{align*} $$

for some $n\in \mathbb {N}$ , $l_1,\ldots ,l_n\in \mathbb {F}_p$ with $l_k\ne l_{k+1}$ for $1\le k\le n-1$ , and $j_1,\ldots ,j_n\in \mathbb {F}_p^*$ with $\sum _{k=1}^n j_k= t$ in $\mathbb {F}_p$ .

Now, for $u\in X$ , we have $\varphi _u(g) = \varphi _u(b^{a^{l_1}})^{j_1}\varphi _u(b^{a^{l_2}})^{j_2}\cdots \varphi _u(b^{a^{l_n}})^{j_n}$ , and because

$$ \begin{align*}\varphi_u(b^{a^{l}}) = \begin{cases} a^{e_{u-l}} & \text{ if } l\ne u, \\ b & \text{ if } l=u, \end{cases}\end{align*} $$

we have $\varphi _u(g)\equiv _{{G}^{\prime }} a^{n_u}b^{m_u}$ with

$$ \begin{align*} m_u= \sum j_k, \end{align*} $$

where the sum is taken over all k such that $l_k=u$ , and

(5.1) $$ \begin{align} n_u=\sum_{\substack{i=0 \\ i\ne u}}^{p-1} e_{u-i}\,m_i = \sum_{j=1}^{p-1} e_j\,m_{u-j}. \end{align} $$

It is clear that Cases 1 and 2 are mutually exclusive. Thus, it suffices to show that if Case 1 does not hold, then Case 2 does. So we assume now that Case 1 does not hold. Suppose that $n_u = \lambda m_u$ for all $u\in X$ . Then,

$$ \begin{align*} \sum_{j=1}^{p-1} e_j\,m_{u-j} -\lambda m_u=0 \quad\text{for all }u\in X. \end{align*} $$

This is equivalent to

$$ \begin{align*} (e_1, \ldots , e_{p-1},-\lambda) \cdot \left(\begin{array}{c} m_{p}\\ m_{p-1}\\ \vdots \\ m_{2}\\ m_1 \end{array}\right) = (e_1, \ldots , e_{p-1},-\lambda) \cdot \left(\begin{array}{c} m_{p-1}\\ m_{p-2}\\ \vdots \\ m_{1}\\ m_{p} \end{array}\right) =\cdots \end{align*} $$

$$ \begin{align*} \cdots = (e_1, \ldots , e_{p-1},-\lambda) \cdot \left(\begin{array}{c} m_{1}\\ m_{p}\\ \vdots \\ m_{3}\\ m_{2} \end{array}\right)= 0. \end{align*} $$

In other words,

(5.2) $$ \begin{align} \left(\begin{array}{c} m_{p}\\ m_{p-1}\\ \vdots \\ m_{2}\\ m_1 \end{array}\right), \left(\begin{array}{c} m_{p-1}\\ m_{p-2}\\ \vdots \\ m_1\\ m_{p} \end{array}\right), \ldots, \left(\begin{array}{c} m_{1}\\ m_{p}\\ \vdots \\ m_{3}\\ m_{2} \end{array}\right) \end{align} $$

are all in the subspace orthogonal to $(e_1 , \ldots , e_{p-1} ,-\lambda )$ . However, the vectors in (5.2) form the rows of a circulant matrix. The rank of this circulant matrix is less than p if and only if $\sum _{i=1}^p m_i = 0$ (cf. [Reference Fernández-Alcober and Zugadi-Reizabal5, Lemma 2.7(i)]). However, $\sum _{i=1}^p m_i = \sum _{k=1}^n j_k= t\ne 0$ . Thus, there is at least one vertex $u_1\in X$ such that $n_{u_1}\ne \lambda m_{u_1}$ .

Suppose that there is only one such $u_1$ . Then, $n_u= \lambda m_u$ for all $u\ne u_1$ . Notice that we have

$$ \begin{align*} \sum_{u\in X}n_u &= \sum_{u\in X}\sum_{j=1}^{p-1}e_jm_{u-j} = \sum_{j=1}^{p-1}\sum_{u\in X}e_jm_{u-j}= \sum_{j=1}^{p-1} e_j \sum_{u\in X} m_u=\lambda \sum_{u\in X} m_u. \end{align*} $$

Hence, we have

$$ \begin{align*} n_{u_1}+\lambda \sum_{u\ne u_1} m_u&= \lambda\sum_{u\in X}m_u, \end{align*} $$

which yields $n_{u_1} = \lambda m_{u_1}$ , a contradiction.

For the final statement, clearly $m_u\ne 0$ for at least one $u\in X$ , because $\sum _{u\in X} m_u= t\ne 0$ . Because we are not in Case 1, it follows that $n_u\ne 0$ . From (5.1), the result follows.▪

Proof Suppose for the sake of contradiction that this is not the case. By Lemma 3.2 and using the fact that x is in Case 2 of Lemma 5.1, there exist two vertices $v_1,v_2\in X$ such that $|x_{v_1}|=|x_{v_2}|=\frac {|x|}{2}$ and $x_u=1$ for all $u\in X\backslash \{v_1,v_2\}$ . This means that

$$ \begin{align*} x=\left(b^{i_1}\right)^{a^{v_1}}\left(b^{j_1}\right)^{a^{v_2}}\cdots \left(b^{i_{\mu}}\right)^{a^{v_1}}\left(b^{j_\mu}\right)^{a^{v_2}}, \end{align*} $$

for some $i_1,\ldots , i_{\mu }, j_1, \ldots , j_{\mu }\in \mathbb {F}_p$ . Because the result is true for x if and only if it is true for $x^{a^{-v_1}}$ , we can suppose without loss of generality that

$$ \begin{align*} x=b^{i_1}\left(b^{j_1}\right)^{a^{v}}\cdots b^{i_{\mu}}\left(b^{j_\mu}\right)^{a^{v}}, \end{align*} $$

for some $v\in X$ . Let us set $i=\sum _{k=1}^{\mu }i_k$ and $j=\sum _{k=1}^{\mu }j_k$ . Notice that $i+j=t$ . Furthermore, because x is in Case 2 of Lemma 5.1, there must exist two vertices with nonzero powers of b, and our assumptions then imply that i and j are both nonzero.

For all $u\in X\backslash \{p,v\}$ , the elements $\varphi _u(b)$ and $\varphi _u(b^{a^v})$ are both powers of a, and thus commute. Therefore, for $u\in X\backslash \{p,v\}$ , we have

$$ \begin{align*} x_u=\varphi_u(x)=\varphi_u(b^i(b^j)^{a^v})=a^{ie_u + je_{u-v}}. \end{align*} $$

Because $x_u=1$ for all $u\in X\backslash \{p,v\}$ , we have $ie_u+je_{u-v}=0$ for all $u\in X\backslash \{p,v\}$ . In other words, for all $2\leq k \leq p-1$ , we have $ie_{kv}+je_{(k-1)v}=0$ . By induction, we see that for all $1\leq k \leq p-1$ , we have $e_{kv} = r^{k-1}e_{v}$ , where $r=-\frac {j}{i}$ , which is well defined, because $i\ne 0$ . Notice that $r\ne 0$ , because $j\ne 0$ , and $r\ne 1$ , because $i+j=t\ne 0$ . Therefore, we have

$$ \begin{align*}\lambda = \sum_{k=1}^{p-1}e_k = \sum_{k=1}^{p-1}e_{kv} = \sum_{k=1}^{p-1}r^{k-1}e_v = e_v\sum_{k=0}^{p-2}r^k = e_v\frac{r^{p-1}-1}{r-1} =0,\end{align*} $$

a contradiction. The conclusion follows.▪

We will now establish the following in several steps.

First, let us consider the set

$$ \begin{align*} U=\left\{g\in \bigsqcup_{v\in X^*} H_v \mid g\equiv_{G^{\prime}} a^{ \lambda i}b^i \text{ for some }i\ne 0 \right\}. \end{align*} $$

Because H is a prodense subgroup of G and $G^{\prime }$ is a nontrivial normal subgroup in G, the set U is nonempty. Let $y\in U$ be an element of minimal length in U. Let $v\in X^*$ be such that $y\in H_v$ . By Proposition 4.1, the subgroup $H_v$ is prodense in $G_v=G$ . Hence, if we prove the above proposition for $H_v$ , we will also have proved it for H. Thus, without loss of generality, we may assume that v is the root of the tree $X^*$ . Furthermore, in light of Lemma 5.3, we have “y,” that is, an element of U of minimal length, everywhere.

Now, let us consider the set

$$ \begin{align*} V=\left\{g\in \bigsqcup_{v\in X^*} H_v \mid g\equiv_{G^{\prime}} b^i \text{ for some }i\ne 0 \right\}, \end{align*} $$

and let $x\in V$ be an element of minimal length in V. Likewise, such an element exists as V is nonempty by the prodensity of H. Let $w\in X^*$ be such that $x\in H_w$ . As before, we may assume that w is the root of the tree. Hence, using the fact that we get y everywhere, we have y and x at the root.

If $|x|=1$ , then $x=a^{-j}b^ia^j$ with $i\ne 0$ and $j\in \{1,\ldots ,p\}$ . Thus, we have that $\varphi _j(x)=b^i$ . As i is invertible modulo p, we obtain $b\in H_j$ .

We will now show that $|x|$ must be equal to 1. The case $|x|=0$ is not possible, so we assume that $|x|>1$ . In this case, there cannot exist $u\in X$ such that $\varphi _u(x)\equiv _{G^{\prime }} b^j$ with $j\ne 0$ . Indeed, because $|\varphi _u(x)|<|x|$ , this would contradict the minimality of x (cf. Lemma 3.2). Therefore, according to Lemma 5.1, there exists $u\in X$ such that $\varphi _u(x)\equiv _{G^{\prime }} a^{i_u}b^{j_u}$ with $i_u\ne 0$ and $i_u\ne \lambda j_u$ . For this $u\in X$ , we write $x_u=\varphi _u(x)$ .

Furthermore, also by Lemma 5.1, there exist two vertices $u_1,u_2\in X$ such that $j_{u_1},j_{u_2}\ne 0$ with $i_{u_1},i_{u_2}\ne 0$ . Now, by Lemma 5.3, we see that $\varphi _v(x_{u_i}^{\,p})\in U$ for $i\in \{1,2\}$ and $v\in X$ , and thus

$$ \begin{align*} |y|\le |\varphi_v(x_{u_i}^{\,p})|\le |x_{u_i}|. \end{align*} $$

Hence, using $\sum _{u\in X}|x_u|\le |x|$ , it follows that

(5.3) $$ \begin{align} 2|y|\le |x|. \end{align} $$

Remark 5.5 We observe that if there is a $v\in X^*$ with $a^n\in H_v$ for some $n\ne 0$ , then writing $y\equiv _{G^{\prime }} a^{\lambda j}b^j$ for some $j\ne 0$ , the product of y with $a^{-\lambda j}$ yields

$$ \begin{align*} |x|\le |y|, \end{align*} $$

which contradicts (5.3), and so we are done in this case. Hence, in what follows, we will assume freely without special mention that $a\not \in H_v$ for all $v\in X^*$ .

Proof Without loss of generality, we may assume that $z\notin \text {St}_G(1)$ . Indeed, otherwise, because z is nontrivial, there must exist some $w\in X^*$ such that $z\in \text {st}_G(w)$ and $\varphi _w(z)\notin \text {St}_G(1)$ . As we have $|z|\geq |\varphi _w(z)|$ , it is sufficient to show that $|\varphi _w(z)|\geq |y|$ . Thus, we may assume that $z\notin \text {St}_G(1)$ . Hence, we have $z\equiv _{G^{\prime }} a^{i_z}b^{j_z}$ for some $i_z,j_z\in \mathbb {F}_p$ with $i_z\ne 0$ .

Notice that for all $u\in X$ , we have $\varphi _u(z^p)\equiv _{G^{\prime }} a^{\lambda j_z}b^{j_z}$ by Lemma 5.3. In particular, if $j_z\neq 0$ , by the minimality of $|y|$ , we must have $|\varphi _u(z^p)| \geq |y|$ . Because Lemma 5.3 also says that $|z|\geq |\varphi _u(z^p)|$ , the conclusion follows in this case.

Thus, it only remains to treat the case where $j_z=0$ . Let us assume, for the sake of contradiction, that we have $|z|<|y|$ , and let $j_y\ne 0$ be such that $y\equiv _{G^{\prime }}a^{\lambda j_y}b^{j_y}$ . Because $j_z=0$ , there exists $k\in \{1,2,\ldots , p-1\}$ such that $z^ky\equiv _{G^{\prime }} b^{j_y}$ . By Lemmas 3.2 and 5.3 together with (3.2), for all $u\in X$ , we have $|\varphi _u(z^ky)|\leq |z|+\frac {|y|+1}{2} < 2|y|$ . If there were some $u\in X$ with $\varphi _u(z^ky)\equiv _{G^{\prime }}b^l$ for some $l\ne 0$ , by the minimality of $|x|$ , we would have that $|x|\leq |\varphi _u(z^ky)|<2|y|$ , a contradiction to (5.3). Thus, by Lemma 5.1, there must exist two vertices $u_1,u_2\in X$ such that $\varphi _{u_l}(z^ky)\equiv _{G^{\prime }} a^{i_{u_l}}b^{j_{u_l}}$ with $i_{u_l}, j_{u_l}\ne 0$ , for $l\in \{1,2\}$ .

Let us write $y=a^{i_y} \cdot \psi ^{-1}((y_1,\ldots ,y_p))$ and $z=a^{i_z} \cdot \psi ^{-1}((z_1,\ldots ,z_p))$ . For any $u\in X$ , we have, remembering that we act on the tree T on the right,

$$ \begin{align*} &\varphi_u(z^ky)\\ &\quad=\varphi_u\Big( \psi^{-1}\big((z_1,\ldots,z_p)^{a^{-i_z}} (z_1,\ldots,z_p)^{a^{-2i_z}} \cdots (z_1,\ldots,z_p)^{a^{-ki_z }} (y_1,\ldots,y_p)\big)\Big)\\ &\quad= z_{u+i_z}z_{u+2i_z}\cdots z_{u+ki_z}y_u. \end{align*} $$

We also have

$$ \begin{align*} &\varphi_u(z^{k-p}y)\\ &\quad=\varphi_u\Big( \psi^{-1}\big((z_1^{-1},\ldots,z_p^{-1}) (z_1^{-1},\ldots,z_p^{-1})^{a^{i_z}} \cdots (z_1^{-1},\ldots,z_p^{-1})^{a^{(p-k-1)i_z }} (y_1,\ldots,y_p)\big)\Big)\\ &\quad= z^{-1}_{u}z^{-1}_{u-i_z}\cdots z^{-1}_{u+(k+1)i_z}y_u. \end{align*} $$

As $\varphi _u(z^p)\equiv _{G^{\prime }} 1$ , we have

$$ \begin{align*}\varphi_u(z^{k-p}y)\equiv_{G^{\prime}} \varphi_u(z^p)\varphi_u(z^{k-p}y) \equiv_{G^{\prime}} \varphi_u(z^ky).\end{align*} $$

In particular, this means that for $u\in \{u_1,u_2\}$ , we have

$$ \begin{align*}\varphi_{u}(z^{k-p}y)\equiv_{G^{\prime}} \varphi_{u}(z^{k}y) \equiv_{G^{\prime}} a^{i_{u}}b^{j_{u}}, \end{align*} $$

with $i_{u}, j_{u}\ne 0$ . We have seen above that this implies that $|\varphi _{u}(z^{k-p}y)|, |\varphi _{u}(z^ky)|\geq |y|$ .

Because $\sum _{n=1}^{p}|y_n|\leq |y|$ , there must exist some $u\in \{u_1,u_2\}$ such that $|y_{u}|\leq \frac {|y|}{2}$ . For this u, we have

$$ \begin{align*} |\varphi_{u}(z^ky)| \leq |y_u|+ \sum_{n=1}^{k}|z_{u+ni_z}|, \end{align*} $$

and

$$ \begin{align*} |\varphi_{u}(z^{k-p}y)| \leq |y_u|+ \sum_{n=0}^{p-k-1}|z_{u-ni_z}|. \end{align*} $$

Because $\sum _{n=0}^{p-1}|z_{u-ni_z}|\leq |z|$ , we must have that one of $\sum _{n=1}^{k}|z_{u+ni_z}|$ or $\sum _{n=0}^{p-k-1}|z_{u-ni_z}|$ is at most $\frac {|z|}{2}$ . Consequently, we either have

$$ \begin{align*} |\varphi_{u}(z^ky)| \leq |y_u| + \frac{|z|}{2} \leq \frac{|y|+|z|}{2}< |y|, \end{align*} $$

or $|\varphi _{u}(z^{k-p}y)|<|y|$ . In either case, we get a contradiction, as required.▪

The following result is essential. First, let us write $y=a^{i_y} \cdot \psi ^{-1}((y_1,\ldots ,y_p))\equiv _{G^{\prime }} a^{\lambda j_y}b^{j_y} $ , for some $j_y\ne 0$ and $i_y=\lambda j_y$ . By abuse of notation, we will still write y for $\varphi _u(y^p)$ for any $u\in X$ (cf. Lemma 5.3). In addition, for notational convenience, we sometimes write $z_v$ or $(z_1,\ldots ,z_p)_v$ for $\varphi _v(z)$ , where $z=\psi ^{-1}((z_1,\ldots ,z_p))\in \text {st}_G(v)$ and $v\in X^*\backslash \{\varnothing \}$ .

Proof Let

$$ \begin{align*} W=\left\{g\in \bigsqcup_{v\in X^*} H_v \mid g\equiv_{G^{\prime}} a^n b^m \text{ for some }m,n \text{ with } n\ne 0, n\ne \lambda m \right\}, \end{align*} $$

and let $w\in W$ be an element of minimal length. We need to show that $|w|=|y|$ . Let us first remark that $|w|\leq \frac {|x|}{2}$ . Indeed, because $|x|>1$ , it follows from Lemma 5.1 that there exist $u_1,u_2\in X$ such that $x_{u_1},x_{u_2}\in W$ . Because $|x_{u_1}|+|x_{u_2}|\leq |x|$ , we conclude that $|w|\leq \min \{|x_{u_1}|, |x_{u_2}|\}\leq \frac {|x|}{2}$ .

Let us write $w=a^n\cdot \psi ^{-1}((w_1,\ldots ,w_p))$ . There exists $k\in \{1,2,\ldots ,p-1\}$ with $y^k\equiv _{G^{\prime }} y^{k-p}\equiv _{G^{\prime }} a^n b^{kj_y}$ . Using y, or more precisely the projection of some power of y to the vertex where w lives, to cancel $a^n$ in w yields

(5.4) $$ \begin{align} |(y^{-k}w)_v|&= \lvert\big( (y_1^{-1},\ldots,y_p^{-1}) (y_1^{-1},\ldots,y_p^{-1})^{a^{i_y}} \cdots (y_1^{-1},\ldots,y_p^{-1})^{a^{(k-1)i_y}} (w_1,\ldots,w_p)\big)_v\rvert\nonumber\\ &= | y^{-1}_{v}y^{-1}_{v-i_y}\cdots y^{-1}_{v-(k-1)i_y}w_v| \nonumber\\ &\le |y^{-1}_{v}y^{-1}_{v-i_y}\cdots y^{-1}_{v-(k-1)i_y}|+|w_v|\nonumber\\ &\le \sum_{d=0}^{k-1} |y_{v-di_y}|+|w_v|, \end{align} $$

and

(5.5) $$ \begin{align} |(y^{p-k}w)_v|&= |y_{v-(p-1)i_y}\cdots y_{v-i_y}y_{v} y^{-1}_{v}y^{-1}_{v-i_y}\cdots y^{-1}_{v-(k-1)i_y}w_v| \nonumber\\ &\le |y_{v-(p-1)i_y}\cdots y_{v-ki_y}|+|w_v|\nonumber\\ &\le \sum_{d=k}^{p-1} |y_{v-di_y}|+|w_v|, \end{align} $$

for $v\in X$ .

Because $\sum _{d=0}^{p-1} |y_{v-di}| \leq |y|$ for $v\in X$ , we have

$$ \begin{align*} \min{\left\{|(y^{-k}w)_v|,|(y^{p-k}w)_v|\right\}} \leq \frac{|y|}{2}+|w_v| \leq \frac{|y|}{2} + \frac{|x|}{4}+\frac{1}{2}\leq\frac{|x|+1}{2}, \end{align*} $$

and

$$ \begin{align*} \max{\left\{|(y^{-k}w)_v|,|(y^{p-k}w)_v|\right\}} \leq |y|+|w_v| \leq \frac{|x|}{2} + \frac{|x|}{4}+\frac{1}{2}\leq\frac{3|x|+2}{4}, \end{align*} $$

making use of (5.3).

As we will see, this implies that there exists a $w_1\in \left \{(y^{-k}w)_v, (y^{p-k}w)_v \mid v\in X\right \}$ such that $w_1\in W$ and $|w_1|\leq \frac {|y|+|w|}{2}$ .

Indeed, if $y^{-k}w$ is in Case 1 of Lemma 5.1, then there exists $v\in X$ such that $(y^{-k}w)_v \equiv _{G^{\prime }} b^{t^{\prime }}$ for some $t^{\prime }\ne 0$ . By the minimality of $|x|$ and the above inequalities, we must have $|x|\leq |(y^{-k}w)_v|\leq \frac {3|x|+2}{4}$ . This is only possible if $|x|\leq 2$ , and because we assume that $|x|>1$ , this implies $|x|=2$ , and so $|y|=1$ by (5.3). Notice that because $|x|=2$ , we cannot have $|x|\leq |(y^{-k}w)_v|\leq \frac {|x|+1}{2}$ , and thus, by the inequalities above, we must have $|(y^{p-k}w)_v|\leq \frac {|x|+1}{2} = \frac {3}{2}$ . Because the length must be an integer, we have $|(y^{p-k}w)_v| \leq 1 \leq \frac {|y|+|w|}{2}$ , where the last inequality follows from Lemma 5.6 applied to $w\ne 1$ . Notice that $(y^{p-k}w)_v = (y^p)_v(y^{-k}w)_v \equiv _{G^{\prime }} a^{\lambda j_y}b^{j_y+t^{\prime }}$ with $\lambda j_y \ne 0$ . Thus, in this case, $w_1=(y^{p-k}w)_v$ has the required properties.

By symmetry, if $y^{p-k}w$ is in Case 1 of Lemma 5.1, then we can also find some $w_1$ of the required form. Thus, it only remains to check the case when both $y^{-k}w$ and $y^{p-k}w$ are in Case 2 of Lemma 5.1. In this case, there exist at least two vertices $v_1,v_2\in X$ such that $(y^{- k}w)_{v_1}$ and $(y^ {-k}w)_{v_2}$ have nonzero total a-exponent differing from $\lambda $ times their total b-exponent, and it follows from Lemma 5.3 that $(y^{p-k}w)_{v_1}$ and $(y^ {p-k}w)_{v_2}$ satisfy the same property. Because $|w_{v_1}|+|w_{v_2}|\leq |w|$ , we can assume without loss of generality that $|w_{v_1}|\leq \frac {|w|}{2}$ . Then, we have

$$ \begin{align*}\min{\left\{|(y^{-k}w)_{v_1}|,|(y^{p-k}w)_{v_1}|\right\}} \leq \frac{|y|}{2}+|w_{v_1}| \leq \frac{|y|+|w|}{2},\end{align*} $$

and we set $w_1$ as the smallest of these two elements.

We have thus found some $w_1\in W$ with $|w_1|\leq \frac {|y|+|w|}{2}$ . By the minimality of $|w|$ , we have $|w|\leq |w_1|$ , and so $|w|\leq |y|$ . As we have $|y|\leq |w|$ from Lemma 5.6, we conclude that $|w|=|y|$ .▪

Finally, to prove Proposition 5.4, we return to our assumption that $|x|>1$ . We will now obtain a contradiction using the above results. Hence, $|x|=1$ , as required.

We will use w from Lemma 5.7 to show that there exists a $g \in \bigsqcup _{v\in X^*}H_v$ with $0<|g|<|y|$ , which will contradict Lemma 5.6.

As before, we write $y=a^{i_y} \cdot \psi ^{-1}((y_1,\ldots ,y_p))\equiv _{G^{\prime }} a^{\lambda j_y}b^{j_y} $ for some $j_y\ne 0$ and so $i_y=\lambda j_y$ . From Lemma 5.7, we have $w\in H_v$ for some $v\in X^*$ with $w\equiv _{G^{\prime }} a^nb^m$ for $n\ne \lambda m$ , $n\ne 0$ and $|w|=|y|$ . Let $k\in \{1,2,\ldots ,p-1\}$ be such that $y^k\equiv _{G^{\prime }} a^{n}b^{kj_y}$ . There must exist at least two vertices $v_1,v_2\in X$ such that $(y^{-k}w)_{v_1}$ and $(y^{-k}w)_{v_2}$ have nonzero total a-exponent differing from $\lambda $ times their total b-exponent. Indeed, as we saw in the proof of the previous lemma, both $y^{-k}w$ and $y^{p-k}w$ must be in Case 2 of Lemma 5.1, unless possibly when $|x|=2$ and $|y|=1$ . As we will now see, this last case is impossible. Certainly, if $|x|=2$ , $|y|=1$ and one of $y^{-k}w$ or $y^{p-k}w$ is in Case 1 of Lemma 5.1, then there exists $u\in X$ such that $(y^{-k}w)_u\equiv _{G^{\prime }} b^{t^{\prime }}$ or $(y^{p-k}w)_u\equiv _{G^{\prime }} b^{t^{\prime }}$ for some $t^{\prime }\ne 0$ . For concreteness, let us assume that $(y^{-k}w)_u\equiv _{G^{\prime }} b^{t^{\prime }}$ ; the other case is similar. Because $|y|=|w|=1$ , there exist $i_1,i_2,i_3,i_4\in \{0,1,\ldots , p-1\}$ such that $y=a^{i_1}b^{j_y}a^{i_2}$ and $w=a^{i_3}b^{m}a^{i_4}$ (in particular, notice that we have $i_1+i_2=i_y$ and $i_3+i_4=n$ , where these equations are taken modulo p). A direct computation then yields that

$$ \begin{align*} y^{-k}w &= a^{-i_2}(b^{-j_y})(b^{-j_y})^{a^{i_y}} \cdots (b^{-j_y})^{a^{(k-1)i_y}}a^{i_2-ki_y}a^{i_3}b^ma^{i_4}\\ &= (b^{-j_y})^{a^{i_2}}(b^{-j_y})^{a^{i_2+i_y}} \cdots (b^{-j_y})^{a^{i_2+(k-1)i_y}}(b^m)^{a^{n-i_3}}. \end{align*} $$

Therefore, we have

$$ \begin{align*}(y^{-k}w)_u = (b^{-j_y})_{u-i_2}(b^{-j_y})_{u-i_2-i_y} \cdots (b^{-j_y})_{u-i_2-(k-1)i_y}(b^m)_{u-n+i_3}.\end{align*} $$

Using the fact that for $v\in X$ , we have

$$ \begin{align*}b_v=\begin{cases} a^{e_v} & \text{ if }v\ne p,\\ b & \text{ if } v=p, \end{cases}\end{align*} $$

and that $u-i_2, u-i_2-i_y,\ldots , u-i_2-(k-1)i_y$ are all different vertices, we see that there are only four different possible forms for $(y^{-kw})_u$ : $a^*$ , $a^*b^{-j_y}a^*$ , $a^*b^{m}$ , or $a^*b^{-j_y}a^*b^{m}$ , where the stars represent unimportant (possibly zero) powers of a. Because we assumed that $(y^{-k}w)_u\equiv _{G^{\prime }} b^{t^{\prime }}$ , we must have $2=|x|\leq |(y^{-k}w)_u|$ , and thus only the last form is possible, because the other forms yield elements of length at most one. In particular, notice that we have $t^{\prime }=m-j_y$ . Let us now consider the element $(y^{p-k}w^{1-p})_u = (y^p)_u(y^{-k}w)_u(w^{-p})_u$ . Because $(y^p)_u\equiv _{G^{\prime }} a^{\lambda j_y}b^{j_y}$ and $w^{-p}\equiv _{G^{\prime }} a^{-\lambda m}b^{-m}$ , we have $(y^{p-k}w^{1-p})_u \equiv _{G^{\prime }} a^{\lambda (j_y-m)}b^{t^{\prime }+j_y-m} = a^{-\lambda t^{\prime }}$ . Considerations of length allow us to conclude that in fact, we must have $(y^{p-k}w^{1-p})_u = a^{-\lambda t^{\prime }}$ . Indeed, in exactly the same way as above, we find that

$$ \begin{align*}(y^{p-k}w^{1-p})_u = \prod_{d=1}^{p-k}(b^{j_y})_{u-i_2+di_y}\prod_{v\in X\backslash \{u-n+i_3\}}(b^{-m})_v.\end{align*} $$

As the sets $\{u-i_2+i_y, u-i_2+2i_y,\ldots , u-i_2+(p-k)i_y\}$ and $X\backslash \{u-n+i_3\}$ are disjoint from $\{u-i_2, u-i_2-i_y, \ldots , u-i_2-(k-1)i_y\}$ and $\{u-n+i_3\}$ , respectively, and that the latter two sets both contain p by previous considerations, we conclude that all the elements in the above product are powers of a, and thus that $(y^{p-k}w^{1-p})_u=a^{-\lambda t^{\prime }}$ . As $t^{\prime }\ne 0$ , this element is nontrivial, but it is of length $0$ , a contradiction with Lemma 5.6. A similar argument yields a contradiction if we assume that $y^{p-k}w$ is in Case 1 of Lemma 5.1. We conclude that both $y^{-k}w$ and $y^{p-k}w$ must be in Case 2 of Lemma 5.1.

Let $u\in \{v_1,v_2\}$ . We may assume that $|w_u|=\frac {|w|}{2}$ , else we are done. Indeed, otherwise, one of $|w_{v_1}|$ or $|w_{v_2}|$ must be strictly smaller than $\frac {|w|}{2}$ . Let us suppose without loss of generality that $|w_{v_2}|<\frac {|w|}{2}$ . Then, either $|(y^{-k}w)_{v_2}|<\frac {|y|}{2}+\frac {|w|}{2}=|y|$ or $|(y^{p-k}w)_{v_2}|<\frac {|y|}{2}+\frac {|w|}{2}=|y|$ . As this contradicts Lemma 5.6, we have that

(5.6) $$ \begin{align} |w_{v_1}|=|w_{v_2}|=\frac{|w|}{2}, \end{align} $$

and hence $|y|=|w|=2\mu $ for some $\mu \in \mathbb {N}$ .

By symmetry, that is, by considering $w^{k^{\prime }}y$ for some $k^{\prime }$ , it likewise follows that only two first-level sections of y are of nonzero length and that the sum of their length must be $|y|$ . Let $i_1,i_2\in X$ be such that

$$ \begin{align*} |y_{i_1}|=|y_{i_2}|=\frac{|y|}{2}=\mu. \end{align*} $$

If $k=1$ , then $|x|\le |y^{-1}w|\leq |y|+|w|=2|y|$ and hence $|y^{-1}w|=2|y|$ by (5.3). Therefore, it follows from Lemma 5.2 that there exists $u\in X$ such that $(y^{-1}w)_u\ne 1$ and $|(y^{-1}w)_u|<|y|$ . As this contradicts Lemma 5.6, we conclude that the case $k=1$ is impossible. Likewise, we see that the case $k=p-1$ is impossible by looking at $y^{p-(p-1)}w=yw$ .

Let us now assume that $1<k<p-1$ . For $v\in X$ , recall that we have

$$ \begin{align*}|(y^{-k}w)_v|=|y^{-1}_{v}y^{-1}_{v-i_y}\cdots y^{-1}_{v-(k-1)i_y}w_v|\leq \sum_{d=0}^{k-1} |y_{v-di_y}|+|w_v|,\end{align*} $$

and

$$ \begin{align*}|(y^{p-k}w)_v|=|(y_{v-ki_y}^{-1}y_{v-(k+1)i_y}^{-1}\cdots y_{v-(p-1)i_y}^{-1})^{-1}w_v|\leq \sum_{d=k}^{p-1} |y_{v-di_y}|+|w_v|.\end{align*} $$

Because we assume that $|w_{v_1}|=|w_{v_2}|=\frac {|y|}{2}$ , it follows that unless $\sum _{d=0}^{k-1} |y_{v_l-di_y}|=\sum _{d=k}^{p-1} |y_{v_l-di_y}|=\frac {|y|}{2}$ for $l\in \{1,2\}$ , then one of $|(y^{-k}w)_{v_l}|$ or $|(y^{p-k}w)_{v_l}|$ is strictly smaller than $|y|$ , which is impossible.

Notice also that if there exists $v\in X\backslash \{v_1,v_2\}$ such that $0<|\prod _{d=0}^{k-1}y_{v-di_y}^{-1}|<|y|$ , then $(y^{-k}w)_{v}$ is a nontrivial element of length strictly smaller than $|y|$ , because we must have $|w_v|=0$ when v is different from $v_1$ or $v_2$ . As this contradicts Lemma 5.6, for all $v\in X\backslash \{v_1,v_2\}$ , we must have either $|\prod _{d=0}^{k-1}y_{v-di_y}^{-1}|=0$ or $|\prod _{d=0}^{k-1}y_{v-di_y}^{-1}|=|y|$ . Likewise, we must also have either $|\prod _{d=k}^{p-1}y_{v-di_y}^{-1}|=0$ or $|\prod _{d=k}^{p-1}y_{v-di_y}^{-1}|=|y|$ .

This implies strong restrictions on the form of y. Recall from above that there exist $i_1,i_2\in X$ such that $|y_{i_1}|=|y_{i_2}|=\frac {|y|}{2}$ . This implies that, up to renaming $i_1$ and $i_2$ , we must have

(5.7) $$ \begin{align} y=a^{i_y}(b^{s_1})^{a^{i_1}}(b^{s_2})^{a^{i_2}}\cdots (b^{s_{2\mu-1}})^{a^{i_1}}(b^{s_{2\mu}})^{a^{i_2}}, \end{align} $$

where $s_1+\cdots +s_{2\mu }=j_y$ . In particular, for all $u\in X\backslash \{i_1,i_2\}$ , we have $|y_u|=0$ . This implies that if, for some $v\in X$ , the set $\{v-di_y \mid 0\leq d\leq k-1\}$ contains exactly one of either $i_1$ or $i_2$ , then $|\prod _{d=0}^{k-1}y_{v-di}^{-1}|=\frac {|y|}{2}$ . By the above considerations, there can be only two such sets, namely when $v=v_1$ or $v=v_2$ . As the next lemma shows, this can only be the case if $i_2=i_1+i_y$ or $i_2=i_1-i_y$ .

Lemma 5.8 Let $i, k, i_1,i_2\in \mathbb {F}_p$ be four elements of $\mathbb {F}_p$ , with $i\ne 0$ , $i_1\ne i_2$ , and $1<k<p-1$ . For all $v\in \mathbb {F}_p$ , let

$$ \begin{align*} I_v = \{v-di\mid 0\leq d \leq k-1\}. \end{align*} $$

If there exist two elements $v_1, v_2\in \mathbb {F}_p$ such that

$$ \begin{align*} |I_{v}\cap\{i_1,i_2\}| = 1 \quad\Longleftrightarrow \quad v\in \{v_1, v_2\}, \end{align*} $$

then either $i_2=i_1+i$ or $i_2=i_1-i$ .

Proof Let $f\colon \mathbb {F}_p\rightarrow \mathbb {F}_p$ be the map defined by $f(x)=i^{-1}(x-i_1)$ . As f is a bijection, we have $|f(I_v)\cap \{f(i_1), f(i_2)\}|= |I_{v}\cap \{i_1,i_2\}|$ , and because we have

$$ \begin{align*}f(I_v) = \{f(v)-d \mid 0\leq d \leq k-1\},\end{align*} $$

for all $v\in \mathbb {F}_p$ , it suffices to prove the result for $i=1$ and $i_1=0$ .

Suppose for the sake of contradiction that $i_2\ne 1, -1$ , and let $v_1\in \{k,k+1,\ldots , {p-1}\}$ be the smallest element, with respect to the standard order on $\{k,k+1,\ldots , p-1\}$ , such that $i_2\in I_{v_1}$ . Note that $v_1$ exists, because $\bigcup _{v=k}^{p-1}I_v=\{1,2,\ldots , p-1\}$ . Note also that because $v_1\in \{k, k+1, \ldots , p-1\}$ , we know that $0\notin I_{v_1}$ . We claim that $i_2\in I_{v_1+1}$ and that $0\notin I_{v_1+1}$ . Indeed, we have $I_{v_1+1}\backslash I_{v_1}=\{v_1+1\}$ . Thus, if we had $0\in I_{v_1+1}$ , this would imply that $v_1=p-1$ . Because $k<p-1$ , by the minimality of $v_1$ , this would then imply that $i_2\notin I_{p-2}$ , and therefore $i_2=p-1$ , a contradiction. Similarly, because $I_{v_1}\backslash I_{v_1+1}=\{v_1-k+1\}$ , if we had $i_2\notin I_{v_1+1}$ , this would imply that $i_2=v_1-k+1$ . Because $k>1$ , we have $i_2<v_1$ , and so $i_2\in I_{v_1-1}$ . By the minimality of $v_1$ , this implies that $v_1=k$ . However, this then means that $i_2=1$ , a contradiction.

We conclude that $i_2\in I_{v_1}, I_{v_1+1}$ and that $0\notin I_{v_1}, I_{v_1+1}$ . Likewise, we can also find $v_2\in \{i_2+k,i_2+k+1,\ldots ,i_2+p-1\}$ such that $0\in I_{v_2}, I_{v_2+1}$ and $i_2\notin I_{v_2}, I_{v_2+1}$ . Clearly, $v_1, v_1+1, v_2, v_2+1$ are four different elements, and we have $|I_{v}\cap \{0, i_2\}|=1$ for all $v\in \{v_1, v_1+1, v_2, v_2+1\}$ , which contradicts our assumptions. The result follows.▪

Coming back to our considerations on the form of y, we have

$$ \begin{align*}y=a^{i_y}(b^{s_1})^{a^{i_1}}(b^{s_2})^{a^{i_2}}\cdots (b^{s_{2\mu-1}})^{a^{i_1}}(b^{s_{2\mu}})^{a^{i_2}},\end{align*} $$

where $i_2=i_1+i_y$ or $i_2=i_1-i_y$ by the previous lemma. If we had $i_2=i_1-i_y$ , then we would have

$$ \begin{align*}\varphi_{i_2-i_y}(y^p) =y_{i_2}y_{i_1} y_{i_2+2i_y}\cdots y_{i_2+(p-1)i_y}.\end{align*} $$

From the form of y above, we see that $y_{i_2}$ ends with $b^{s_{2\mu }}$ and $y_{i_1}$ begins with $b^{s_1}$ , which implies that $|y_{i_2}y_{i_1}|\leq |y_{i_2}|+|y_{i_1}|-1$ . Consequently, we have $|\varphi _{i_2-i_y}(y^p)|<|y|$ , a contradiction to the minimality of y by Lemma 5.3. We conclude that we must have $i_2=i_1+i_y$ .

To finish the proof, it suffices to show that if y is of the above form, then one of $(y^{p})_{v_1}$ or $(y^p)_{v_2}$ is not. Indeed, we can then repeat the whole argument above with $y^{\prime }=(y^p)_{v}$ and $w^{\prime }=(y^{-k}w)_v$ for the corresponding $v\in \{v_1,v_2\}$ and we will reach a contradiction.

Let us then prove this last claim. Suppose that

(5.8) $$ \begin{align} y=a^{i_y}(b^{s_1})^{a^{i_1}}(b^{s_2})^{a^{i_1+i_y}}\cdots (b^{s_{2\mu-1}})^{a^{i_1}}(b^{s_{2\mu}})^{a^{i_1+i_y}}. \end{align} $$

For $u\in \{v_1,v_2\}$ , we have

$$ \begin{align*}(y^p)_{u} = y_{u+i_y}y_{u+2i_y}\cdots y_{u+pi_y},\end{align*} $$

and because we have supposed above that $\sum _{d=0}^{k-1} |y_{u-di_y}|=\frac {|y|}{2}$ and that $|y_{i_1}|=|y_{i_1+i_y}|=\frac {|y|}{2}$ , we conclude that either $u=i_1$ or $u-ki_y=i_1$ . Thus, up to renaming $v_1$ and $v_2$ , we can assume that $v_1=i_1+ki_y$ and $v_2=i_1$ . It follows that

(5.9) $$ \begin{align} (y^p)_{v_1} = a^{k_1}y_{i_1}y_{i_1+i_y}a^{k_2}, \end{align} $$

where $a^{k_1} = y_{v_1+i_y}\cdots y_{v_1+(p-k-1)i_y}$ and $a^{k_2}= y_{v_{1}+(p-k+2)i_y}\cdots y_{v_1+pi_y}$ , and that

(5.10) $$ \begin{align} (y^p)_{v_2} = y_{i_1+i_y}a^{k_2+k_1}y_{i_1}. \end{align} $$

From (5.8) and (5.9), we see that

$$ \begin{align*} (y^p)_{v_1} &= a^{k_1}b^{t_1}a^{r_1}b^{t_2}a^{r_2}\cdots a^{r_{2\mu-1}}b^{t_{2\mu}}a^{k_2}\\ &= a^{k_1+\sum_{d=1}^{2\mu-1}r_d+k_2}(b^{t_1})^{a^{\sum_{d=1}^{2\mu-1}r_d+k_2}}(b^{t_2})^{a^{\sum_{d=2}^{2\mu-1}r_d+k_2}}\cdots (b^{t_{2\mu-1}})^{a^{r_{2\mu-1}+k_2}}(b^{t_{2\mu}})^{a^{k_2}}, \end{align*} $$

for some $t_1,\ldots , t_{2\mu }$ and $r_1, \ldots , r_{2\mu -1}$ in $\mathbb {F}_p$ . Suppose that $(y^p)_{v_1}$ is of the same form as y, namely that there exists $i_3\in X$ such that

$$ \begin{align*} (y^p)_{v_1}=a^{i_y}(b^{t_1})^{a^{i_3}}(b^{t_2})^{a^{i_3+i_y}}\cdots (b^{t_{2\mu-1}})^{a^{i_3}}(b^{t_{2\mu}})^{a^{i_3+i_y}}. \end{align*} $$

This implies that $r_d=-i_y$ if d is odd and $r_d=i_y$ if d is even. Therefore, we obtain $\sum _{d=1}^{2\mu -1}r_d=-i_y$ . Because we know from Lemma 5.3 that $k_1+k_2+\sum _{d=1}^{2\mu -1}r_d=i_y$ , we conclude that $k_1+k_2=2i_y$ .

If we now turn our attention to $(y^p)_{v_2}$ , it follows from (5.8) and (5.10) that

$$ \begin{align*}(y^p)_{v_2} = a^{r^{\prime}_1}b^{t^{\prime}_1}\cdots a^{r^{\prime}_{\mu}}b^{t^{\prime}_{\mu}} a^{k_1+k_2} b^{t^{\prime}_{\mu+1}}a^{r^{\prime}_{\mu+1}}\cdots b^{t^{\prime}_{2\mu}}a^{r^{\prime}_{2\mu}},\end{align*} $$

for some $t^{\prime }_1,\ldots , t^{\prime }_{2\mu }$ and $r^{\prime }_1, \ldots , r^{\prime }_{2\mu }$ in $\mathbb {F}_p$ . Using the same reasoning as above, for $(y^p)_{v_2}$ to be of the same form as y, we would need either $k_1+k_2=i_y$ or $k_1+k_2=-i_y$ , depending on the parity of $\mu $ .

We conclude that $(y^p)_{v_1}$ and $(y^p)_{v_2}$ cannot both be in the same form as y. Indeed, this would imply that either $i_y=2i_y$ or $-i_y=2i_y$ . Because $i_y\ne 0$ , the first equation is impossible, and the second can only be satisfied if $p=3$ . However, we assumed that there existed some $1<k<p-1$ , which is impossible if $p=3$ .

Therefore, there is some $u\in \{v_1, v_2\}$ such that $y^{\prime }=(y^p)_{u}$ is not in the form of (5.8). Setting $w^{\prime }=(y^{-k}w)_u$ and repeating the whole argument above with $y^{\prime }$ and $w^{\prime }$ will thus necessarily yield a contradiction. We conclude that $|x|=1$ , and the result follows.

The first statement of Theorem 1.1 is now proved. We show the second.

For the constant GGS-group $\mathcal {G}$ , the situation is different.

Proof By [Reference Fernández-Alcober, Garrido and Uria-Albizuri4, Proposition 3.4] and using the notation of Section 3.2, we have $\mathcal {G}/K^{\prime } \cong \left (\mathbb {Z}/p\mathbb {Z}\right ) \ltimes \mathbb {Z}^{p-1}$ where the action of $\mathbb {Z}/p\mathbb {Z}$ on $\mathbb {Z}^{p-1}$ is given by the matrix

We see that for all $q\in \mathbb {N}$ , the subgroup $(q\mathbb {Z})^{p-1}\leq \mathbb {Z}^{p-1}$ is invariant under the action of $\mathbb {Z}/p\mathbb {Z}$ . Therefore, for all $q\in \mathbb {N}$ different from $0$ or $1$ , we can consider the subgroup $\left (\mathbb {Z}/p\mathbb {Z}\right ) \ltimes (q\mathbb {Z})^{p-1}$ , which is nontrivial and proper in $\left (\mathbb {Z}/p\mathbb {Z}\right ) \ltimes \mathbb {Z}^{p-1}$ . As $\left (\mathbb {Z}/p\mathbb {Z}\right ) \ltimes \mathbb {Z}^{p-1}$ is finitely generated, this subgroup is contained in a maximal subgroup $M_q\leq \left (\mathbb {Z}/p\mathbb {Z}\right ) \ltimes \mathbb {Z}^{p-1}$ .

We notice that if $q_1, q_2\in \mathbb {N}$ are two different prime numbers, then $M_{q_1} \ne M_{q_2}$ . Indeed, otherwise, we would have $(q_1\mathbb {Z})^{p-1}, (q_2\mathbb {Z})^{p-1}\leq M_{q_1}$ , which would imply $\mathbb {Z}^{p-1}\leq M_{q_1}$ , because $q_1$ and $q_2$ are coprime. As we already had $\left (\mathbb {Z}/p\mathbb {Z}\right ) \ltimes (q_1\mathbb {Z})^{p-1}\leq M_{q_1}$ , this means that $M_{q_1}=\left (\mathbb {Z}/p\mathbb {Z}\right ) \ltimes \mathbb {Z}^{p-1}$ , which is absurd.

We have thus shown that we have an infinite number of maximal subgroups of $\mathcal {G}/K^{\prime }$ which are all pairwise distinct. By the correspondence theorem, the same is true for $\mathcal {G}$ . This immediately implies that $\mathcal {G}$ has maximal subgroups that are not of index p, because there can only be finitely many such subgroups. It also implies that $\mathcal {G}$ admits maximal subgroups that are not normal. Indeed, otherwise, the Frattini subgroup of $\mathcal {G}$ would contain $\mathcal {G}^{\prime }$ , because the quotient of any group by a normal maximal subgroup must be a cyclic group of prime order, and thus abelian, but $\mathcal {G}/\mathcal {G^{\prime }}$ is a p-group, which would imply that every maximal subgroup is of index p.▪