Let us first prove the inclusion $\mathcal{K}(\mathrm{BS}(m, n)) \supseteq \mathcal{J}$. It suffices to show that no element of $\mathcal{J}$ is isolated in $\mathcal{J}$. Recall the definition of the topology in terms of pointed actions, see Section 2.2 and in particular Equation (2.7). Let us fix a pointed transitive action $(\alpha_0, v)$ representing an element of $\mathcal{J}$ and a radius $R \ge 0$. We will show that the basic neighborhood $\mathcal{N}([\alpha_0, v], R)$ contains at least two distinct elements of $\mathcal{J}$.

Let $(\beta, \tau)$ be the pre-action obtained by restricting $\alpha_0$ to the union of the $b$-orbits of the vertices of the ball of radius $R+1$ centered at $v$ in the Schreier graph of $\alpha_0$. The Bass-Serre graph of $(\beta, \tau)$ is the projection in $\mathbf{BS}(\alpha_0)$ (see Definition 3.10) of this ball, hence is finite. Since $\mathbf{BS}(\alpha_0)$ is infinite, the pre-action $(\beta, \tau)$ is not saturated.

We now build two $(m, n)$-graphs $G_1, G_2$ that extend the finite non-saturated Bass-Serre graph $\mathcal{G}$ of $(\beta, \tau)$ in two different ways. First, let $G_1$ be a forest-saturation of $\mathcal{G}$ given by Lemma 4.22. In particular, the subgraph induced in $G_1$ by $V(G_1) \setminus V(\mathcal{G})$ is a forest whose vertices have degree at least $3 \le 1 + \min(|m|, |n|)$ in $G_1$.

We then construct $G_2$ by modifying $G_1$. Let us pick a vertex $v \in V(G_1) \setminus V(\mathcal{G})$. The subgraph induced in $G_1$ by $V(G_1) \setminus \{v\}$ has at least 3 connected components. Choose two connected components disjoint from $\mathcal{G}$ and remove them. In the resulting $(m, n)$-graph $G_1'$, the vertex $v$ is the only one that is not saturated: two edges are missing.

Theorem 4.17 gives us an $(m, n)$-graph which is a simple edge path $\mathcal{P}$ whose extremities have the same label as $v$ and for which the orientations of the end edges are compatible with that of the missing edges of $v$. We then apply twice the welding lemma, Lemma 4.16, so as to weld the two extremities of $\mathcal{P}$ to $v$. We eventually define $G_2$ to be a forest-saturation of the graph that we obtained. Observe that $G_1$ is not isomorphic to $G_2$ since the fundamental groups of their underlying graphs are free groups of distinct ranks.

Finally, we extend $(\beta, \tau)$ to pre-actions $\alpha_1$ and $\alpha_2$ whose Bass-Serre graphs are $G_1$ and $G_2$ respectively, thanks to Proposition 3.23. Since $G_1, G_2$ are saturated, $\alpha_1, \alpha_2$ are actually actions by Example 3.21. We already remarked that $G_1$ is not isomorphic to $G_2$, so the pointed transitive actions $(\alpha_1, v)$ and $(\alpha_2, v)$ are not isomorphic: $[\alpha_1, v] \ne [\alpha_2, v]$. Moreover, the balls of radius $R$ centered at the basepoints in the Schreier graphs of $\alpha_0, \alpha_1, \alpha_2$ all coincide by construction with that of $(\beta, \tau)$, so $[\alpha_1, v]$ and $[\alpha_2, v]$ are both in $\mathcal{N}([\alpha_0, v], R)$.

Let us now turn to the inclusion $\mathcal{K}(\mathrm{BS}(m, n)) \subseteq \mathcal{J}$. Let us pick a sub-
It easily follows from the definitions that if $\mathbf{Ph}(\Lambda) = \mathbf{Ph}(\Lambda')$ then $\mathbf{Ph}(\Lambda) = \mathbf{Ph}(\Lambda \cap \Lambda')$, see Remark 4.4.

**Remark 5.7.** The phenotype of $\Lambda \leq BS(m, n)$ is the phenotype of the index of $\Lambda \cap \langle b \rangle$ in $\langle b \rangle$ since this index is the cardinal of the $b$-orbit of the point $\Lambda \in \Lambda \setminus BS(m, n)$. In other words, given $\Lambda \leq BS(m, n)$, we have:

$$ \mathbf{Ph}(\Lambda) = \mathbf{Ph}(\Lambda \cap \langle b \rangle) = \mathrm{Ph}_{m,n}(\langle b \rangle : \Lambda \cap \langle b \rangle). $$

In particular $\mathbf{Ph}(\langle b^k \rangle) = \mathrm{Ph}_{m,n}(k)$ for $k \in \mathbb{Z}_{\geq 1}$ and the phenotype of the trivial subgroup is infinite.

**Proposition 5.8.** In the partition of the space of subgroups of $BS(m, n)$ according to their phenotype

$$ \mathrm{Sub}(\mathrm{BS}(m, n)) = \bigsqcup_{q \in \mathcal{Q}_{m,n}} \mathbf{Ph}^{-1}(q), $$

the pieces are non-empty and satisfy:

1. For every finite $q \in \mathcal{Q}_{m,n}$, the piece $\mathbf{Ph}^{-1}(q)$ is open; it is also closed if and only if $|m| = |n|$.

2. For $q = \infty$, the piece $\mathbf{Ph}^{-1}(\infty)$ is closed and not open.

In particular, the function $\mathbf{Ph}$: $\mathrm{Sub}(\mathrm{BS}(m, n)) \to \mathbb{Z}_{\geq 1} \cup \{+\infty\}$ is Borel. It is continuous if and only if $|m| = |n|$.

*Proof.* Given $k \in \mathbb{Z}_{\geq 1}$, we set

$$ A_k := \{ \Lambda \in \mathrm{Sub}(\mathrm{BS}(m, n)) : \Lambda \cap \langle b \rangle = \langle b^k \rangle \}. $$

Writing $A_k$ as

$$ A_k = \{ \Lambda \in \mathrm{Sub}(\mathrm{BS}(m, n)) : b^k \in \Lambda, b^i \notin \Lambda \text{ for every } 1 \leq i < k \} $$

makes it clear that $A_k$ is clopen for every $k \in \mathbb{Z}_{\geq 1}$. Moreover $\langle b^k \rangle \in A_k$, so in particular $A_k$ is not empty. Now, Remark 5.7 implies that for every $q \in \mathbb{Z}_{\geq 1}$

$$ \mathbf{Ph}^{-1}(q) = \bigsqcup_{k \in \mathrm{Ph}_{m,n}^{-1}(q)} A_k. \qquad (5.9) $$

Hence $\mathbf{Ph}^{-1}(q)$ is open for every finite $q$ and, by taking the complement, $\mathbf{Ph}^{-1}(\infty)$ is closed.
group $\Lambda \in \text{Sub}(\text{BS}(m, n)) \setminus \mathcal{J}(m, n)$ and let us prove that it is not in the perfect kernel.

If $|m| \neq |n|$, then $\Lambda$ has finite index in $\text{BS}(m, n)$ by Remark 5.2, hence it is isolated in $\text{Sub}(\text{BS}(m, n))$.

If $|m| = |n|$, then $\pi(\Lambda)$ has finite index in $\text{BS}(m, n)/\langle b^m \rangle$ by Remark 5.2, hence it is finitely generated. Therefore, the set

$$ \mathcal{V} := \{ \Lambda' \in \text{Sub}(\text{BS}(m, n)) : \pi(\Lambda') \geq \pi(\Lambda) \} $$

is a neighborhood of $\Lambda$, since it contains the basic neighborhood $\mathcal{V}(S, \emptyset) = \{ \Lambda' \in \text{Sub}(\text{BS}(m, n)) : S \subseteq \Lambda' \}$ where $S \subseteq \Lambda$ is a finite set such that $\pi(S)$ generates $\pi(\Lambda)$.

Now, for any $\Lambda' \in \mathcal{V}$, the subgroup $\pi(\Lambda')$ has finite index in $\text{BS}(m, m)/\langle b^m \rangle$. Hence $\pi(\Lambda')$ is finitely generated, so $\Lambda'$ itself is finitely generated since it is written as an extension with cyclic kernel:

$$ 1 \to \langle b^m \rangle \cap \Lambda' \to \Lambda' \to \pi(\Lambda') \to 1. $$

Therefore all subgroups of $\mathcal{V}$ are finitely generated, which implies that $\mathcal{V}$ is countable and hence $\Lambda$ is not in $\mathcal{K}(\text{BS}(m, n))$. $\square$

**Corollary 5.4.** If $|m| \geq 2$, $|n| \geq 2$ and $|m| \neq |n|$, then

$$ \text{Ph}^{-1}(\infty) \subseteq \mathcal{K}(\text{BS}(m, n)); $$

in other words, every infinite phenotype subgroup is in the perfect kernel.

*Proof.* Any subgroup with infinite phenotype has infinite index and hence it is in $\mathcal{K}(\text{BS}(m, n))$ according to Theorem 5.3. $\square$

## 5.2 Phenotypical decomposition of the perfect kernel

Let us now turn to a description of the internal structure of $\mathcal{K}(\text{BS}(m, n))$.

**Notation 5.5.** Let $m, n \in \mathbb{Z} \setminus \{-1, 0, 1\}$. We denote by $\mathcal{Q}_{m,n}$ the set of all possible $(m, n)$-phenotypes, that is, $\mathcal{Q}_{m,n} := \text{Ph}_{m,n}(\mathbb{Z}_{\geq 1} \cup \{\infty\})$.

**Definition 5.6.** The phenotype of a subgroup $\Lambda \leq \text{BS}(m, n)$ is the $(m, n)$-phenotype of the index of $\Lambda \cap \langle b \rangle$ in $\langle b \rangle$

$$ \text{Ph}(\Lambda) = \text{Ph}(\Lambda \cap \langle b \rangle) := \text{Ph}_{m,n}(\langle b \rangle : \Lambda \cap \langle b \rangle). $$

This yields a function $\text{Ph}: \text{Sub}(\text{BS}(m, n)) \to \mathcal{Q}_{m,n} \subseteq \mathbb{Z}_{\geq 1} \cup \{\infty\}$.
Take a sequence of positive integers $(k_i)_{i \in \mathbb{N}}$ tending to $\infty$. Observe that the subgroups $\{\langle b^{k_i} \rangle\}_{i}$ have finite phenotype and converge to the trivial subgroup which has infinite phenotype. Therefore $\mathbf{Ph}^{-1}(\infty)$ is not open. Moreover, if $\mathbf{Ph}_{m,n}^{-1}(q)$ is not finite, we can choose all the $k_i$'s with phenotype $q$; the same argument shows that $\mathbf{Ph}^{-1}(q)$ is not closed. Finally, the clopen decomposition (5.9) shows that $\mathbf{Ph}^{-1}(q)$ is closed as soon as $\mathbf{Ph}_{m,n}^{-1}(q)$ is finite. By Lemma 4.5, $\mathbf{Ph}_{m,n}^{-1}(q)$ is finite exactly when $|m| = |n|$. $\square$

We now restrict the above partition to the perfect kernel

$$ \mathcal{K}(\mathrm{BS}(m, n)) = \bigsqcup_{q \in \mathcal{Q}_{m,n}} \mathcal{K}_q(\mathrm{BS}(m, n)), \qquad (5.10) $$

where

$$ \mathcal{K}_q(\mathrm{BS}(m, n)) := \mathcal{K}(\mathrm{BS}(m, n)) \cap \mathbf{Ph}_{m,n}^{-1}(q). \qquad (5.11) $$

**Remark 5.12.** Observe that each $\mathcal{K}_q(\mathrm{BS}(m, n))$ is not empty: indeed it contains $\langle b^q \rangle$ which belongs to the perfect kernel by Theorem 5.1. Moreover, in the proof of Theorem 5.1 the $(m, n)$-graphs we construct have the same phenotype, so every element of $\mathcal{K}_q(\mathrm{BS}(m, n))$ is actually a non-trivial limit of elements of $\mathcal{K}_q(\mathrm{BS}(m, n))$. We conclude that $\mathcal{K}_q(\mathrm{BS}(m, n))$ is equal to the perfect kernel of $\mathbf{Ph}_{m,n}^{-1}(q)$.

Let us show that the action of $\mathrm{BS}(m, n)$ by conjugation on each term is topologically transitive in the following sense.

**Definition 5.13.** An action by homeomorphisms of a group $\Gamma$ on a topological space $X$ is called **topologically transitive** if for every nonempty open sets $U$ and $V$, there is a point whose $\Gamma$-orbit intersects both $U$ and $V$.

**Theorem 5.14.** Let $m, n$ be integers such that $|m|, |n| \ge 2$. Then for every phenotype $q \in \mathcal{Q}_{m,n}$, the action by conjugation of $\mathrm{BS}(m, n)$ on the invariant subspace $\mathcal{K}_q(\mathrm{BS}(m, n))$ is topologically transitive.

*Proof.* We again use the definition of the topology in terms of pointed actions, see Section 2.2 and in particular Equation (2.7). So let us fix two pointed actions $(\alpha_1, v_1)$ and $(\alpha_2, v_2)$ in $\mathcal{K}_q(\mathrm{BS}(m, n))$, take $R > 0$, and consider the basic open sets $\mathcal{N}([\alpha_1, v_1], R)$ and $\mathcal{N}([\alpha_2, v_2], R)$. We need to construct a pointed action whose orbit meets both open sets.

As in the proof of Theorem 5.1, we let $(\beta_i, \tau_i)$, for $i = 1, 2$, be the pre-action obtained by restricting $\alpha_i$ to the reunion of the $b$-orbits of the vertices
In particular $\mathbf{Ph}(\langle b^k \rangle) = \text{Ph}_{m,n}(k)$ for $k \in \mathbb{Z}_{\ge 1}$ and the phenotype of the trivial subgroup is infinite.

**Remark 5.7.** The index $[\langle b \rangle : \Lambda \cap \langle b \rangle]$ is the cardinal of the $\langle b \rangle$-orbit of the point $\Lambda$ in the action $\Lambda \backslash \text{BS}(m, n) \hookrightarrow \text{BS}(m, n)$. Hence $\mathbf{Ph}(\Lambda)$ is the phenotype of this action (as given in Definition 4.9). Since the latter doesn't depend on the basepoint, the function $\mathbf{Ph}$ is invariant under conjugation.

It easily follows from the definitions that if $\mathbf{Ph}(\Lambda) = \mathbf{Ph}(\Lambda')$ then $\mathbf{Ph}(\Lambda) = \mathbf{Ph}(\Lambda \cap \Lambda')$, see Remark 4.4.

**Proposition 5.8.** In the partition of the space of subgroups of $\text{BS}(m, n)$ according to their phenotype

$$ \text{Sub}(\text{BS}(m, n)) = \bigsqcup_{q \in \mathcal{Q}_{m,n}} \mathbf{Ph}^{-1}(q), $$

the pieces are non-empty and satisfy:

1. For every finite $q \in \mathcal{Q}_{m,n}$, the piece $\mathbf{Ph}^{-1}(q)$ is open; it is also closed if and only if $|m| = |n|$.

2. For $q = \infty$, the piece $\mathbf{Ph}^{-1}(\infty)$ is closed and not open.

In particular, the function $\mathbf{Ph}: \text{Sub}(\text{BS}(m, n)) \to \mathbb{Z}_{\ge 1} \cup \{+\infty\}$ is Borel. It is continuous if and only if $|m| = |n|$.

*Proof.* Given $k \in \mathbb{Z}_{\ge 1}$, we set

$$ A_k := \{ \Lambda \in \text{Sub}(\text{BS}(m, n)) : \Lambda \cap \langle b \rangle = \langle b^k \rangle \}. $$

Writing $A_k$ as

$$ A_k = \{ \Lambda \in \text{Sub}(\text{BS}(m, n)) : b^k \in \Lambda, b^i \notin \Lambda \text{ for every } 1 \le i < k \} $$

makes it clear that $A_k$ is clopen for every $k \in \mathbb{Z}_{\ge 1}$. Moreover $\langle b^k \rangle \in A_k$, so in particular $A_k$ is not empty. Now, by definition, for every $q \in \mathbb{Z}_{\ge 1}$ we have

$$ \mathbf{Ph}^{-1}(q) = \bigsqcup_{k \in \text{Ph}_{m,n}^{-1}(q)} A_k. \qquad (5.9) $$

Hence $\mathbf{Ph}^{-1}(q)$ is open for every finite $q$ and, by taking the complement, $\mathbf{Ph}^{-1}(\infty)$ is closed.
of the balls $B(v_i, R + 1)$ of radius $R + 1$ centered at $v_i$ in the Schreier graph of $\alpha_i$. The Bass-Serre graph of $(\beta_i, \tau_i)$ is finite. Since $\mathbf{BS}(\alpha_i)$ is infinite, the pre-action $(\beta_i, \tau_i)$ is not saturated.

Moreover $(\beta_1, \tau_1)$ and $(\beta_2, \tau_2)$ have the same phenotype, so we can apply the merging machine (Theorem 4.12) to obtain an action $\alpha$ whose Schreier graph contains (copies of) the balls $B(v_i, R + 1)$.

Pointing $\alpha$ at the copy of $v_1$ that we denote by $v$, we have $(\mathbf{Sch}(\alpha), v) \simeq_R (\mathbf{Sch}(\alpha_1), v_1)$. By transitivity of $\alpha$, there is $\gamma \in \mathrm{BS}(m, n)$ such that $v\alpha(\gamma)$ is the copy of $v_2$, and thus $(\mathbf{Sch}(\alpha), v\alpha(\gamma)) \simeq_R (\mathbf{Sch}(\alpha_2), v_2)$. In particular, the orbit of $[\alpha, v]$ meets both $\mathcal{N}([\alpha_1, v_1], R)$ and $\mathcal{N}([\alpha_2, v_2], R)$. $\square$

**Corollary 5.15.** Let $m, n$ be integers such that $|m|, |n| \ge 2$. Then for every $q \in \mathcal{Q}_{m,n}$, there is a dense $G_\delta$ subset of $\mathcal{K}_q(\mathrm{BS}(m, n))$ consisting of subgroups with dense conjugacy class in $\mathcal{K}_q(\mathrm{BS}(m, n))$.

*Proof of Corollary 5.15.* By Proposition 5.8, each $\mathcal{K}_q(\mathrm{BS}(m, n))$ is Polish as an open or a closed subset of the Polish space $\mathcal{K}(\mathrm{BS}(m, n))$.

The corollary now follows from a well-known characterization of topological transitivity in Polish spaces: if $(U_i)$ is a countable base of non-empty open subsets, then the set $\cap_{i \in \mathbb{N}} U_i \Gamma$ of points with dense orbit is a dense $G_\delta$ by the Baire theorem. $\square$

## 5.3 Closed invariant subsets with a fixed finite phenotype

Given a finite phenotype $q$, we will show that there is a largest closed invariant subset inside the (open but non closed when $|m| \neq |n|$) set of subgroups of phenotype $q$. The following lemma is key.

**Lemma 5.16.** Let $|m| \neq |n|$, and let $L \in \mathbb{Z}_{\ge 1}$ satisfying:

$$ \exists p \in \mathcal{P}, |m|_p \neq |n|_p \text{ and } |L|_p > \min(|m|_p, |n|_p). $$

Then for any saturated $(m, n)$-graph which contains $L$ as a label, the range of the label map is unbounded.

*Proof.* By symmetry, we may as well assume that $|n|_p < |m|_p$ for a fixed prime $p$, and so $|L|_p > |n|_p$. Let $v_0 \in V(\mathcal{G})$ have label $L$. Since our Bass-Serre graph $\mathcal{G}$ is saturated, every vertex has at least one outgoing edge. We can thus build inductively an infinite path $(e_k)_{k \in \mathbb{N}}$ consisting of positive edges with $s(e_0) = v_0$. The conclusion then follows directly from Lemma 3.30. $\square$
Take a sequence of positive integers $(k_i)_{i \in \mathbb{N}}$ tending to $\infty$. Observe that the subgroups $\{\langle b^{k_i} \rangle\}_{i}$ have finite phenotype and converge to the trivial subgroup which has infinite phenotype. Therefore $\mathbf{Ph}^{-1}(\infty)$ is not open. Moreover, if $\mathbf{Ph}_{m,n}^{-1}(q)$ is not finite, we can choose all the $k_i$'s with phenotype $q$; the same argument shows that $\mathbf{Ph}^{-1}(q)$ is not closed. Finally, the clopen decomposition (5.9) shows that $\mathbf{Ph}^{-1}(q)$ is closed as soon as $\mathbf{Ph}_{m,n}^{-1}(q)$ is finite. By Lemma 4.5, $\mathbf{Ph}_{m,n}^{-1}(q)$ is finite exactly when $|m| = |n|$. $\square$

We now restrict the above partition to the perfect kernel

$$ \mathcal{K}(\mathrm{BS}(m, n)) = \bigsqcup_{q \in \mathcal{Q}_{m,n}} \mathcal{K}_q(\mathrm{BS}(m, n)), \qquad (5.10) $$

where

$$ \mathcal{K}_q(\mathrm{BS}(m, n)) := \mathcal{K}(\mathrm{BS}(m, n)) \cap \mathbf{Ph}_{m,n}^{-1}(q). \qquad (5.11) $$

**Remark 5.12.** Observe that each $\mathcal{K}_q(\mathrm{BS}(m, n))$ is not empty: indeed it contains $\langle b^q \rangle$ which belongs to the perfect kernel by Theorem 5.1. Moreover, in the proof of Theorem 5.1 the $(m, n)$-graphs we construct have the same phenotype, so every element of $\mathcal{K}_q(\mathrm{BS}(m, n))$ is actually a non-trivial limit of elements of $\mathcal{K}_q(\mathrm{BS}(m, n))$. We conclude that $\mathcal{K}_q(\mathrm{BS}(m, n))$ is equal to the perfect kernel of $\mathbf{Ph}_{m,n}^{-1}(q)$.

Let us show that the action of $\mathrm{BS}(m, n)$ by conjugation on each term is topologically transitive in the following sense.

**Definition 5.13.** An action by homeomorphisms of a group $\Gamma$ on a topological space $X$ is called **topologically transitive** if for every nonempty open sets $U$ and $V$, there is a point whose $\Gamma$-orbit intersects both $U$ and $V$.

**Theorem 5.14.** Let $m, n$ be integers such that $|m|, |n| \ge 2$. Then for every phenotype $q \in \mathcal{Q}_{m,n}$, the action by conjugation of $\mathrm{BS}(m, n)$ on the invariant subspace $\mathcal{K}_q(\mathrm{BS}(m, n))$ is topologically transitive.

*Proof.* We again use the definition of the topology in terms of pointed actions, see Section 2.2 and in particular Equation (2.7). So let us fix two pointed actions $(\alpha_1, v_1)$ and $(\alpha_2, v_2)$ in $\mathcal{K}_q(\mathrm{BS}(m, n))$, take $R > 0$, and consider the basic open sets $\mathcal{N}([\alpha_1, v_1], R)$ and $\mathcal{N}([\alpha_2, v_2], R)$. We need to construct a pointed action whose orbit meets both open sets.

As in the proof of Theorem 5.1, we let $(\beta_i, \tau_i)$, for $i = 1, 2$, be the pre-action obtained by restricting $\alpha_i$ to the union of the $b$-orbits of the vertices
**Remark 5.17.** When $|n| = |m|$, the lemma fails because labels are bounded: if $L_0$ is a label then all labels in the same connected component must satisfy $|L|_p \le \max(|L_0|_p, |m|_p, |n|_p)$ for all prime $p$ by Equation (3.29) and the discussion that precedes it.

Let $q$ be a finite $(m, n)$-phenotype. In order to describe which saturated $(m, n)$-graphs have unbounded labels, we now define

$$s(q, m, n) := q \cdot \prod_{\substack{p \in \mathcal{P} \\ |q|_p = 0 \\ |m|_p = |n|_p > 0}} p^{|m|_p} \cdot \prod_{\substack{p \in \mathcal{P} \\ |m|_p \ne |n|_p}} p^{\min\{|n|_p, |m|_p\}}. \quad (5.18)$$

**Remark 5.19.** The definition is motivated by the fact that $s(q, m, n)$ is the largest label of phenotype $q$ which does not satisfy the hypothesis of Lemma 5.16. As we will see in the proof of Theorem 5.20, a saturated $(m, n)$-graph with phenotype $q$ has unbounded labels if and only if one of its labels does not divide $s(q, m, n)$.

Proposition 5.8 implies that every subgroup (or pointed action) adherent to the set of subgroups of phenotype $q$ has phenotype $q$ or $\infty$, and phenotype $\infty$ can occur only when $|m| \ne |n|$. We can now characterize the subgroups $\Lambda$ with phenotype $q$ whose orbit approaches subgroups with infinite phenotype.

**Theorem 5.20.** Let $m, n$ be integers such that $|m|, |n| \ge 2$ and denote by $q \in Q_{m,n} \setminus \{\infty\}$ a finite $(m, n)$-phenotype. Let $s = s(q, m, n)$ as in Equation (5.18). Then the space

$$MC_q := \mathbf{Ph}^{-1}(q) \cap \{\Lambda \in \mathrm{Sub}(\mathrm{BS}(m, n)) : \Lambda \ge \langle\langle b^s \rangle\rangle\}$$

of subgroups of phenotype $q$ containing the normal subgroup $\langle\langle b^s \rangle\rangle$ satisfies the following properties:

(1) $MC_q$ is the largest closed $\mathrm{BS}(m, n)$-invariant subset of $\mathrm{Sub}(\mathrm{BS}(m, n))$ contained in $\mathbf{Ph}^{-1}(q)$; in particular, all normal subgroups of phenotype $q$ and all finite index subgroups of phenotype $q$ contain $\langle\langle b^s \rangle\rangle$.

(2) For every $\Lambda \in \mathbf{Ph}^{-1}(q) \setminus MC_q$, the orbit of $\Lambda$ accumulates to $\mathbf{Ph}^{-1}(\infty)$;

(3) If $|m| = |n|$, then $MC_q = \mathbf{Ph}^{-1}(q)$.

(4) If $|m| \ne |n|$, then $MC_q \cap K_q(\mathrm{BS}(m, n))$ has empty interior in $K_q(\mathrm{BS}(m, n))$.

(5) If $\mathrm{gcd}(m, n) = 1$, then $s = q$ and $MC_q \cap K(\mathrm{BS}(m, n)) = \{\langle\langle b^q \rangle\rangle\}$; in particular $\langle\langle b^q \rangle\rangle$ is the unique normal subgroup of phenotype $q$ of infinite index.
of the balls $B(v_i, R + 1)$ of radius $R + 1$ centered at $v_i$ in the Schreier graph of $\alpha_i$. The Bass-Serre graph of $(\beta_i, \tau_i)$ is finite. Since $\mathbf{BS}(\alpha_i)$ is infinite, the pre-action $(\beta_i, \tau_i)$ is not saturated.

Moreover $(\beta_1, \tau_1)$ and $(\beta_2, \tau_2)$ have the same phenotype, so we can apply the merging machine (Theorem 4.12) to obtain an action $\alpha$ whose Schreier graph contains (copies of) the balls $B(v_i, R + 1)$.

Pointing $\alpha$ at the copy of $v_1$ that we denote by $v$, we have $(\mathbf{Sch}(\alpha), v) \simeq_R (\mathbf{Sch}(\alpha_1), v_1)$. By transitivity of $\alpha$, there is $\gamma \in \mathrm{BS}(m, n)$ such that $v\alpha(\gamma)$ is the copy of $v_2$, and thus $(\mathbf{Sch}(\alpha), v\alpha(\gamma)) \simeq_R (\mathbf{Sch}(\alpha_2), v_2)$. In particular, the orbit of $[\alpha, v]$ meets both $\mathcal{N}([\alpha_1, v_1], R)$ and $\mathcal{N}([\alpha_2, v_2], R)$. $\square$

**Corollary 5.15.** Let $m, n$ be integers such that $|m|, |n| \ge 2$. Then for every $q \in Q_{m,n}$, there is a dense $G_\delta$ subset of $K_q(\mathrm{BS}(m, n))$ consisting of subgroups with dense conjugacy class in $K_q(\mathrm{BS}(m, n))$.

*Proof of Corollary 5.15.* By Proposition 5.8, each $K_q(\mathrm{BS}(m, n))$ is Polish as an open or a closed subset of the Polish space $K(\mathrm{BS}(m, n))$.

The corollary now follows from a well-known characterization of topological transitivity in Polish spaces: if $(U_i)$ is a countable base of non-empty open subsets, then the set $\cap_{i \in \mathbb{N}} U_i \Gamma$ of points with dense orbit is a dense $G_\delta$ by the Baire theorem. $\square$

## 5.3 Closed invariant subsets with a fixed finite phenotype

Given a finite phenotype $q$, we will show that there is a largest closed invariant subset inside the (open but non closed when $|m| \ne |n|$) set of subgroups of phenotype $q$. The following lemma is key.

**Lemma 5.16.** Let $|m| \ne |n|$, and let $L \in \mathbb{Z}_{\ge 1}$ satisfying:

$$ \exists p \in \mathcal{P}, |m|_p \ne |n|_p \text{ and } |L|_p > \min(|m|_p, |n|_p). $$

Then for any saturated $(m, n)$-graph which contains $L$ as a label, the range of the label map is unbounded.

*Proof.* By symmetry, we may as well assume that $|n|_p < |m|_p$ for a fixed prime $p$, and so $|L|_p > |n|_p$. Let $v_0 \in V(\mathcal{G})$ have label $L$. Since our Bass-Serre graph $\mathcal{G}$ is saturated, every vertex has at least one outgoing edge. We can thus build inductively an infinite path $(e_k)_{k \in \mathbb{N}}$ consisting of positive edges with $s(e_0) = v_0$. The conclusion then follows directly from Lemma 3.30. $\square$
*Proof of Theorem 5.20.* Let us first show (3) and (2). Observe that a subgroup $\Lambda$ contains $\langle\langle b^s \rangle\rangle$ if and only if all the $b$-orbits of the corresponding action $\Lambda \setminus \text{BS}(m, n) \hookrightarrow \text{BS}(m, n)$ have cardinality which divides $s$.

Let $\Lambda \in \mathbf{Ph}^{-1}(q)\mathcal{MC}_q$. Its Bass-Serre graph admits a label $L$ that does not divide $s$ but has phenotype $q$. We fix a prime $p$ such that $|L|_p > |s|_p$.

Let us show by contradiction that $|m|_p \neq |n|_p$. So assume that $|m|_p = |n|_p$. Then $|s|_p \geq |m|_p = |n|_p$: by the definition of $s$ (5.18),

* if $p$ divides $q = \text{Ph}_{m,n}(s)$, then $|s|_p = |q|_p > |m|_p = |n|_p$;

* if $p$ does not divide $q = \text{Ph}_{m,n}(s)$, then $|s|_p = |m|_p = |n|_p$.

Thus, we have $|L|_p > |m|_p = |n|_p$, in other words $p \in \mathcal{P}_{m,n}(L)$ (see Definition 4.1). Hence, we have $|\text{Ph}_{m,n}(L)|_p = |L|_p > |s|_p \geq |\text{Ph}_{m,n}(s)|_p$. This is a contradiction since both phenotypes are equal to $q$. Therefore $|m|_p \neq |n|_p$. In particular, $|n| \neq |m|$ when $\mathbf{Ph}^{-1}(q) \neq \mathcal{MC}_q$. This proves (3).

From Equation (5.18) again, $|s|_p = \min(|m|_p, |n|_p)$, so $|L|_p > \min(|m|_p, |n|_p)$. Lemma 5.16 thus applies, and so there is a sequence of vertices in the Bass-Serre graph of $\Lambda$ whose labels tend to $+\infty$. In other words, there is a sequence $(\gamma_i)_{i \geq 0}$ such that the index of $\gamma_i \Lambda \gamma_i^{-1} \cap \langle b \rangle$ in $\langle b \rangle$ tends to $+\infty$. By compactness, we may assume that this sequence converges, and its limit $\Delta$ cannot contain a non-zero power of $b$ since $[\langle b \rangle : \gamma_i \Lambda \gamma_i^{-1} \cap \langle b \rangle] \to +\infty$. Hence $\Delta$ has infinite phenotype, which proves (2).

We now prove (1). We first claim that $\mathcal{MC}_q$ is closed in $\text{Sub}(\text{BS}(m, n))$. Indeed the set

$$ \mathcal{B}_s := \{ \Lambda \in \text{Sub}(\text{BS}(m, n)) : \Lambda \geq \langle\langle b^s \rangle\rangle \} $$

is a countable intersection of basic clopen sets and hence it is closed. Then, notice that $\mathcal{B}_s$ intersects only finitely many sets $\mathbf{Ph}^{-1}(q')$, since $q'$ must be finite and divides $s$. Proposition 5.8 claims that $\mathbf{Ph}^{-1}(q')$ are open, hence

$$ \mathcal{MC}_q = \mathcal{B}_s \setminus \bigcup_{\substack{q' \neq q \\ q' \text{ divides } s}} \mathbf{Ph}^{-1}(q') $$

is closed. Also note that $\mathcal{MC}_q$ is obviously $\text{BS}(m, n)$-invariant. Finally Item (2) implies that no larger $\text{BS}(m, n)$-invariant subset of $\mathbf{Ph}^{-1}(q)$ can be closed in $\text{Sub}(\text{BS}(m, n))$. This proves that $\mathcal{MC}_q$ is the largest closed $\text{BS}(m, n)$-invariant subset of $\text{Sub}(\text{BS}(m, n))$ contained in $\mathbf{Ph}^{-1}(q)$. Since all normal subgroups and all finite index subgroups have finite (hence closed) orbits, the remaining statement in Item (1) is clear.
**Remark 5.17.** When $|n| = |m|$, the lemma fails because labels are bounded: if $L_0$ is a label then all labels in the same connected component must satisfy $|L|_p \le \max(|L_0|_p, |m|_p, |n|_p)$ for all primes $p$ by Equation (3.29) and the discussion that precedes it.

Let $q$ be a finite $(m, n)$-phenotype. In order to describe which saturated $(m, n)$-graphs have unbounded labels, we now define

$$s(q, m, n) := q \cdot \prod_{\substack{p \in \mathcal{P} \\ |q|_p = 0 \\ |m|_p = |n|_p > 0}} p^{|m|_p} \cdot \prod_{\substack{p \in \mathcal{P} \\ |m|_p \ne |n|_p}} p^{\min\{|n|_p, |m|_p\}}. \quad (5.18)$$

**Remark 5.19.** The definition is motivated by the fact that $s(q, m, n)$ is the largest label of phenotype $q$ which does not satisfy the hypothesis of Lemma 5.16. As we will see in the proof of Theorem 5.20, a saturated $(m, n)$-graph with phenotype $q$ has unbounded labels if and only if one of its labels does not divide $s(q, m, n)$.

Proposition 5.8 implies that every subgroup (or pointed action) that lies in the closure of the set of subgroups of phenotype $q$ has phenotype either $q$ or $\infty$, and phenotype $\infty$ can occur only when $|m| \ne |n|$. We can now characterize the subgroups $\Lambda$ with phenotype $q$ whose orbit approaches subgroups with infinite phenotype.

**Theorem 5.20.** Let $m, n$ be integers such that $|m|, |n| \ge 2$ and denote by $q \in Q_{m,n} \setminus \{\infty\}$ a finite $(m, n)$-phenotype. Let $s = s(q, m, n)$ as in Equation (5.18). Then the space

$$MC_q := \mathbf{Ph}^{-1}(q) \cap \{\Lambda \in \mathrm{Sub}(\mathrm{BS}(m, n)) : \Lambda \ge \langle\langle b^s \rangle\rangle\}$$

of subgroups of phenotype $q$ containing the normal subgroup $\langle\langle b^s \rangle\rangle$ satisfies the following properties:

(1) $MC_q$ is the largest closed $\mathrm{BS}(m, n)$-invariant subset of $\mathrm{Sub}(\mathrm{BS}(m, n))$ contained in $\mathbf{Ph}^{-1}(q)$; in particular, all normal subgroups of phenotype $q$ and all finite index subgroups of phenotype $q$ contain $\langle\langle b^s \rangle\rangle$;

(2) If $|m| = |n|$, then $MC_q = \mathbf{Ph}^{-1}(q)$;

(3) For every $\Lambda \in \mathbf{Ph}^{-1}(q) \setminus MC_q$, the orbit of $\Lambda$ accumulates to $\mathbf{Ph}^{-1}(\infty)$;

(4) If $|m| \ne |n|$, then $MC_q \cap K_q(\mathrm{BS}(m, n))$ has empty interior in $K_q(\mathrm{BS}(m, n))$;

(5) If $\mathrm{gcd}(m, n) = 1$, then $s = q$ and $MC_q \cap K(\mathrm{BS}(m, n)) = \{\langle\langle b^q \rangle\rangle\}$; in particular $\langle\langle b^q \rangle\rangle$ is the unique normal subgroup of phenotype $q$ of infinite index.
Let us prove Item (4). Suppose $|n| \neq |m|$; let $p$ be a prime number such that $|m|_p \neq |n|_p$. By definition $\mathrm{Ph}_{m,n}(sp) = \mathrm{Ph}_{m,n}(s) = q$, so that $\langle b^{sp} \rangle \in \mathcal{K}_q(\mathrm{BS}(m,n))\backslash\mathcal{MC}_q$. Consider a subgroup $\Lambda \in \mathcal{K}_q(\mathrm{BS}(m,n))$ whose orbit is dense in $\mathcal{K}_q(\mathrm{BS}(m,n))$, as provided by Corollary 5.15. Since the orbit of $\Lambda$ accumulates to $\langle b^{sp} \rangle \notin \mathcal{MC}_q$ and $\mathcal{MC}_q$ is invariant and closed, the latter does not contain any point of that orbit. Hence the complement $\mathcal{K}_q(\mathrm{BS}(m,n))\backslash\mathcal{MC}_q$ contains the dense orbit of $\Lambda$. We conclude that $\mathcal{MC}_q \cap \mathcal{K}_q(\mathrm{BS}(m,n))$ has empty interior in $\mathcal{K}_q(\mathrm{BS}(m,n))$.

We finally prove Item (5). The equality $s = q$ follows immediately from Formula (5.18) for $s(q, m, n)$. We have the presentation

$$ \mathrm{BS}(m, n)/\langle\langle b^q \rangle\rangle = \langle \bar{b}, \bar{t} : \bar{t}\bar{b}^m\bar{t}^{-1} = \bar{b}^n, \bar{b}^q = 1 \rangle. $$

Since $\mathrm{gcd}(q, m) = \mathrm{gcd}(q, n) = 1$, the elements $\bar{b}^m$ and $\bar{b}^n$ both generate $\langle \bar{b} \rangle$ in $\mathrm{BS}(m, n)/\langle\langle b^q \rangle\rangle$. We thus have a natural semi-direct product decomposition

$$ \mathrm{BS}(m, n)/\langle\langle b^q \rangle\rangle \cong \mathbb{Z}/q\mathbb{Z} \times \mathbb{Z} = \langle \bar{b} \rangle \times \langle \bar{t} \rangle $$

Consider $\Lambda \in \mathcal{MC}_q$ in the perfect kernel; it contains $\langle\langle b^q \rangle\rangle$. It suffices to prove that the image $\Lambda_q := \Lambda/\langle\langle b^q \rangle\rangle$ of $\Lambda$ in $\langle \bar{b} \rangle \times \langle \bar{t} \rangle$ is trivial. Since $\mathrm{Ph}(\Lambda) = q$, the index $[\langle b \rangle : \Lambda \cap \langle b \rangle]$ is a multiple of $q$, so we have $\Lambda_q \cap \langle \bar{b} \rangle = \{\mathrm{id}\}$. Thus $\Lambda_q$ is mapped injectively in the quotient $\langle \bar{b} \rangle \times \langle \bar{t} \rangle / \langle \bar{b} \rangle \simeq \mathbb{Z}$. If this image were not $\{0\}$, then $\Lambda$ would have finite index in $\mathrm{BS}(m, n)$, contradicting that $\Lambda$ is in the perfect kernel. The group $\Lambda_q$ is thus trivial as wanted. $\square$

**Remark 5.21.** In terms of actions, $\mathcal{MC}_q$ is the set of classes $[\alpha, v]$ all of whose cardinalities of $b$-orbits divide $s$ and have phenotype $q$.

**Proposition 5.22.** Let $m, n \in \mathbb{Z}\setminus\{0\}$ and $k \in \mathbb{Z}_{\ge 1}$. Let

$$ G_{m,n,k} := \mathrm{BS}(m, n)/\langle\langle b^k \rangle\rangle = \langle \bar{t}, \bar{b} \mid \bar{t}\bar{b}^m\bar{t}^{-1} = \bar{b}^n, \bar{b}^k = 1 \rangle $$

and let

$$ r(k) := \max\{r' \in \mathbb{N} : r' \text{ divides } k \text{ and } \mathrm{gcd}(r', m) = \mathrm{gcd}(r', n)\}. $$

Then:

1. $b$ has order $r(k)$ in the quotient $G_{m,n,k}$; in particular $\langle\langle b^k \rangle\rangle = \langle\langle b^{r(k)} \rangle\rangle$;

2. the group $G_{m,n,k} = G_{m,n,r(k)}$ is the HNN extension of $\mathbb{Z}/r(k)\mathbb{Z} = \langle \bar{b} \rangle$ with respect to the relation $\bar{t}\bar{b}^m\bar{t}^{-1} = \bar{b}^n$.
Proof of Theorem 5.20. The proofs of (2) and (3) rely on the following claim.

**Claim.** For any $\Lambda \in \mathbf{Ph}^{-1}(q) \setminus \mathcal{MC}_q$, there is a prime $p$ such that $|m|_p \neq |n|_p$ and a vertex label $L$ in the Bass-Serre graph of $\Lambda$ such that $|L|_p > |s|_p$.

*Proof of the claim.* Observe that a subgroup $\Lambda$ contains $\langle\langle b^s \rangle\rangle$ if and only if all the $b$-orbits of the corresponding action $\Lambda \setminus \text{BS}(m, n) \hookrightarrow \text{BS}(m, n)$ have cardinality which divides $s$. So if $\Lambda \in \mathbf{Ph}^{-1}(q) \setminus \mathcal{MC}_q$, we can fix a prime $p$ such that $|L|_p > |s|_p$, and we will prove that $|m|_p \neq |n|_p$.

Assume by contradiction that $|m|_p = |n|_p$. Then $|s|_p \geq |m|_p = |n|_p$; if $|m|_p = 0$ then the inequality clearly holds, otherwise by Equation (5.18),

* if $p$ divides $q = \text{Ph}_{m,n}(s)$, then $|s|_p = |q|_p > |m|_p = |n|_p$;

* if $p$ does not divide $q = \text{Ph}_{m,n}(s)$, then $|s|_p = |m|_p = |n|_p$.

Thus, we have $|L|_p > |m|_p = |n|_p$, in other words $p \in \mathcal{P}_{m,n}(L)$ (see Definition 4.1). Hence, we have $|\text{Ph}_{m,n}(L)|_p = |L|_p > |s|_p \geq |\text{Ph}_{m,n}(s)|_p$. This is a contradiction since both phenotypes are equal to $q$. □claim

We can now easily prove (2) by the contrapositive: by the above claim if $\mathcal{MC}_q \neq \mathbf{Ph}^{-1}(q)$ then there is a prime $p$ such that $|m|_p \neq |n|_p$, in particular $|m| \neq |n|$.

Let us prove (3). Let $\Lambda \in \mathbf{Ph}^{-1}(q) \setminus \mathcal{MC}_q$. The claim above provides a prime $p$ such that $|m|_p \neq |n|_p$ and the Bass-Serre graph of $\Lambda$ admits a vertex label $L$ such that $|L|_p > |s|_p$. It follows from Equation (5.18) that $|s|_p = \min(|m|_p, |n|_p)$, so $|L|_p > \min(|m|_p, |n|_p)$. Lemma 5.16 thus applies, and so there is a sequence of vertices in the Bass-Serre graph of $\Lambda$ whose labels tend to $+\infty$. In other words, there is a sequence $(\gamma_i)_{i \geq 0}$ such that the index of $\gamma_i \Lambda \gamma_i^{-1} \cap \langle b \rangle$ in $\langle b \rangle$ tends to $+\infty$. By compactness, we may assume that this sequence converges, and its limit $\Delta$ cannot contain a non-zero power of $b$ since $[\langle b \rangle : \gamma_i \Lambda \gamma_i^{-1} \cap \langle b \rangle] \to +\infty$. Hence $\Delta$ has infinite phenotype, which proves (3).

We now prove (1). We first claim that $\mathcal{MC}_q$ is closed in $\text{Sub}(\text{BS}(m, n))$. Indeed the set

$$ \mathcal{B}_s := \{ \Lambda \in \text{Sub}(\text{BS}(m, n)) : \Lambda \geq \langle\langle b^s \rangle\rangle \} $$

is a countable intersection of basic clopen sets and hence it is closed. Then, notice that $\mathcal{B}_s$ intersects only finitely many sets $\mathbf{Ph}^{-1}(q')$, since $q'$ must be finite and divide $s$. Proposition 5.8 claims that the $\mathbf{Ph}^{-1}(q')$ are open, hence

$$ \mathcal{MC}_q = \mathcal{B}_s \setminus \bigcup_{\substack{q' \neq q \\ q' \text{ divides } s}} \mathbf{Ph}^{-1}(q') $$
3. $\mathrm{Ph}_{m,n}(k) = \mathrm{Ph}_{m,n}(r(k)) = \mathbf{Ph}(\langle\langle b^k \rangle\rangle)$.

It is a routine computation, working prime number by prime number, to check that

$$r(k) = \prod_{\substack{p \in \mathcal{P} \\ |m|_p = |n|_p}} p^{|k|_p} \cdot \prod_{\substack{p \in \mathcal{P} \\ |m|_p \neq |n|_p}} p^{\min(|k|_p, |m|_p, |n|_p)} \quad (5.23)$$

In particular, $r(k)$ is a multiple of all the $r$'s which divide $k$ and satisfy $\gcd(r', m) = \gcd(r', n)$.

*Proof.* Set $r := r(k)$. Since $\bar{b}^m$ and $\bar{b}^n$ are conjugate in $G_{m,n,k}$, they have the same order:

$$\frac{\mathrm{ord}(\bar{b})}{\mathrm{gcd}(\mathrm{ord}(\bar{b}), m)} = \mathrm{ord}(\bar{b}^m) = \mathrm{ord}(\bar{b}^n) = \frac{\mathrm{ord}(\bar{b})}{\mathrm{gcd}(\mathrm{ord}(\bar{b}), n)}.$$

Thus $\mathrm{gcd}(\mathrm{ord}(\bar{b}), m) = \mathrm{gcd}(\mathrm{ord}(\bar{b}), n)$. Moreover $\mathrm{ord}(\bar{b})$ divides $k$. So by the definition of $r$, the order $\mathrm{ord}(\bar{b})$ divides $r$ and hence $b^r \in \langle\langle b^k \rangle\rangle$. On the other hand $b^k \in \langle b^r \rangle$, so that $\langle\langle b^r \rangle\rangle = \langle\langle b^k \rangle\rangle$ and $G_{m,n,k} = G_{m,n,r}$.

Since $\mathrm{gcd}(r, m) = \mathrm{gcd}(r, n)$, the subgroups generated by $\tilde{b}^m$ and $\tilde{b}^n$ in the group $\mathbb{Z}/r\mathbb{Z} = \langle \tilde{b} : \tilde{b}^r = 1 \rangle$ are isomorphic. We can thus consider the HNN-extension of $\mathbb{Z}/r\mathbb{Z} = \langle \tilde{b} : \tilde{b}^r = 1 \rangle$ with the relation $\tilde{t}\tilde{b}^m\tilde{t}^{-1} = \tilde{b}^n$. It admits the presentation $\langle \tilde{t}, \tilde{b} \mid \tilde{t}\tilde{b}^m\tilde{t}^{-1} = \tilde{b}^n, \tilde{b}^r = 1 \rangle$ and it is hence isomorphic to $G_{m,n,r}$.

By the Normal Form Theorem for HNN-extensions, the vertex group injects, i.e., $\bar{b}$ has order exactly $r$. Finally Remark 5.7 and Formula (5.23) imply that $\mathrm{Ph}_{m,n}(k) = \mathrm{Ph}_{m,n}(r) = \mathbf{Ph}(\langle\langle b^r \rangle\rangle)$. $\square$

**Theorem 5.24.** Let $m, n \in \mathbb{Z}\{0\}$ and $q$ be a finite phenotype.

(1) If $\mathrm{gcd}(m, n) = 1$, then the perfect kernel contains a unique normal subgroup of phenotype $q$, namely $\langle\langle b^q \rangle\rangle$.

(2) If $\mathrm{gcd}(m, n) \neq 1$, then the perfect kernel contains continuum many normal subgroups of phenotype $q$.

*Proof.* The case $\mathrm{gcd}(m, n) = 1$ follows from Item (5) of Theorem 5.20. Therefore let us assume that $\mathrm{gcd}(m, n) \neq 1$.

Consider a prime $p$ which divides both $m$ and $n$. Then either $|q|_p \neq 0$ and we set $k := q$ otherwise set $k := qp$. In both cases, remark that $\mathrm{Ph}_{m,n}(k) = q$, that $\mathrm{gcd}(k, m) = \mathrm{gcd}(k, n)$ and hence $r(k) = k$. Then Proposition 5.22 yields that $\bar{b}$ has order $k$ in $G_{m,n,k}$. Furthermore since $k_0 := \mathrm{gcd}(k, m) = \mathrm{gcd}(k, n) > 1$, the elements $\bar{b}^n$ and $\bar{b}^m$ are not generators of the subgroup $\langle\bar{b}\rangle$: the group
is closed. Also note that $\mathcal{M}\mathcal{C}_q$ is obviously $\text{BS}(m, n)$-invariant. Finally Item (3) implies that every closed $\text{BS}(m, n)$-invariant subset of $\mathbf{Ph}^{-1}(q)$ is contained in $\mathcal{M}\mathcal{C}_q$. This proves that $\mathcal{M}\mathcal{C}_q$ is the largest closed $\text{BS}(m, n)$-invariant subset of $\text{Sub}(\text{BS}(m, n))$ contained in $\mathbf{Ph}^{-1}(q)$. Since all normal subgroups and all finite index subgroups have finite (hence closed) orbits, the remaining statement in Item (1) is clear.

Let us prove Item (4). Suppose $|n| \neq |m|$; let $p$ be a prime number such that $|m|_p \neq |n|_p$. By definition $\text{Ph}_{m,n}(sp) = \text{Ph}_{m,n}(s) = q$, so $\langle b^{sp} \rangle \in \mathcal{K}_q(\text{BS}(m, n)) \setminus \mathcal{M}\mathcal{C}_q$. Consider a subgroup $\Lambda \in \mathcal{K}_q(\text{BS}(m, n))$ whose orbit is dense in $\mathcal{K}_q(\text{BS}(m, n))$, as provided by Corollary 5.15. Since the orbit of $\Lambda$ accumulates to $\langle b^{sp} \rangle \notin \mathcal{M}\mathcal{C}_q$ and $\mathcal{M}\mathcal{C}_q$ is invariant and closed, the latter does not contain any point of that orbit. Hence the complement $\mathcal{K}_q(\text{BS}(m, n)) \setminus \mathcal{M}\mathcal{C}_q$ contains the dense orbit of $\Lambda$. We conclude that $\mathcal{M}\mathcal{C}_q \cap \mathcal{K}_q(\text{BS}(m, n))$ has empty interior in $\mathcal{K}_q(\text{BS}(m, n))$.

We finally prove Item (5). The equality $s = q$ follows immediately from Formula (5.18) for $s(q, m, n)$. We have the presentation

$$ \text{BS}(m, n)/\langle\langle b^q \rangle\rangle = \langle \bar{b}, \bar{t} : \bar{t}\bar{b}^m\bar{t}^{-1} = \bar{b}^n, \bar{b}^q = 1 \rangle. $$

Since $\text{gcd}(q, m) = \text{gcd}(q, n) = 1$, the elements $\bar{b}^m$ and $\bar{b}^n$ both generate $\langle \bar{b} \rangle$ in $\text{BS}(m, n)/\langle\langle b^q \rangle\rangle$. We thus have a natural semi-direct product decomposition

$$ \text{BS}(m, n)/\langle\langle b^q \rangle\rangle \cong \mathbb{Z}/q\mathbb{Z} \times \mathbb{Z} = \langle \bar{b} \rangle \times \langle \bar{t} \rangle $$

Consider $\Lambda \in \mathcal{M}\mathcal{C}_q$ in the perfect kernel; it contains $\langle\langle b^q \rangle\rangle$. It suffices to prove that the image $\Lambda_q := \Lambda/\langle\langle b^q \rangle\rangle$ of $\Lambda$ in $\langle \bar{b} \rangle \times \langle \bar{t} \rangle$ is trivial. Since $\mathbf{Ph}(\Lambda) = q$, the index $[\langle b \rangle : \Lambda \cap \langle b \rangle]$ is a multiple of $q$, so we have $\Lambda_q \cap \langle \bar{b} \rangle = \{\text{id}\}$. Thus $\Lambda_q$ is mapped injectively in the quotient $\langle \bar{b} \rangle \times \langle \bar{t} \rangle / \langle \bar{b} \rangle \simeq \mathbb{Z}$. If this image were not $\{0\}$, then $\Lambda$ would have finite index in $\text{BS}(m, n)$, contradicting that $\Lambda$ is in the perfect kernel. The group $\Lambda_q$ is thus trivial as wanted. $\square$

**Remark 5.21.** In terms of actions, $\mathcal{M}\mathcal{C}_q$ is the set of classes $[\alpha, v]$ all of whose cardinalities of $b$-orbits divide $s$ and have phenotype $q$.

**Proposition 5.22.** Let $m, n \in \mathbb{Z} \setminus \{0\}$ and $k \in \mathbb{Z}_{\ge 1}$. Let

$$ G_{m,n,k} := \text{BS}(m, n)/\langle\langle b^k \rangle\rangle = \langle \bar{t}, \bar{b} \mid \bar{t}\bar{b}^m\bar{t}^{-1} = \bar{b}^n, \bar{b}^k = 1 \rangle $$

and let

$$ r(k) := \max\{r' \in \mathbb{N} : r' \text{ divides } k \text{ and } \text{gcd}(r', m) = \text{gcd}(r', n)\}. $$

Then:
$G_{m,n,k}$ is not a semi-direct product. We claim that $G_{m,n,k}$ is not amenable. Indeed, we can write the group $G_{m,n,k}$ as the amalgamated free product

$$G_{m,n,k} = \langle \bar{t}, \bar{c} \mid \bar{t}(\bar{c})^{\frac{m}{k_0}} \bar{t}^{-1} = (\bar{c})^{\frac{n}{k_0}}, (\bar{c})^{\frac{k}{k_0}} = 1 \rangle *_{\bar{c}=\bar{b}^{k_0}} \langle \bar{b} \mid \bar{b}^k \rangle$$

and one can easily check that $G_{m,n,k}$ admits as a quotient the non-amenable free product $\langle \bar{t} \rangle * \langle \bar{b} \mid \bar{b}^{k_0} \rangle$.

Since $G_{m,n,k}$ is the fundamental group of a finite graph of finite groups, it admits a finite index normal subgroup $F$ which is a finitely generated free group [Ser80, Prop. 11 p. 120]. Since $G_{m,n,k}$ is non-amenable, this normal free subgroup is not amenable.

Every characteristic subgroup $N$ of $F$ is itself normal in $G_{m,n,k}$. Thus the pull-back under the quotient map $\text{BS}(m, n) \to G_{m,n,k}$ is a normal subgroup $\tilde{N} \subset \text{BS}(m, n)$. Since the intersection of $F$ with the finite group $\langle \bar{b} \rangle$ is trivial, the same holds for its characteristic subgroups: $N \cap \langle \bar{b} \rangle = \{\text{id}\}$. Therefore the order of the image of $b$ in $G_{m,n,k}/N = \text{BS}(m, n)/\tilde{N}$ is the same as in $G_{m,n,k}$, namely $k$. In other words, $\tilde{N} \cap \langle \bar{b} \rangle = \langle \bar{b}^k \rangle$. By Remark 5.7,

$$\mathbf{Ph}(\tilde{N}) = \text{Ph}_{m,n}(\langle \bar{b} \rangle : \tilde{N} \cap \langle \bar{b} \rangle] = \text{Ph}_{m,n}(k) = q.$$

There are continuum many characteristic subgroups $N$ in the finitely generated free subgroup $F$ [Bry74] (see also [BGK17]). At most countably many of them lie outside the perfect kernel, so the theorem follows. $\square$

# 6 Limits of finite phenotype subgroups

In this section, we characterize the subgroups of infinite phenotype of $\text{BS}(m, n)$ which arise as limits of finite phenotype subgroups. We will use a version of the straightforward fact that finitely generated subgroups always form a dense set in the space of subgroups.

**Lemma 6.1.** Let $m, n \in \mathbb{Z}\setminus\{0\}$. For every phenotype $q \in \mathcal{Q}_{m,n}$, the finitely generated subgroups of phenotype $q$ are dense in $\mathbf{Ph}^{-1}(q)$.

*Proof.* Let $\Lambda$ be a non finitely generated subgroup of phenotype $q$. Let $k \in \mathbb{Z}_{\ge 0}$ such that $\Lambda \cap \langle \bar{b} \rangle = \langle \bar{b}^k \rangle$. The group $\Lambda$ can be written as the increasing union of finitely generated subgroups all containing $\bar{b}^k$. They have the same phenotype as $\Lambda$. $\square$
1. b has order $r(k)$ in the quotient $G_{m,n,k}$; in particular $\langle\langle b^k \rangle\rangle = \langle\langle b^{r(k)} \rangle\rangle$;

2. the group $G_{m,n,k} = G_{m,n,r(k)}$ is the HNN extension of $\mathbb{Z}/r(k)\mathbb{Z} = \langle \bar{b} \rangle$ with respect to the relation $\tilde{b}\tilde{b}^m\tilde{t}^{-1} = \bar{b}^n$.

3. $\mathrm{Ph}_{m,n}(k) = \mathrm{Ph}_{m,n}(r(k)) = \mathbf{Ph}(\langle\langle b^k \rangle\rangle)$.

**Remark 5.23.** It follows from Item 1 in the above proposition that $r(k) = [\langle b \rangle : \langle\langle b^k \rangle\rangle \cap \langle b \rangle]$.

It is a routine computation, working prime number by prime number, to check that

$$r(k) = \prod_{\substack{p \in \mathcal{P} \\ |m|_p = |n|_p}} p^{|k|_p} \cdot \prod_{\substack{p \in \mathcal{P} \\ |m|_p \neq |n|_p}} p^{\min(|k|_p, |m|_p, |n|_p)} \quad (5.24)$$

In particular, $r(k)$ is a multiple of all the $r'$s which divide $k$ and satisfy $\gcd(r', m) = \gcd(r', n)$. Moreover $r(r(k)) = r(k)$.

**Remark 5.25.** It also follows from Item 1 and 3 of the above proposition that the set of integers $k$ of phenotype $q$ such that $r(k) = k$ parametrizes the normal subgroups of the form $\langle\langle b^{k'} \rangle\rangle$ of phenotype $q$. Comparing Equations (5.24) and (5.18), one can check that this is exactly the set of integers $k$ that are multiple of $q$ and that divide $s(q, m, n)$, i.e. $k = q \cdot j$ where

1. $|j|_p = 0$ for $p \in \mathcal{P}$ such that $|m|_p = |n|_p = 0$;

2. $|j|_p \le |m|_p$ for $p \in \mathcal{P}$ such that $|m|_p = |n|_p > 0$ and $|q|_p = 0$;

3. $|j|_p \le \min\{|n|_p, |m|_p\}$ for $p \in \mathcal{P}$ such that $|m|_p \ne |n|_p$.

**Proof of Proposition 5.22.** Set $r := r(k)$. Since $\bar{b}^m$ and $\bar{b}^n$ are conjugate in $G_{m,n,k}$, they have the same order:

$$\frac{\mathrm{ord}(\bar{b})}{\gcd(\mathrm{ord}(\bar{b}), m)} = \mathrm{ord}(\bar{b}^m) = \mathrm{ord}(\bar{b}^n) = \frac{\mathrm{ord}(\bar{b})}{\gcd(\mathrm{ord}(\bar{b}), n)}.$$

Thus $\gcd(\mathrm{ord}(\bar{b}), m) = \gcd(\mathrm{ord}(\bar{b}), n)$. Moreover $\mathrm{ord}(\bar{b})$ divides $k$. So by the definition of $r$, the order $\mathrm{ord}(\bar{b})$ divides $r$ and hence $b^r \in \langle\langle b^k \rangle\rangle$. On the other hand $b^k \in \langle b^r \rangle$, so $\langle\langle b^r \rangle\rangle = \langle\langle b^k \rangle\rangle$ and $G_{m,n,k} = G_{m,n,r}$.

Since $\gcd(r, m) = \gcd(r, n)$, the subgroups generated by $\tilde{b}^m$ and $\tilde{b}^n$ in the group $\mathbb{Z}/r\mathbb{Z} = \langle \tilde{b} : \tilde{b}^r = 1 \rangle$ are isomorphic. We can thus consider the HNN-extension of $\mathbb{Z}/r\mathbb{Z} = \langle \tilde{b} : \tilde{b}^r = 1 \rangle$ with the relation $\tilde{t}\tilde{b}^m\tilde{t}^{-1} = \tilde{b}^n$. It admits the presentation $\langle \tilde{t}, \tilde{b} | \tilde{t}\tilde{b}^m\tilde{t}^{-1} = \tilde{b}^n, \tilde{b}^r = 1 \rangle$ and it is hence isomorphic to $G_{m,n,r}$.

By the Normal Form Theorem for HNN-extensions, the vertex group in- jects, i.e., $\bar{b}$ has order exactly $r$. Finally Formula (5.24) imply that $\mathrm{Ph}_{m,n}(k) = \mathrm{Ph}_{m,n}(r) = \mathbf{Ph}(\langle\langle b^r \rangle\rangle)$. $\square$
## 6.1 Limits of subgroups with fixed finite phenotype

Recall from Proposition 5.8 that, for $q$ finite, $\mathbf{Ph}^{-1}(q)$ is open while $\mathbf{Ph}^{-1}(\infty)$ is closed, and from Theorem 5.20 (2) that the orbit of any $\Lambda \in \mathbf{Ph}^{-1}(q) \setminus \mathcal{MC}_q$ accumulates to $\mathbf{Ph}^{-1}(\infty)$. We now determine the set of such accumulation points in $\mathbf{Ph}^{-1}(\infty)$: this is exactly the set of subgroups contained in the normal closure $\langle\!\langle b\rangle\!\rangle$ of $b$ but having trivial intersection with $\langle\!\langle b\rangle\!\rangle$ itself.

**Theorem 6.2.** Suppose $|m| \neq |n|$ and let $q$ be a finite phenotype. Then

$$ \overline{\mathbf{Ph}^{-1}(q)} \cap \mathbf{Ph}^{-1}(\infty) = \{\Lambda \in \mathbf{Ph}^{-1}(\infty) : \Lambda \le \langle\!\langle b\rangle\!\rangle\}. $$

We need two preparatory lemmas. We start with an easy consequence of the defining relation $tb^m = b^nt$ of $\mathrm{BS}(m, n)$.

**Notation 6.3.** Given $\gamma \in \mathrm{BS}(m, n)$, let us denote:

* by $\kappa_\gamma$ the $t$-length of $\gamma$, namely the number of occurrences of $t^{\pm 1}$ in the normal form of $\gamma$;

* by $\Sigma_\gamma$ the number of occurrences of $t$ minus the number of occurrences of $t^{-1}$ in the normal form of $\gamma$, which is often called the $t$-height of $\gamma$.

Remark that $\Sigma_\gamma$ is the image of $\gamma$ in $\mathrm{BS}(m, n)/\langle\!\langle b\rangle\!\rangle \cong \mathbb{Z}$. In particular $\Sigma_\gamma = 0$ if and only if $\gamma \in \langle\!\langle b\rangle\!\rangle$.

**Lemma 6.4.** Fix $\gamma \in \mathrm{BS}(m, n)$. Let $A \in \mathbb{Z}$ be such that for all primes $p \in \mathcal{P}$

* if $|m|_p = |n|_p$ then $|A|_p \ge |m|_p$;

* otherwise $|A|_p \ge \kappa_\gamma |m|_p$ and $|A|_p \ge \kappa_\gamma |n|_p$.

Then there is $B \in \mathbb{Z}$, such that $\gamma b^A = b^B \gamma$, where $|B|$ is determined by:

$$ |B|_p = |A|_p + \Sigma_\gamma (|n|_p - |m|_p) \text{ for all } p \in \mathcal{P}. $$

*Proof.* This follows from a straightforward induction on $\kappa_\gamma$ using the relation $tb^m = b^nt$. We leave the details to the reader. $\square$

The proof of the inclusion in Theorem 6.2 from left to right relies on the following lemma.

**Lemma 6.5.** Fix $\gamma \notin \langle\!\langle b\rangle\!\rangle$ and let $q$ be a finite phenotype. There is an integer $R = R(q, \gamma)$ such that every subgroup $\Lambda$ of phenotype $q$ containing $\gamma$ must also contain $b^R$.
**Theorem 5.26.** Let $m, n \in \mathbb{Z} \setminus \{0\}$ and $q$ be a finite phenotype.

(1) If $\gcd(m, n) = 1$, then the perfect kernel contains a unique normal subgroup of phenotype $q$, namely $\langle\langle b^q \rangle\rangle$.

(2) If $\gcd(m, n) \neq 1$, then the perfect kernel contains continuum many normal subgroups of phenotype $q$.

*Proof.* The case $\gcd(m, n) = 1$ follows from Item (5) of Theorem 5.20. Therefore let us assume that $\gcd(m, n) \neq 1$.

Consider a prime $p$ which divides both $m$ and $n$. Then either $|q|_p \neq 0$ and we set $k := q$ otherwise set $k := qp$. In both cases, remark that $\mathrm{Ph}_{m,n}(k) = q$, that $\mathrm{gcd}(k, m) = \mathrm{gcd}(k, n)$ and hence $r(k) = k$. Then Proposition 5.22 yields that $\bar{b}$ has order $k$ in $G_{m,n,k}$. Furthermore since $k_0 := \mathrm{gcd}(k, m) = \mathrm{gcd}(k, n) > 1$, the elements $\bar{b}^n$ and $\bar{b}^m$ are not generators of the subgroup $\langle\bar{b}\rangle$: the group $G_{m,n,k}$ is not a semi-direct product. We claim that $G_{m,n,k}$ is not amenable. Indeed, we can write the group $G_{m,n,k}$ as the amalgamated free product

$$G_{m,n,k} = \langle \bar{t}, \bar{c} \mid \bar{t}(\bar{c})^{\frac{m}{k_0}} \bar{t}^{-1} = (\bar{c})^{\frac{n}{k_0}}, (\bar{c})^{\frac{k}{k_0}} = 1 \rangle *_{\bar{c}=\bar{b}^{k_0}} \langle \bar{b} \mid \bar{b}^k \rangle$$

and one can easily check that $G_{m,n,k}$ admits as a quotient the non-amenable free product $\langle\bar{t}\rangle * \langle\bar{b} \mid \bar{b}^{k_0}\rangle$.

Since $G_{m,n,k}$ is the fundamental group of a finite graph of finite groups, it admits a finite index normal subgroup $F$ which is a finitely generated free group [Ser80, Prop. 11 p. 120]. Since $G_{m,n,k}$ is non-amenable, this normal free subgroup is not amenable.

Every characteristic subgroup $N$ of $F$ is itself normal in $G_{m,n,k}$. Thus the pull-back under the quotient map $\mathrm{BS}(m, n) \to G_{m,n,k}$ is a normal subgroup $\tilde{N} \subsetneq \mathrm{BS}(m, n)$. Since the intersection of $F$ with the finite group $\langle\bar{b}\rangle$ is trivial, the same holds for its characteristic subgroups: $N \cap \langle\bar{b}\rangle = \{\mathrm{id}\}$. Therefore the order of the image of $b$ in $G_{m,n,k}/\tilde{N} = \mathrm{BS}(m, n)/\tilde{N}$ is the same as in $G_{m,n,k}$, namely $k$. In other words, $\tilde{N} \cap \langle b \rangle = \langle b^k \rangle$. By definition,

$$\mathrm{Ph}(\tilde{N}) = \mathrm{Ph}_{m,n}(\langle b \rangle : \tilde{N} \cap \langle b \rangle] = \mathrm{Ph}_{m,n}(k) = q.$$

There are continuum many characteristic subgroups $N$ in the finitely generated free subgroup $F$ [Bry74] (see also [BGK17]). At most countably many of them lie outside the perfect kernel, so the theorem follows. $\square$
Proof. Up to replacing $\gamma$ by its inverse, let us assume $\Sigma_{\gamma} > 0$. We define $M := \max\{|m|_p, |n|_p : p \in \mathcal{P}\}$, and then

$$R := q \left( \prod_{\substack{p \in \mathcal{P} \\ |m|_p + |n|_p > 0}} p \right)^{\kappa_{\gamma} M}.$$

Fix $\Lambda$ of phenotype $q$. Since $q$ is finite, we have $\langle b \rangle \cap \Lambda = \langle b^N \rangle$ with $N > 0$. We have to show that $N$ divides $R$. Notice that $\mathrm{Ph}_{m,n}(N) = q$, thus $N$ decomposes as

$$N = q \cdot p_1^{l_1} \cdots p_k^{l_k} p_{k+1}^{l_{k+1}} \cdots p_r^{l_r},$$

where $r \ge 0$ and $l_1, \dots, l_r \ge 1$, while the $p_i$ are distinct prime numbers co-prime with $q$, see Definition 4.1. Moreover, we order them so that $p_1, \dots, p_k \in \mathcal{P}_{m,n} \setminus \mathcal{P}_{m,n}(N)$ and $p_{k+1}, \dots, p_r \in \mathcal{P} \setminus \mathcal{P}_{m,n}$.

Observe that $|m|_{p_i} = |n|_{p_i} \ge |N|_{p_i} = l_i \ge 1$ when $p_i \in \mathcal{P}_{m,n} \setminus \mathcal{P}_{m,n}(N)$ and $|m|_{p_i} \ne |n|_{p_i}$ when $p_i \in \mathcal{P} \setminus \mathcal{P}_{m,n}$. Hence, $|m|_{p_i} + |n|_{p_i} > 0$ for every $i \in \{1, \dots, r\}$. Consequently, to establish that $N$ divides $R$, it suffices to prove

$$\forall i \in \{1, \dots, r\}, \quad l_i \le \kappa_{\gamma} M. \qquad (6.6)$$

Observe that $\kappa_{\gamma} \ge 1$ since $\gamma \notin \langle\langle b \rangle\rangle$. For $i \in \{1, \dots, k\}$, Equation (6.6) holds since $p_i \in \mathcal{P}_{m,n} \setminus \mathcal{P}_{m,n}(N)$, thus

$$l_i \le |m|_{p_i} = |n|_{p_i} \le M \le \kappa_{\gamma} M.$$

Let us hence fix $i \in \{k+1, \dots, r\}$ and suppose by contradiction that $l_i > \kappa_{\gamma} M$. Consider

$$N' = N \times (p_1 \cdots p_k)^M (p_{k+1} \cdots \hat{p}_i \cdots p_r)^{\kappa_{\gamma} M}$$

where by $\hat{p}_i$ we mean that the factor $p_i$ is removed from the product. Clearly $b^{N'} \in \Lambda$ and $|N'|_{p_i} = l_i$. Put

$$\varepsilon := \mathrm{sign}(|m|_{p_i} - |n|_{p_i}).$$

Note that $p_i \notin \mathcal{P}_{m,n}$, hence $|m|_{p_i} \ne |n|_{p_i}$, so $\varepsilon \ne 0$. Since we assumed $|N|_{p_i} = l_i \ge \kappa_{\gamma} M$, we also have $|N'|_{p_i} \ge \kappa_{\gamma} M$. It is then clear that $N'$
# 6 Limits of finite phenotype subgroups

In this section, we characterize the subgroups of infinite phenotype of $BS(m, n)$ which arise as limits of finite phenotype subgroups. We will use a version of the straightforward fact that finitely generated subgroups always form a dense set in the space of subgroups.

**Lemma 6.1.** Let $m, n \in \mathbb{Z} \setminus \{0\}$. For every phenotype $q \in Q_{m,n}$, the finitely generated subgroups of phenotype $q$ are dense in $\mathbf{Ph}^{-1}(q)$.

*Proof.* Let $\Lambda$ be a non finitely generated subgroup of phenotype $q$. Let $k \in \mathbb{Z}_{\ge 0}$ such that $\Lambda \cap \langle b \rangle = \langle b^k \rangle$. The group $\Lambda$ can be written as the increasing union of finitely generated subgroups all containing $b^k$. They have the same phenotype as $\Lambda$. $\square$

## 6.1 Limits of subgroups with fixed finite phenotype

Recall from Proposition 5.8 that, for $q$ finite, $\mathbf{Ph}^{-1}(q)$ is open while $\mathbf{Ph}^{-1}(\infty)$ is closed, and from Theorem 5.20 (3) that the orbit of any $\Lambda \in \mathbf{Ph}^{-1}(q) \setminus \mathcal{MC}_q$ accumulates to $\mathbf{Ph}^{-1}(\infty)$. We now determine the set of such accumulation points in $\mathbf{Ph}^{-1}(\infty)$: this is exactly the set of subgroups contained in the normal closure $\langle\!\langle b \rangle\!\rangle$ of $\langle b \rangle$ but having trivial intersection with $\langle b \rangle$ itself (since they belong to $\mathbf{Ph}^{-1}(\infty)$).

**Theorem 6.2.** Suppose $|m| \ne |n|$ and let $q$ be a finite phenotype. Then

$$ \overline{\mathbf{Ph}^{-1}(q)} \cap \mathbf{Ph}^{-1}(\infty) = \{\Lambda \in \mathbf{Ph}^{-1}(\infty) : \Lambda \le \langle\!\langle b \rangle\!\rangle\}. $$

We need two preparatory lemmas. We start with an easy consequence of the defining relation $tb^m = b^n t$ of $BS(m, n)$.

**Notation 6.3.** Given $\gamma \in BS(m, n)$, let us denote:

* by $\kappa_\gamma$ the $t$-length of $\gamma$, namely the number of occurrences of $t^{\pm 1}$ in the normal form of $\gamma$;

* by $\Sigma_\gamma$ the number of occurrences of $t$ minus the number of occurrences of $t^{-1}$ in the normal form of $\gamma$, which is often called the $t$-height of $\gamma$.

Remark that $\Sigma_\gamma$ is the image of $\gamma$ in $BS(m, n)/\langle\!\langle b \rangle\!\rangle \cong \mathbb{Z}$. In particular $\Sigma_\gamma = 0$ if and only if $\gamma \in \langle\!\langle b \rangle\!\rangle$.

**Lemma 6.4.** Fix $\gamma \in BS(m, n)$. Let $A \in \mathbb{Z}$ be such that for all primes $p \in \mathcal{P}$
satisfies the assumption of Lemma 6.4, so $\gamma^\epsilon b^{N'}\gamma^{-\epsilon} = b^{N''}$, where

$$ \begin{aligned} |N''|_{p_i} &= l_i + \sum_{\gamma^\epsilon} (|n|_p - |m|_p) = l_i + \sum_{\gamma^\epsilon} (|n|_p - |m|_p) \\ &= l_i - \sum_{\gamma} |m|_{p_i} - |n|_{p_i} < l_i. \end{aligned} $$

Clearly $b^{N''} \in \Lambda$, hence $b^{N''} \in \langle b^N \rangle$. But $|N''|_{p_i} < |N|_{p_i}$, a contradiction. We thus have established Equation (6.6), which finishes the proof. □

*Proof of Theorem 6.2.* Set

$$ \mathcal{L} := \{ \Lambda \in \mathbf{Ph}^{-1}(\infty) : \Lambda \le \langle\langle b \rangle\rangle \}. $$

We first show the inclusion $\overline{\mathbf{Ph}^{-1}(q)} \cap \mathbf{Ph}^{-1}(\infty) \subseteq \mathcal{L}$. Take $\Delta \in \mathbf{Ph}^{-1}(\infty) \setminus \mathcal{L}$ and $\gamma \in \Delta \setminus \langle\langle b \rangle\rangle$. By Lemma 6.5, there is an $R$ such that every subgroup $\Lambda$ of phenotype $q$ containing $\gamma$ also contains $b^R$. Thus the clopen neighborhood of $\Delta$ given by

$$ \mathcal{O} := \{ \Lambda \in \mathrm{Sub}(\mathrm{BS}(m, n)) : \gamma \in \Lambda, b^R \notin \Lambda \} $$

does not intersect $\mathbf{Ph}^{-1}(q)$. Thus $\Delta$ is not in the closure of $\mathbf{Ph}^{-1}(q)$.

We now show the reverse inclusion $\mathcal{L} \subseteq \overline{\mathbf{Ph}^{-1}(q)} \cap \mathbf{Ph}^{-1}(\infty)$. Remark that as in Lemma 6.1, the finitely generated elements of $\mathcal{L}$ are dense in $\mathcal{L}$: every element of $\mathcal{L}$ is an increasing union of finitely generated subgroups which have to be in $\mathcal{L}$ as well. So take $\Lambda = \langle S \rangle \in \mathcal{L}$ where $S$ is finite; we will show that $\Lambda$ is limit of subgroups with phenotype $q$. Set $\kappa := \max_{\gamma \in S} \kappa_\gamma$, where $\kappa_\gamma$ is the $t$-length of $\gamma$ (see Notation 6.3). Set $M := \max\{|m|_p, |n|_p : p \in \mathcal{P}\}$. Note that $\mathcal{P} \setminus \mathcal{P}_{m,n}$ is finite, since it is composed of primes $p$ such that $|m|_p + |n|_p > 0$, and that $|m|_p = 0$ for all but finitely many primes $p$. Hence, for $j \ge 1$, we can define the integer

$$ N_j := q \cdot \prod_{p \in \mathcal{P}_{m,n} \setminus \mathcal{P}_{m,n}(q)} p^{|m|_p} \cdot \prod_{p \in \mathcal{P} \setminus \mathcal{P}_{m,n}} p^{|\kappa M|} $$

Observe that $\mathrm{Ph}_{m,n}(N_j) = q$.

Since $\Lambda \le \langle\langle b \rangle\rangle$, the height $\Sigma_\gamma$ is zero (see Notation 6.3) for every $\gamma \in S$, whence, for every $\gamma \in S$ and every $j$, Lemma 6.4 gives $\gamma b^{N_j} = b^{\pm N_j} \gamma$. Thus, $\Lambda = \langle S \rangle$ normalizes $\langle b^{N_j} \rangle$. Moreover, $\Lambda$ has trivial intersection with $\langle b^{N_j} \rangle$ because it has infinite phenotype. In particular for $j = 1$, we have a natural isomorphism

$$ \Phi : \Lambda \times \langle b^{N_1} \rangle \to \langle \lambda, b^{N_1} \rangle. $$
* if $|m|_p = |n|_p$ then $|A|_p \ge |m|_p$;

* otherwise $|A|_p \ge \kappa_\gamma |m|_p$ and $|A|_p \ge \kappa_\gamma |n|_p$.

Then there is $B \in \mathbb{Z}$, such that $\gamma b^A = b^B \gamma$, where $|B|$ is determined by:

$$|B|_p = |A|_p + \Sigma_\gamma (|n|_p - |m|_p) \text{ for all } p \in \mathcal{P}.$$

*Proof.* This follows from a straightforward induction on $\kappa_\gamma$ using the relation $tb^m = b^n t$. We leave the details to the reader. □

The proof of the inclusion in Theorem 6.2 from left to right relies on the following lemma.

**Lemma 6.5.** Fix $\gamma \notin \langle\langle b \rangle\rangle$ and let $q$ be a finite phenotype. There is an integer $R = R(q, \gamma)$ such that every subgroup $\Lambda$ of phenotype $q$ containing $\gamma$ must also contain $b^R$.

*Proof.* Up to replacing $\gamma$ by its inverse, let us assume $\Sigma_\gamma > 0$. We define $M := \max\{|m|_p, |n|_p : p \in \mathcal{P}\}$, and then

$$R := q \left( \prod_{\substack{p \in \mathcal{P} \\ |m|_p + |n|_p > 0}} p \right)^{\kappa_\gamma M}.$$

Fix $\Lambda$ of phenotype $q$. Since $q$ is finite, we have $\langle b \rangle \cap \Lambda = \langle b^N \rangle$ with $N > 0$. We have to show that $N$ divides $R$. Notice that $\text{Ph}_{m,n}(N) = q$, thus $N$ decomposes as

$$N = q \cdot p_1^{l_1} \cdots p_k^{l_k} p_{k+1}^{l_{k+1}} \cdots p_r^{l_r},$$

where $r \ge 0$ and $l_1, \dots, l_r \ge 1$, while the $p_i$ are distinct prime numbers co-prime with $q$, see Definition 4.1. Moreover, we order them so that $p_1, \dots, p_k \in \mathcal{P}_{m,n} \setminus \mathcal{P}_{m,n}(N)$ and $p_{k+1}, \dots, p_r \in \mathcal{P} \setminus \mathcal{P}_{m,n}$.

Observe that $|m|_{p_i} = |n|_{p_i} \ge |N|_{p_i} = l_i \ge 1$ when $p_i \in \mathcal{P}_{m,n} \setminus \mathcal{P}_{m,n}(N)$ and $|m|_{p_i} \ne |n|_{p_i}$ when $p_i \in \mathcal{P} \setminus \mathcal{P}_{m,n}$. Hence, $|m|_{p_i} + |n|_{p_i} > 0$ for every $i \in \{1, \dots, r\}$. Consequently, to establish that $N$ divides $R$, it suffices to prove

$$\forall i \in \{1, \dots, r\}, \quad l_i \le \kappa_\gamma M. \qquad (6.6)$$

Observe that $\kappa_\gamma \ge 1$ since $\gamma \notin \langle\langle b \rangle\rangle$. For $i \in \{1, \dots, k\}$, Equation (6.6) holds since $p_i \in \mathcal{P}_{m,n} \setminus \mathcal{P}_{m,n}(N)$, thus

$$l_i \le |m|_{p_i} = |n|_{p_i} \le M \le \kappa_\gamma M.$$
Since $N_1$ divides $N_j$, we get

$$ \Phi(\Lambda \times \langle b^{N_j} \rangle) = \langle \lambda, b^{N_j} \rangle. $$

Observe that $\Phi$ induces a homeomorphism

$$ \mathrm{Sub}(\Lambda \times \langle b^{N_1} \rangle) \to \mathrm{Sub}(\langle \lambda, b^{N_1} \rangle) \subseteq \mathrm{Sub}(\mathrm{BS}(m, n)), $$

and that the sequence of subgroups $(\Lambda \times \langle b^{N_j} \rangle)_{j \ge 1}$ converges to $\Lambda \times \{e\}$.

Therefore we have that $\langle \lambda, b^{N_j} \rangle$ converges to $\Lambda$. Since $\mathbf{Ph}(\langle \lambda, b^{N_j} \rangle) = \mathrm{Ph}_{m,n}(N_j) = q$, the group $\Lambda$ is the limit of a sequence of elements of phenotype $q$ as wanted. $\square$

## 6.2 Limits of subgroups with finite phenotype

In Theorem 6.2, we showed that $\overline{\mathbf{Ph}^{-1}(q)} \cap \mathbf{Ph}^{-1}(\infty)$ does not depend on the finite phenotype $q$. We will now consider the closure of all subgroups with finite phenotype and we will first analyse what happens if $|m| = |n|$.

**Proposition 6.7.** Let $m, n$ be integers such that $|m| = |n| \ge 2$. Then

$$ \mathbf{Ph}^{-1}(\infty) \subseteq \overline{\bigcup_{q \text{ finite}} \mathbf{Ph}^{-1}(q)}. $$

In other words, every subgroup with infinite phenotype is a limit of subgroups with finite (variable) phenotypes.

*Proof.* Let us fix $\Lambda \in \mathbf{Ph}^{-1}(\infty)$. Note that $\langle b^n \rangle$ is normalized by $\Lambda$ thanks to the relation $tb^n t^{-1} = b^{\pm n}$. We now proceed as in the second part of the proof of Theorem 6.2: the group $\langle \lambda, b^{jn} \rangle$ has finite phenotype, it is isomorphic to $\Lambda \times \langle b^{jn} \rangle$ and the sequence of subgroups $(\langle \lambda, b^{jn} \rangle)_{j \ge 1}$ converges to $\Lambda$. $\square$

The situation is completely different in the case $|m| \ne |n|$.

**Proposition 6.8.** Let $m, n$ be integers such that $|m| \ne |n|$ and $|m|, |n| \ge 2$.

Then

$$ \mathbf{Ph}^{-1}(\infty) \notin \overline{\bigcup_{q \text{ finite}} \mathbf{Ph}^{-1}(q)}. $$

In other words, there are subgroups with infinite phenotype that are not limits of subgroups with finite (variable) phenotypes.
Let us hence fix $i \in \{k+1, \dots, r\}$ and suppose by contradiction that $l_i > \kappa_\gamma M$. Consider

$$N' = N \times (p_1 \cdots p_k)^M (p_{k+1} \cdots \hat{p}_i \cdots p_r)^{\kappa_\gamma M}$$

where by $\hat{p}_i$ we mean that the factor $p_i$ is removed from the product. Clearly $b^{N'} \in \Lambda$ and $|N'|_{p_i} = l_i$. Put

$$\varepsilon := \operatorname{sign}(|m|_{p_i} - |n|_{p_i}).$$

Note that $p_i \notin \mathcal{P}_{m,n}$, hence $|m|_{p_i} \neq |n|_{p_i}$, so $\varepsilon \neq 0$. Since we assumed $|N|_{p_i} = l_i \ge \kappa_\gamma M$, we also have $|N'|_{p_i} \ge \kappa_\gamma M$. It is then clear that $N'$ satisfies the assumption of Lemma 6.4, so $\gamma^\varepsilon b^{N'} \gamma^{-\varepsilon} = b^{N''}$, where

$$\begin{aligned} |N''|_{p_i} &= l_i + \sum_{\gamma^\varepsilon} (|n|_{p_i} - |m|_{p_i}) = l_i + \varepsilon \sum_{\gamma} (|n|_{p_i} - |m|_{p_i}) \\ &= l_i - \sum_{\gamma} |m|_{p_i} - |n|_{p_i} < l_i. \end{aligned}$$

Clearly $b^{N''} \in \Lambda$, hence $b^{N''} \in \langle b^N \rangle$. But $|N''|_{p_i} < |N|_{p_i}$, a contradiction. We thus have established Equation (6.6), which finishes the proof. $\square$

*Proof of Theorem 6.2.* Set

$$\mathcal{L} := \{ \Lambda \in \mathbf{Ph}^{-1}(\infty) : \Lambda \le \langle \! \langle b \rangle \! \rangle \}.$$

We first show the inclusion $\overline{\mathbf{Ph}^{-1}(q)} \cap \mathbf{Ph}^{-1}(\infty) \subseteq \mathcal{L}$. Take $\Delta \in \mathbf{Ph}^{-1}(\infty) \setminus \mathcal{L}$ and $\gamma \in \Delta \setminus \langle \! \langle b \rangle \! \rangle$. By Lemma 6.5, there is an $R$ such that every subgroup $\Lambda$ of phenotype $q$ containing $\gamma$ also contains $b^R$. Thus the clopen neighborhood of $\Delta$ given by

$$\mathcal{O} := \{ \Lambda \in \mathrm{Sub}(\mathrm{BS}(m, n)) : \gamma \in \Lambda, b^R \notin \Lambda \}$$

does not intersect $\mathbf{Ph}^{-1}(q)$. Thus $\Delta$ is not in the closure of $\mathbf{Ph}^{-1}(q)$.

We now show the reverse inclusion $\mathcal{L} \subseteq \overline{\mathbf{Ph}^{-1}(q)} \cap \mathbf{Ph}^{-1}(\infty)$. Remark that as in Lemma 6.1, the finitely generated elements of $\mathcal{L}$ are dense in $\mathcal{L}$: every element of $\mathcal{L}$ is an increasing union of finitely generated subgroups which have to be in $\mathcal{L}$ as well. So take $\Lambda = \langle S \rangle \in \mathcal{L}$ where $S$ is finite; we will show that $\Lambda$ is a limit of subgroups with phenotype $q$. Set $\kappa := \max_{\gamma \in S} \kappa_\gamma$, where $\kappa_\gamma$ is the $t$-length of $\gamma$ (see Notation 6.3). Set $M := \max\{|m|_p, |n|_p : p \in \mathcal{P}\}$. Note that $\mathcal{P} \setminus \mathcal{P}_{m,n}$ is finite, since it is composed of primes $p$ such that
Let us recall from Corollary 5.4 that $Ph^{-1}(\infty) = K_{\infty}(BS(m, n))$ whenever $|m| \neq |n|$. Hence, the subgroups given by the proposition lie in fact in $K_{\infty}(BS(m, n))$.

In the proof of Proposition 6.8, we will need a lemma and a proposition.

**Lemma 6.9.** Let $m, n$ be integers such that $|m| \neq |n|$ and $|m|, |n| \ge 2$. Let $k := \gcd(m, n)$. Let $\Lambda \le BS(m, n)$ be a subgroup containing the following elements

$$t, btb^{-1}, \dots, b^{k-1}tb^{-(k-1)}$$

If $\Lambda$ has finite phenotype, then $\Lambda$ has finite index in $BS(m, n)$.

*Proof.* Let $\alpha$ be the action $\Lambda \backslash BS(m, n) \hookrightarrow BS(m, n)$. Since the phenotype is finite, it is sufficient to show that the Bass-Serre graph $BS(\alpha)$ is finite (see Remark 4.10).

Since $\Lambda$ contains $t$, there is a loop in $BS(\alpha)$ at the vertex $v := \Lambda \langle b \rangle$. In particular, Equation (3.13) gives $\frac{L(v)}{\gcd(L(v),m)} = \frac{L(v)}{\gcd(L(v),n)}$. As $\Lambda$ has finite phenotype, $L(v)$ is finite, so that $\gcd(L(v),m) = \gcd(L(v),n)$. Moreover, as $BS(\alpha)$ is a saturated $(m, n)$-graph, we obtain

$$\deg_{\text{in}}(v) = \gcd(L(v), m) = \gcd(L(v), n) = \deg_{\text{out}}(v).$$

This number, that we will denote $d$, is the greatest common divisor of $m, n$ and $L(v)$. Hence $d$ divides $k = \gcd(m, n)$.

The $d$ outgoing edges at $v$ are exactly $\Lambda \langle b^n \rangle, \Lambda b \langle b^n \rangle, \dots, \Lambda b^{d-1} \langle b^n \rangle$. As $d \le k$, the subgroup $\Lambda$ contains $t, btb^{-1}, \dots, b^{d-1}tb^{-(d-1)}$. Since $\Lambda b^j t = (\Lambda b^j t b^{-j}) b^j = \Lambda b^j$, the element $t$ fixes all the points $\Lambda, \Lambda b, \dots, \Lambda b^{d-1} \in \Lambda \backslash BS(m, n)$. The terminal vertex of the edge $\Lambda b^j \langle b^n \rangle$ is precisely the vertex $\Lambda b^j t \langle b \rangle = \Lambda b^j \langle b \rangle = v$ (see Definition 3.5), so that all outgoing edges at $v$ are loops.

Since the outgoing degree at $v$ is equal to the ingoing degree, all ingoing edges at $v$ are loops as well. Therefore $BS(\alpha)$ consists only of the vertex $v$ and $d$ loops. It is thus finite as wanted. $\square$

**Proposition 6.10.** Let $m, n$ be integers with $|m|, |n| \ge 2$. Let $\Lambda$ be a finitely generated subgroup of infinite phenotype and infinite Bass-Serre graph. Then there is a sequence of conjugates of $\Lambda$ which converges to $\{\text{id}\}$. In particular, such a subgroup does not contain any non-trivial normal subgroup of $BS(m, n)$.
$|m|_p + |n|_p > 0$, and that $|m|_p = 0$ for all but finitely many primes $p$. Hence, for $j \ge 1$, we can define the integer

$$N_j := q \cdot \prod_{p \in \mathcal{P}_{m,n} \setminus \mathcal{P}_{m,n}(q)} p^{|m|_p} \cdot \prod_{p \in \mathcal{P} \setminus \mathcal{P}_{m,n}} p^{j\kappa M}.$$

Observe that $\mathrm{Ph}_{m,n}(N_j) = q$.

Since $\Lambda \le \langle\langle b \rangle\rangle$, the height $\Sigma_\gamma$ is zero (see Notation 6.3) for every $\gamma \in S$, whence, for every $\gamma \in S$ and every $j$, Lemma 6.4 gives $\gamma b^{N_j} = b^{\pm N_j} \gamma$. Thus, $\Lambda = \langle S \rangle$ normalizes $\langle b^{N_j} \rangle$. Moreover, $\Lambda$ has trivial intersection with $\langle b^{N_j} \rangle$ because it has infinite phenotype. In particular for $j = 1$, we have a natural isomorphism

$$\Phi: \Lambda \times \langle b^{N_1} \rangle \to \langle \Lambda, b^{N_1} \rangle.$$

Since $N_1$ divides $N_j$, we get

$$\Phi(\Lambda \times \langle b^{N_j} \rangle) = \langle \Lambda, b^{N_j} \rangle.$$

Observe that $\Phi$ induces a homeomorphism

$$\mathrm{Sub}(\Lambda \times \langle b^{N_1} \rangle) \to \mathrm{Sub}(\langle \Lambda, b^{N_1} \rangle) \subseteq \mathrm{Sub}(\mathrm{BS}(m, n)),$$

and that the sequence of subgroups $(\Lambda \times \langle b^{N_j} \rangle)_{j \ge 1}$ converges to $\Lambda \times \{\mathrm{id}\}$. Therefore we have that $\langle \Lambda, b^{N_j} \rangle$ converges to $\Lambda$. Since $\mathbf{Ph}(\langle \Lambda, b^{N_j} \rangle) = \mathrm{Ph}_{m,n}(N_j) = q$, the group $\Lambda$ is the limit of a sequence of elements of phenotype $q$ as wanted. $\square$

## 6.2 Limits of subgroups with varying finite phenotype

In Theorem 6.2, we showed that $\overline{\mathbf{Ph}^{-1}(q)} \cap \mathbf{Ph}^{-1}(\infty)$ does not depend on the finite phenotype $q$. We will now consider the closure of all subgroups with finite phenotype and we will first analyse what happens if $|m| = |n|$.

**Proposition 6.7.** Let $m, n$ be integers such that $|m| = |n| \ge 2$. Then

$$\mathbf{Ph}^{-1}(\infty) \subseteq \overline{\bigcup_{q \text{ finite}} \mathbf{Ph}^{-1}(q)}.$$

In other words, every subgroup with infinite phenotype is a limit of subgroups with finite (variable) phenotypes.
*Proof.* First recall that $\Lambda$ is free. Indeed, having infinite phenotype, it acts freely on the Bass-Serre tree $\mathcal{T}$ of $\text{BS}(m, n)$. Taking the class $\langle b \rangle$ as a base point in $\mathcal{T}$, the subgroup $\Lambda$ is the fundamental group of the quotient graph $\Lambda \backslash \mathcal{T}$ based at $\Lambda \langle b \rangle$. This quotient graph is equal to the Bass-Serre graph of $\Lambda$, see Section 3.6, so it is infinite. Since moreover $\Lambda$ is finitely generated, it consists of a finite graph to which are attached finitely many infinite trees. Moving the basepoint along one of those infinite trees toward infinity amounts to conjugating $\Lambda$ by a certain sequence of elements $\gamma_i$ of $\text{BS}(m, n)$ for which we claim that $\gamma_i \Lambda \gamma_i^{-1} \to \{\text{id}\}$. Indeed, each non-trivial element of $\gamma_i \Lambda \gamma_i^{-1}$ is represented by a long path in the tree, followed by a closed path in the finite graph and the long path back to the new basepoint. All such elements have a uniformly large $t$-length which tends to $+\infty$ with $i$: their $t$-length is bounded below by twice the $t$-length of $\gamma_i$ minus the diameter of the finite graph. In particular, for any finite set $F \subset \Gamma \setminus \{\text{id}\}$ and large enough $n$, all the elements of $\gamma_i \Lambda \gamma_i^{-1}$ have $t$-length larger than all those of $F$; so $\gamma_i \Lambda \gamma_i^{-1} \cap F = \emptyset$. This proves that $\gamma_i \Lambda \gamma_i^{-1} \to \{\text{id}\}$ as wanted. $\square$

*Proof of Proposition 6.8.* Consider the group $\Lambda := \langle t, btb^{-1}, \dots, b^{k-1}tb^{-(k-1)} \rangle$. Observe that by Britton's Lemma (see e.g. [LS01, Chapter IV.2]), it is a free group freely generated by $t, btb^{-1}, \dots, b^{k-1}tb^{-(k-1)}$. Every non-trivial element of $\Lambda$ contains at least one $t^{\pm 1}$ in its normal form, in particular $\Lambda \cap \langle b \rangle = \{\text{id}\}$: the phenotype of $\Lambda$ is infinite. We claim that

$$ \Lambda \notin \overline{\bigcup_{q \text{ finite}} \mathbf{Ph}^{-1}(q)}. $$

Suppose that $(\Lambda_i)_{i \ge 0}$ is a sequence of subgroups of finite (variable) phenotypes converging to $\Lambda$. For $i$ large enough, we have $t, btb^{-1}, \dots, b^{k-1}tb^{-(k-1)} \in \Lambda_i$, and thus the subgroup $\Lambda_i$ has finite index by Lemma 6.9. However, recall that since $|m| \ne |n|$, the group $\text{BS}(m, n)$ is not residually finite [Mes72]. Therefore there is a non-trivial normal subgroup $N \le \text{BS}(m, n)$ contained in every finite index subgroup, and we have $N \le \Lambda$ since $\Lambda_i \to \Lambda$. This is impossible by Proposition 6.10. $\square$

**Corollary 6.11.** Let $m, n$ be integers such that $|m| \ne |n|$ and $|m|, |n| \ge 2$. Then

$$ \overline{\bigcup_{q \text{ finite}} \mathbf{Ph}^{-1}(q) \cap \mathbf{Ph}^{-1}(\infty)} $$

has empty interior in $\mathbf{Ph}^{-1}(\infty)$.
Proof. Let us fix $\Lambda \in \mathbf{Ph}^{-1}(\infty)$. Note that $\langle b^n \rangle$ is normalized by $\Lambda$ thanks to the relation $tb^n t^{-1} = b^{\pm n}$. We now proceed as in the second part of the proof of Theorem 6.2: the group $\langle \Lambda, b^{jn} \rangle$ has finite phenotype, it is isomorphic to $\Lambda \times \langle b^{jn} \rangle$ and the sequence of subgroups $(\langle \Lambda, b^{jn} \rangle)_{j \ge 1}$ converges to $\Lambda$. $\square$

The situation is completely different in the case $|m| \ne |n|$.

**Proposition 6.8.** Let $m, n$ be integers such that $|m| \ne |n|$ and $|m|, |n| \ge 2$. Then

$$ \mathbf{Ph}^{-1}(\infty) \notin \overline{\bigcup_{q \text{ finite}} \mathbf{Ph}^{-1}(q)}. $$

In other words, there are subgroups with infinite phenotype that are not limits of subgroups with finite (variable) phenotypes.

Let us recall from Corollary 5.4 that $\mathbf{Ph}^{-1}(\infty) = \mathcal{K}_{\infty}(\mathrm{BS}(m, n))$ whenever $|m| \ne |n|$. Hence, the subgroups given by the proposition lie in fact in $\mathcal{K}_{\infty}(\mathrm{BS}(m, n))$.

In the proof of Proposition 6.8, we will need a lemma and a proposition.

**Lemma 6.9.** Let $m, n$ be integers such that $|m| \ne |n|$ and $|m|, |n| \ge 2$. Let $k := \gcd(m, n)$. Let $\Lambda \le \mathrm{BS}(m, n)$ be a subgroup containing the following elements

$$ t, btb^{-1}, \dots, b^{k-1}tb^{-(k-1)}. $$

If $\Lambda$ has finite phenotype, then $\Lambda$ has finite index in $\mathrm{BS}(m, n)$.

Proof. Let $\alpha$ be the action $\Lambda \backslash \mathrm{BS}(m, n) \hookrightarrow \mathrm{BS}(m, n)$. Since the phenotype is finite, it is sufficient to show that the Bass-Serre graph $\mathbf{BS}(\alpha)$ is finite (see Remark 4.10).

Since $\Lambda$ contains $t$, there is a loop in $\mathbf{BS}(\alpha)$ at the vertex $v := \Lambda \langle b \rangle$. In particular, Equation (3.13) gives $\frac{L(v)}{\gcd(L(v),m)} = \frac{L(v)}{\gcd(L(v),n)}$. As $\Lambda$ has finite phenotype, $L(v)$ is finite, so $\gcd(L(v),m) = \gcd(L(v),n)$. Moreover, as $\mathbf{BS}(\alpha)$ is a saturated $(m, n)$-graph, we obtain

$$ \deg_{\mathrm{in}}(v) = \gcd(L(v), m) = \gcd(L(v), n) = \deg_{\mathrm{out}}(v). $$

This number, that we will denote $d$, is the greatest common divisor of $m, n$ and $L(v)$. Hence $d$ divides $k = \gcd(m, n)$.

The $d$ outgoing edges at $v$ are exactly $\Lambda \langle b^n \rangle, \Lambda b \langle b^n \rangle, \dots, \Lambda b^{d-1} \langle b^n \rangle$. As $d \le k$, the subgroup $\Lambda$ contains $t, btb^{-1}, \dots, b^{d-1}tb^{-(d-1)}$. Since $\Lambda b^j t =$
*Proof.* Recall again that $\mathbf{Ph}^{-1}(\infty) = \mathcal{K}_{\infty}(\mathrm{BS}(m, n))$, see Corollary 5.4. In this space, the subset $\mathcal{K}_{\infty}(\mathrm{BS}(m, n)) \setminus \cup_{q \text{ finite}} \mathbf{Ph}^{-1}(q)$ is open and Proposition 6.8 implies that it is non-empty. By Corollary 5.15, this open subset contains a subgroup $\Lambda$ whose orbit is dense in $\mathcal{K}_{\infty}(\mathrm{BS}(m, n))$. Therefore $\cup_{q \text{ finite}} \mathbf{Ph}^{-1}(q)$ has empty interior in $\mathcal{K}_{\infty}(\mathrm{BS}(m, n))$. $\square$

**Proposition 6.12.** Let $m, n$ be integers such that $|m|, |n| \ge 2$. For any finite phenotype $q_0$, the following inclusion is strict:

$$ \overline{\mathbf{Ph}^{-1}(q_0)} \cap \mathbf{Ph}^{-1}(\infty) \subsetneq \overline{\bigcup_{q \text{ finite}} \mathbf{Ph}^{-1}(q)} \cap \mathbf{Ph}^{-1}(\infty). $$

Observe that Proposition 6.12 is trivially true if $|m| = |n|$. Indeed, Proposition 6.7 implies that the right hand side is equal to $\mathbf{Ph}^{-1}(\infty)$. Since Proposition 5.8 yields that $\mathbf{Ph}^{-1}(q_0)$ is closed, the left hand side is empty.

*Proof of Proposition 6.12.* For a prime $p$ which divides neither $m$ nor $m$, define $\Lambda_p := \langle b^p, t \rangle$. Then $\Lambda_p$ clearly has phenotype $p$ (and index $p$ in $\mathrm{BS}(m, n)$). Let $\Lambda$ be an accumulation point of the sequence $(\Lambda_p)$, then by construction $\Lambda$ has infinite phenotype, so it is in the set $\overline{\cup_{q \text{ finite}} \mathbf{Ph}^{-1}(q)} \cap \mathbf{Ph}^{-1}(\infty)$. However, it contains $t \notin \langle\langle b \rangle\rangle$ so it is not in $\overline{\mathbf{Ph}^{-1}(q_0)} \cap \overline{\mathbf{Ph}^{-1}(q_0)}$ by Theorem 6.2. $\square$

**Corollary 6.13.** Let $m, n$ be integers such that $|m|, |n| \ge 2$. The following inclusion is strict:

$$ \bigcup_{q \text{ finite}} \overline{\mathbf{Ph}^{-1}(q)} \cap \mathbf{Ph}^{-1}(\infty) \subsetneq \overline{\bigcup_{q \text{ finite}} \mathbf{Ph}^{-1}(q)} \cap \mathbf{Ph}^{-1}(\infty). $$

*Proof.* If $|m| = |n|$, then as already remarked the left hand side is empty.

If $|m| \ne |n|$, recall from Theorem 6.2 that $\overline{\mathbf{Ph}^{-1}(q_0)} \cap \mathbf{Ph}^{-1}(\infty) = \{\Lambda \in \mathbf{Ph}^{-1}(\infty) : \Lambda \le \langle\langle b \rangle\rangle\}$ and hence it is independent of $q_0$. $\square$

We can also give a statement analogous to Proposition 6.12 in the perfect kernel, which is less easy to obtain.

**Theorem 6.14.** Let $m, n$ be integers such that $|m|, |n| \ge 2$. For any finite phenotype $q_0$, the following inclusion is strict:

$$ \overline{\mathcal{K}_{q_0}(\mathrm{BS}(m, n))} \cap \mathcal{K}_{\infty}(\mathrm{BS}(m, n)) \subsetneq \overline{\bigcup_{q \text{ finite}} \mathcal{K}_q(\mathrm{BS}(m, n))} \cap \mathcal{K}_{\infty}(\mathrm{BS}(m, n)). $$
$(\Lambda b^j t b^{-j}) b^j = \Lambda b^j$, the element $t$ fixes all the points $\Lambda, \Lambda b, \dots, \Lambda b^{d-1} \in \Lambda \setminus \text{BS}(m, n)$. The terminal vertex of the edge $\Lambda b^j \langle b^n \rangle$ is precisely the vertex $\Lambda b^j t \langle b \rangle = \Lambda b^j \langle b \rangle = v$ (see Definition 3.5), so all outgoing edges at $v$ are loops.

Since the outgoing degree at $v$ is equal to the incoming degree, all incoming edges at $v$ are loops as well. Therefore $\text{BS}(\alpha)$ consists only of the vertex $v$ and $d$ loops. It is thus finite as wanted. $\square$

**Proposition 6.10.** Let $m, n$ be integers with $|m|, |n| \ge 2$. Let $\Lambda$ be a finitely generated subgroup of infinite phenotype and infinite Bass-Serre graph. Then there is a sequence of conjugates of $\Lambda$ which converges to $\{\text{id}\}$. In particular, $\Lambda$ does not contain any non-trivial normal subgroup of $\text{BS}(m, n)$.

*Proof.* First recall that $\Lambda$ is free. Indeed, having infinite phenotype, it acts freely on the Bass-Serre tree $\mathcal{T}$ of $\text{BS}(m, n)$. Taking the class $\langle b \rangle$ as a base point in $\mathcal{T}$, the subgroup $\Lambda$ is the fundamental group of the quotient graph $\Lambda \setminus \mathcal{T}$ based at $\Lambda \langle b \rangle$. This quotient graph is equal to the Bass-Serre graph of $\Lambda$, see Section 3.6, so it is infinite. Since moreover $\Lambda$ is finitely generated, it consists of a finite graph to which are attached finitely many infinite trees. Moving the basepoint along one of those infinite trees toward infinity amounts to conjugating $\Lambda$ by a certain sequence of elements $\gamma_i$ of $\text{BS}(m, n)$ for which we claim that $\gamma_i \Lambda \gamma_i^{-1} \to \{\text{id}\}$. Indeed, each non-trivial element of $\gamma_i \Lambda \gamma_i^{-1}$ is represented by a long path in the tree, followed by a closed path in the finite graph and the long path back to the new basepoint. All such elements have a uniformly large $t$-length which tends to $+\infty$ with $i$: their $t$-length is bounded below by twice the $t$-length of $\gamma_i$ minus the diameter of the finite graph. In particular, for any finite set $F \subset \Gamma \setminus \{\text{id}\}$ and large enough $i$, all the elements of $\gamma_i \Lambda \gamma_i^{-1}$ have $t$-length larger than all those of $F$; so $\gamma_i \Lambda \gamma_i^{-1} \cap F = \emptyset$. This proves that $\gamma_i \Lambda \gamma_i^{-1} \to \{\text{id}\}$ as wanted. $\square$

*Proof of Proposition 6.8.* Consider the group $\Lambda := \langle t, btb^{-1}, \dots, b^{k-1}tb^{-(k-1)} \rangle$. Observe that by Britton's Lemma (see e.g. [LS01, Chapter IV.2]), it is a free group freely generated by $t, btb^{-1}, \dots, b^{k-1}tb^{-(k-1)}$. Every non-trivial element of $\Lambda$ contains at least one $t^{\pm 1}$ in its normal form, in particular $\Lambda \cap \langle b \rangle = \{\text{id}\}$: the phenotype of $\Lambda$ is infinite. We claim that

$$ \Lambda \notin \bigcup_{q \text{ finite}} \mathbf{Ph}^{-1}(q). $$
Proof. For a fixed prime $p$ which divides neither $m$ nor $n$, let us define a pre-action $(\beta_p, \tau_p)$ as follows. Consider three $\beta_p$-cycles say $o_1, o_2$ and $o_3$, of cardinals $pn, p$ and $pm$ respectively. Then fix basepoints $y_i \in o_i$ for $i = 1, 2, 3$. Remark that $o_1$ splits into $|n| \ge 2 \beta_p^n$-orbits of size $p$ and that $o_3$ splits into $|m| \ge 2 \beta_p^m$-orbits of size $p$. Therefore we can define $\tau_p$ by setting

$$y_1\beta_p^{jn}\tau_p := y_2\beta_p^{jm}, \quad y_2\beta_p^{jn}\tau_p := y_3\beta_p^{jm} \quad \text{and} \quad y_1\beta_p^{-1+jn}\tau_p := y_3\beta_p^{1+jm}.$$

Clearly the phenotype of such a pre-action is $p$ and the associated Bass-Serre graph $\mathcal{G}_{0,p} := \mathbf{BS}(\beta_p, \tau_p)$ is a triangle. Set $x_p := y_1$ and note that for every $p$, we have

$$x_p\tau_p\tau_p\beta_p\tau_p^{-1}\beta_p = x_p.$$

By Lemma 4.21, we can then extend $\mathcal{G}_{0,p}$ to a saturated $(m, n)$-graph $\mathcal{G}_p$, see Figure 5, and by Proposition 3.23 we can extend the pre-action $(\beta_p, \tau_p)$ to an action $\alpha_p$ whose Bass-Serre graph is $\mathcal{G}_p$.

<figure><img src="image_8.png" /><figcaption>Figure 5: A (2,3)-graph $\mathcal{G}_p$, where $m = 2$ and $n = 3$.</figcaption></figure>

Define $\Lambda_p$ to be the stabilizer of the action $\alpha_p$ at $x_p$ and remark that $t^2bt^{-1}b \in \Lambda_p$. Moreover by construction $\mathbf{Ph}(\Lambda_p) = p$.

By compactness, we find an accumulation point $\Lambda$ of the sequence $(\Lambda_p)_p$. Since $\mathbf{Ph}(\Lambda_p) = p$, the subgroup $\Lambda$ has infinite phenotype. Since $t^2bt^{-1}b \in \Lambda_p$ for every $p$, we have that $t^2bt^{-1}b \in \Lambda$. Moreover $t^2bt^{-1}b \notin \langle\langle b \rangle\rangle$ so $\Lambda \notin \mathbf{Ph}^{-1}(q_0)$ by Theorem 6.2. Therefore the proof is completed. $\square$
Suppose that $(\Lambda_i)_{i \ge 0}$ is a sequence of subgroups of finite (variable) phenotypes converging to $\Lambda$. For $i$ large enough, we have $t, btb^{-1}, \dots, b^{k-1}tb^{-(k-1)} \in \Lambda_i$, and thus the subgroup $\Lambda_i$ has finite index by Lemma 6.9. However, recall that since $|m| \ne |n|$, the group $\text{BS}(m, n)$ is not residually finite [Mes72]. Therefore there is a non-trivial normal subgroup $N \le \text{BS}(m, n)$ contained in every finite index subgroup, and we have $N \le \Lambda$ since $\Lambda_i \to \Lambda$. This is impossible by Proposition 6.10. $\square$

**Corollary 6.11.** Let $m, n$ be integers such that $|m| \ne |n|$ and $|m|, |n| \ge 2$. Then

$$ \bigcup_{q \text{ finite}} \overline{\mathbf{Ph}^{-1}(q)} \cap \mathbf{Ph}^{-1}(\infty) $$

has empty interior in $\mathbf{Ph}^{-1}(\infty)$.

*Proof.* Recall again that $\mathbf{Ph}^{-1}(\infty) = \mathcal{K}_{\infty}(\text{BS}(m, n))$, see Corollary 5.4. In this space, the subset $\mathcal{K}_{\infty}(\text{BS}(m, n)) \setminus \bigcup_{q \text{ finite}} \overline{\mathbf{Ph}^{-1}(q)}$ is open and Proposition 6.8 implies that it is non-empty. By Corollary 5.15, this open subset contains a subgroup $\Lambda$ whose orbit is dense in $\mathcal{K}_{\infty}(\text{BS}(m, n))$. Therefore $\bigcup_{q \text{ finite}} \overline{\mathbf{Ph}^{-1}(q)}$ has empty interior in $\mathcal{K}_{\infty}(\text{BS}(m, n))$. $\square$

**Proposition 6.12.** Let $m, n$ be integers such that $|m|, |n| \ge 2$. For any finite phenotype $q_0$, the following inclusion is strict:

$$ \overline{\mathbf{Ph}^{-1}(q_0)} \cap \mathbf{Ph}^{-1}(\infty) \subset \bigcup_{q \text{ finite}} \overline{\mathbf{Ph}^{-1}(q)} \cap \mathbf{Ph}^{-1}(\infty). $$

Observe that Proposition 6.12 is trivially true if $|m| = |n|$. Indeed, Proposition 6.7 implies that the right hand side is equal to $\mathbf{Ph}^{-1}(\infty)$. Since Proposition 5.8 yields that $\mathbf{Ph}^{-1}(q_0)$ is closed, the left hand side is empty.

*Proof of Proposition 6.12.* For a prime $p$ which divides neither $m$ nor $n$, define $\Lambda_p := \langle b^p, t \rangle$. Then $\Lambda_p$ clearly has phenotype $p$ (and index $p$ in $\text{BS}(m, n)$). Let $\Lambda$ be an accumulation point of the sequence $(\Lambda_p)$, then by construction $\Lambda$ has infinite phenotype, so it is in the set $\bigcup_{q \text{ finite}} \overline{\mathbf{Ph}^{-1}(q)} \cap \mathbf{Ph}^{-1}(\infty)$. However, it contains $t \notin \langle\langle b \rangle\rangle$ so it is not in $\overline{\mathbf{Ph}^{-1}(q_0)}$ by Theorem 6.2. $\square$

**Corollary 6.13.** Let $m, n$ be integers such that $|m|, |n| \ge 2$. The following inclusion is strict:

$$ \bigcup_{q \text{ finite}} \overline{\mathbf{Ph}^{-1}(q)} \cap \mathbf{Ph}^{-1}(\infty) \subset \bigcup_{q \text{ finite}} \overline{\mathbf{Ph}^{-1}(q)} \cap \mathbf{Ph}^{-1}(\infty). $$
References

[AG22] Pénélope Azuelos and Damien Gaboriau. Some subgroup spaces with maximal perfect kernel. In preparation. 5

[Ben1883] Ivar O. Bendixson. Quelques théorèmes de la théorie des ensembles de points. *Acta Math.*, 2:415–429, 1883. 13

[BGK15] Lewis Bowen, Rostislav Grigorchuk, and Rostislav Kravchenko. Invariant random subgroups of lamplighter groups. *Israel Journal of Mathematics*, 207(2):763–782, April 2015. 5

[BGK17] Lewis Bowen, Rostislav Grigorchuk, and Rostislav Kravchenko. Characteristic random subgroups of geometric groups and free abelian groups of infinite rank. *Trans. Amer. Math. Soc.*, 369(2):755–781, 2017. 49

[BLT19] Oren Becker, Alexander Lubotzky, and Andreas Thom. Stability and invariant random subgroups. *Duke Mathematical Journal*, 168(12):2207–2234, September 2019. Publisher: Duke University Press. 4, 14, 38

[Bry74] Roger M. Bryant. Characteristic Subgroups of Free Groups. In M. F. Newman, editor, *Proceedings of the Second International Conference on the Theory of Groups*, Lecture Notes in Mathematics, pages 141–149, Berlin, Heidelberg, 1974. Springer. 49

[BS62] Gilbert Baumslag and Donald Solitar. Some two-generator one-relator non-Hopfian groups. *Bull. Amer. Math. Soc.*, 68:199–201, 1962. 2

[Can1884] Georg Cantor. Ueber unendliche, lineare punktmannichfaltigkeiten, part 6. *Mathematische Annalen*, 23(4):453–488, 1884. 13

[CGLM22] Alessandro Carderi, Damien Gaboriau, and François Le Maître. On dense totipotent free subgroups in full groups. *To appear in Geometry & Topology*, 2022. 4, 5

[CGLMS23] Alessandro Carderi, Damien Gaboriau, François Le Maître, and Yves Stalder. On the space of subgroups of Baumslag-Solitar groups II: the Cantor-Bendixson rank. in preparation, 2023. 5
Proof. If $|m| = |n|$, then as already remarked the left hand side is empty.

If $|m| \neq |n|$, recall from Theorem 6.2 that $\overline{\mathbf{Ph}^{-1}(q)} \cap \mathbf{Ph}^{-1}(\infty)$ is independent of $q$. The corollary thus follows from Proposition 6.12. $\square$

We can also give a statement analogous to Proposition 6.12 in the perfect kernel, which is less easy to obtain.

**Theorem 6.14.** Let $m, n$ be integers such that $|m|, |n| \ge 2$. For any finite phenotype $q_0$, the following inclusion is strict:

$$ \overline{\mathcal{K}_{q_0}(\mathrm{BS}(m, n))} \cap \mathcal{K}_{\infty}(\mathrm{BS}(m, n)) \subset \overline{\bigcup_{q \text{ finite}} \mathcal{K}_q(\mathrm{BS}(m, n))} \cap \mathcal{K}_{\infty}(\mathrm{BS}(m, n)). $$

Proof. For a fixed prime $p$ which divides neither $m$ nor $n$, let us define a pre-action $(\beta_p, \tau_p)$ as follows. Consider three $\beta_p$-cycles say $o_1, o_2$ and $o_3$, of cardinals $pn, p$ and $pm$ respectively. Then fix basepoints $y_i \in o_i$ for $i = 1, 2, 3$. Remark that $o_1$ splits into $|n| \ge 2 \beta_p^n$-orbits of size $p$ and that $o_3$ splits into $|m| \ge 2 \beta_p^m$-orbits of size $p$. Therefore we can define $\tau_p$ by setting

$$ y_1\beta_p^{jn}\tau_p := y_2\beta_p^{jm}, \quad y_2\beta_p^{jn}\tau_p := y_3\beta_p^{jm} \quad \text{and} \quad y_1\beta_p^{-1+jn}\tau_p := y_3\beta_p^{1+jm}. $$

Clearly the phenotype of such a pre-action is $p$ and the associated Bass-Serre graph $\mathcal{G}_{0,p} := \mathbf{BS}(\beta_p, \tau_p)$ is a triangle. Set $x_p := y_1$ and note that for every $p$, we have

$$ x_p\tau_p\tau_p\beta_p\tau_p^{-1}\beta_p = x_p. $$

By Lemma 4.22, we can then extend $\mathcal{G}_{0,p}$ to a saturated $(m, n)$-graph $\mathcal{G}_p$, see Figure 5, and by Proposition 3.23 we can extend the pre-action $(\beta_p, \tau_p)$ to an action $\alpha_p$ whose Bass-Serre graph is $\mathcal{G}_p$.

<figure><img src="image_6.png" /><figcaption>Figure 5: A (2,3)-graph $\mathcal{G}_p$, where $m = 2$ and $n = 3$.</figcaption></figure>
[Dud09] Fedor. A. Dudkin. Subgroups of Baumslag-Solitar groups. *Algebra Logika*, 48(1):3–30, 150, 152, 2009. 3

[FMMS20] Pierre Fima, François Le Maître, Soyoung Moon, and Yves Stalder. A characterization of high transitivity for groups acting on trees. *Discrete Analysis*, 2022:8, 63 pp. www.discreteanalysisjournal.com 10

[For07] Max Forester. Splittings of generalized Baumslag-Solitar groups. *Geometriae Dedicata*, 121(1):43–59, January 2007. 16

[GJ03] Światosław R. Gal and Tadeusz Januszkiewicz New a-Tmenable HNN-extensions *Journal of Lie Theory*, 13(2):383–385, 2003 3

[Gel05] Efraim Gelman. Subgroup growth of Baumslag-Solitar groups. *J. Group Theory*, 8(6):801–806, 2005. 3

[GKM16] Yair Glasner, Daniel Kitroser, and Julien Melleray. From isolated subgroups to generic permutation representations. *J. Lond. Math. Soc. (2)*, 94(3):688–708, 2016. 5

[Kec95] Alexander S. Kechris. *Classical descriptive set theory*, volume 156 of *Graduate Texts in Mathematics*. Springer-Verlag, New York, 1995. 13, 14

[Lev07] Gilbert Levitt. On the automorphism group of generalized Baumslag-Solitar groups. *Geometry & Topology*, 11(1):473–515, March 2007. Publisher: Mathematical Sciences Publishers. 16

[Lev15] Gilbert Levitt. Quotients and subgroups of Baumslag-Solitar groups. *J. Group Theory*, 18(1):1–43, 2015. 3

[LS01] Roger C. Lyndon and Paul E. Schupp. *Combinatorial Group Theory*, volume 89 of *Classics in Mathematics*. Springer, Berlin, Heidelberg, 2001. 55

[Mes72] Stephen Meskin. Nonresidually finite one-relator groups. *Trans. Amer. Math. Soc.*, 164:105–114, 1972. 3, 55

[Ser80] Jean-Pierre Serre. *Trees*. Springer, Berlin, Heidelberg, 1980. 11, 12, 15, 16, 49
Define $\Lambda_p$ to be the stabilizer of the action $\alpha_p$ at $x_p$ and remark that $t^2bt^{-1}b \in \Lambda_p$. Moreover by construction $\mathbf{Ph}(\Lambda_p) = p$.

By compactness, we find an accumulation point $\Lambda$ of the sequence $(\Lambda_p)_p$. Since $\mathbf{Ph}(\Lambda_p) = p$, the subgroup $\Lambda$ has infinite phenotype. Since $t^2bt^{-1}b \in \Lambda_p$ for every $p$, we have that $t^2bt^{-1}b \in \Lambda$. Moreover $t^2bt^{-1}b \notin \langle\!\langle b\rangle\!\rangle$ so $\Lambda \notin \overline{\mathbf{Ph}^{-1}(q_0)}$ by Theorem 6.2. Therefore the proof is completed. $\square$

## References

[AG23] Pénélape Azuelos and Damien Gaboriau. Perfect kernel and dynamics: from Bass-Serre theory to hyperbolic groups. to appear in *Mathematische Annalen*, 2023. arXiv:2308.05954. 5

[Ben83] Ivar Bendixson. Quelques théorèmes de la théorie des ensembles de points. *Acta Mathematica*, 2(1):415–429, 1883. doi:10.1007/BF02612172. 15

[BGK15] Lewis Bowen, Rostislav Grigorchuk, and Rostyslav Kravchenko. Invariant random subgroups of lamplighter groups. *Israel Journal of Mathematics*, 207(2):763–782, 2015. doi:10.1007/s11856-015-1160-1. 5

[BGK17] Lewis Bowen, Rostislav Grigorchuk, and Rostyslav Kravchenko. Characteristic random subgroups of geometric groups and free abelian groups of infinite rank. *Transactions of the American Mathematical Society*, 369(2), 2017. doi:10.1090/tran/6695.51

[BLT19] Oren Becker, Alexander Lubotzky, and Andreas Thom. Stability and invariant random subgroups. *Duke Mathematical Journal*, 168(12):2207–2234, 2019. doi:10.1215/00127094-2019-0024. 4, 16, 40

[Bon24] Sasha Bontemps. Perfect kernel of generalized Baumslag-Solitar groups. *arXiv preprint*, 2024. doi:10.48550/arXiv.2411.03221. 12

[Bry74] Roger M. Bryant. Characteristic Subgroups of Free Groups. In M. F. Newman, editor, *Proceedings of the Second International*
[Sta06b] Yves Stalder. Moyennabilité intérieure et extensions HNN. *Ann. Inst. Fourier (Grenoble)*, 56(2):309-323, 2006. 3

[SW20] Rachel Skipper and Phillip Wesolek. On the Cantor-Bendixson rank of the Grigorchuk group and the Gupta-Sidki 3 group. *J. Algebra*, 555:386-405, 2020. 5

[WebTool] Alessandro Carderi, Damien Gaboriau, François Le Maître, and Yves Stalder. *How to build (m,n)-graphs*. Zenodo, 2022. doi:10.5281/zenodo.7225585 23

A.C., INSTITUT FÜR ALGEBRA UND GEOMETRIE, KARLSRUHE INSTITUTE OF TECHNOLOGY, 76128 KARLSRUHE, GERMANY
alessandro.carderi@kit.edu

D.G., CNRS, ENS-LYON, UNITÉ DE MATHÉMATIQUES PURES ET APPLIQUÉES, 69007 LYON, FRANCE
damien.gaboriau@ens-lyon.fr

F.L.M., UNIVERSITÉ PARIS CITÉ, SORBONNE UNIVERSITÉ, CNRS, INSTITUT DE MATHÉMATIQUES DE JUSSIEU-PARIS RIVE GAUCHE, F-75013 PARIS, FRANCE
francois.le-maitre@imj-prg.fr

Y.S., UNIVERSITÉ CLERMONT AUVERGNE, CNRS, LMBP, F-63000 CLERMONT-FERRAND, FRANCE
yves.stalder@uca.fr
*Conference on the Theory of Groups*, Lecture Notes in Mathematics, pages 141–149, Berlin, Heidelberg, 1974. Springer. doi:10.1007/978-3-662-21571-5_11. 51

[BS62] Gilbert Baumslag and Donald Solitar. Some two-generator one-relator non-Hopfian groups. *Bulletin of the American Mathematical Society*, 68(3):199–201, 1962. doi:10.1090/S0002-9904-1962-10745-9. 3

[Can84] Georg Cantor. Ueber unendliche, lineare Punktmannichfaltigkeiten, part 6. *Mathematische Annalen*, 23(4):453–488, 1884. doi:10.1007/BF01446598. 15

[CGLM23] Alessandro Carderi, Damien Gaboriau, and François Le Maître. On dense totipotent free subgroups in full groups. *Geometry & Topology*, 27(6):2297–2318, 2023. doi:10.2140/gt.2023.27. 2297. 4, 5

[CGLMS] Alessandro Carderi, Damien Gaboriau, François Le Maître, and Yves Stalder. On the space of subgroups of Baumslag-Solitar groups III: The Cantor-Bendixson rank. In preparation. 5

[CGLMS22] Alessandro Carderi, Damien Gaboriau, François Le Maître, and Yves Stalder. How to build (m,n)-graphs. Zenodo, 2022. doi:10.5281/zenodo.7225585. 25

[Dud09] Fedor A. Dudkin. Subgroups of Baumslag–Solitar groups. *Algebra and Logic*, 48(1):1–19, 2009. doi:10.1007/s10469-009-9038-0. 3

[FLMMS22] Pierre Fima, François Le Maître, Soyoung Moon, and Yves Stalder. A characterization of high transitivity for groups acting on trees. *Discrete Analysis*, (8), 2022. doi:10.19086/da.37645. 11

[For06] Max Forester. Splittings of generalized Baumslag–Solitar groups. *Geometriae Dedicata*, 121(1):43–59, 2006. doi:10.1007/s10711-006-9085-9. 17
[Gel05] Efraim Gelman. Subgroup growth of Baumslag–Solitar groups. *Journal of Group Theory*, 8(6):801–806, 2005. doi:10.1515/jgth.2005.8.6.801. 3

[GJ03] Światosław R. Gal and Tadeusz Januszkiewicz. New a-Tmenable HNN-extensions. *Journal of Lie Theory*, 13(2):383–385, 2003. URL: https://www.emis.de/journals/JLT/vol.13_no.2/5.html. 3

[GKM16] Yair Glasner, Daniel Kitroser, and Julien Melleray. From isolated subgroups to generic permutation representations. *Journal of the London Mathematical Society*, 94(3):688–708, 2016. doi:10.1112/jlms/jdw054. 5

[GLMS24] Damien Gaboriau, François Le Maître, and Yves Stalder. On the space of subgroups of Baumslag-Solitar groups II: High transitivity. *arXiv preprint*, 2024. doi:10.48550/arXiv.2410.23224. 6, 12, 39

[Kec95] Alexander S. Kechris. *Classical Descriptive Set Theory*, volume 156 of *Graduate Texts in Mathematics*. Springer-Verlag, New York, 1995. doi:10.1007/978-1-4612-4190-4. 5, 15

[Lev07] Gilbert Levitt. On the automorphism group of generalized Baumslag–Solitar groups. *Geometry & Topology*, 11(1):473–515, 2007. doi:10.2140/gt.2007.11.473. 17

[Lev15] Gilbert Levitt. Quotients and subgroups of Baumslag–Solitar groups. *Journal of Group Theory*, 18(1):1–43, 2015. doi:10.1515/jgth-2014-0028. 3

[LS01] Roger C. Lyndon and Paul E. Schupp. *Combinatorial Group Theory*, volume 89 of *Classics in Mathematics*. Springer, Berlin, Heidelberg, 2001. doi:10.1007/978-3-642-61896-3. 57

[Mes72] Stephen Meskin. Nonresidually finite one-relator groups. *Transactions of the American Mathematical Society*, 164:105–114, 1972. doi:10.2307/1995962. 3, 58

[Ser80] Jean-Pierre Serre. *Trees*. Springer-Verlag, Berlin Heidelberg, 1980. doi:10.1007/978-3-642-61856-7. 13, 17, 18, 51
[Sta06] Yves Stalder. Moyennabilité intérieure et extensions HNN. *Anales de l'institut Fourier*, 56(2):309-323, 2006. doi:10.5802/aif.2183.3

[SW20] Rachel Skipper and Phillip Wesolek. On the Cantor-Bendixson rank of the Grigorchuk group and the Gupta-Sidki 3 group. *Journal of Algebra*, 555:386-405, 2020. doi:10.1016/j.jalgebra.2020.02.034.5

A.C., INSTITUT FÜR ALGEBRA UND GEOMETRIE, KARLSRUHE INSTITUTE OF TECHNOLOGY, 76128 KARLSRUHE, GERMANY
alessandro.carderi@kit.edu

D.G., CNRS, ENS-LYON, UNITÉ DE MATHÉMATIQUES PURES ET APPLIQUÉES, 69007 LYON, FRANCE
damien.gaboriau@ens-lyon.fr

F.L.M., UNIVERSITÉ PARIS CITÉ, SORBONNE UNIVERSITÉ, CNRS, INSTITUT DE MATHÉMATIQUES DE JUSSIEU-PARIS RIVE GAUCHE, F-75013 PARIS, FRANCE
francois.le-maitre@imj-prg.fr

Y.S., UNIVERSITÉ CLERMONT AUVERGNE, CNRS, LMBP, F-63000 CLERMONT-FERRAND, FRANCE
yves.stalder@uca.fr