The class $\mathcal{Q}^{\prime}$ contains all non-abelian limit groups
([BHMS09, Corollary 2.2]). Another important class of groups contained in
$\mathcal{Q}^{\prime}$ is that of graphs of free groups with cyclic edges such
that at least one vertex group is non-cyclic.

The following is a corollary of Theorems 5.9 and 5.11.

######  Corollary 5.23.

Let $G$ be any group in the class $\mathcal{Q}$. Then
$\operatorname{\mathcal{K}}(G)=\operatorname{Sub}_{[\infty]}(G)$. Moreover,
the conclusions of Theorem 5.11 apply. In particular, if
$G\in\mathcal{Q}^{\prime}$, then the action
$G\curvearrowright\operatorname{\mathcal{K}}(G)$ highly topologically
transitive.

We postpone the proof of this corollary until after the following two
necessary propositions and after their proofs (which mimic that of [BHMS09,
Proposition 5.2]). Observe that the assumptions of the propositions both
repeat those defining the class $\mathcal{Q}$ each with one condition missing
(the third and the fourth respectively).

######  Proposition 5.24.

Let $G$ be the fundamental group of a finite graph of finitely generated
groups, where all of the edge groups are finitely generated virtually abelian
and the associated Bass-Serre tree $T$ admits a pair $(f_{1},f_{2})$ of edges
whose stabilisers have finite intersection. Let $\Lambda,C\leq G$ be finitely
generated subgroups where $C$ is virtually abelian. If the set of double
classes $\Lambda\backslash G/C$ is finite, then the quotient graph
$\Lambda\backslash T$ is finite.

This proposition applies in particular to any $G\in\mathcal{Q}$ and its Bass-
Serre tree.

######  Proposition 5.25.

Let $G$ be the fundamental group of a finite graph of finitely generated
groups, where all of the edge groups are virtually abelian. Assume that the
associated Bass-Serre tree $T$ admits a vertex $w$ satisfying
$\operatorname{Stab}_{G}(w)\in\mathcal{Q}_{n}$, for $n\geq 0$. Then the
quotient graph $\Lambda\backslash T$ is finite if and only if the index
$[G:\Lambda]$ is finite.

###### Proof of Proposition 5.24.

We consider two cases.
Case 1. Suppose that $C$ fixes a vertex $v$ of $T$. Then, by the double-coset
hypothesis, the $G$-orbit of $v$ consists of only finitely many
$\Lambda$-orbits $\Lambda\cdot v_{i}$. Since the action of $G$ on T is
cocompact, every point in $T$ is at a bounded distance from $G\cdot v$, thus
$\Lambda\backslash T$ has finite diameter. The fundamental group of the
topological graph underlying $\Lambda\backslash T$ is finitely generated since
$\Lambda$ is finitely generated and maps onto it. By minimality of
$G\curvearrowright T$, there are no valency 1 edges in $T$. Moreover, the
valency of a vertex $z$ in the tree $T$ is the sum of the indices
$[\operatorname{Stab}_{\Lambda}(z):\operatorname{Stab}_{\Lambda}(e)]$ over a
subset of the edges $e$ incident to $z$ containing a representative from each
$\Lambda$-orbit. It follows that if $z$ is a vertex with
$\operatorname{Stab}_{\Lambda}(z)=\operatorname{Stab}_{\Lambda}(e)$ for some
incident edge $e$ then the image of $z$ in $\Lambda\backslash T$ does not have
valency 1. By finite generation of $\Lambda$, only finitely many valency 1
vertices of $\Lambda\backslash T$ can have a vertex group larger than the
group of their incident edge. It follows that $\Lambda\backslash T$ is finite.

Case 2. Suppose that $C$ does not fix a vertex.  Since $C$ is finitely
generated, it admits a hyperbolic element $c\in C$ [Ser80, Corollary 3, p.
65]. Let $A$ be the axis of $c$. Let $C^{\prime}\leq C$ be a finite index
abelian subgroup. For some integer $j$, we have $c^{\prime}:=c^{j}\in
C^{\prime}$, and $A$ is also the axis of $c^{\prime}$. Let $a\in C$. The image
$a\cdot A$ is the axis of the hyperbolic element $ac^{\prime}a^{-1}$. In case
$a\in C^{\prime}$ then $ac^{\prime}a^{-1}=c^{\prime}$, so the line $A$ is
$C^{\prime}$-invariant and $A$ is the axis of all hyperbolic elements of
$C^{\prime}$. If $a\in C$, a certain power $k$ satisfies $(aca^{-1})^{k}\in
C^{\prime}$, thus the axis of $aca^{-1}$ is also $A$. It follows that $A$ is
$C$-invariant and is the minimal subtree of $C$.

Let $e\in E(T)$ be an edge that belongs to $A$. Contract to one point each
connected component of $T\smallsetminus G\cdot e$ in order to produce the
$G$-equivariant factor map $P:T\to S$ to the edge-transitive $G$-tree $S$. Up
to enlarging the segment $L\coloneqq[f_{1},f_{2}]$ of $T$ so as to start and
finish with edges in $G\cdot e$, the image $P(L)=[P(f_{1}),P(f_{2})]$
satisfies $|\operatorname{Stab}_{G}(P(L))|<\infty$. Let $\bar{e}:=P(e)$. The
minimal tree of $C$ in $S$ is the image $\bar{A}:=P(A)=C\cdot\bar{e}$ of the
axis $A$. The two maps $(G\to E(S)\>;\>g\mapsto g\cdot\bar{e})$ and
$(\Lambda\backslash G\to E(\Lambda\backslash S)\>;\>\Lambda
g\mapsto\Lambda(g\cdot\bar{e}))$ are onto and the finiteness of the double
cosets $\Lambda\backslash G/C$ thus implies the following:
Claim: The quotient graph $\Lambda\backslash S$ is covered by finitely many
images $\Lambda g_{i}C\cdot\bar{e}$, i.e. by finitely many images
$Q_{\Lambda}(g_{i}(\bar{A}))$ of lines under the quotient map
$Q_{\Lambda}:S\to\Lambda\backslash S$. $\diamond$

Let $S_{\Lambda}\subseteq S$ be a minimal $\Lambda$-invariant subtree. Assume
for a contradiction that $S_{\Lambda}\not=S$. Let $\varepsilon$ be an edge of
$S\smallsetminus S_{\Lambda}$ and $Z_{1}$ the half-tree of
$S\smallsetminus\\{\varepsilon\\}$ that does not contain $S_{\Lambda}$. By
Lemma 5.2, there is $\gamma_{2}\in G$ such that
$\gamma_{2}([P(f_{1}),P(f_{2})])\subseteq Z_{1}$ and $\gamma_{2}(P(f_{1}))$
separates $\gamma_{2}(P(f_{2}))$ from $\varepsilon$. We show that the map
$Q_{\Lambda}$ is boundedly finite-to-one in restriction to the half-tree
$Z_{2}$ of $S\smallsetminus\\{\gamma_{2}(P(f_{2}))\\}$ that does not contain
$S_{\Lambda}$. Indeed an element $\lambda\in\Lambda$ sending a vertex $u$ of
$Z_{2}$ inside $Z_{2}$ must fix the projection $v$ of $u$ on $S_{\Lambda}$ and
thus the segment $[v,\gamma_{2}(P(f_{2}))]$. But this segment contains the
segment $\gamma_{2}(P(L))$. Thus $\lambda$ belongs to the
$\gamma_{2}$-conjugate
$\gamma_{2}\operatorname{Stab}_{G}(P(L))\gamma_{2}^{-1}$ of the point-wise
stabiliser of $P(L)$, which is finite. The map $Z_{2}\to
Q_{\Lambda}(Z_{2})=(\gamma_{2}\operatorname{Stab}_{G}(P(L))\gamma_{2}^{-1}\cap\Lambda)\backslash
Z_{2}$ has fibers of uniformly bounded size, thus $Q_{\Lambda}(Z_{2})$ has
infinitely many ends. The graph $\Lambda\backslash S$ contains
$Q_{\Lambda}(Z_{2})$, thus it has infinitely many ends, and it cannot be
covered by finitely many images of lines, contrarily to the above Claim. Thus
$S_{\Lambda}=S$.

Since $\Lambda$ is finitely generated, $\Lambda\backslash S$ is finite (Remark
5.6). It follows that $\Lambda\backslash T$ is finite. Proposition 5.24 is
proved. ∎

###### Proof of Proposition 5.25.

We will argue by induction on $n$.

The initialisation of the induction is that groups in $\mathcal{Q}_{0}$
satisfy the conclusion of Proposition 5.25: If $G\in\mathcal{Q}_{0}$, i.e. its
Stallings tree is an irreducible edge transitive action $G\curvearrowright T$
with a finite edge stabiliser $K$, then for any $\Lambda\leq G$, the quotient
$\Lambda\backslash T$ is finite if and only if $\Lambda\backslash G/K$ is
finite, if and only if $\Lambda\backslash G$ is finite.

Assume the conclusion of Proposition 5.25 holds for all groups satisfying the
assumptions up to some $n-1$. In particular, this conclusion holds for all
groups in $\mathcal{Q}_{n}$.

Let now $G$ be a group satisfying the assumptions for $n$ with
$\operatorname{Stab}_{\Lambda}(w)\in\mathcal{Q}_{n}$. Let $\Lambda\leq G$ and
assume $\Lambda\backslash T$ is finite. Observe that
$\operatorname{Stab}_{\Lambda}(w)$ is also finitely generated since (a)
subgroups of finitely generated (virtually) abelian groups are finitely
generated and (b) a finitely generated graph of groups (such as $\Lambda$)
with finitely generated edge groups must have finitely generated vertex groups
by [Coh89, p.43, Propositions 29 and 35].

Let $e$ be some edge incident at $w$. Then
$|\operatorname{Stab}_{\Lambda}(w)\backslash\operatorname{Stab}_{G}(w)/\operatorname{Stab}_{G}(e)|$
is bounded above by the finite number of edges of $\Lambda\backslash T$ that
are incident at $\Lambda\cdot w\in\Lambda\backslash T$. Since
$\operatorname{Stab}_{G}(w)\in\mathcal{Q}_{n}$, since
$\operatorname{Stab}_{\Lambda}(w)$ and $\operatorname{Stab}_{G}(e)$ are
finitely generated and $\operatorname{Stab}_{G}(e)$ is virtually abelian, then
applying Proposition 5.24 to the action of $\operatorname{Stab}_{G}(w)$ on its
own Bass-Serre tree $T^{\prime}$ implies that
$\operatorname{Stab}_{\Lambda}(w)\backslash T^{\prime}$ is finite. Then, by
the induction hypothesis (including the initialisation when $n=0$),
$\operatorname{Stab}_{\Lambda}(w)$ has finite index in
$\operatorname{Stab}_{G}(w)$. And similarly, for each $g\in G$, the subgroup
$\operatorname{Stab}_{\Lambda}(gw)$ has finite index in
$\operatorname{Stab}_{G}(gw)$: Indeed
$\operatorname{Stab}_{G}(gw)\simeq\operatorname{Stab}_{G}(w)\in\mathcal{Q}_{n}$
and $gw$ could have played the role of $w$.

Consider the action of $\operatorname{Stab}_{G}(w)$ by right multiplication on
$\Lambda\backslash G$: the orbits are the double cosets $\Lambda\backslash
G/\operatorname{Stab}_{G}(w)$ and hence are finite in number because they
index a subset of the vertices of $\Lambda\backslash T$ ; moreover, the
stabilizer of $\Lambda g$ is $g^{-1}\Lambda
g\cap\operatorname{Stab}_{G}(w)=g^{-1}(\Lambda\cap
g\operatorname{Stab}_{G}(w)g^{-1})g=g^{-1}\operatorname{Stab}_{\Lambda}(gw)g$,
which we have just seen has finite index in
$g^{-1}\operatorname{Stab}_{G}(gw)g=\operatorname{Stab}_{G}(w)$. Thus
$\Lambda\backslash G$ is finite. The converse is clear: if $\Lambda\backslash
G$ is finite, the quotient $\Lambda\backslash T$ is finite since $G\backslash
T$ is finite. Proposition 5.25 is proved. ∎

###### Proof of Corollary 5.23.

An element $G\in\mathcal{Q}$ is equipped with its defining Bass-Serre $G$-tree
$T$. Observe that the action $G\curvearrowright T$ can be assumed to be
minimal and irreducible: take the minimal $j$ such that $G\in\mathcal{Q}_{j}$
and note that the stabiliser of $w$ is not virtually abelian, thus the
stabiliser of any edge incident to $w$ has infinite index in
$\operatorname{Stab}_{G}(w)$. Irreducibility of the action then follows
Proposition 5.1. By Theorem 5.9 $\operatorname{Sub}_{|\bullet\backslash
T|\infty}(G)\subseteq\operatorname{\mathcal{K}}(G)\subseteq\operatorname{Sub}_{[\infty]}(G)$.
It is enough to check that $\overline{\operatorname{Sub}_{|\bullet\backslash
T|\infty}(G)}\supseteq\operatorname{Sub}_{[\infty]}(G)$.

If $\Lambda\in\operatorname{Sub}_{[\infty]}(G)$ is finitely generated then
$\Lambda\backslash T$ is infinite: if $n=0$ this is because the action of $G$
on the edges of its Stallings tree is transitive with finite stabilisers, and
otherwise this follows from Proposition 5.25. Thus
$\Lambda\in\operatorname{Sub}_{|\bullet\backslash T|\infty}(G)$. If
$\Lambda\in\operatorname{Sub}_{[\infty]}(G)$ is not finitely generated, it is
a limit of proper finitely generated subgroups $\Lambda_{n}\leq\Lambda$, which
are thus also of infinite index, so that
$\Lambda_{n}\in\operatorname{Sub}_{|\bullet\backslash T|\infty}(G)$ and
$\Lambda\in\overline{\operatorname{Sub}_{|\bullet\backslash T|\infty}(G)}$.
Then Theorem 5.11 applies concerning the dynamics. ∎

##  6 Hyperbolic groups

###  6.1 Generalities on the space of subgroups of hyperbolic groups

Recall from [Gro87, 5.3] that a subset $Y$ of a geodesic metric space $X$ is
quasiconvex if its convex hull is contained within a bounded neighbourhood of
$Y$.

Let $G$ be a hyperbolic (aka word-hyperbolic) group. A subgroup $H\leq G$ is
quasiconvex if it is a quasiconvex subset of the Cayley graph $X$ of $G$
associated with a finite generating set of $G$. Note that this does not depend
on the choice of Cayley graph as quasiconvexity in hyperbolic spaces is
invariant under quasi-isometry (this is not true for metric spaces in
general). In fact $H\leq G$ is quasiconvex if and only if $H$ admits a finite
generating set $T$ and the inclusion $H\hookrightarrow X$ is a quasi-isometric
embedding with respect to the word metric defined by $T$.

Given a hyperbolic group $G$, let

$\operatorname{Sub}_{\mathrm{QC}[\infty]}(G)\subseteq\operatorname{Sub}_{[\infty]}(G)$

denote the set of quasiconvex subgroups of $G$ with infinite index. Observe
that $\operatorname{Sub}_{\mathrm{QC}[\infty]}(G)$ is invariant under the
$G$-action by conjugation. If $G$ is non-elementary, we prove that this subset
is in fact a $G$-invariant subspace of the perfect kernel of $G$.

######  Theorem 6.1.

Let $G$ be a non-elementary hyperbolic group. Then

$\overline{\operatorname{Sub}_{\mathrm{QC}[\infty]}(G)}\subseteq\operatorname{\mathcal{K}}(G).$

Recall that a non-elementary hyperbolic group $G$ has a unique maximal finite
normal subgroup $N\unlhd G$ and by Proposition 3.2 the set
$\mathrm{Conj}(N)^{G}$ of $G$-conjugacy classes of subgroups of $N$ defines a
finite $G$-invariant clopen partition:
$\operatorname{Sub}(G)=\bigsqcup_{C\in\mathrm{Conj}(N)^{G}}\\{H\leq G\colon
H\cap N\in C\\}$. In analogy with Theorem 5.11, we denote for each
$C\in\mathrm{Conj}(N)^{G}$ the (which will be shown to be perfect) subspace

$\mathcal{F}_{C}:=\overline{\operatorname{Sub}_{\mathrm{QC}[\infty]}(G)}\cap\\{H\leq
G\colon H\cap N\in C\\}$ (18)

so that the clopen partition of $\operatorname{Sub}(G)$ induces a
$G$-invariant partition of
$\overline{\operatorname{Sub}_{\mathrm{QC}[\infty]}(G)}$:

$\overline{\operatorname{Sub}_{\mathrm{QC}[\infty]}(G)}=\bigsqcup_{C\in\mathrm{Conj}(N)^{G}}\mathcal{F}_{C}$

whose dynamics satisfy:

######  Theorem 6.2.

Let $G$ be a non-elementary hyperbolic group and fix
$C\in\mathrm{Conj}(N)^{G}$. Then

  * (i)

the action of $G$ on $\mathcal{F}_{C}$ is topologically transitive, and

  * (ii)

if $|C|=1$ then the action $G\curvearrowright\mathcal{F}_{C}$ is highly
topologically transitive.

In particular, if $N=\\{\mathrm{id}\\}$ then the action
$G\curvearrowright\overline{\operatorname{Sub}_{\mathrm{QC}[\infty]}(G)}$ is
highly topologically transitive.

Note that the assumption that $|C|=1$ is a necessary condition for a
topologically $r$-transitive action on $\mathcal{F}_{C}$ whenever $r\geq 2$
(see Remark 5.13).

We delay the proofs of Theorems 6.1 and 6.2 until the next section. The
following is an immediate consequence.

A hyperbolic group is locally quasiconvex if every finitely generated subgroup
of $G$ is quasiconvex.

######  Corollary 6.3.

If a group $G$ is non-elementary hyperbolic and locally quasiconvex, then

$\operatorname{\mathcal{K}}(G)=\operatorname{Sub}_{[\infty]}(G)=\overline{\operatorname{Sub}_{\mathrm{QC}[\infty]}(G)}.$

If, in addition, the maximal finite normal subgroup of $G$ is trivial then the
action of $G$ on $\operatorname{\mathcal{K}}(G)$ is highly topologically
transitive.

######  Example 6.4.

Hyperbolic limit groups are locally quasiconvex by [Dah03, Proposition 4.6]
and torsion free. Therefore, the above corollary provides another proof that
the perfect kernel of a hyperbolic limit group $G\not=\mathbb{Z}$ is the space
$\operatorname{Sub}_{[\infty]}(G)$ and that the $G$-action on
$\operatorname{\mathcal{K}}(G)$ is highly topologically transitive.

######  Example 6.5.

If $G$ is non-elementary hyperbolic and admits a small hierarchy then by
[BW13, Theorem 3.6] it is locally quasiconvex and Corollary 6.3 applies.

For instance, if $G$ is not virtually cyclic and splits non-trivially as a
graph of groups where each vertex group is a free group, each edge group is
cyclic and $G$ does not contain any subgroup which is a Baumslag-Solitar group
then $G$ is non-elementary hyperbolic by [BF92] and locally quasiconvex by the
above. Compare with the drastically different behaviour when $G$ is a
Baumslag-Solitar group [CGLMS22].

######  Example 6.6.

Let $S$ be a finite set with $|S|\geq 2$ and consider the one-relator group
$G=\langle S\>|\>w^{m}\rangle$ where $w$ is a freely and cyclically reduced
word in $S\cup S^{-1}$ and $m\geq|w|_{S}$. Such one-relator groups with
torsion are non-elementary hyperbolic and have no non-trivial normal subgroups
[New68, Theorems 3 and 2], and they are locally quasiconvex by [HW01, Theorem
1.2] so, by Corollary 6.3,
$\operatorname{\mathcal{K}}(G)=\operatorname{Sub}_{[\infty]}(G)$ and the
action $G\curvearrowright\operatorname{\mathcal{K}}(G)$ is highly
topologically transitive.

######  Example 6.7.

Let $G$ be the Coxeter group $\langle
s_{1},\dots,s_{n}|s_{i}^{2},(s_{i}s_{j})^{m_{i,j}}\;\forall\;i<j\rangle$ where
$\infty\geq m_{i,j}>n\geq 3$. Then $G$ is hyperbolic [Mou88, Theorem 17.1] and
locally quasiconvex [MW08, Theorem 12.2], so
$\operatorname{\mathcal{K}}(G)=\operatorname{Sub}_{[\infty]}(G)$. Moreover, it
follows from [PV23] that $G$ has no non-trivial finite normal subgroups so the
action $G\curvearrowright\operatorname{\mathcal{K}}(G)$ is highly
topologically transitive.

######  Remark 6.8.

It is likely that the results of this section can be extended to include
relatively hyperbolic groups with suitable parabolic subgroups, for instance
finitely generated abelian groups.

###  6.2 Proof of Theorems 6.1 and 6.2

The proofs of both the main results of this section depend on the fact that,
in a non-elementary hyperbolic group $G$, any infinite index quasiconvex
subgroup $H$ is a factor in a splitting of a quasiconvex subgroup $L$ of $G$
over a finite subgroup (see Proposition 6.11). Moreover, there is a great deal
of freedom in the choice of $L$. In the torsion-free case, this result was
first observed by Gromov in [Gro87, 5.3] and was developed in a detailed and
effective manner by Arzhantseva in [Arz01]. Other combination theorems
allowing torsion were proved by Gitik in [Git99] and were generalised to
relatively quasiconvex subgroups of relatively hyperbolic groups by Martínez-
Pedroza in [MP09]. In fact, even when restricted to hyperbolic groups, the
results of Martínez-Pedroza improve upon Gitik’s results in some respects (see
for instance Theorems 6.9 and 6.10 below). These improvements are needed for
various aspects of our proofs, and they allow us to obtain Proposition 6.11
for hyperbolic groups with torsion.

The Gromov boundary $\partial G$ of a non-elementary hyperbolic group $G$ is a
perfect compact metrisable space; it is equipped with a $G$-action that
extends the left multiplication action of $G$ on itself. Given a subgroup $H$,
let $\Lambda(H)\subseteq\partial G$ denote the limit set of $H$. If $H$ is
virtually infinite cyclic then $\Lambda(H)$ consists of precisely two points
(the limit points of any infinite order element of $H$). Let $g\in G$ be an
infinite order element and $P=\operatorname{Stab}(\Lambda(\langle g\rangle))$.
Then $\langle g\rangle$ is a finite index subgroup of $P$ [GH90, Theorem 30]
which implies that $P$ is quasiconvex [Gro87, 8.1D]. Moreover if $a\in
G\smallsetminus P$ then $P\cap aPa^{-1}=\operatorname{Stab}(\Lambda(\langle
g\rangle)\cap\operatorname{Stab}(\Lambda(\langle ag\rangle))$ where
$\Lambda(\langle g\rangle)\neq\Lambda(\langle ag\rangle)$. It follows that
there is a finite index subgroup $Q$ of $P\cap aPa^{-1}$ which fixes three
points of $\partial G$; this implies that $Q$ fixes a bounded subset of $G$
and is therefore finite. Thus $P$ is almost malnormal (i.e. $P\cap P^{a}$ is
finite for all $a\in G\smallsetminus P$; here and in the sequel, $P^{a}$
denotes the conjugate $aPa^{-1}$). By [Bow12, Theorem 7.11] this implies that
$G$ is hyperbolic relative to $P$. Since subgroups of conjugates of $P$ are
virtually cyclic and therefore quasiconvex it follows from [MP12, Theorem 1.1]
that quasiconvexity in $G$ is equivalent to quasiconvexity relative to $P$. In
particular, the following theorems are special cases of the results of [MP09].

######  Theorem 6.9 (Martínez-Pedroza, [MP09, Theorem 1.1]).

Let $G$ be a hyperbolic group and fix a finite generating set $S$ for $G$. Let
$g\in G$ be an infinite order element and
$P=\operatorname{Stab}(\Lambda(\langle g\rangle))$. Let $H\leq G$ be a
quasiconvex subgroup. Then there is a constant $C_{1}\geq 0$ depending only on
$H$, $P$ and $S$ such that the following holds. If $Q\leq P$ is a subgroup
such that

  1. 1.

$H\cap P\leq Q$ and

  2. 2.

$|k|_{S}\geq C_{1}$ for all $k\in Q\smallsetminus H$

then $\langle H,Q\rangle$ splits as the amalgamated free product $\langle
H,Q\rangle=H\ast_{H\cap Q}Q$ and $\langle H,Q\rangle$ is quasiconvex in $G$.

######  Theorem 6.10 (Martínez-Pedroza, [MP09, Theorem 1.2]).

Let $G$ be a hyperbolic group and fix a finite generating set $S$ for $G$. Let
$g\in G$ be an infinite order element and
$P=\operatorname{Stab}(\Lambda(\langle g\rangle))$. Let $H_{1},H_{2}\leq G$ be
quasiconvex subgroups such that $M:=H_{1}\cap P=H_{2}\cap P$. Then there
exists a constant $C_{2}\geq 0$ depending only on $H_{1},H_{2},P$ and $S$ such
that the following holds. If $p\in P$ is such that

  1. 1.

$pMp^{-1}=M$

  2. 2.

$|k|_{S}\geq C_{2}$ for all $k\in MpM$

then $\langle H_{1},pH_{2}p^{-1}\rangle$ splits as the amalgamated free
product $\langle H_{1},pH_{2}p^{-1}\rangle=H_{1}\ast_{M}pH_{2}p^{-1}$ and
$\langle H_{1},pH_{2}p^{-1}\rangle$ is quasiconvex in $G$.

We are now in a position to prove the combination result we need. In order to
produce the relevant subgroup $P\leq G$, we adapt a standard argument to
produce a suitable hyperbolic element of $G$ by taking advantage of the
“holes” in the boundary of $G$ left by the subgroups $H_{i}$.

######  Proposition 6.11.

Let $G$ be a non-elementary hyperbolic group and $H_{1},\dots,H_{r}\leq G$ be
quasiconvex subgroups with infinite index. Let $N\unlhd G$ be the unique
maximal finite normal subgroup of $G$ and suppose that $M:=H_{1}\cap
N=H_{i}\cap N$ for each $i$. Then the following hold.

  1. 1.

For each $i$, the subset $\Lambda(H_{i})\subseteq\partial G$ is closed with
empty interior. Thus the complement $\partial
G\smallsetminus\cup_{i=1}^{r}\Lambda(H_{i})$ is a dense open set.

  2. 2.

For any non-empty open set $Y\subseteq\partial G$, there exists an infinite
order element $p\in G$ whose limit points are contained in $Y$ and such that,
if we denote $P=\operatorname{Stab}(\Lambda(\langle p\rangle))$, then $P\cap
H_{i}=M$ for each $i$.

  3. 3.

If $u\in G$ is a sufficiently high power of $p$ then, for each
$n\in\mathbb{N}$, the subgroup $Q_{n}:=\langle u^{n},M\rangle$ generated by
$u^{n}$ and $M$ splits as the direct product $Q_{n}=\langle u^{n}\rangle\times
M$, and for each $i\in\\{1,\dots,r\\}$, the subgroup $\langle
H_{i},Q_{n}\rangle$ splits as the amalgamated free product

$\langle H_{i},Q_{n}\rangle=H_{i}\ast_{M}Q_{n}$

and is quasiconvex and has infinite index in $G$.

###### Proof.

Recall that the unique maximal finite normal subgroup $N\unlhd G$ is the
kernel of the action of $G$ on $\partial G$ (see e.g. [DK18, Lemma 11.130]).
Let $H\leq G$ be an infinite index quasiconvex subgroup. Then the limit set
$\Lambda(H)$ is a (possibly empty, if $H$ is finite) closed proper subspace of
$\partial G$. Suppose for a contradiction that there is an open subset
$U\subseteq\partial G$ such that $U\subseteq\Lambda(H)$. Then by [Gro87,
8.2G], there is an infinite order element $h\in H$ such that $\Lambda(\langle
h\rangle)\subseteq U$. Since $h$ acts on $\partial G$ with north-south
dynamics ([Gro87, 8.1G]), a certain power $h^{n}$ satisfies $h^{n}U\cup
U=\partial G$. But $h^{n}U\cup U\subseteq
h^{n}\Lambda(H)\cup\Lambda(H)=\Lambda(H)$. Thus $\Lambda(H)=\partial G$ which
contradicts the fact that $H$ has infinite index in $G$. Thus $\Lambda(H)$ has
empty interior. The same holds for a finite union of closed sets with empty
interior. This proves part 1 of the lemma.

To prove part 2, we first show that, for any $g\in G\smallsetminus N$, the
fixed point set $\operatorname{Fix}(g)\coloneqq\\{x\in\partial G:gx=x\\}$ is
closed with empty interior. Fix an element $g\in G$. First,
$\operatorname{Fix}(g)$ is closed. Suppose that there is a non-empty open set
$W\subseteq\operatorname{Fix}(g)$ and let $L\leq G$ be the pointwise
stabiliser of $W$. Since $\partial G$ is perfect, $W$ contains more than two
points so every element of $L$ has finite order. By [GH90, Corollaire 36,
Chapitre 8] this implies that $L$ is finite. Let $h\in G$ be an infinite order
element whose limit points $h^{+},h^{-}$ are in $W$ and let
$W^{+},W^{-}\subseteq W$ be open neighbourhoods of $h^{+},h^{-}$ respectively
such that $W^{+}\cap W^{-}=\emptyset$. Let $L^{-}$ be the pointwise stabiliser
of $W^{-}$ and note that, as above, $L^{-}$ is finite. Moreover, for
sufficiently large $n\in\mathbb{N}$, $h^{n}W^{-}\cup W^{+}=\partial G$ and in
particular $W^{-}\subseteq h^{n}W^{-}$, so $h^{n}L^{-}h^{-n}\leq L^{-}$. Since
$L^{-}$ is finite, this implies that $h^{n}L^{-}h^{-n}=L^{-}$. But then $g$ is
in the pointwise stabiliser of $h^{n}W^{-}$ and of $W^{+}$ so $g\in N$.

Recall that any subgroup $H\leq G$ acts properly discontinuously on $\partial
G\smallsetminus\Lambda(H)$ ([Coo89]. Let $Y\subseteq\partial G$ be non-empty
and open and let $x_{0}\in Y^{\prime}:=Y\cap(\partial
G\smallsetminus\cup_{i=1}^{r}\Lambda(H_{i}))$. Then for each $i$ there is an
open neighbourhood $x_{0}\in U_{i}\subseteq Y^{\prime}$ such that $|\\{h\in
H_{i}:hU_{i}\cap U_{i}\neq\emptyset\\}|<\infty$. Let
$U\coloneqq\cap_{i=1}^{r}U_{i}$ and note that $U$ is non-empty and open. Let
$A=\\{h\in(\cup_{i=1}^{r}H_{i})\smallsetminus M:hU\cap U\neq\emptyset\\}$.
Since, for each $h\in A$, the fixed point set $\operatorname{Fix}(h)$ is
closed with empty interior, the complement $V:=U\smallsetminus(\cup_{h\in
A}\operatorname{Fix}(h))$ is non-empty and open. Now enumerate
$A=\\{a_{1},\dots,a_{l}\\}$. If $a_{1}V\cap V=\emptyset$, let $V_{1}:=V$.
Otherwise there exists $x_{1}\in V$ such that $a_{1}x_{1}\in V$ and, since
$x_{1}\notin\operatorname{Fix}(a_{1})$, we have $x_{1}\neq a_{1}x_{1}$. It
follows that there exists an open neighbourhood $V_{1}\subseteq V$ of $x_{1}$
such that $a_{1}V_{1}\cap V_{1}=\emptyset$. Iterate this process to find a
sequence of non-empty open sets $V_{l}\subseteq\dots\subseteq V_{1}\subseteq
V$ such that for each $1\leq k\leq l$ and each $1\leq i\leq k$ we have
$a_{i}V_{k}\cap V_{k}=\emptyset$. In particular $aV_{l}\cap V_{l}=\emptyset$
for all $a\in A$.

Let $p\in G$ be an infinite order element whose limit points are contained in
$V_{l}\subseteq Y$ and let $P=\operatorname{Stab}(\Lambda(\langle p\rangle))$.
Then $P\cap H_{i}=M$ for each $i$. This proves part 2 of the lemma.

Let $u$ be a power of $p$ which commutes with $M$. For each $n\in\mathbb{N}$
let $Q_{n}:=\langle u^{n},M\rangle=\langle u^{n}\rangle\times M\leq P$. Then
$Q_{n}\cap H_{i}=M$ for each $i$ and, since $P$ is quasi-isometrically
embedded in $G$, for sufficiently large $n$ and $C_{1}\geq 0$ as in Theorem
6.9, $|k|_{S}\geq C_{1}$ for all $k\in Q_{n}\smallsetminus M$. Therefore, by
Theorem 6.9, for sufficiently large $n$,

$\langle H_{i},Q_{n}\rangle=H_{i}\ast_{M}Q_{n}\leq G$ (19)

is quasiconvex. Moreover, up to replacing $u$ with $u^{n_{0}}$ for some
sufficiently large $n_{0}$, each $\langle H_{i},Q_{n}\rangle$ is an infinite
index quasiconvex subgroup of $G$ satisfying (19). Indeed, if
$H_{i}\ast_{M}Q_{n_{0}}$ has finite index, then the splitting is non-trivial
(if it were trivial, i.e. $H_{i}=M$, and $G$ would be elementary), thus
$\langle H_{i},Q_{2n_{0}}\rangle$ has infinite index in $\langle
H_{i},Q_{n_{0}}\rangle$ and therefore in $G$. This proves part 3 of the lemma.
∎

###### Proof of Theorem 6.1.

Let $H\leq G$ be an infinite index quasiconvex subgroup, $N\unlhd G$ be the
unique maximal finite normal subgroup of $G$ and $M=H\cap N$. By Proposition
6.11, there exists an infinite order element $u\in G$ such that, for each
$n\in\mathbb{N}$, $Q_{n}:=\langle u^{n},M\rangle=\langle u^{n}\rangle\times M$
and

$H_{n}:=\langle
H,Q_{n}\rangle=H\ast_{M}Q_{n}\in\operatorname{Sub}_{QC[\infty]}(G).$

Let $T$ be a finite generating set for $H$ and $\lambda\geq 1$ be such that
the natural inclusion of the first term of the sequence $H_{1}\hookrightarrow
G$ is a $\lambda$-quasi-isometric embedding with respect to the generating set
$S$ for $G$ and the generating set $T\cup\\{u\\}$ for $H_{1}$. For any
$n\in\mathbb{N}$, if $o\in H_{n}\smallsetminus H$ then the reduced spelling of
$o$ must contain the subword $u^{n}$, and since $o\in H_{1}\smallsetminus H$
we have

$|o|_{S}\geq\frac{1}{\lambda}|o|_{T\cup\\{u\\}}-\lambda\geq\frac{n}{\lambda}-\lambda.$

Hence, given a basic open neighbourhood $\mathcal{V}(\mathcal{I},\mathcal{O})$
of $H$ in $\operatorname{Sub}(G)$, for all sufficiently large $n$, the
subgroup $H_{n}$ contains $H$ but does not contain $o\in\mathcal{O}$, thus
$H_{n}\in\mathcal{V}(\mathcal{I},\mathcal{O})\cap\operatorname{Sub}_{QC[\infty]}(G)$
so $H$ is not isolated in $\operatorname{Sub}_{\mathrm{QC}[\infty]}(G)$. ∎

###### Proof of Theorem 6.2.

Fix $C\in\mathrm{Conj}(N)^{G}$. To prove part (i) (i.e. that the action of $G$
on $\mathcal{F}_{C}$ is topologically transitive) we will show the following.
Given two non-empty open subsets $U,V\subseteq\mathcal{F}_{C}$, there exists
$g\in G$ such that $(g\cdot U)\cap V=gUg^{-1}\cap V\neq\emptyset$. As in the
proof of Theorem 5.11, we can assume, up to replacing $V$ with its image by
some element of $G$, that there exist finitely generated subgroups $H\in U$
and $K\in V$ such that $M:=H\cap N=K\cap N$.

To prove part (ii), we need to show that, if $C=\\{M\\}$ (i.e. $M$ is normal
in $G$) then, for all $r$ and for all non-empty open sets
$U_{1},\dots,U_{r},V_{1},\dots,V_{r}\subseteq\mathcal{F}_{C}$, there exists
$g\in G$ such that $gU_{i}\cap V_{i}\neq\emptyset$ for each $i$. Since
finitely generated subgroups are dense and, for all $H_{i}\in U_{i}$ and
$K_{i}\in V_{i}$, we have $H_{i}\cap N=K_{i}\cap N=M$, both part (i) and part
(ii) follow from the following statement:

Fix $C\in\mathrm{Conj}(N)^{C}$ and $M\in C$ (we don’t assume that $|C|=1$).
Let $r\geq 1$ and $U_{1},\dots,U_{r},V_{1},\dots
V_{r}\subseteq\mathcal{F}_{C}$ be open subsets such that, for each
$i\in\\{1,\dots,r\\}$, there exist finitely generated subgroups $H_{i}\in
U_{i}$ and $K_{i}\in V_{i}$ where $H_{i}\cap N=K_{i}\cap N=M$. Then there is a
common element $g\in G$ such that $gU_{i}\cap V_{i}\neq\emptyset$ for all $i$.

Let us prove that this statement indeed holds.

######  Claim 1.

There exists an infinite order element $p\in G$ which commutes with $M$ and
such that, for each $i$, $\langle
H_{i},K_{i}^{p}\rangle=H_{i}\ast_{M}K_{i}^{p}\leq G$ is quasiconvex and has
infinite index in $G$.

###### Proof of Claim 1.

If $G$ has infinitely many ends, then it admits a Stallings tree, a minimal
irreducible $G$-tree $T$ with finite edge stabilisers. Note that the kernel of
the action on $T$ is the maximal finite normal subgroup $N$. The infinite
index subgroups $H_{i},K_{i}$ have proper minimal subtrees
$T_{H_{i}},T_{K_{i}}$ in $T$; let $Z_{e}$ be the half-tree separated from the
union $\cup_{i}T_{H_{i}}\cup\cup_{i}T_{K_{i}}$ of these minimal subtrees by
some edge $e$ such that
$\operatorname{Stab}_{H_{i}}(e)=\operatorname{Stab}_{K_{i}}(e)=M$ for each $i$
(see Claim 4 in the proof of Theorem 5.11 Take a vertex $x$ of $T$. The map
$G\ni g\mapsto g\cdot x$ extends to a $G$-equivariant map $\Psi:G\cup\partial
G\to T\cup\partial T$. By Lemma 5.2 there is a hyperbolic (with respect to the
action on $T$) element $p\in G$ whose axis $\mathrm{Axis}(p)$ is contained in
$Z_{e}$. Since it is fixed by $\langle p\rangle$, the image
$\Psi(\Lambda(\langle p\rangle)$ is the set of limit points of
$\mathrm{Axis}(p)$. Thus, the stabiliser
$P:=\operatorname{Stab}_{G}(\Lambda(\langle p\rangle))$ stabilises globally
$\mathrm{Axis}(p)$. It follows that any $g\in P\cap H_{i}$ (respectively of
$P\cap K_{i}$), for any $i$, must stabilise $\mathrm{Axis}(p)$ and $T_{H_{i}}$
(respectively $T_{K_{i}}$), therefore
$g\in\operatorname{Stab}_{H_{i}}(e)=\operatorname{Stab}_{K_{i}}(e)=M$.

Otherwise, if $G$ has only one end, let $Y$ any open subset of $\partial G$.
By Proposition 6.11 there exists an infinite order element $p\in G$ whose
limit points are contained in $Y$ such that, if
$P\coloneqq\operatorname{Stab}(\Lambda(\langle p\rangle))$, then $P\cap
H_{i}=P\cap K_{i}=M$ for each $i$.

In either case, replace $p$ with a proper power so that it commutes with $M$.

Now let $C_{2}\geq 0$ be as in Theorem 6.10. Since $\langle p\rangle$ is
quasi-isometrically embedded in $G$ and $M$ is finite, we can replace $p$ with
a proper power so that $|pm|\geq|p|-|m|\geq C_{2}$ for all $pm\in pM=MpM$.
Thus by Theorem 6.10, for each $i$, the subgroup generated by $H_{i}$ and
$K_{i}^{p}$ splits as

$\langle H_{i},K_{i}^{p}\rangle=H_{i}\ast_{M}K_{i}^{p},$

and is quasiconvex in $G$.

Moreover, up to taking a proper power of $p$, we can assume that for each $i$,
$H_{i}\ast_{M}K_{i}^{p}$ has infinite index in $G$. Indeed, suppose that
$H_{i}\ast_{M}K_{i}^{p}$ has finite index in $G$ for some $i$. Then the
splitting is non-trivial (if $M$ is equal to one of the factors then the other
factor has finite index in $G$) so $G$ has at least two, thus infinitely many,
ends. In this case, $p$ acts hyperbolically on the Stallings tree with axis
$\mathrm{Axis}(p)$ contained in $Z_{e}$. Let $n\geq 2$ and let $\pi(e)$ be the
projection of $e$ on $\mathrm{Axis}(p)$. Then any edge
$e^{\prime}\not\subset\mathrm{Axis}(p)$ but adjacent to $\mathrm{Axis}(p)$ at
a vertex strictly between $\pi(e)$ and $p^{n}\cdot\pi(e)$ defines a half-tree
$Z^{\prime}$ (the one that does not contain $e$) which lies outside any
minimal invariant $H_{i}\ast_{M}K_{i}^{p^{n}}$-subtree of $T$. Thus the
quotient $H_{i}\ast_{M}K_{i}^{p^{n}}\backslash T$ is infinite. Since the
action of $G$ on $T$ is minimal and $G$ is finitely generated, the quotient
$G\backslash T$ is compact (see Remark 5.6) so this implies that
$H_{i}\ast_{M}K_{i}^{p^{n}}$ has infinite index in $G$. ∎

By Proposition 6.11-(3), there exists an infinite order element $u\in G$ which
commutes with $M$ and such that, for all $i$,

$\displaystyle A^{(i)}:=\langle
H_{i},K_{i}^{p},u\rangle=(H_{i}\ast_{M}K_{i}^{p})\ast_{M}\langle
u,M\rangle\leq G\quad\text{and}$ $\displaystyle
B^{(i)}:=(A^{(i)})^{p^{-1}}=\langle
H_{i}^{p^{-1}},K_{i},u^{p^{-1}}\rangle=(H_{i}^{p^{-1}}\ast_{M}K_{i})\ast_{M}\langle
u^{p^{-1}},M\rangle\leq G$

are quasiconvex and have infinite index in $G$. For ease of notation, we now
fix $i\in\\{1,\dots,r\\}$ and denote $A=A^{(i)}$ and $B=B^{(i)}$.

######  Claim 2.

For all large enough $n$, we have $A_{n}:=H_{i}\ast_{M}(K_{i})^{pu^{n}}\in
U_{i}$ while
$B_{n}=(A_{n})^{u^{-n}p^{-1}}=(H_{i})^{u^{-n}p^{-1}}\ast_{M}K_{i}\in V_{i}$.

###### Proof of Claim 2.

First observe, for each $n$, that $A_{n}\leq A$, $B_{n}\leq B$ and the
inclusions $A_{n}\hookrightarrow A$ and $B_{n}\hookrightarrow B$ are quasi-
isometric embeddings so $A_{n},B_{n}$ are quasiconvex in $G$ and have infinite
index. Moreover $A_{n}\cap N=B_{n}\cap N=M$ so
$A_{n},B_{n}\in\mathcal{F}_{C}$. Now let
$\mathcal{V}(\mathcal{I},\mathcal{O})\subseteq\operatorname{Sub}(G)$ be a
basic neighbourhood of $H_{i}$ such that
$\mathcal{V}(\mathcal{I},\mathcal{O})\cap\mathcal{F}_{C}\subseteq U_{i}$. Let
$T,T^{\prime}$ be finite generating sets for $H_{i},K_{i}$ respectively and
let $\lambda\geq 1$ be such that $A\hookrightarrow G$ is a $\lambda$-quasi-
isometry with respect to the word metric on $G$ defined by $S$ and the word
metric on $A$ defined by $S_{A}:=T\cup pT^{\prime}p^{-1}\cup\\{u\\}$. Then for
all $a\in A_{n}\smallsetminus H_{i}$, the reduced spelling of $a$ must contain
a subword $pu^{n}t^{\prime}u^{-n}p^{-1}$, thus

$|a|_{S}\geq\frac{1}{\lambda}|a|_{S_{A}}-\lambda\geq\frac{2n}{\lambda}-\lambda.$

Therefore if $n$ is sufficiently large then $|a|_{S}>|o|_{S}$ for all $a\in
A_{n}\smallsetminus H_{i}$ and $o\in\mathcal{O}$. Since $\mathcal{I}\subseteq
H_{i}\subseteq A_{n}$, this implies that
$A_{n}\in\mathcal{V}(\mathcal{I},\mathcal{O})\cap\mathcal{F}_{C}\subseteq
U_{i}$. By a symmetric argument, $(u^{-n}p^{-1})\cdot A_{n}=B_{n}\in V_{i}$
for sufficiently large $n$. ∎

Hence, by applying Claim 2 to each $i\in\\{1,\dots,r\\}$, we find
$N\in\mathbb{N}$ such that, for each $i$, the intersection $(u^{-N}p^{-1}\cdot
U_{i})\cap V_{i}$ is non-empty. ∎

###  6.3 Hyperbolic 3-manifold groups

In the case of hyperbolic 3-manifold groups the results of the previous
section, together with celebrated results about 3-manifolds, allow us to
describe the perfect kernel completely.

The following dichotomy, which is a consequence of Canary’s Covering Theorem
[Can96] and the Tameness Theorem [Ago04, CG06] (see [AFW15, Theorem 5.2] for a
precise statement) allows us to understand the status of non-quasiconvex
subgroups in the space of subgroups of a hyperbolic 3-manifold groups. If
$G=\pi_{1}(M)$, where $M$ is a hyperbolic 3-manifold, and $H\leq G$ is
finitely generated then one (and only one) of the following holds:

  1. (a)

$H$ is quasiconvex or

  2. (b)

$H$ is a virtual fiber, i.e. there exists a finite index subgroup
$G^{\prime}\leq G$ such that $H^{\prime}\coloneqq G^{\prime}\cap H$ is a
hyperbolic surface group and $G^{\prime}$ fits in a short exact sequence

${1}$${H^{\prime}}$${G^{\prime}}$${\mathbb{Z}}$${1.}$

######  Theorem 6.12.

Let $M$ be a closed hyperbolic 3-manifold and $G=\pi_{1}(M)$. Then

$\operatorname{\mathcal{K}}(G)=\overline{\operatorname{Sub}_{\mathrm{QC}[\infty]}(G)}.$

More precisely, the perfect kernel of $G$ is the set of infinite index
subgroups $H\leq G$ such that either $H$ is quasiconvex or $H$ is not finitely
generated.
Moreover, the action of $G$ on $\operatorname{\mathcal{K}}(G)$ is highly
topologically transitive and the Cantor-Bendixson rank of $G$ is either 2 or
3.

###### Proof.

By Theorem 6.1 and since $G$ is finitely generated
$\overline{\operatorname{Sub}_{\mathrm{QC}[\infty]}(G)}\subseteq\operatorname{\mathcal{K}}(G)\subseteq\operatorname{Sub}_{[\infty]}(G)$.
Let us show that the first inclusion is in fact an equality.

If $H\leq G$ is a virtual fiber subgroup then, by Corollary 3.6,
$H\notin\operatorname{\mathcal{K}}(G)$.

It remains to show that every non-finitely generated subgroup of $G$ is the
limit of a sequence of quasiconvex subgroups with infinite index.

It follows from Proposition 3.15 that a non-finitely generated subgroup does
not contain any virtual fiber subgroup. Indeed, for any tower of subgroups
$K_{1}\leq K_{2}\leq G$ where $K_{1}$ is a virtual fiber subgroup, either
$[K_{1}:K_{2}]<\infty$ or $[K_{2}:G]<\infty$ and in either case $K_{2}$ is
finitely generated. Thus if $H=\langle s_{n}:n\in\mathbb{N}\rangle\leq G$ is
not finitely generated, it is the limit of the non-stationary sequence
$(\langle s_{1},\dots,s_{n}\rangle)_{n}$ of infinite index subgroups and each
$\langle s_{1},\dots,s_{n}\rangle$ is quasiconvex.

To see that $\operatorname{rk_{CB}}(G)=2$ or 3, first note that the Cantor-
Bendixson rank of $G$ is the supremum of the Cantor-Bendixson erasing ranks of
its finite index and virtual fiber subgroups. But virtual fiber subgroups are
separable. Indeed if $G^{\prime}\leq G$, $H^{\prime}\leq H$ are finite index
subgroups such that $G^{\prime}/H^{\prime}=\mathbb{Z}$ then
$H^{\prime}=\cap_{n\in\mathbb{N}}\pi^{-1}(n\mathbb{Z})$ where
$\pi:G^{\prime}\rightarrow\mathbb{Z}$ is the quotient map. For each
$n\in\mathbb{N}$, $\pi^{-1}(n\mathbb{Z})$ has finite index in $G^{\prime}$ and
therefore in $G$, so $H^{\prime}$ is closed with respect to the profinite
topology on $G$. Since $H$ is the union of finitely many cosets of
$H^{\prime}$, it follows that $H$ is also closed. Thus
$\operatorname{rk^{\times}_{CB}}(H;G)\geq 2$. Since by [Ago13, Theorem 9.2]
every closed hyperbolic 3-manifold group contains a virtual fiber subgroup,
this implies that $\operatorname{rk_{CB}}(G)$ is in fact the supremum of the
Cantor-Bendixson erasing ranks of the virtual fiber subgroups. For any such
subgroup $H\leq G$, any Schreier graph of $H$ is quasi-isometric to
$\mathbb{Z}$ so it follows from Theorem 3.10 that
$\operatorname{rk^{\times}_{CB}}(H;G)\leq 3$. ∎

Note that there exist hyperbolic 3-manifold groups with Cantor-Bendixson rank
3. Indeed, it was pointed out to us by Ian Biringer that there are hyperbolic
3-manifold groups which algebraically fiber over the infinite dihedral group
$D_{\infty}$ and the kernel of this fibering has Cantor-Bendixson erasing rank
$3$. Recall that a group $G$ algebraically fibers over $Q$ if it fits into a
short exact sequence $1\to H\to G\to Q\to 1$ where $H$ is finitely generated.
He moreover convinced us (personal communication) that if a virtual fiber
subgroup is not virtually of this form then its Cantor-Bendixson erasing rank
is $2$.

It follows that the Cantor-Bendixson rank of a hyperbolic 3-manifold group $G$
is 3 if and only if $G$ virtually fibers over $D_{\infty}$, otherwise it is 2.
This leads us to the following question.

######  Question 6.13.

Does there exist a closed hyperbolic 3-manifold $M$ such that no finite index
subgroup of $\pi_{1}(M)$ algebraically fibers over $D_{\infty}$? Equivalently,
does there exist a hyperbolic 3-manifold group $G$ with
$\operatorname{rk_{CB}}(G)=2$?

##  7 Some facts and more questions

Let us consider the supremum of the Cantor-Bendixson ranks among various
classes of groups

$\displaystyle\alpha_{\text{fp}}$
$\displaystyle\coloneqq\sup\\{\operatorname{rk_{CB}}(G)\mid G\text{ finitely
presented}\\}$ $\displaystyle\alpha_{\text{fg}}$
$\displaystyle\coloneqq\sup\\{\operatorname{rk_{CB}}(G)\mid G\text{ finitely
generated}\\}$ $\displaystyle\alpha_{\text{c}}$
$\displaystyle\coloneqq\sup\\{\operatorname{rk_{CB}}(G)\mid G\text{
countable}\\}$

######  Question 7.1.

Clearly $\alpha_{\text{fp}}\leq\alpha_{\text{fg}}\leq\alpha_{\text{c}}$.

  1. 1.

Which inequalities are strict?

  2. 2.

What are the values of $\alpha_{\text{fp}},\alpha_{\text{fg}}$ and
$\alpha_{\text{c}}$?

  3. 3.

What are the possible values of Cantor-Bendixson ranks for the various classes
of groups?

Recall the Rips construction and the Haglund-Wise construction:
Given any finitely presented group $Q$ there exists a specific group $G$ and a
finitely generated normal subgroup $N$ with a short exact sequence

${1}$${N}$${G}$${Q}$${1,}$$\scriptstyle{\phi}$

where the group $G$ can be chosen to be
– a hyperbolic group [Rip82], or
– a finitely presented subgroup of $\mathrm{SL}(n,\mathbb{Z})$ for some $n$
[HW08, Th. 1.6].

It follows that:

######  Proposition 7.2.

$\displaystyle\alpha_{\text{\rm fp}}$
$\displaystyle=\sup\\{\operatorname{rk_{CB}}(G)\mid G\text{\rm\
hyperbolic}\\}.$ $\displaystyle\alpha_{\text{\rm fp}}$
$\displaystyle=\sup\\{\operatorname{rk_{CB}}(G)\mid G\text{\rm\ finitely
presented linear}\\}.$

Recall from Proposition 3.5 that given an exact sequence $1\rightarrow
N\rightarrow G\overset{\phi}{\rightarrow}Q\rightarrow 1$ where $N$ is finitely
generated, then $\operatorname{Sub}_{N\leq}(G)$ is a $G$-invariant clopen
subset of $\operatorname{Sub}(G)$ and $\phi$ induces a homeomorphism
$\operatorname{Sub}_{N\leq}(G)\simeq\operatorname{Sub}(Q)$ that is
$G$-equivariant for the $G$-actions, where $G$ acts on $\operatorname{Sub}(Q)$
via $\phi$, so that
$\operatorname{\mathcal{K}}(G)\cap\operatorname{Sub}_{N\leq}(G)\simeq\operatorname{\mathcal{K}}(Q)$
and $\operatorname{rk}(G)\geq\operatorname{rk}(Q)$ (Proposition 3.5 (4)).

If follows from the Rips and Haglund-Wise constructions that for any finitely
presented $Q$ we can embed the action
$Q\curvearrowright\operatorname{Sub}_{[\infty]}(Q)$, regardless of its
complexity, inside the space of non quasiconvex subgroups (see Corollary 7.4)
of some hyperbolic group, or alternatively inside the space of subgroups of a
finitely presented linear group.

######  Question 7.3.

These observations give rise to the questions:

  1. 1.

How “pathological” can the action $H\curvearrowright\operatorname{Sub}(H)$ be
for a finitely presented group $H$?

  2. 2.

Which bounds can be set for the Cantor-Bendixson ranks of finitely presented
groups?

Observe that the lamplighter groups, the Grigorchuk group and the Gupta-Sidki
$3$-group mentioned in the introduction are not finitely presented. The first
example of a finitely presented group with infinite Cantor-Bendixson rank
(namely the Baumslag group) has been announced by Bontemps, Frisch and the
second author [BFG23]. This implies in particular the existence of hyperbolic
groups with infinite Cantor-Bendixson rank.

######  Corollary 7.4.

If $G$ is produced using the Rips construction above with a non-hyperbolic
$Q$, then

$\operatorname{Sub}_{N\leq}(G)\cap\overline{\operatorname{Sub}_{\mathrm{QC}[\infty]}(G)}=\emptyset.$

Thus, if $\operatorname{\mathcal{K}}(Q)\not=\emptyset$, the reverse inclusion
$\overline{\operatorname{Sub}_{\mathrm{QC}[\infty]}(G)}\supseteq\operatorname{\mathcal{K}}(G)$
of Theorem 6.1 does not hold since

$\operatorname{\mathcal{K}}(Q)\simeq\operatorname{\mathcal{K}}(G)\cap\operatorname{Sub}_{\leq
N}(G)\subseteq\operatorname{\mathcal{K}}(G)\smallsetminus\overline{\operatorname{Sub}_{\mathrm{QC}[\infty]}(G)}.$

The corollary follows from the following result of Mihalik and Towle [MT94,
Theorem 1]:
If a subgroup of a hyperbolic group $G$ has infinite index and is quasiconvex,
then it does not contain any infinite normal subgroup of $G$. The fact that
$Q$ is not hyperbolic ensures that $N$ is infinite.

The Theorems 5.9, 5.11, 6.1, 6.2 all use the property that the subgroups we
consider fix a proper quasiconvex subset of an ambient hyperbolic space. This
leads to the following question, which we have not pursued in depth:

######  Question 7.5.

Is there a common generalisation of our results in the context of group
actions on hyperbolic spaces with suitable constraints?

##  8 An application: amenable faithful transitive actions

The class $\mathcal{A}$ is the class of countably infinite groups that admit a
faithful, transitive and amenable action.

Recall from [GM07] that if a triple of groups $K_{1}\leq K_{2}\leq K_{3}$
satisfies that $K_{i}$ is co-amenable in $K_{i+1}$, then $K_{1}$ is co-
amenable in $K_{3}$ and if $K_{1}$ belongs to $\mathcal{A}$, then $K_{3}$
belongs to $\mathcal{A}$. In particular, if $H\in\mathcal{A}$ and $G\geq H$ is
a finite index supergroup, then $G\in\mathcal{A}$. Moreover, if
$H_{1},H_{2}\in\mathcal{A}$ then $H_{1}\times H_{2}\in\mathcal{A}$.

Moon observed that Baumslag-Solitar groups belong $\mathcal{A}$ since they
admit a co-amenable free subgroup (or they are amenable, depending on the
parameters).

######  Proposition 8.1.

Let $G$ be a countable group and $\operatorname{Sub}_{[\infty]}(G)$ its space
of infinite index subgroups. Assume that there is

  1. (a)

a subgroup $H_{1}\in\operatorname{Sub}_{[\infty]}(G)$ such that
$G\curvearrowright G/H_{1}$ is faithful;

  2. (b)

a co-amenable subgroup $H_{2}\in\operatorname{Sub}_{[\infty]}(G)$ (or more
generally a family $(H_{2}^{i})_{i\in\mathbb{N}}$ of subgroups
$H_{2}^{i}\in\operatorname{Sub}_{[\infty]}(G)$ such that the action
$G\curvearrowright G/H_{2}^{1}\sqcup G/H_{2}^{2}\sqcup\cdots\sqcup
G/H_{2}^{i}\sqcup\cdots$ is amenable);

  3. (c)

a subgroup $H_{0}\in\operatorname{Sub}_{[\infty]}(G)$ whose orbit under the
$G$-action by conjugation on $\operatorname{Sub}_{[\infty]}(G)$ accumulates on
both $H_{1}$ and $H_{2}$ (or more generally on $H_{1}$ and all the
$H_{2}^{i}$’s).

Then the action $G\curvearrowright G/H_{0}$ is faithful, transitive and
amenable. In particular, $G\in\mathcal{A}$.

Moreover, if $R\subseteq\operatorname{Sub}(G)$ is a $G$-invariant closed
subspace containing both $H_{1}$ and $H_{2}$ (resp. $H_{1}$ and all the
$H_{2}^{i}$) together with a dense orbit, then the set of $H_{0}$ satisfying
(c) is a dense Gδ subset of $R$.

###### Proof of Corollary I.

First observe that the group $G$ is infinite, thus
$H_{1}\coloneqq\\{\mathrm{id}\\}\in\operatorname{Sub}_{[\infty]}(G)$. The
amenability condition on $G$ implies that it admits subgroups satisfying
Condition (b) of Proposition 8.1, and thus its conclusion. ∎

###### Proof of Proposition 8.1.

It is enough to show that $G\curvearrowright G/H_{0}$ is faithful and
amenable. Let $\gamma\in G\smallsetminus\\{\mathrm{id}\\}$. By faithfulness of
$G\curvearrowright G/H_{1}$, there is $g_{1}\in G$ such that $g_{1}^{-1}\gamma
g_{1}\not\in H_{1}$. Let $R$ be the closure of the orbit of $H_{0}$ and let
$V$ be the open neighbourhood of $H_{1}$ in $R$ consisting of the subgroups
that do not contain $g_{1}^{-1}\gamma g_{1}$. By density, there is $g\in G$
such that $gH_{0}g^{-1}\in V$, i.e. $\gamma\not\in g_{1}gH_{0}(g_{1}g)^{-1}$.
Thus $G\curvearrowright G/H_{0}$ is faithful.

Consider a Følner sequence $(A_{i})_{i}$ with $A_{i}\subseteq G/H_{2}^{i}$ (or
$A_{i}\subseteq G/H_{2}$). More precisely, one can (up to re-indexing) write
$G$ as an increasing union of finite subsets
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6.3mu\mkern-0.3mu}\,}\slimits@_{i}S_{i}$ such that for each $i$ and $\gamma\in
S_{i}$ we have $|\gamma A_{i}\triangle A_{i}|\leq\frac{1}{i}|A_{i}|$ with
$A_{i}=\\{g_{j}^{i}H_{2}^{i}\colon j=1,2,\cdots,|A_{i}|\\}$. Let
$|A_{i}|=n_{i}$ and $J_{i}=\\{1,2,\cdots,n_{i}\\}$. The intersection $\gamma
A_{i}\cap A_{i}$ is encoded by the equations $(g_{k}^{i})^{-1}\gamma
g_{j}^{i}\in H_{2}^{i}$ (for $j,k\in J_{i}$ and $\gamma\in S_{i}$) while
$|A_{i}|$ is encoded by $(g_{k}^{i})^{-1}g_{j}^{i}\not\in H_{2}^{i}$ (for
$j,k\in J_{i}$, $j\not=k$). These conditions define an open neighbourhood
$V_{i}$ of $H_{2}^{i}$ in $R$ and by density of the orbit of $H_{0}$, there is
$k_{i}\in G$ such that $k_{i}H_{0}k_{i}^{-1}\in V_{i}$, i.e.
$(g_{k}^{i})^{-1}\gamma g_{j}^{i}\in H_{2}^{i}$ $\implies$
$(g_{k}^{i})^{-1}\gamma g_{j}^{i}\in k_{i}H_{0}k_{i}^{-1}$ and
$(g_{k}^{i})^{-1}g_{j}^{i}\not\in k_{i}H_{0}k_{i}^{-1}$. It follows that the
$B_{i}\colon=\\{g_{j}^{i}k_{i}H_{0}\colon j\in J_{i}\\}$ satisfy $|\gamma
B_{i}\triangle B_{i}|\leq\frac{1}{i}|B_{i}|$. Thus the sequence $(B_{i})_{i}$
is a Følner sequence for the action $G\curvearrowright G/H_{0}$, which is
therefore amenable.

Concerning the “moreover” part, since $R$ admits a countable basis
$(U_{\ell})_{\ell\in\mathbb{N}}$ of open sets, the set $\cap_{\ell}G\cdot
U_{\ell}$ of points with dense orbits in $R$ is a Gδ. Since their orbit
accumulates to some infinite index subgroup (namely $H_{1}$), they have
infinite index themselves. ∎

######  Corollary 8.2.

Any finite graph of free groups with cyclic edges with at least one non-cyclic
vertex group is in $\mathcal{A}$.

This generalises a result of Moon [Moo11a].

###### Proof.

We will apply Proposition 8.1. By Example 5.22, $G$ belongs to
$\mathcal{Q}^{\prime}$. Thus,
$G\curvearrowright\operatorname{Sub}_{[\infty]}(G)=:R\ni\\{\mathrm{id}\\}$ is
topologically transitive. It remains to find a co-amenable infinite index
subgroup $H_{2}\in R$.

If the graph is not a tree, then $G$ surjects onto $\mathbb{Z}$. Take $H_{2}$
to be the kernel of this morphism. Otherwise, if one of the leaves $v$ of the
finite graph has a non-cyclic stabiliser $G_{v}$, then $G$ surjects onto the
abelianisation of $G_{v}$ modulo the group $G_{e}$ of the incident edge $e$.
Eventually, if $v_{1},v_{2}$ are two leaves of the finite graph, with cyclic
stabilisers $G_{v_{1}},G_{v_{2}}$ (and the stabilisers $G_{e_{1}}\lneqq
G_{v_{1}}$ and $G_{e_{2}}\lneqq G_{v_{2}}$ of the incident edges $e_{1},e_{2}$
are proper subgroups). Then $G$ maps onto the non-trivial free product
$G_{v_{1}}/G_{e_{1}}*G_{v_{2}}/G_{e_{2}}$, that itself maps virtually onto
$\mathbb{Z}$. Thus $G$ virtually maps onto $\mathbb{Z}$. The kernel of this
virtual morphism is co-amenable and has infinite index. ∎

######  Corollary 8.3.

Let $G$ be a countable group which admits a minimal and irreducible action
$G\curvearrowright T$ on a tree. Suppose that there are two edges
$f_{1},f_{2}$ of $T$ with
$\operatorname{Stab}_{G}(f_{1})\cap\operatorname{Stab}_{G}(f_{2})=N_{0}$ (the
kernel of the $G$-action on $T$), and assume moreover that $G$ admits a co-
amenable subgroup $G_{0}$ (possibly a finite index subgroup) such that
$G_{0}\cap N_{0}=\\{\mathrm{id}\\}$ and such that the graph $G_{0}\backslash
T$ is not a finite tree. Then $G\in\mathcal{A}$.

###### Proof of Corollary 8.3.

If $G_{0}\backslash T$ is not a tree, $G_{0}$ admits a homomorphism onto
$\mathbb{Z}$ whose kernel $N$ gives rise to a infinite quotient $N\backslash
T$: indeed, pick a simple closed loop $c$ in the graph $G_{0}\backslash T$,
then the fundamental group $G_{0}$ of the graph of groups associated with
$G_{0}\curvearrowright T$ maps onto $\pi_{1}(c)\simeq\mathbb{Z}$ with kernel
$N\leq G_{0}$. The graph of groups associated with $N\curvearrowright T$
contains an embedded copy of the universal cover of the loop $c$. It follows
that $H_{2}\coloneqq N$ is co-amenable in $G_{0}$ and thus in $G$ and that
$H_{2}\in\overline{\operatorname{Sub}_{|\bullet\backslash T|\infty}(G)}$ and
$H_{2}\cap N_{0}=\\{\mathrm{id}\\}$ (i.e.
$H_{2}\in\mathcal{F}_{\\{\\{\mathrm{id}\\}\\}}^{T}$ in the terminology of
Equation 13 and Theorem 5.11). In case $G_{0}\backslash T$ is infinite, then
simply set $H_{2}\coloneqq G_{0}$. By Theorem 5.11,
$R\coloneqq\mathcal{F}_{\\{\\{\mathrm{id}\\}\\}}^{T}$ and $H_{2}$ satisfy the
assumptions of Proposition 8.1. ∎

######  Corollary 8.4.

Let $G$ be a group which decomposes non-trivially as the fundamental group of
a graph of groups where the graph is not a finite tree. Assume that, for some
edge $e$ and incident vertex $v$, the associated edge group $G_{e}$ and vertex
group $G_{v}$ satisfy: $\exists g\in G_{v}$ such that $gG_{e}g^{-1}\cap
G_{e}=\\{\mathrm{id}\\}$. Then $G\in\mathcal{A}$.

###### Proof.

The condition $gG_{e}g^{-1}\cap G_{e}=\\{\mathrm{id}\\}$ ensures that in the
associated Bass-Serre tree $T$, the edge $f_{1}$ stabilised by $G_{e}\leq G$
and its image by $g\in G_{v}$ have stabilisers that intersect trivially. Thus
the kernel $N_{0}$ of the action is trivial. The corollary follows by applying
Corollary 8.3 with $G_{0}\coloneqq G$. ∎

Our final application is to show that the class of (virtually) compact special
groups introduced by Haglund and Wise in [HW08] is contained in $\mathcal{A}$.
These are groups $G$ which (virtually) act co-specially, freely and
cocompactly on a CAT(0) cube complex $X$: a cell complex constructed by
attaching $n$-cubes $[0,1]^{n}$ isometrically along faces, which is CAT(0)
when equipped with the piecewise $\ell^{2}$-metric. The action
$G\curvearrowright X$ is co-special if the quotient $G\backslash X$ is special
in the sense of [Wis21, Chapter 6, p. 167].

We recall some key definitions and properties related to CAT(0) cube complexes
and co-special actions.

Let $X$ be a CAT(0) cube complex. The dimension of $X$ is the supremum of the
set of $n\in\mathbb{N}$ such that $X$ contains an $n$-cube. We say that $X$ is
reducible if it splits non-trivially as a product $X=X_{1}\times X_{2}$ and
$X$ is irreducible otherwise.

A hyperplane $h\subseteq X$ is a non-empty connected subspace of $X$ whose
intersection with each $n$-cube $\sigma$ of $X$ is either empty or a midcube
of $\sigma$ (the space obtained from $[0,1]^{n}$ by restricting one coordinate
to $1/2$). It was proved in [Sag95] that each hyperplane of $X$ separates $X$
into two connected components, called halfspaces; we denote these $h^{+}$ and
$h^{-}$. Given two subspaces $U,V\subseteq X$, we say that $h$ separates $U$
from $V$ if $U$ and $V$ lie in distinct connected components of $X\setminus
h$. Given two hyperplanes $h,k\subseteq X$, we say that $h$ and $k$ are
strongly separated if no hyperplane in $X$ intersects both $h$ and $k$.

Let $d$ denote the piecewise $\ell^{1}$-metric on $X$. The distance between
any two vertices of $X$ is precisely the number of hyperplanes separating
them; this coincides with the restriction to the 0-skeleton of the path metric
on the 1-skeleton of $X$. There is a well-defined gate map
$\mathfrak{g}_{h}:X\rightarrow h$ which maps each vertex of $X$ to its
(unique) closest point projection in $h$ (see [BHS17, Section 2.1]). This map
has the property that if $x,y$ are vertices of $X$ and $k$ is a hyperplane
then $k$ separates $\mathfrak{g}_{h}(x)$ and $\mathfrak{g}_{h}(y)$ if and only
if $k$ intersects $h$ and separates $x$ from $y$. In particular, if $h$ and
$k$ are strongly separated hyperplanes then $\mathfrak{g}_{h}(k)$ is a single
point.

We consider actions by cellular automorphisms, these are isometries with
respect to both $d$ and the CAT(0) metric (i.e. the piecewise
$\ell^{2}$-metric). The group of all cellular automorphisms of $X$ is denoted
by $\operatorname{Aut}(X)$. An action $G\curvearrowright X$ on a CAT(0) cube
complex is said to be proper if the stabiliser of each $n$-cube is finite. It
is essential if for any orbit $G\cdot v$ and for each halfspace $h^{+}$
bounded by a hyperplane $h$, there are points in $(G\cdot v)\cap h^{+}$
arbitrarily far from $h$.

We recall the Double Skewering Lemma of [CS11]: Suppose that $G$ is a group
acting essentially on a finite dimensional CAT(0) cube complex without a fixed
point at infinity. Then for any halfspaces $h^{+}\subseteq k^{+}$, there is an
element $\gamma\in G$ such that $\gamma k^{+}\subsetneq h^{+}\subseteq k^{+}$.

Suppose that $h\subseteq X$ is a hyperplane such that the set of hyperplanes
$G\cdot h$ is pairwise disjoint (i.e. for all $g\in G$ either $gh=h$ or
$gh\cap h=\emptyset$) and such that $G$ does not invert $h$ (i.e. if
$g\in\operatorname{Stab}_{G}(h)$ then $gh^{+}=h^{+}$). Let $N(h)$ be the open
carrier of $h$; that is, the union of all open cubes which intersect $h$. Let
$T_{h}$ be the graph whose vertex set is the set of connected components of
$X\smallsetminus(G\cdot N(h))$ and where $v,w\in V(T)$ are connected by an
edge if they are both bounded by a common hyperplane carrier in $G\cdot N(h)$.
This defines a continuous surjection $\pi_{h}:X\rightarrow T_{h}$ which maps
each connected component of $X\smallsetminus(G\cdot N(h))$ to a vertex and
each open hyperplane carrier $N(k)$ in $G\cdot N(h)$ to an open edge whose
midpoint is $\pi_{h}(k)$. Note that edges of $T_{h}$ separate the space into
two connected components so $T_{h}$ is a tree. We will refer to $T_{h}$ as the
dual tree of $h$. The action $G\curvearrowright(G\cdot N(h))$ induces an
action $G\curvearrowright T_{h}$ via $\pi_{h}$. Since $G$ does not invert
hyperplanes in the orbit of $h$, this action preserves the orientation of the
edges, and is therefore an “action on a graph” in the sense of Section 5.

Recall that if $G\curvearrowright X$ is co-special then $G$ acts with pairwise
disjoint hyperplane orbits and no hyperplane inversions. By the above, this
implies that $G$ acts on the dual tree $T_{h}$ for any hyperplane $h$ of $X$.

######  Lemma 8.5.

Let $G$ be a countable group acting on an irreducible CAT(0) cube complex $X$,
where $X$ is not quasi-isometric to $\mathbb{R}$ and the automorphism group
$\operatorname{Aut}(X)$ acts cocompactly. Suppose that the action is proper,
essential and does not fix a point in the visual boundary
$\partial_{\infty}X$. Suppose moreover that there is a hyperplane $h\subseteq
X$ such that the set of hyperplanes $G\cdot h$ is pairwise disjoint and such
that $G$ does not invert $h$. Let $T_{h}$ be the corresponding dual tree.
Under these conditions, the action $G\curvearrowright T_{h}$ is minimal and
irreducible and there are edges $f_{1},f_{2}\in E(T)$ such that
$\operatorname{Stab}_{G}(f_{1})\cap\operatorname{Stab}_{G}(f_{2})$ is finite.

###### Proof.

Since the action of $\operatorname{Aut}(X)$ is cocompact, $X$ is finite
dimensional.

We first show that there is a pair of strongly separated hyperplanes in the
orbit $G\cdot h$. Since $X$ is irreducible and finite dimensional and since
$G$ acts essentially without a fixed point at infinity, there is a pair of
strongly separated hyperplanes $k_{1},k_{2}\subseteq X$ by [CS11, Theorem
5.1]. Since $h$ does not intersect both $k_{1}$ and $k_{2}$, then up to
relabelling, we can assume that $h^{-}\subseteq k_{2}^{-}$ and that
$k_{2}^{+}\subseteq k_{1}^{+}$ where $k_{i}^{+},k_{i}^{-}$ are the halfspaces
of $k_{i}$. By the Double Skewering Lemma, there exists $\gamma,\psi\in G$,
such that $\gamma k_{2}^{-}\subsetneq k_{1}^{-}$ and $\psi h^{+}\subsetneq
k_{2}^{+}$. Thus $\gamma h\subseteq k_{1}^{-}$ and $\psi h\subseteq
k_{2}^{+}$. If a hyperplane intersects both $\gamma h$ and $\psi h$ then it
must intersect $k_{1}$ and $k_{2}$ so no such hyperplane exists. Therefore
$h_{1}\coloneqq\gamma h$ and $h_{2}\coloneqq\psi h$ are strongly separated.

Recall that the image $\mathfrak{g}_{h_{1}}(h_{2})$ is a single point, say
$x_{1}$. If
$g\in\operatorname{Stab}_{G}(h_{1})\cap\operatorname{Stab}_{G}(h_{2})$ then
$d(gx_{1},h_{2})=d(x_{1},h_{2})$ and, since $gx_{1}\in h_{1}$ and $x_{1}$ is
the unique point in $\mathfrak{g}_{h_{1}}(h_{2})$, we have $gx_{1}=x_{1}$.
Since the action of $G$ on $X$ is proper, this implies that
$\operatorname{Stab}_{G}(h_{1})\cap\operatorname{Stab}_{G}(h_{2})$ is finite.
Thus if $f_{i}$ is the edge of $T_{h}$ with midpoint $\pi_{h}(h_{i})$ for
$i=1,2$ then
$\operatorname{Stab}_{G}(f_{1})\cap\operatorname{Stab}_{G}(f_{2})$ is finite,
as required.

Let us now show that the action $G\curvearrowright T_{h}$ is minimal and
irreducible. Firstly, by the Double Skewering Lemma, $T_{h}$ is unbounded and
the $G$-action has no global fixed point. Since the action has a single orbit
of edges, it is minimal.

Since $X$ is irreducible and not a quasi-line and since the action of
$\operatorname{Aut}(X)$ is cocompact, [CS11, Lemma 7.1] implies that
$\operatorname{Aut}(X)$ does not stabilise any Euclidean flat $\mathbb{R}^{n}$
in $X$. By [CS11, Theorem 7.2], this implies that there is a facing triple of
hyperplanes in $X$: i.e. a pairwise disjoint triple of hyperplanes
$k_{0},k_{1},k_{2}$ such that no $k_{j}$ separates the remaining two
hyperplanes. Suppose that for each $j$, $k_{j}^{+}$ is the halfspace
containing the remaining $k_{\ell}$’s. Since $h_{1},h_{2}$ are strongly
separated, there is $i\in\\{1,2\\}$ and $j\in\mathbb{Z}/3\mathbb{Z}$ such that
$h_{i}\cap k_{j}=\emptyset$. By the Double Skewering Lemma, there exists $g\in
G$ such that $gh_{i}\subseteq k_{j}^{-}$ (set $g=\mathrm{id}$ if this was
already the case). Then, using the Double Skewering Lemma again, there are
$g_{j+1},g_{j+2}\in G$ such that $g_{j+1}gh_{i}\subseteq k_{j+1}^{-}$ and
$g_{j+2}gh_{i}\subseteq k_{j+2}^{-}$. The set
$\\{gh_{i},g_{j+1}gh_{i},g_{j+2}gh_{i}\\}$ is a facing triple in the $G$-orbit
of $h$. Thus there is a facing triple of hyperplanes (in this case mid-points
of edges) in $T_{h}$, so $T_{h}$ cannot be a line. By minimality of the
action, $G$ cannot fix a line in $T_{h}$.

It remains to show that $G$ does not fix an end. Let
$(e_{n})_{n\in\mathbb{N}}$ be a geodesic ray such that $e_{1}$ is the edge
whose midpoint is $\pi_{h}(h)$. There is a halfspace of $h$ which contains
$\pi_{h}^{-1}(e_{n})$ for all $n\geq 2$, suppose without loss of generality
that this halfspace is $h^{+}$. By the Flipping Lemma of [CS11] there is an
element $\gamma\in G$ such that $\gamma h^{+}\subsetneq h^{-}$. Thus the paths
$(e_{n})_{n\geq 2}$ and $(\gamma e_{n})_{n\geq 2}$ are contained in distinct
components of $T_{h}\setminus\\{e_{1}\\}$ and therefore represent distinct
ends of $T_{h}$. ∎

######  Theorem 8.6.

If $G$ is an infinite virtually compact special group then $G\in\mathcal{A}$.

Examples of virtually compact special groups include right-angled Artin groups
(essentially by definition); right-angled and hyperbolic Coxeter groups [NR03,
HW10]; limit groups [Wis21, Theorem 18.6]; one-relator groups with torsion
[Wis21, Corollary 19.2]. Many groups are known to act properly and cocompactly
on CAT(0) cube complexes. By [Ago13], those that are moreover hyperbolic are
then virtually compact special. This includes C’(1/6) small cancellation
groups [Wis04]; random groups at density $d<1/6$ [OW11], hyperbolic-by-cyclic
groups which are themselves hyperbolic [DMM23].

###### Proof.

Recall that class $\mathcal{A}$ contains every infinite amenable group and is
closed under taking direct products. Moreover if $G$ is virtually in
$\mathcal{A}$ then $G$ itself is in $\mathcal{A}$. Thus we can assume that $G$
is compact special. Let $X$ be a CAT(0) cube complex on which $G$ acts freely,
cocompactly and co-specially. In particular, $G$ is torsion-free. By [CS11,
Proposition 3.5], there is a convex subcomplex of $X$, called the essential
core, on which $G$ acts essentially. The action of $G$ on this subcomplex is
free and cocompact and co-special. Thus by replacing $X$ with its essential
core and $G$ with a finite index subgroup, we can assume that the action
$G\curvearrowright X$ is free, cocompact, co-special and essential.

If $X$ splits non-trivially as a product $X=X_{1}\times X_{2}$ then by [Zal23,
Lemma 3.8] there is a finite index subgroup $G^{\prime}\leq G$ which splits as
a direct product $G^{\prime}=G_{1}\times G_{2}$, where
$G_{1}=\operatorname{Stab}_{G^{\prime}}(X_{1}\times\\{x_{2}\\})$ and
$G_{2}=\operatorname{Stab}_{G^{\prime}}(\\{x_{1}\\}\times X_{2})$ for any
$x_{i}\in X_{i}$. The actions $G_{i}\curvearrowright X_{i}$ are free,
cocompact and co-special. Moreover, for each $i$, the dimension of $X_{i}$ is
strictly inferior to that of $X$.

By induction on the dimension, if we iterate the process of passing to the
essential core, then applying [Zal23, Lemma 3.8] if the subspace is reducible,
we arrive at a set of subgroups $H_{1},\dots,H_{k}$ of $G$ such that:
– $G$ can be constructed from $H_{1},\dots,H_{k}$ by taking direct products
and finite index supergroups and
– each $H_{i}$ acts on an irreducible CAT(0) cube complex $X_{i}$ such that
the action is free, cocompact, co-special and essential.
We can therefore assume that $X$ is irreducible and $G\curvearrowright X$ has
the above properties.

Let $\rho:X\rightarrow X\backslash G$ be the quotient map. Fix a hyperplane
$h\subseteq X$ and let $T_{h}$ be the corresponding dual tree. By [Wis21,
Lemma 14.6] we can replace $G$ with a finite index subgroup so that $\rho(h)$
does not separate $G\backslash X$. This implies that $G\backslash T_{h}$ is a
loop.

By [CS11, Corollary 4.9], and since $X$ is irreducible, either $X$ is quasi-
isometric to a line or there is no fixed point at infinity. In the former
case, $G$ is virtually cyclic (in fact, cyclic since it is torsion free) so
$G\in\mathcal{A}$ since it is amenable. On the other hand, if $G$ does not fix
a point in $\partial_{\infty}X$ then it follows from Lemma 8.5 that the action
of $G$ on $T_{h}$ is minimal and irreducible, and that there are edges
$f_{1},f_{2}\in E(T_{h})$ such that
$\operatorname{Stab}_{G}(f_{1})\cap\operatorname{Stab}_{G}(f_{2})$ is finite.
Since $G$ is torsion-tree this intersection is in fact trivial. Thus
$G\in\mathcal{A}$ by Corollary 8.3. ∎

##  Appendix A Appendix: A list of examples

A list of examples of finitely generated groups for which
$\operatorname{\mathcal{K}}(G)=\operatorname{Sub}_{[\infty]}(G)$, thus
$\operatorname{rk_{CB}}(G)=1$, and for which
$G\curvearrowright\operatorname{\mathcal{K}}(G)$ is highly topologically
transitive.

  1. 1.

Non-abelian limit groups (Corollary 5.23 and Example 5.22).

  2. 2.

Non-elementary free products $A*B$ and groups with infinitely many ends and no
non-trivial finite normal subgroup (Corollary 5.17). As for
$\mathrm{SL}(2,\mathbb{Z})$, see Example 5.18.

  3. 3.

Graphs of free groups with cyclic edges such that at least one vertex group is
non-cyclic (Corollary 5.23 and Example 5.22).

  4. 4.

Groups in the class $\mathcal{Q}^{\prime}$ (Corollary 5.23).

  5. 5.

Non-elementary and locally quasiconvex hyperbolic groups with trivial finite
normal subgroup (Corollary 6.3). For instance, non-elementary hyperbolic
groups with a small hierarchy and trivial finite normal subgroup (Example
6.5).

  6. 6.

One-relator group $G=\langle S\>|\>w^{m}\rangle$ where $w$ is a cyclically
reduced word in $S\cup S^{-1}$ and $m\geq|w|_{S}$ (Example 6.6).

  7. 7.

Coxeter groups $G=\langle
s_{1},\dots,s_{n}|s_{i}^{2},(s_{i}s_{j})^{m_{i,j}}\;\forall\;i<j\rangle$ where
$\infty\geq m_{i,j}>n\geq 3$ (Example 6.7).

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P.A., University of Bristol, School of Mathematics, Bristol, UK

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D.G., CNRS, ENS-Lyon, Unité de Mathématiques Pures et Appliquées, 69007 Lyon,
France

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