# Perfect kernel and dynamics: from Bass-Serre theory to hyperbolic groups

Pénélope Azuelos and Damien Gaboriau

(October 3, 2023)

###### Abstract

We introduce several approaches to studying the Cantor-Bendixson decomposition
of and the dynamics on the (topological) space of subgroups for various
families of countable groups. In particular, we uncover the perfect kernel and
the Cantor-Bendixson rank of the space of subgroups of many new groups,
including for instance infinitely ended groups, limit groups, hyperbolic
3-manifold groups and many graphs of groups. We also study the topological
dynamics of the conjugation action on the perfect kernel, establishing the
conditions for topological transitivity and higher topological transitivity.
As an application, we obtain many new examples of groups in the class
$\mathcal{A}$ of Glasner and Monod, i.e. admitting faithful transitive
amenable actions. This includes for example right-angled Artin groups, limit
groups, C’(1/6) small cancellation groups, random groups at density $d<1/6$,
and more generally all virtually compact special groups.

Keywords: Space of subgroups; perfect kernel; topologically transitive
actions; hyperbolic groups, subgroup separable groups, Bass-Serre theory.

MSC-classification: 37B; 20E; 20F67; 20F65; 37B20

###### Contents

  1. 1 Introduction
  2. 2 Notation and preliminaries
  3. 3 Generalities on the topological structure of $\operatorname{Sub}(G)$
    1. 3.1 Normal subgroups and invariant subsets
    2. 3.2 Examples
    3. 3.3 Subgroups with Schreier graph quasi-isometric to $\mathbb{Z}$
  4. 4 Subgroup separable groups and the first $\ell^{2}$-Betti number
  5. 5 Actions on trees
    1. 5.1 Main technical dendrological lemma
    2. 5.2 Main dendrological theorems
    3. 5.3 Applications
    4. 5.4 A rich class of groups with maximal perfect kernel
  6. 6 Hyperbolic groups
    1. 6.1 Generalities on the space of subgroups of hyperbolic groups
    2. 6.2 Proof of Theorems 6.1 and 6.2
    3. 6.3 Hyperbolic 3-manifold groups
  7. 7 Some facts and more questions
  8. 8 An application: amenable faithful transitive actions
  9. A Appendix: A list of examples

##  1 Introduction

Let $G$ be a countable group and $\operatorname{Sub}(G)$ be the space of all
subgroups of $G$. There has been a recent increase in studies of the action by
conjugation of $G$ on this space. It is the crucible where certain properties
of the non-free actions of G boil down, whether they are of a topological or
measured nature.

When endowed with the Chabauty topology, $\operatorname{Sub}(G)$ is a compact
metrizable totally disconnected space, thus satisfying the Baire property. The
Cantor–Bendixson structure theory of compact spaces leads us to consider the
unique decomposition
$\operatorname{Sub}(G)=\operatorname{\mathcal{K}}(G)\sqcup C$ where
$\operatorname{\mathcal{K}}(G)$, the perfect kernel of $G$, is the maximal
perfect subspace of $\operatorname{Sub}(G)$ and $C$ is a countable set.
Associated to this decomposition is an ordinal, called the Cantor-Bendixson
rank of $G$ and denoted $\operatorname{rk_{CB}}(G)$. It captures after how
many steps the iterative process of removing isolated points from
$\operatorname{Sub}(G)$ results in the stable subspace
$\operatorname{\mathcal{K}}(G)$. Precise definitions are given in Section 2.

The action of $G$ on its space of subgroups by conjugation is continuous and
the perfect kernel is invariant under this action. The interesting uniformly
recurrent subgroups (URS) and invariant random subgroups (IRS) are supported
on the perfect kernel.

It is an instructive exercise to check that
$\operatorname{\mathcal{K}}(\mathbb{Z}^{d})=\emptyset$ and that
$\operatorname{rk_{CB}}(\mathbb{Z}^{d})=d+1$. In fact, since a non-empty
perfect Polish space contains a Cantor set,
$\operatorname{\mathcal{K}}(G)=\emptyset$ if and only if $G$ admits only
countably many subgroups. This is the case for instance for all finitely
generated nilpotent groups. More generally, this happens for all groups whose
subgroups are all finitely generated (the so-called Noetherian groups), for
instance for the Tarski Monsters: those non-abelian groups whose proper
subgroups are all cyclic.

If $G$ is the $\mathfrak{R}$-free group of countably infinite rank of a (non-
trivial) variety $\mathfrak{R}$ of groups (such as the free group
$\mathbf{F}_{\infty}$, $\oplus_{\mathbb{N}}\mathbb{Z}$,
$\oplus_{\mathbb{N}}\mathbb{Z}/m\mathbb{Z}$, the free Burnside group
$B(\mathbb{N},p)=\langle(s_{i})_{i\in\mathbb{N}}|g^{p}=1,\text{ for all
}g\rangle$ for $p\geq 2$,…), then
$\operatorname{\mathcal{K}}(G)=\operatorname{Sub}(G)$ (see Proposition 3.8)
and $\operatorname{rk_{CB}}(G)=0$.

In general, if $G$ is finitely generated then its finite index subgroups are
isolated, so

$\operatorname{\mathcal{K}}(G)\subseteq\operatorname{Sub}_{[\infty]}(G)\text{
where }\operatorname{Sub}_{[\infty]}(G):=\\{H\leq G\text{ s.t.
}[G:H]=\infty\\}.$

In the case where
$\operatorname{\mathcal{K}}(G)=\operatorname{Sub}_{[\infty]}(G)$, note that
$\operatorname{rk_{CB}}(G)=1$. Observe that $\operatorname{Sub}(G)$ may
contain infinite index isolated points. There are even examples of infinite
groups for which the trivial subgroup $\\{\mathrm{id}\\}$ is isolated:
consider for instance the Prüfer $p$-group, or see the more interesting
finitely generated Example 3.7.

It is not hard to check that the equality holds for all finitely generated
non-cyclic free groups $\mathbf{F}_{r}$:

$\operatorname{\mathcal{K}}(\mathbf{F}_{r})=\operatorname{Sub}_{[\infty]}(\mathbf{F}_{r}),$

see e.g. [CGLM20, Proposition 2.1]. Moreover, the action of $\mathbf{F}_{r}$
on $\operatorname{\mathcal{K}}(\mathbf{F}_{r})$ is highly topologically
transitive (see Corollary 5.17 or Corollary 6.3). Recall that a continuous
action $G\curvearrowright X$ on a compact metrizable space of a countable
group $G$ is called topologically $r$-transitive for a positive integer $r$ if
the diagonal action $G\curvearrowright X^{r}$ on the $r$-fold space is
topologically transitive; i.e. if for every $2r$-tuple
$(V_{1},V_{2},\cdots,V_{2r})$ of non-empty open sets, there is an element
$g\in G$ such that $g\cdot V_{i}\cap V_{r+i}\not=\emptyset$ for all
$i=1,2,\cdots,r$ (see Section 2). The action is called highly topologically
transitive if it is topologically $r$-transitive for all $r\geq 1$.

One purpose of this article is to extend the validity of these statements to
several new groups.

In recent years, the space of subgroups of several classes of groups have been
studied by various authors. Cornulier, Guyot and Pitsch [CGP10] classified the
spaces of subgroups of all countable abelian groups up to homeomorphism and
computed their Cantor-Bendixson rank. In [BGK15], Bowen, Grigorchuk and
Kravchenko found the perfect kernel and Cantor-Bendixson rank of the
lamplighter groups
$(\mathbb{Z}/k\mathbb{Z})^{n}\wr\mathbb{Z}=\oplus_{\mathbb{Z}}(\mathbb{Z}/k\mathbb{Z})^{n}\rtimes\mathbb{Z}$,
i.e.
$\operatorname{\mathcal{K}}((\mathbb{Z}/k\mathbb{Z})^{n}\wr\mathbb{Z})=\operatorname{Sub}(\oplus_{\mathbb{Z}}(\mathbb{Z}/k\mathbb{Z})^{n})$
and
$\operatorname{rk_{CB}}((\mathbb{Z}/k\mathbb{Z})^{n}\wr\mathbb{Z})=\omega$,
the first infinite ordinal. Skipper and Wesolek described in [SW20] the
perfect kernels of the Grigorchuk group, the Gupta-Sidki $3$-group and some
other families of groups of automorphisms of infinite regular rooted trees;
they also established that $\operatorname{rk_{CB}}(G)=\omega$ for these
groups.

Recently Carderi, Le Maître, Stalder and the second author described the
perfect kernel and the dynamics for the Baumslag-Solitar groups
$\mathrm{BS}(m,n)=\langle b,t|tb^{m}t^{-1}=b^{n}\rangle$ depending on the
parameters $m,n$ in [CGLMS22]. If $\lvert m\lvert,|n|>1$, then
$\operatorname{\mathcal{K}}(\mathrm{BS}(m,n))$ is the space of subgroups whose
action on the standard Bass-Serre tree of $\mathrm{BS}(m,n)$ has infinite
quotient. Furthermore, the perfect kernel admits a countably infinite
partition into $\mathrm{BS}(m,n)$-invariant subspaces. One of them is closed
and not open; all the others are open (and also closed exactly when
$|m|=|n|$). Moreover, the $\mathrm{BS}(m,n)$-action on each piece is
topologically transitive.

Intrinsically, this theory is less interested in the algebraic structure of
subgroups so much as the “geometry of their embedding” into $G$. A result
which exemplifies this is the following:
If $H\leq G$ are finitely generated groups and the quotient $H\backslash G$ is
quasi-isometric to $\mathbb{Z}$ then $H$ vanishes by the third Cantor-
Bendixson derivative of $\operatorname{Sub}(G)$ (Theorem 3.10).

#### Hyperbolic groups

An important class of finitely generated groups is that of hyperbolic groups,
introduced by Gromov [Gro87]. Describing the space of subgroups of a
hyperbolic group is in general a difficult task, largely due to the fact that
finitely generated subgroups of hyperbolic groups need not respect the
geometry of the group. There are however a range of tools available to study
those subgroups that do respect the geometry: namely the quasiconvex
subgroups.

Given a hyperbolic group $G$, we denote by
$\operatorname{Sub}_{\mathrm{QC}[\infty]}(G)$ the set of quasiconvex subgroups
of $G$ with infinite index.

######  Theorem A (Theorems 6.1 and 6.2).

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

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

If, in addition, the maximal finite normal subgroup of $G$ is trivial then the
action of $G$ on $\overline{\operatorname{Sub}_{\mathrm{QC}[\infty]}(G)}$ is
highly topologically transitive.

A hyperbolic group is said to be locally quasiconvex if each of its finitely
generated subgroups is quasiconvex. There are many examples of hyperbolic
groups which satisfy this property, including (but not limited to): hyperbolic
surface groups, hyperbolic limit groups, certain one-relator groups with
torsion, certain small cancellation groups and graphs of free groups with
cyclic edge groups which do not contain any Baumslag-Solitar subgroups (see
Examples 6.4, 6.5, 6.6, and 6.7, and the references therein).

######  Corollary B (Corollary 6.3).

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

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

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.

Another class of hyperbolic groups of particular interest is that of
hyperbolic 3-manifold groups. While these are not locally quasiconvex, their
finitely generated subgroups are well understood. As a result, we are able to
describe their perfect kernels in their entirety.

######  Theorem C (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 $2$ or $3$.

The above theorem makes use of significant discoveries which have been made
about 3-manifolds over the past few decades. In particular, we use a
consequence of the Tameness Theorem [Ago04], [CG06] and Canary’s Covering
Theorem [Can96] as well as the fact that closed hyperbolic 3-manifolds are
virtually fibered [Ago13] (by virtue of being virtually special in the sense
of [HW08]).

In general, the inclusion
$\overline{\operatorname{Sub}_{QC[\infty]}(G)}\subseteq\operatorname{\mathcal{K}}(G)$
is not an equality. Indeed, Rips’ construction [Rip82] allows us to find a
hyperbolic group $G$ so that we can embed a $G$-invariant open subset $U$ into
the complement
$\operatorname{\mathcal{K}}(G)\smallsetminus\overline{\operatorname{Sub}_{QC[\infty]}(G)}$.
Moreover, one can ensure that the action of $G$ on $U$ is as pathological as
that of any finitely presented group on its perfect kernel. We discuss this in
more detail in Section 7.

#### Subgroup separability

A group $G$ is subgroup separable (also referred to as LERF or locally
extended residually finite) if every finitely generated subgroup of $G$ can be
expressed as an intersection of finite index subgroups. The class of groups
with this property includes free groups [Hal49] and surface groups [Sco78].

Subgroup separability is equivalent to the density in $\operatorname{Sub}(G)$
of the set of finite index subgroups, as observed in [GKM16] (see Section 4).
This does not provide any further information on the perfect kernel of a
group, except when we can relate the rank and the index of finite index
subgroups. For this purpose, we require that the first $\ell^{2}$-Betti number
be positive, we will see that in many cases this condition is in fact optimal
(see Remark 4.2).

######  Theorem D (Theorem 4.1).

Let $G$ be a countable and subgroup separable group. If $G$ is not finitely
generated then $\operatorname{Sub}(G)$ is a perfect space. If $G$ is finitely
generated with positive first $\ell^{2}$-Betti number then
$\mathcal{K}(G)=\operatorname{Sub}_{[\infty]}(G)$ and
$\operatorname{rk_{CB}}(G)=1$.

As mentioned above, free groups and surface groups are subgroup separable.
Moreover, hyperbolic surface groups have positive first $\ell^{2}$-Betti
number, as do non-abelian free groups. Thus the above theorem implies that the
perfect kernel of a finite rank non-abelian free group or a hyperbolic compact
surface group coincides with its set of infinite index subgroups, and its
Cantor-Bendixson rank is $1$. Note that the remaining (non-hyperbolic) compact
surface groups are virtually abelian so their perfect kernel is empty.

More generally, we have the following corollary for Sela’s limit groups.
Recall that a group $G$ is a limit group, or $\omega$-residually free, if for
any finite set $S\subseteq G\smallsetminus\\{\mathrm{id}\\}$ there exists a
normal subgroup $N\unlhd G$ such that the quotient $G/N$ is free and $S\cap
N=\emptyset$. Equivalently, limit groups are the limits of free groups in the
space of marked groups [CG05, Theorem 1.1].

######  Corollary E (Limit groups (Corollaries 4.4 and 5.23)).

If $G$ is a finitely generated non-abelian limit group, then
$\operatorname{\mathcal{K}}(G)=\operatorname{Sub}_{[\infty]}(G)$ and the
action $G\curvearrowright\operatorname{\mathcal{K}}(G)$ is highly
topologically transitive.

#### Groups with infinitely many ends and actions on trees

A fundamental notion in geometric group theory is the notion of ends of a
group. Recall that a finitely generated group has either zero, one, two or
infinitely many ends. Stallings’ Theorem [Sta68, Sta71] states that a group
has infinitely many ends if and only if it acts non-trivially and irreducibly
on a tree with finite edge stabilisers. This theorem is invaluable in the
study of infinitely ended groups and, in the context of spaces of subgroups,
it allows us to show the following.

######  Theorem F (Infinitely many ends (Corollary 5.17)).

If $G$ is a finitely generated group with infinitely many ends, then the
perfect kernel of $G$ consists of its subgroups with infinite index:

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

Moreover, if $G$ has no non-trivial finite normal subgroups then the action of
$G$ on $\operatorname{\mathcal{K}}(G)$ is highly topologically transitive.

It is also interesting to consider actions on trees where stabilisers of
individual edges are not necessarily finite or trivial but rather where the
stabilisers of sufficiently long paths are trivial. These actions are known as
acylindrical. Examples of groups admitting such actions include amalgamated
free products $A\ast_{C}B$ where $C$ is malnormal in $A$ or $B$ (See Section
5.3).

######  Corollary G (Acylindrical actions (see Corollary 5.15)).

Let $G\curvearrowright T$ be a minimal irreducible and acylindrical action of
a countable group on a tree. If $H\leq G$ is a subgroup such that the quotient
graph $H\backslash T$ is infinite then $H\in\operatorname{\mathcal{K}}(G)$.

An acylindrical action on a tree only provides information on those subgroups
whose quotient is infinite. We construct examples of finitely generated groups
$G$ which act acylindrically on trees and which contain infinite index
subgroups $H$ whose quotients are finite. Sometimes these subgroups $H$ are in
the perfect kernel and sometimes they are not (see Remark 5.16). More
information about the group $G$ is therefore required to describe the perfect
kernel fully.

The conclusion of Corollary G in fact holds under a weaker condition than
acylindricity of the action. Given a group $G$ acting on a tree $T$, define
the set

$\operatorname{Sub}_{|\bullet\backslash
T|\infty}(G):=\\{H\in\operatorname{Sub}(G)\colon\left|H\backslash
T\right|=\infty\\},$

of subgroups that act on $T$ with infinite quotient graph.

######  Theorem H (Theorems 5.9 and 5.11).

Assume $G\curvearrowright T$ is a minimal and irreducible action on a tree.
Assume that there are two edges $f_{1},f_{2}$ of $T$ such that
$\operatorname{Stab}_{G}(f_{1})\cap\operatorname{Stab}_{G}(f_{2})$ is finite.
Then

$\overline{\operatorname{Sub}_{|\bullet\backslash
T|\infty}(G)}\subseteq\operatorname{\mathcal{K}}(G).$

If, in addition, the kernel of the action $G\curvearrowright T$ is trivial
then the action of $G$ on $\overline{\operatorname{Sub}_{|\bullet\backslash
T|\infty}(G)}$ is highly topologically transitive.

This result allows us to establish that
$\operatorname{\mathcal{K}}(G)=\operatorname{Sub}_{[\infty]}(G)$ for a large
family of groups (namely the class $\mathcal{Q}$, see Definition 5.21), which
includes graphs of free groups with cyclic edge groups such that at least one
vertex group is non-cyclic. It also allows us to deduce the topological
transitivity properties of the action
$G\curvearrowright\operatorname{\mathcal{K}}(G)$ (see Corollary 5.23).

The reader will probably have noticed that our results above concerning the
dynamics of the action $G\curvearrowright\operatorname{\mathcal{K}}(G)$
exclude groups $G$ that contain a non-trivial finite normal subgroup $N_{0}$.
They may have wondered what happens in this case. In fact, the action of $G$
on $\operatorname{Sub}(N_{0})$ by conjugation induces a finite $G$-invariant
clopen partition of $\operatorname{Sub}(G)$ and of
$\operatorname{\mathcal{K}}(G)$ (see Proposition 3.2). When we allow the
existence of such an $N_{0}$, under the conditions of Theorems A and H, the
action on each piece of this clopen partition is topologically transitive (see
Theorem 6.2 and Theorem 5.11). However, in general the action on the pieces is
not topologically $2$-transitive, see Remark 5.13.

#### An application to transitive amenable actions

As a by-product of our investigations, we obtain information on transitive
actions. Following Glasner-Monod [GM07], let $\mathcal{A}$ be the class of
countable infinite groups that admit a faithful transitive and amenable
action.

This class contains all free groups and most free products (see [GM07]),
surface groups and hyperbolic $3$-manifolds groups (see [Moo10]) and some
instances of amalgamated free products and HNN-extension over finite or
amenable groups (see [Moo11a, Moo11b, Fim14]).

Our results allow us to greatly enlarge the collection of examples of groups
in the class $\mathcal{A}$. This is done by singling out two subgroups
$H_{1},H_{2}\in\operatorname{Sub}_{[\infty]}(G)$ such that $G\curvearrowright
G/H_{1}$ is faithful while $G\curvearrowright G/H_{2}$ is amenable (thus
isolating two separate conditions: faithfulness and co-amenability) and by
finding (thanks to topological transitivity) subgroups whose orbit in
$\operatorname{Sub}(G)$ accumulates on both $H_{1}$ and $H_{2}$ (see
Proposition 8.1).

Our Corollary 5.17 (on groups with infinitely many ends) admits as a corollary
Theorem 3.4 of [GM07]. The following generalisation was suggested to us by
François Le Maître, following the circulation of a previous version of this
paper.

######  Corollary I (of Proposition 8.1).

Assume that the action $G\curvearrowright\operatorname{Sub}_{[\infty]}(G)$ is
topologically transitive. Assume that $G$ admits an amenable action with only
infinite orbits. Then $G\in\mathcal{A}$.

Here is a (non-exhaustive) list of groups which we can prove belong to the
class $\mathcal{A}$.

  * •

Any infinitely ended group without a non-trivial finite normal subgroup and
with some amenable action with infinite orbits (Corollary 5.17 and Proposition
8.1). For instance, any HNN-extension $G$ of any infinite group $H$ above a
pair of finite subgroups with trivial intersection.

  * •

The fundamental group of any non-tree graph of groups where some edge group is
malnormal in an incident vertex group (See Corollary 8.4).

  * •

Virtually compact special groups (see Theorem 8.6). In particular:
– Right-angled Artin groups;
– C’(1/6) small cancellation groups;
– Right-angled and hyperbolic Coxeter groups;
– One-relator groups with torsion;
– Limit groups;
– Random groups at density $d<1/6$;
– Hyperbolic-by-cyclic groups which are themselves hyperbolic.

  * •

Groups in the class $\mathcal{Q}^{\prime}$ (Definition 5.21) that admit an
infinite index co-amenable subgroup (for instance if the defining graph of
groups is not a tree) (Corollary 5.23 and Proposition 8.1). In particular,
finite graphs of free groups with cyclic edge groups such that at least one
vertex group is non-cyclic (Corollary 8.2).

#### Acknowledgments

We are extremely grateful to Mark Hagen for his helpful comments, for many
enlightening discussions on hyperbolic groups and for asking us about the
space of subgroups of hyperbolic 3-manifold groups. The first author is
particularly grateful for his ongoing support.
We would like to thank Ian Biringer for pointing out an example of a
hyperbolic 3-manifold group with Cantor-Bendixson rank 3 and for a discussion
which motivated us to include and improve upon the results of Section 3.3.
We thank Alessandro Carderi, François Le Maître and Yves Stalder for many
enlightening discussions and for drawing our attention to the example of the
group $G$ in Example 3.7.
We also thank François Dahmani for several discussions on hyperbolic groups.
We are grateful to Eli Glasner for asking us whether our previous results
about topologically transitive actions could be upgraded to weakly mixing
actions and for giving us a few basics on the subject. This led us to consider
highly topologically transitive actions. This question was asked during the
wonderful conference “Measured Group Theory, Stochastic Processes on Groups
and Borel Combinatorics” which took place at the CIRM in May 2023.
We are grateful to Matthias Uschold for his comments on a previous version of
this paper.
P. A. is supported by a University of Bristol PhD scholarship. She was
supported by the LABEX MILYON (ANR-10-LABX-0070) of Université de Lyon, within
the program “Investissements d’Avenir” (ANR-11-IDEX-0007) operated by the
French National Research Agency (ANR) during the academic year 2021-2022, when
we initiated this project.
D. G. is supported by the C.N.R.S.

##  2 Notation and preliminaries

Let $G$ be a countable group. The space of subgroups $\operatorname{Sub}(G)$
of $G$ is a subspace of the set $\\{0,1\\}^{G}$, where each subset of $G$ is
identified with an element of $\\{0,1\\}^{G}$ via the characteristic map. The
Chabauty topology on $\operatorname{Sub}(G)$ is the subspace topology induced
by the product topology on $\\{0,1\\}^{G}$. It follows that
$\operatorname{Sub}(G)$ is compact, metrisable, separable and totally
disconnected. In particular, it satisfies the Baire property and it is a
Polish space. The collection of subsets

$\mathcal{V}_{G}(\mathcal{I},\mathcal{O}):=\\{H\leq
G\colon\mathcal{I}\subseteq
H,H\cap\mathcal{O}=\emptyset\\}\subseteq\operatorname{Sub}(G)$

for all finite subsets $\mathcal{I},\mathcal{O}\subseteq G$ forms a basis of
clopen sets for the Chabauty topology on $\operatorname{Sub}(G)$. When there
is no risk of confusion we will write $\mathcal{V}(\mathcal{I},\mathcal{O})$
instead of $\mathcal{V}_{G}(\mathcal{I},\mathcal{O})$. The action of $G$ by
conjugation ($g\cdot H=gHg^{-1}$) on $\operatorname{Sub}(G)$ is continuous.

#### The Cantor-Bendixson decomposition

Let $X$ be a Polish space. The Cantor-Bendixson derivatives of $X$ are defined
by transfinite induction. The derived set of $X$ is the complement
$X^{\prime}:=X\smallsetminus\operatorname{Is}(X)$ of the set
$\operatorname{Is}(X)$ of isolated points of $X$. Let $X^{(0)}=X$. The Cantor-
Bendixson derivatives of $X$ are defined by repeatedly applying the derived
set operation using transfinite recursion: Given an ordinal $\alpha$ such that
the $\alpha$-th Cantor-Bendixson derivative $X^{(\alpha)}$ is defined, let
$X^{(\alpha+1)}:=\left(X^{(\alpha)}\right)^{\prime}$. If $\beta$ is a limit
ordinal such that $X^{(\alpha)}$ is defined for each $\alpha<\beta$, then let
$X^{(\beta)}:=\cap_{\alpha<\beta}X^{(\alpha)}$. The Cantor-Bendixson theorem
[Can84, Ben00] implies that there is an, at most countable, ordinal $\alpha$
such that $X^{(\alpha)}=X^{(\alpha+1)}$. In this case, $X^{(\alpha)}$ is the
(unique) maximal perfect subspace of $X$, called the perfect kernel of $X$; we
denote it $\operatorname{\mathcal{K}}(X):=X^{(\alpha)}$. The minimal ordinal
$\alpha$ such that $X^{(\alpha)}=\operatorname{\mathcal{K}}(X)$ is called the
Cantor-Bendixson rank of $X$ and is denoted by $\operatorname{rk_{CB}}(X)$. A
point $x\in X$ is a condensation point if every open neighbourhood of $x$ is
uncountable. It follows from the proof of the Cantor-Bendixson theorem (see
e.g. [Kec95] for an accessible account) that
$x\in\operatorname{\mathcal{K}}(X)$ if and only if $x$ is a condensation
point.

When $G$ is a countable group, we define

$\displaystyle\operatorname{\mathcal{K}}(G)$ $\displaystyle:=$
$\displaystyle\operatorname{\mathcal{K}}(\operatorname{Sub}(G))\text{ the
{perfect kernel of $G$}, and}$ $\displaystyle\operatorname{rk_{CB}}(G)$
$\displaystyle:=$
$\displaystyle\operatorname{rk_{CB}}(\operatorname{Sub}(G))\text{ the {Cantor-
Bendixson rank of $G$}.}$

Be careful not to confuse this with the Cantor-Bendixson rank $\mathrm{cb}(G)$
that some authors define for a group $G$ as the supremum of the ordinals
$\alpha$ such that $\\{\mathrm{id}\\}\in\mathcal{N(G)}^{(\alpha)}$ (where
$\mathcal{N}(G)$ denotes the space of normal subgroups of $G$) when such an
$\alpha$ exists and $\mathfrak{C}$ otherwise, i.e. when
$\\{\mathrm{id}\\}\in\operatorname{\mathcal{K}}(\mathcal{N}(G))$ (see for
instance [Cor11]).

Let $H\leq G$ be a subgroup. The Cantor-Bendixson erasing rank
$\operatorname{rk^{\times}_{CB}}(H;G)$ of $H$ in $G$ is, when
$H\notin\operatorname{\mathcal{K}}(G)$, the infimum of the ordinals $\alpha$
such that the derivative $\operatorname{Sub}(G)^{(\alpha)}$ does not contain
$H$; and we write $\operatorname{rk^{\times}_{CB}}(H;G)=\infty$ when
$H\in\operatorname{\mathcal{K}}(G)$. We make the convention that $\infty$ is
larger than all the ordinals. Observe that
$\operatorname{rk^{\times}_{CB}}(H;G)$ is not a limit ordinal:

$\operatorname{rk^{\times}_{CB}}(H;G)=\begin{cases}\infty&\text{ if
}H\in\operatorname{\mathcal{K}}(G)\\\ \alpha+1&\text{ if
}H\in\operatorname{Sub}(G)^{(\alpha)}\text{ but
}H\not\in\operatorname{Sub}(G)^{(\alpha+1)}.\end{cases}$

Moreover, $\operatorname{rk_{CB}}(G)=0$ if and only if $\operatorname{Sub}(G)$
is perfect and
$\operatorname{rk_{CB}}(G)=\sup\\{\operatorname{rk^{\times}_{CB}}(H;G)\colon
H\in\operatorname{Sub}(G)\smallsetminus\operatorname{\mathcal{K}}(G)\\}$
otherwise.

######  Example 2.1.

For instance, when $G=\mathbb{Z}^{d}$, then
$\operatorname{rk^{\times}_{CB}}(H;\mathbb{Z}^{d})=d-\operatorname{rk}(H)+1$
where $\operatorname{rk}(H)$ is the minimal number of generators of the
abelian torsion-free group $H\leq\mathbb{Z}^{d}$. In particular,
$\operatorname{rk^{\times}_{CB}}(\\{\mathrm{id}\\};\mathbb{Z}^{d})=d+1$,
$\operatorname{rk_{CB}}(\mathbb{Z}^{d})=d+1$, and
$\operatorname{\mathcal{K}}(\mathbb{Z}^{d})=\emptyset$.

#### Topological transitivity

A continuous action $G\curvearrowright X$ of a countable discrete group $G$ on
a compact metrizable space is called topologically transitive if it satisfies
one of the following equivalent conditions:

  1. 1.

For every pair $(V_{1},V_{2})$ of non-empty open subsets of $X$, there is an
element $g\in G$ such that $g\cdot V_{1}\cap V_{2}\not=\emptyset$;

  2. 2.

There is a dense orbit, i.e. a point $x\in X$ whose orbit $G\cdot x$ is dense
in $X$;

  3. 3.

There is a Gδ subset of $X$ consisting of dense orbits.

A continuous action $G\curvearrowright X$ on a compact metrizable space of a
countable group $G$ is called topologically $r$-transitive for a positive
integer $r$ if the diagonal action $G\curvearrowright X^{r}$ on the $r$-fold
space is topologically transitive; i.e. if for every $2r$-tuple
$(V_{1},V_{2},\cdots,V_{2r})$ of non-empty open sets, there is an element
$g\in G$ such that $g.V_{i}\cap V_{r+i}\not=\emptyset$ for all
$i=1,2,\cdots,r$. A topologically $2$-transitive action is also called
topologically weakly mixing. If an action is topologically $r$-transitive for
all $r\geq 1$ we say that it is highly topologically transitive.

Contrarily to the analogous concept in ergodic theory, topological
$2$-transitivity does not imply high topological transitivity when the group
$G$ is not abelian.

##  3 Generalities on the topological structure of $\operatorname{Sub}(G)$

In this section we highlight the effects of various properties of individual
subgroups, and of their quotients, on the Cantor-Bendixson decomposition of
the space of subgroups and the dynamics of the action by conjugation.

######  Remark 3.1.

If $H\leq G$ is a subgroup of the countable group $G$, then the natural
continuous embedding $\operatorname{Sub}(H)\subseteq\operatorname{Sub}(G)$
induces a continuous embedding
$\operatorname{\mathcal{K}}(H)\subseteq\operatorname{\mathcal{K}}(G)$. In
fact, the topology on $\operatorname{Sub}(H)$ is induced from that on
$\operatorname{Sub}(G)$. Condensation points of $\operatorname{Sub}(H)$ are
thus condensation points of $\operatorname{Sub}(G)$.

###  3.1 Normal subgroups and invariant subsets

A normal subgroup $N\trianglelefteq G$ is clearly a fixed point of the
$G$-action on $\operatorname{Sub}(G)$ by conjugation. In the case where $N$ is
finite, this property induces an invariant partition of the entire space.

######  Proposition 3.2 (Finite normal subgroup and its clopen partition).

If $N\trianglelefteq G$ is a finite normal subgroup $N$. The set
$\mathrm{Conj}(N)^{G}$ of $G$-conjugacy classes of subgroups of $N$ induces a
finite $G$-invariant clopen partition:

$\operatorname{Sub}(G)=\bigsqcup_{C\in\mathrm{Conj}(N)^{G}}\\{H\in\operatorname{Sub}(G)\colon
H\cap N\in C\\}.$

###### Proof.

By normality and finiteness, $G$ acts by conjugation on
$\operatorname{Sub}(N)$ with finite orbits. These orbits form the set
$\mathrm{Conj}(N)^{G}=\\{\\{\gamma M\gamma^{-1}\colon\gamma\in G\\}\colon
M\in\operatorname{Sub}(N)\\}$. Since $\gamma H\gamma^{-1}\cap N=\gamma(H\cap
N)\gamma^{-1}$, the $G$-orbit of $H\cap N\in\operatorname{Sub}(N)$ is an
invariant of the $G$-orbit of $H\in\operatorname{Sub}(G)$. In addition, for
$C\in\mathrm{Conj}(N)^{G}$, the condition $H\cap N\in C$ is a clopen condition
in $\operatorname{Sub}(G)$: it defines a union of basic clopen sets:
$\\{H\in\operatorname{Sub}(G)\colon H\cap N\in C\\}=\cup_{M\in
C}\mathcal{V}(M,N\smallsetminus M)$. ∎

If $G^{\prime}$ is a subgroup of the countable group $G$, then the map

$P:\left(\begin{array}[]{ccc}\operatorname{Sub}(G)&\longrightarrow&\operatorname{Sub}(G^{\prime})\\\
H&\longmapsto&H\cap G^{\prime}\end{array}\right)$ (1)

is continuous: if
$\mathcal{V}_{G^{\prime}}(\mathcal{I}^{\prime},\mathcal{O}^{\prime})$ is a
basic open set in $\operatorname{Sub}(G^{\prime})$ (with
$\mathcal{I}^{\prime},\mathcal{O}^{\prime}$ finite subsets of $G^{\prime}$),
then
$P^{-1}(\mathcal{V}_{G^{\prime}}(\mathcal{I}^{\prime},\mathcal{O}^{\prime}))=\mathcal{V}_{G}(\mathcal{I}^{\prime},\mathcal{O}^{\prime})$.

######  Proposition 3.3.

Let $G^{\prime}\leq G$ be a finite index subgroup and
$P:\operatorname{Sub}(G)\rightarrow\operatorname{Sub}(G^{\prime})$ be the
continuous map from (1). Then each fiber of $P$ is countable. It follows that
the natural embedding
$\operatorname{\mathcal{K}}(G^{\prime})\subseteq\operatorname{\mathcal{K}}(G)\cap\operatorname{Sub}(G^{\prime})$
is an equality:
$\operatorname{\mathcal{K}}(G^{\prime})=\operatorname{\mathcal{K}}(G)\cap\operatorname{Sub}(G^{\prime})$.

###### Proof of Proposition 3.3.

Fix a subgroup $G_{0}^{\prime}\leq G^{\prime}$. Let $k:=[G:G^{\prime}]$ and
let $g_{1},\dots,g_{k}\in G$ be such that $G=\sqcup_{i=1}^{k}g_{i}G^{\prime}$.
If $H\in P^{-1}(G_{0}^{\prime})$ let $I\subseteq\\{1,\dots k\\}$ be such that
$H\cap g_{i}G^{\prime}\neq\emptyset$ if and only if $i\in I$. Then, for each
$i\in I$, there is $h_{i}\in g_{i}G^{\prime}$ such that $H\cap
g_{i}G^{\prime}=h_{i}G_{0}^{\prime}$. Thus $H=\sqcup_{i\in
I}h_{i}G_{0}^{\prime}$. It follows that

$P^{-1}(G_{0}^{\prime})\subseteq\bigcup_{I\subseteq\\{1,\dots,k\\}}\Big{\\{}\bigsqcup_{i\in
I}h_{i}G_{0}^{\prime}:h_{i}\in g_{i}G^{\prime}\Big{\\}}.$

Thus $P^{-1}(G_{0}^{\prime})$ is contained in a finite union of countable sets
and is therefore countable.

To see that
$\operatorname{\mathcal{K}}(G)\cap\operatorname{Sub}({G^{\prime}})\subseteq\operatorname{\mathcal{K}}({G^{\prime}})$,
let
$H^{\prime}\in\operatorname{Sub}(G^{\prime})\smallsetminus\operatorname{\mathcal{K}}(G^{\prime})$.
Pick a countable neighbourhood $\mathcal{V}^{\prime}$ of $H^{\prime}$ in
$\operatorname{Sub}(G^{\prime})$. Its pre-image $P^{-1}(\mathcal{V}^{\prime})$
is a countable neighbourhood of $H^{\prime}$ in $\operatorname{Sub}(G)$ so
$H^{\prime}\notin\operatorname{\mathcal{K}}(G)$. ∎

######  Proposition 3.4.

Let $N\trianglelefteq G$ be a finite normal subgroup of the countable group
$G$. Assume $G$ contains a torsion-free subgroup $H$ such that
$\\{\mathrm{id}\\}\in\operatorname{\mathcal{K}}(H)$. Then
$\operatorname{Sub}(N)\subseteq\operatorname{\mathcal{K}}(G)$.

###### Proof.

The subgroup $H$ acting by conjugation on the finite subgroup $N$ admits a
finite index subgroup $H^{\prime}$ that acts trivially; i.e. $H^{\prime}$
commutes with $N$, and since it is torsion-free, it intersects $N$ trivially,
thus $K=H^{\prime}\times N$ injects naturally as a subgroup of $G$. If
$M=\\{\mathrm{id}\\}\times M$ is a subgroup of $N=\\{\mathrm{id}\\}\times N$
and $\mathcal{V}=\mathcal{V}_{K}(M,\mathcal{O})$ is any basic neighbourhood of
$M$ in $\operatorname{Sub}(K)$, with
$\mathcal{O}=\\{(x_{1},n_{1})(x_{2},n_{2}),\cdots,(x_{j},n_{j})\\}$ where
$x_{i}\in H^{\prime}$ and $n_{i}\in N$, then
$\mathcal{V}^{\prime}:=\mathcal{V}_{H^{\prime}}(\emptyset,\\{x_{1},x_{2},\cdots,x_{j}\\})$
is a basic neighbourhood of $\\{\mathrm{id}\\}$ in
$\operatorname{Sub}(H^{\prime})$. By Proposition 3.3,
$\\{\mathrm{id}\\}\in\operatorname{\mathcal{K}}(H^{\prime})$. Thus
$\mathcal{V}^{\prime}$ is uncountable and the uncountable set
$\\{H^{\prime}\times M\colon H^{\prime}\in\mathcal{V}^{\prime}\\}$ is
contained in $\mathcal{V}$. It follows that
$M\in\operatorname{\mathcal{K}}(K)$, and so
$M\in\operatorname{\mathcal{K}}(G)$ by Remark 3.1. ∎

Let $H\leq G$ be a subgroup. We denote by

$\operatorname{Sub}_{H\leq}(G)\\!:=\\{L\leq G\colon H\leq L\\}$

the closed subspace of $\operatorname{Sub}(G)$ consisting of subgroups that
contain $H$. In the case where $H$ is generated by a finite set $S$, then
$\operatorname{Sub}_{H\leq}(G)=\mathcal{V}(S,\emptyset)$ is a clopen
neighbourhood of $H$ in $\operatorname{Sub}(G)$. If $H$ is normal, then the
action of $G$ on $\operatorname{Sub}(G)$ by conjugation induces natural
actions on the spaces of subgroups
$G\curvearrowright\operatorname{Sub}_{H\leq}(G)$ and
$G\curvearrowright\operatorname{Sub}(G/H)$.

######  Proposition 3.5.

Let $1\to N\to G\overset{\phi}{\to}Q\to 1$ be a short exact sequence.

  1. 1.

Then the bijection

$q:\left(\begin{array}[]{ccc}\operatorname{Sub}_{N\leq}(G)&\to&\operatorname{Sub}(Q)\\\
H&\mapsto&H/N\end{array}\right)$

is a $G$-equivariant homeomorphism with respect to the subspace topology on
$\operatorname{Sub}_{N\leq}(G)$. In particular,

$q^{-1}(\operatorname{\mathcal{K}}(Q))\subseteq\operatorname{\mathcal{K}}(G).$
(2)

  2. 2.

Assume that $N$ is finitely generated. Then $\operatorname{Sub}_{N\leq}(G)$ is
a clopen neighbourhood of $N$ in $\operatorname{Sub}(G)$. It follows that the
restriction

$q_{\restriction}:\operatorname{\mathcal{K}}(G)\cap\operatorname{Sub}_{N\leq}(G)\longrightarrow\operatorname{\mathcal{K}}(Q)$
(3)

is a $G$-equivariant homeomorphism.

If $H\in\operatorname{Sub}_{N\leq}(G)$, then the Cantor-Bendixson erasing
ranks of $H$ in $G$ and of $q(H)=H/N$ in $Q$ are the same:

$\operatorname{rk^{\times}_{CB}}(H;G)=\operatorname{rk^{\times}_{CB}}(H/N;Q).$
(4)

In particular, if $\operatorname{Sub}(Q)$ is countable, and if
$H\in\operatorname{Sub}_{N\leq}(G)$, then

$\operatorname{rk^{\times}_{CB}}(H;G)\leq\operatorname{rk_{CB}}(Q).$ (5)

###### Proof.

Let us show that the bijection
$q:\operatorname{Sub}_{N\leq}(G)\to\operatorname{Sub}(Q)$ is continuous. If
$\mathcal{V}_{Q}(\mathcal{I},\mathcal{O})$ is a basic clopen set in
$\operatorname{Sub}(Q)$, let $\mathcal{I}^{\prime}$ and $\mathcal{O}^{\prime}$
be any (bijective!) $\phi$-sections in $G$ of the finite sets $\mathcal{I}$
and $\mathcal{O}$. Then
$q^{-1}\left(\mathcal{V}_{Q}(\mathcal{I},\mathcal{O})\right)=\mathcal{V}_{G}(\mathcal{I}^{\prime},\mathcal{O}^{\prime})\cap\operatorname{Sub}_{N\leq}(G)$.
To see that (2) holds, recall that if $H\in
q^{-1}(\operatorname{\mathcal{K}}(Q))$ then every open neighbourhood of $q(H)$
in $\operatorname{Sub}(Q)$ is uncountable. Since $q$ is a homeomorphism, it
follows that every open neighbourhood of $H$ in
$\operatorname{Sub}_{N\leq}(G)$ is uncountable and thus so is every open
neighbourhood of $H$ in $\operatorname{Sub}(G)$. Hence
$H\in\operatorname{\mathcal{K}}(G)$.

If $N$ is generated by the finite set $S$, then
$\operatorname{Sub}_{N\leq}(G)$ coincides with the basic clopen set
$\mathcal{V}_{G}(S,\emptyset)$.

If $Y$ is an open subset of a Polish space $X$ then a point $y\in Y$ is
isolated with respect to the subspace topology on $Y$ if and only if $y$ is
isolated in $X$. It follows by transfinite induction that, for every $y\in Y$
and every ordinal $\alpha$, we have $y\in Y^{(\alpha)}$ if and only if $y\in
X^{(\alpha)}$. Thus $Y^{(\alpha)}=Y\cap X^{(\alpha)}$. Again by transfinite
induction, one can show that, if $f:X\rightarrow Y$ is a homeomorphism between
Polish spaces $X,Y$, then $f(X^{(\alpha)})=Y^{(\alpha)}$ for all $\alpha$. In
particular, $f(\operatorname{\mathcal{K}}(X))=\operatorname{\mathcal{K}}(Y)$
and the Cantor-Bendixson erasing rank is invariant under homeomorphism.

Thus
$q(\operatorname{\mathcal{K}}(G)\cap\operatorname{Sub}_{N\leq}(G))=q(\operatorname{\mathcal{K}}(\operatorname{Sub}_{N\leq}(G)))=\operatorname{\mathcal{K}}(Q)$
and the restricted homeomorphism (3) is well-defined. The equality (4) also
follows since, for $H\in\operatorname{Sub}_{N\leq G}(G)$,

$\operatorname{rk^{\times}_{CB}}(H;G)=\infty\Leftrightarrow
H\in\operatorname{\mathcal{K}}(G)\cap\operatorname{Sub}_{N\leq
G}(G)\Leftrightarrow
q(H)\in\operatorname{\mathcal{K}}(Q)\Leftrightarrow\operatorname{rk^{\times}_{CB}}(H/N;Q)=\infty$

and, if $H\notin\operatorname{\mathcal{K}}(G)$ then
$H/N\notin\operatorname{\mathcal{K}}(Q)$ so

$\displaystyle\operatorname{rk^{\times}_{CB}}(H;G)$
$\displaystyle=\inf\left\\{\alpha\geq
1:H\notin\operatorname{Sub}(G)^{(\alpha)}\right\\}$
$\displaystyle=\inf\left\\{\alpha\geq
1:H\notin\operatorname{Sub}_{N\leq}(G)^{(\alpha)}\right\\}$
$\displaystyle=\inf\left\\{\alpha\geq
1:q(H)\notin\operatorname{Sub}(Q)^{(\alpha)}\right\\}$
$\displaystyle=\operatorname{rk^{\times}_{CB}}(H/N;Q).$

If $\operatorname{Sub}(Q)$ is countable, then
$\operatorname{\mathcal{K}}(Q)=\emptyset$ and then
$\operatorname{rk_{CB}}(Q)=\sup\left\\{\operatorname{rk^{\times}_{CB}}(L;Q)\colon
L\in\operatorname{Sub}(Q)\right\\}$. This proves the inequality (5). ∎

Propositions 3.3 and 3.5 imply that virtual algebraic fibering allows one to
find finitely generated, infinite index subgroups which do not belong to the
perfect kernel. More generally:

######  Corollary 3.6.

Let $G$ be a countable group and $H\leq G$ be a finitely generated subgroup.
Suppose that there exists a finite index subgroup $G^{\prime}\leq G$ and a
finitely generated subgroup $H^{\prime}\leq H\cap G^{\prime}$ such that there
is a short exact sequence

${1}$${H^{\prime}}$${G^{\prime}}$${Q}$${1}$

where $\operatorname{Sub}(Q)$ is countable. Then
$H\notin\operatorname{\mathcal{K}}(G)$.

###### Proof.

By Proposition 3.5, $\operatorname{Sub}_{H^{\prime}\leq}(G^{\prime})$ is a
countable neighbourhood of $H^{\prime}$. Let $P:G\rightarrow G^{\prime}$ be
defined by $P(L)=L\cap G^{\prime}$. Then, by Proposition 3.3,
$P^{-1}(\operatorname{Sub}_{H^{\prime}\leq}(G^{\prime}))$ is a countable
neighbourhood of $H$. ∎

###  3.2 Examples

Note that the existence of a short exact sequence with finitely generated
kernel does not in general provide any information about subgroups which do
not contain the kernel. In fact, there exist pairs $N\trianglelefteq G$ with
$N$ finite and normal in the countable group $G$ such that
$\\{\mathrm{id}\\}\in\operatorname{\mathcal{K}}(G/N)$, yet the subgroup
$\\{\mathrm{id}\\}$ is isolated in $\operatorname{Sub}(G)$.

######  Example 3.7 (Example with $\\{\mathrm{id}\\}$ isolated).

In order to allow for the possibility that $\\{\mathrm{id}\\}$ be isolated, it
is necessary that $G$ contains no infinite order element. A very interesting
example is a central extension by $\mathbb{Z}/p\mathbb{Z}$ of the Burnside
group $B(m,p):=\langle h_{1},h_{2},\cdots,h_{m}|g^{p}=1,\forall g\rangle$. Let
$m,p\in\mathbb{N}$, $m\geq 2$ and $p$ a prime number $\geq 665$. There is a
central extension

${1}$${\mathbb{Z}/p\mathbb{Z}}$${G}$${B(m,p)}$${1}$

in which all the elements have order $p^{2}$ except the elements of the center
$\mathbb{Z}/p\mathbb{Z}$. Indeed, first take the Adian group $A(m,p)$ – this
is a torsion free $m$-generated group whose center is an infinite cyclic group
$\langle c\rangle$ and with quotient $A(m,p)/\langle c\rangle\simeq B(m,p)$ –
and set $G=A(m,p)/\langle c^{p}\rangle$, see [Zel11, Example 1.25]. The $p$-th
power of every element belongs to the center. It follows that every non-
trivial subgroup contains the center. The trivial subgroup is thus isolated in
$\operatorname{Sub}(G)$.

On the other hand, by a theorem of Širvanjan [Šir76], when $m\geq 2$ and $p$
is larger than $665$ and odd, $B(m,p)$ contains a copy of the countably
infinite rank free Burnside group $B(\mathbb{N},p)=\langle
s_{i},i\in\mathbb{N}|g^{p}=1\text{ for all }g\rangle$. Since
$\\{\mathrm{id}\\}\in\operatorname{\mathcal{K}}(B(\mathbb{N},p))=\operatorname{Sub}(B(\mathbb{N},p))$,
as shown in Proposition 3.8 below, then
$\\{\mathrm{id}\\}\in\operatorname{\mathcal{K}}(B(m,p))$. By Proposition 3.5,
$\mathbb{Z}/p\mathbb{Z}\in\operatorname{\mathcal{K}}(G)$.

Recall that the $\mathfrak{R}$-free group with generators indexed by $I$ of a
variety $\mathfrak{R}$ of groups is the group
$\mathfrak{F}^{\mathfrak{R}}_{I}=\langle s_{i},i\in I\ |\ \mathfrak{R}\rangle$
where $\mathfrak{R}$ represents a collection of relators of the form
$R(g_{1},g_{2},\cdots,g_{p})=1$ for all $p$-tuples
$(g_{1},g_{2},\cdots,g_{p})$ of elements of the (usual) free group on
$(s_{i})_{i\in I}$. Examples of $\mathfrak{R}$-free group
$\mathfrak{F}^{\mathfrak{R}}_{\mathbb{N}}$ of countably infinite rank include
the free group $\mathbf{F}_{\infty}$, the free abelian group
$\oplus_{\mathbb{N}}\mathbb{Z}$, the free abelian group
$\oplus_{\mathbb{N}}\mathbb{Z}/m\mathbb{Z}$ of exponent $m$, the free
nilpotent group of class $c$, the free Burnside group $B(\mathbb{N},p)=\langle
s_{i},i\in\mathbb{N}|g^{p}=1,\text{ for all }g\rangle$,… By non-trivial
variety, we simply mean that $\mathfrak{F}^{\mathfrak{R}}_{\mathbb{N}}$ is not
finitely generated.

######  Proposition 3.8 (Free group of a variety).

Let $\mathfrak{F}^{\mathfrak{R}}_{\mathbb{N}}=\langle
s_{1},s_{2},\cdots|\mathfrak{R}\rangle$ be the $\mathfrak{R}$-free group of
countably infinite rank of a non-trivial variety $\mathfrak{R}$ of groups.
Then
$\operatorname{\mathcal{K}}(\mathfrak{F}^{\mathfrak{R}}_{\mathbb{N}})=\operatorname{Sub}(\mathfrak{F}^{\mathfrak{R}}_{\mathbb{N}})$.

###### Proof.

For $I\subseteq\mathbb{N}$, consider the subgroup $H_{I}:=\langle s_{i}\colon
i\in I\rangle\leq\mathfrak{F}^{\mathfrak{R}}_{\mathbb{N}}$, and let
$\mathfrak{F}^{\mathfrak{R}}_{I}:=\langle t_{i}\colon i\in
I|\mathfrak{R}\rangle$ be the $\mathfrak{R}$-free group with generators
indexed by $I$.

Consider the natural morphisms, well-defined by the universal property of the
$\mathfrak{R}$-free groups,
$\mathfrak{F}^{\mathfrak{R}}_{\mathbb{N}}\overset{u}{\to}\mathfrak{F}^{\mathfrak{R}}_{I}\overset{v}{\to}\mathfrak{F}^{\mathfrak{R}}_{\mathbb{N}}$
defined by $u(s_{i})=t_{i}$ if $i\in I$ and $u(s_{i})=\mathrm{id}$ otherwise,
and by $v(t_{i})=s_{i}$.

Since $v$ is the inverse of $u$ when restricted to $H_{I}$, this ensures that
$H_{I}\simeq\mathfrak{F}^{\mathfrak{R}}_{I}$ and $w_{I}:=v\circ u$ is a
retraction $\mathfrak{F}^{\mathfrak{R}}_{\mathbb{N}}\to H_{I}$ (i.e. it is the
identity on $H_{I}$). Similarly, when $I\subseteq J$, the obvious morphism
$w_{I,J}:H_{J}\overset{}{\to}H_{I}$ is a retraction.

Let $L$ be a finitely generated subgroup of
$\mathfrak{F}^{\mathfrak{R}}_{\mathbb{N}}$. Then, there is a finite subset
$I\subset\mathbb{N}$ such that $L\leq H_{I}$. For every
$i\in\mathbb{N}\smallsetminus I$, let $L_{i}=\langle L,s_{i}\rangle\leq
H_{I\sqcup\\{i\\}}$. Observe, by non-triviality of the variety, that
$s_{i}\not\in H_{I}$ and thus $L_{i}\not=L$. We claim that the sequence
$(L_{i})_{i}$ tends to $L$ when $i$ tends to $\infty$: If $g\in L_{i}$, then
write it as a word $m$ using elements of $L$ and $s_{i}$. When $j>i$, if $g$
continues to belong to $L_{j}$ then under the retraction
$H_{I\sqcup\\{i,j\\}}\to H_{I\sqcup\\{j\\}}$, we see that the word $m$
continues to represent $g$ even after the erasing of the letters $s_{i}$. Thus
$g$ can be written as a word using only elements of $L$, i.e. $g\in L$. It
follows that $L$ is a non-trivial limit of finitely generated subgroups of
$\mathfrak{F}^{\mathfrak{R}}_{\mathbb{N}}$ and thus all finitely generated
subgroups belong to
$\operatorname{\mathcal{K}}(\mathfrak{F}^{\mathfrak{R}}_{\mathbb{N}})$.
Now, if $L$ is a non finitely generated subgroup of
$\mathfrak{F}^{\mathfrak{R}}_{\mathbb{N}}$, then it is a limit of finitely
generated groups. We have proved that
$\operatorname{\mathcal{K}}(\mathfrak{F}^{\mathfrak{R}}_{\mathbb{N}})=\operatorname{Sub}(\mathfrak{F}^{\mathfrak{R}}_{\mathbb{N}})$.
∎

When $\mathfrak{F}^{\mathfrak{R}}_{\mathbb{N}}$ is abelian, its action on
$\operatorname{Sub}(\mathfrak{F}^{\mathfrak{R}}_{\mathbb{N}})$ is trivial,
thus not topologically transitive, while
$\mathbf{F}_{\infty}\curvearrowright\operatorname{\mathcal{K}}(\mathbf{F}_{\infty})$
is highly topologically transitive (see Corollary 5.19).

######  Question 3.9 (Free Burnside groups and dynamics).

For the free Burnside groups, $B(\mathbb{N},p)$ of countably infinite rank and
$B(m,p)$ of rank $m$:

  1. 1.

Is the action
$B(\mathbb{N},p)\curvearrowright\operatorname{Sub}(B(\mathbb{N},p))$
topologically $r$-transitive for some $r$?

  2. 2.

For $m\geq 2$ and $p$ such that $B(m,p)$ is infinite, what is perfect kernel
$\operatorname{\mathcal{K}}(B(m,p))$? How transitive is the corresponding
action of $B(m,p)$?

###  3.3 Subgroups with Schreier graph quasi-isometric to $\mathbb{Z}$

We now consider subgroups which are not necessarily normal but whose
“quotients” are nonetheless equipped with a strong geometric structure: namely
those with Schreier graph quasi-isometric to $\mathbb{Z}$.

If $G$ is a finitely generated group and $H$ is a subgroup, we say that
$H\backslash G$ is quasi-isometric to $\mathbb{Z}$ if for a (equivalently any)
Cayley graph $X$ of $G$ associated with a finite generating set, the Schreier
graph $H\backslash X$ is quasi-isometric with $\mathbb{Z}$.

######  Theorem 3.10.

If $H\leq G$ be a pair of finitely generated groups such that $H\backslash G$
is quasi-isometric to $\mathbb{Z}$, then the Cantor-Bendixson erasing rank of
$H$ satisfies $\operatorname{rk^{\times}_{CB}}(H;G)\leq 3$.

Note that the above theorem does not hold if one assumes only that
$H\backslash G$ is 2-ended. Indeed, if $G$ is a hyperbolic surface group and
$H\leq G$ is a cyclic subgroup then $H\backslash G$ is 2-ended but
$H\in\operatorname{\mathcal{K}}(G)$. It also does not hold if one removes the
assumption the $H$ is finitely generated: if $G$ is a non-abelian free group
and $H$ is the kernel of a morphism onto $\mathbb{Z}$, then
$H\in\operatorname{\mathcal{K}}(G)$. We are however working on extending this
result to finitely generated subgroups $H$ such that $H\backslash G$ is quasi-
isometric to $\mathbb{Z}^{d}$ for $d\geq 2$.

######  Example 3.11.

The trivial group $\\{\mathrm{id}\\}$ in $\mathbb{Z}$ and in the infinite
dihedral group $D_{\infty}$ furnishes examples of Cantor-Bendixson erasing
rank $2$ and $3$, respectively. This will be useful in Section 6.3.

Let $G$ be a group with finite generating set $S$. Let
$X:=\operatorname{Cay}(G,S)$ be the associated Cayley graph. The ball of
radius $R$ centered at $g$ in $\operatorname{Cay}(G,S)$ is denoted by
$B_{X}(g,R)$.

For every tower of groups $H\leq K\leq G$, we shall consider the corresponding
covering maps

${X\ }$${H\backslash X}$${K\backslash X}$${G\backslash
X.}$$\scriptstyle{p_{H}}$$\scriptstyle{p_{K,H}}$$\scriptstyle{p_{G,K}}$

The idea of the proof is as follows. Given $H\leq G$ such that $H\backslash X$
is quasi-isometric to $\mathbb{Z}$, we will show that, for all but finitely
many intermediate subgroups $H\leq K\leq G$, either $K$ has finite index in
$G$ or the covering map $p_{K,H}$ folds the space $H\backslash X$ over,
meaning that $K\backslash X$ is quasi-isometric to a half-line $\mathbb{N}$.
In the latter case, we show that all but finitely many intermediate subgroups
$K\leq L\leq G$ have finite index in $G$.

######  Proposition 3.12.

Let $H\leq G=\langle S\rangle$ be any subgroup. Let $x_{0}\in H\backslash X$
and let $D>0$ be a positive integer. There are at most
$2^{|B_{X}(\mathrm{id},D)|}$ intermediate subgroups $H\leq K\leq G$ such that
the fiber $p_{K,H}^{-1}(p_{K,H}(x_{0}))$ of $p_{K,H}:H\backslash X\to
K\backslash X$ has diameter $\leq D$.

###### Proof.

Consider any such $K$. The finite set
$p_{K,H}^{-1}(p_{K,H}(x_{0}))=\\{x_{0},x_{1},x_{2},\cdots,x_{n}\\}$ is
contained in the ball $B(x_{0},D)$ of $H\backslash X$. Let $\tilde{x}_{0}\in
X$ be any pre-image of $x_{0}\in H\backslash X$ under the covering map
$p_{H}:X\to H\backslash X$. By lifting under $p_{H}$ some geodesic paths from
$x_{0}$ to $x_{i}$ to geodesic paths from $\tilde{x}_{0}$ to $\tilde{x}_{i}$,
one obtains a family of points
$\\{\tilde{x}_{0},\tilde{x}_{1},\tilde{x}_{2},\cdots,\tilde{x}_{n}\\}$ such
that $p_{H}(\tilde{x}_{i})=x_{i}$. Each $\tilde{x}_{i}$ belongs to the ball
$B_{X}(\tilde{x}_{0},D)$ of $X$.

Let $g_{0}=\mathrm{id},g_{1},g_{2},\cdots,g_{n}\in G$ such that
$g_{i}\tilde{x}_{0}=\tilde{x}_{i}$. We claim that $K=\langle
H,g_{1},g_{2},\cdots,g_{n}\rangle$.
In fact, $p_{K}(g_{i}\tilde{x}_{0})=p_{K}(\tilde{x}_{0})$ implies that
$g_{i}\in K$. Thus $K\geq\langle H,g_{1},g_{2},\cdots,g_{n}\rangle$.
Since $p_{K}=p_{K,H}\circ p_{H}$, it follows that

$K\cdot\tilde{x}_{0}=H\cdot\\{\tilde{x}_{0},\tilde{x}_{1},\tilde{x}_{2},\cdots,\tilde{x}_{n}\\}=\bigcup_{i=0}^{n}H\cdot
g_{i}(\tilde{x}_{0})=(\bigcup_{i=0}^{n}Hg_{i})\cdot\tilde{x}_{0}.$

Since the action $G\curvearrowright X$ is free, it follows that $K\leq\langle
H,g_{1},g_{2},\cdots,g_{n}\rangle$.

Now let $A_{D}:=\\{\gamma_{1},\gamma_{2},\cdots,\gamma_{m}\\}$ be the set of
elements of $G$ such that
$B_{X}(\tilde{x}_{0},D)=\\{\tilde{x}_{0}\\}\cup\bigcup_{i=1}^{m}\gamma_{i}(\tilde{x}_{0})\subseteq
X$. If $K\leq G$ is a subgroup containing $H$ such that the $p_{K,H}$-fiber of
$p_{K,H}(x_{0})$ has diameter $\leq D$, then $K=\langle H,A_{K}\rangle$ for
some subset $A_{K}\subseteq A_{D}$. The number of such subgroups $K$ is thus
bounded by the cardinality $2^{|B_{X}(\mathrm{id},D)|}$ of the power set of
$B_{X}(\mathrm{id},D)$. ∎

######  Lemma 3.13.

Consider subgroups $H\leq K\leq G$. Let $\bar{x}\in K\backslash X$ and
$x_{1},x_{2}\in H\backslash X$ such that
$p_{K,H}(x_{1})=p_{K,H}(x_{2})=\bar{x}$. Let $\bar{\delta}$ be a geodesic path
in $K\backslash X$ starting at $\bar{x}$. Its pull-back $\delta_{1}$ in
$H\backslash X$ starting at $x_{1}$ is again a geodesic. If $\delta_{1}$ meets
the ball $B_{H\backslash X}(x_{2},r)$ then $d(x_{1},x_{2})\leq 2r$.

###### Proof.

Let $z_{2}\in\delta_{1}\cap B_{H\backslash X}(x_{2},r)$. Then, since
(successively) $\delta_{1},\bar{\delta}$ are geodesics,
$p_{K,H}(x_{1})=p_{K,H}(x_{2})$ and $p_{K,H}$ does not increase distances, we
have

$d(x_{1},z_{2})=d(p_{K,H}(x_{1}),p_{K,H}(z_{2}))=d(p_{K,H}(x_{2}),p_{K,H}(z_{2}))\leq
d(x_{2},z_{2})\leq r.$

Thus, $d(x_{1},x_{2})\leq d(x_{1},z_{2})+d(z_{2},x_{2})\leq 2r$. ∎

A connected graph $(\mathcal{Y},d)$ is $C$-quasi-isometric with $\mathbb{Z}$
if there is a quasi-isometry $f:\mathbb{Z}\to\mathcal{Y}$, such that
$\frac{1}{C}|i-j|-C\leq d(f(i),f(j))\leq C|i-j|+C$ for every
$i,j\in\mathbb{Z}$, and $d(z,f(\mathbb{Z}))\leq C$ for all $z\in\mathcal{Y}$.
In this case, there is a constant $C_{1}$ (for instance
$C_{1}:=3C^{3}+C^{2}+3C$) such that the complement of every ball of radius
$\geq C_{1}$ has two infinite connected components.

######  Lemma 3.14.

Consider subgroups $H\leq K\leq G$. Assume $H\backslash X$ is $C$-quasi-
isometric with $\mathbb{Z}$. Assume that $K\backslash X$ is infinite. Let
$\bar{x}\in K\backslash X$ and $x_{1},x_{2}\in H\backslash X$ such that
$p_{K,H}(x_{1})=p_{K,H}(x_{2})=\bar{x}$ and let $\bar{\delta}$ be an infinite
geodesic path in $K\backslash X$ starting at $\bar{x}$. If the pull-backs
$\delta_{1},\delta_{2}$ starting at $x_{1},x_{2}$ (respectively) point towards
the same end $\xi$ of $H\backslash X$, then

$d(\delta_{1}(t),\delta_{2}(t))\leq C_{2}=2C_{1}$

for every $t\in[0,\infty)$ where $\delta_{1},\delta_{2}:[0,\infty)\to
H\backslash X$ are parametrised by arc length. In particular,
$d(x_{1},x_{2})\leq C_{2}$.

###### Proof.

Recall that the balls $B(x_{i},C_{1})$ both separate the space $H\backslash X$
and have two infinite components in their complement.
If $\delta_{2}\cap B(x_{1},C_{1})=\emptyset$ then the end $\xi$ is associated
with the infinite connected component of $(H\backslash X)\smallsetminus
B(x_{1},C_{1})$ that contains $\delta_{2}$ (and thus $x_{2}$ and
$B(x_{2},C_{1})$). Then $B(x_{2},C_{1})$ separates $B(x_{1},C_{1})$ from
$\xi$. Thus $\delta_{1}\cap B(x_{2},C_{1})\not=\emptyset$. In conclusion, we
have either $\delta_{2}\cap B(x_{1},C_{1})\not=\emptyset$ or $\delta_{1}\cap
B(x_{2},C_{1})\not=\emptyset$. By Lemma 3.13, we conclude that
$d(x_{1},x_{2})\leq 2C_{1}$. Then,
$p_{K,H}(\delta_{1}(t))=p_{K,H}(\delta_{2}(t))=\bar{\delta}(t)$ are the
initial point of infinite geodesic paths to which the above applies. ∎

######  Proposition 3.15.

Consider subgroups $H\leq K\leq G$. Assume $H\backslash X$ is $C$-quasi-
isometric with $\mathbb{Z}$. Then either $[K:H]$ is finite or $[G:K]$ is
finite.

###### Proof.

Assume $[G:K]$ is infinite, i.e. $K\backslash X$ is infinite. Let $\bar{x}\in
K\backslash X$ and let $\bar{\delta}$ be an infinite geodesic path in
$K\backslash X$ starting at $\bar{x}$. If all the $p_{K,H}$-pull-backs of
$\bar{\delta}$ point towards the same end of $H\backslash X$, then
$p_{K,H}^{-1}(\bar{x})$ has diameter bounded by $C_{2}$ (by Lemma 3.14).
Assume now that there are two points $u_{1},u_{2}\in H\backslash X$ such that
$p_{K,H}(u_{i})=\bar{x}$ and such that the pull-backs of $\bar{\delta}$
starting at $u_{1},u_{2}$ point towards different ends of $H\backslash X$.
Then for all other points $u\in p_{K,H}^{-1}(\bar{x})$, the pull-back of
$\bar{\delta}$ starting at $u$ point towards the same end as either that
starting at $u_{1}$ or that starting at $u_{2}$. It follows that
$d(u,\\{u_{1},u_{2}\\})\leq C_{2}$ (by Lemma 3.14). In any case,
$p_{K,H}^{-1}(\bar{x})$ is finite. It follows that $[K:H]$ is finite. ∎

######  Proposition 3.16.

Consider subgroups $H\leq K\leq G$. Assume $H\backslash X$ is $C$-quasi-
isometric with $\mathbb{Z}$ and that the index $[K:H]$ is finite, i.e. the
fibers of $p_{K,H}:H\backslash X\to K\backslash X$ have cardinal $[K:H]$.
Assume that a fiber of $p_{K,H}$ has diameter $>C_{2}=2C_{1}$. Then there are
fibers of $p_{K,H}$ of arbitrarily large diameter and there are only finitely
many subgroups $K\leq L\leq G$ such that $[L:K]$ is finite.

###### Proof.

Let $\bar{x}\in K\backslash X$ and let $x_{1},x_{2}\in H\backslash X$ be two
points such that $p_{K,H}(x_{1})=p_{K,H}(x_{2})=\bar{x}$ and
$d(x_{1},x_{2})>C_{2}$. Observe that $K\backslash X$ is an infinite locally
finite graph. Let $\bar{\delta}$ be a geodesic path in $K\backslash X$
starting at $\bar{x}$ and let $\delta_{1},\delta_{2}$ be its lifts in
$H\backslash X$ starting at $x_{1}$ and $x_{2}$ respectively. By Lemma 3.14,
$\delta_{1},\delta_{2}$ point towards different ends $\xi_{1},\xi_{2}$,
respectively, of $H\backslash X$. Then
$p_{K,H}(\delta_{1}(t))=p_{K,H}(\delta_{2}(t))$ for every $t\in[0,\infty)$
while the distance $d(\delta_{1}(t),\delta_{2}(t))$ tends to $\infty$ as $t$
tends to $\infty$ : there is no bound on the diameter of the $p_{K,H}$-fibers.
Then, consider a subgroup $L$ such that $K\leq L\leq G$ and $[L:K]$ is finite,
thus $[G:L]$ is infinite. Consider the covering maps

${H\backslash X}$${K\backslash X}$${L\backslash
X,}$$\scriptstyle{p_{K,H}}$$\scriptstyle{p_{L,K}}$

and let us prove that there is a bound independent of $L$ for the diameter of
$p_{L,K}^{-1}(p_{L,K}(\bar{x}))$.

Let $z:=p_{L,K}(\bar{x})\in L\backslash X$ and let $B:=(p_{L,K}\circ
p_{K,H})^{-1}(z)$ be its pre-image under $p_{L,K}\circ p_{K,H}$. Let $\delta$
be a geodesic path in $L\backslash X$ starting at $z$ and let
$\delta_{1},\delta_{2}$ be its lifts in $H\backslash X$ starting at
$x_{1},x_{2}$ respectively. Consider the partition $B=B_{1}\sqcup B_{2}$ where
$B_{i}$ is the subset consisting of those $b$ such that the lift of $\delta$
in $H\backslash X$ starting at $b$ points towards $\xi_{i}$. Recall that
$x_{i}\in B_{i}$. It follows from Lemma 3.14 that $B_{i}\subseteq
B(x_{i},C_{2})$. Thus the diameter of $B$ is bounded by
$d(x_{1},x_{2})+2C_{2}$. Since $p_{K,H}$ does not increase distances, it
follows that $p_{L,K}^{-1}(p_{L,K}(\bar{x}))=p_{K,H}(B)$ has diameter bounded
by $d(x_{1},x_{2})+2C_{2}$. By Proposition 3.12, there are at most
$2^{|B_{X}(\mathrm{id},d(x_{1},x_{2})+2C_{2})|}$-many such subgroups $L$. ∎

Let us now give the proof of Theorem 3.10.

###### Proof of Theorem 3.10.

Assume that $\operatorname{rk^{\times}_{CB}}(H;G)>2$. Let $(H_{n})_{n}$ be a
non-stationary sequence of subgroups of $G$ that converges to $H$ and such
that $\operatorname{rk^{\times}_{CB}}(H_{n};G)\geq 2$. Since $G$ is finitely
generated, the finite index subgroups of $G$ have erasing Cantor-Bendixson
rank $=1$, thus $[G:H_{n}]=\infty$. Since $H$ is finitely generated, up to
extraction one can assume that each $H_{n}$ contains $H$, and so by
Proposition 3.15, $[H_{n}:H]<\infty$. By Proposition 3.12, there are only
finitely many subgroups $H\leq K\leq G$ for which the fibers of $p_{K,H}$ have
diameter $\leq C_{2}=2C_{1}$. Thus up to extraction one can assume that each
$H_{n}$ gives rise to a $p_{H_{n},H}$-fiber of diameter $>C_{2}$.

Let us now approximate $H_{n}$. Let $(L_{n,j})_{j}$ be a non-stationary
sequence tending to $H_{n}$ when $j$ tends to $\infty$. Since $[H_{n}:H]$ is
finite, $H_{n}$ is finitely generated, thus we have $H\leq H_{n}\leq L_{n,j}$
for large enough $j$. By Proposition 3.16, there are only finitely many
subgroups $L$ such that $H\leq H_{n}\leq L\leq G$ and $[L:H_{n}]<\infty$, thus
up to extraction, one can assume that $[L_{n,j}:H_{n}]=\infty$ for each $j$.
Then, by Proposition 3.15 $[G:L_{n,j}]<\infty$. It follows that
$\operatorname{rk^{\times}_{CB}}(L_{n,j};G)=1$, so
$\operatorname{rk^{\times}_{CB}}(H_{n};G)=2$ for each $n$ and
$\operatorname{rk^{\times}_{CB}}(H;G)=3$. ∎

##  4 Subgroup separable groups and the first $\ell^{2}$-Betti number

Let $G$ be a countable group. A subgroup $H\leq G$ is separable if it is equal
to the intersection of a sequence of finite index subgroups of $G$ or
equivalently if $H$ is a closed subset of $G$ with respect to the profinite
topology. The group $G$ is said to be subgroup separable if every finitely
generated subgroup is separable. Equivalently, $G$ is subgroup separable if
and only if the subspace $\operatorname{Sub}_{[<\infty]}(G)$ of finite index
subgroups of $G$ is dense in $\operatorname{Sub}(G)$, as observed in [GKM16]:
the closure of the finite index subgroups contains all the finitely generated
subgroups and these are dense in $\operatorname{Sub}(G)$. Applied to the
subgroup $\\{\mathrm{id}\\}$ this property implies that $G$ is residually
finite. It is clear from the first definition that subgroup separability
passes to subgroups. In addition it easily follows, for example via the
profinite characterisation, that subgroup separability is an invariant of
commensurability.

Subgroup separability was shown to hold for finitely generated free groups by
M. Hall Jr in [Hal49] and for finitely generated orientable surface groups
(and so all finitely generated Fuchsian groups) by P. Scott [Sco78].

Assuming $G$ is a finitely generated subgroup separable group, the isolated
points of $\operatorname{Sub}(G)$ are exactly the finite index subgroups; i.e.
$H$ has finite index in $G$ if and only if
$\operatorname{rk^{\times}_{CB}}(H;G)=1$. We will show that
$\operatorname{Sub}(G)^{\prime}$ has no isolated points in the case where $G$
has positive first $\ell^{2}$-Betti number. We refer the reader to [CG86] for
the precise definition of this invariant, but note that it verifies the
following properties:

  * •

$\beta_{1}^{(2)}(G)\leq\operatorname{rk}(G)$ (this follows immediately from
the definition).

  * •

If $H\leq G$ is a finite index subgroup, then $\beta_{1}^{(2)}(H)=[G:H]\
\beta_{1}^{(2)}(G)$ (cf. [CG86, Proposition 2.6]).

The reader may observe that, in the proof of the following theorem, the first
$\ell^{2}$-Betti number can be replaced by any group invariant $a(\cdot)$ such
that, for any subgroup $H\leq G$, $a(H)\leq\operatorname{rk}(H)$ and for any
sequence $(H_{n})_{n}$ of finite index subgroups of $G$, if
$[G:H_{n}]\rightarrow\infty$ then $a(G)\rightarrow\infty$. In fact, the
property we exploit here is that, for any sequence $(H_{n})_{n}$ of finite
index subgroups of $G$, if $[G:H_{n}]\rightarrow\infty$ then
$\operatorname{rk}(H_{n})\rightarrow\infty$.

######  Theorem 4.1.

Let $G$ be a subgroup separable group. If $G$ is not finitely generated then
$Sub(G)$ is a perfect space. If $G$ is finitely generated with positive first
$\ell^{2}$-Betti number then $\mathcal{K}(G)=\operatorname{Sub}_{[\infty]}(G)$
and thus $\operatorname{rk_{CB}}(G)=1$.

###### Proof of Theorem 4.1.

First suppose that $G$ is not finitely generated and let $H\leq G$. If $H$ is
finitely generated then $H$ has infinite index in $G$. Moreover, there is a
sequence of finite index subgroups $(K_{n})_{n}$ of $G$ such that
$H=\cap_{n}K_{n}$. In other words, the sequence of finite index subgroups
$H_{i}=\cap_{n\leq i}K_{n}$ converges toward $H$ and is non-stationary.
Therefore $H$ is not isolated in $\operatorname{Sub}(G)$. If $H$ is not
finitely generated then it is a non-stationary limit of a sequence of finitely
generated subgroups, so $H$ is again not isolated. Thus
$\operatorname{Sub}(G)$ is perfect.

Now suppose that $G$ is finitely generated and $\beta_{1}^{(2)}(G)>0$. Every
finite index subgroup of $G$ is an isolated point of $\operatorname{Sub}(G)$
so $\mathcal{K}(G)\subseteq\operatorname{Sub}_{[\infty]}(G)$. Let $H\leq G$ be
a subgroup with infinite index. If $H$ is not finitely generated and
$H=\langle s_{n}:n\in\mathbb{N}\rangle$ then
$H=\lim_{n\rightarrow\infty}\langle s_{1},\dots,s_{n}\rangle$ and, for each
$n$, $\langle s_{1},\dots,s_{n}\rangle\leq H$ has infinite index so $H$ is not
isolated in $\operatorname{Sub}_{[\infty]}(G)$. Next, suppose that $H$ has a
finite generating set of cardinality $r$. By subgroup separability of $G$, the
subgroup $H$ is the intersection of countably many finite index subgroups of
$G$ and, since finite intersections of finite index subgroups have finite
index, $H$ is the limit of a strictly decreasing sequence
$(K_{n})_{n\in\mathbb{N}}$ of finite index subgroups of $G$. For each $n$, let
$H_{n}:=\langle H,k_{n}\rangle$ for some $k_{n}\in K_{n}\smallsetminus H$.
Since $H\lneq H_{n}\leq K_{n}$ for all $n$ it follows that $H_{n}\rightarrow
H$ non trivially. It remains to show that $H_{n}$ has infinite index in $G$
for almost all $n$.

Since $(K_{n})_{n\in\mathbb{N}}$ is a non-stationary nested sequence of finite
index subgroups, the sequence $([G:K_{n}])_{n\in\mathbb{N}}$ of indices must
tend towards infinity. In particular, $\beta_{1}^{(2)}(G)[G:K_{n}]>r+1$, for
almost all $n$. For all such $n$, if $H_{n}$ has finite index in $G$ then

$\beta_{1}^{(2)}(H_{n})=\beta_{1}^{(2)}(G)[G:H_{n}]\geq\beta_{1}^{(2)}(G)[G:K_{n}]>r+1.$

But $H_{n}$ has a generating set of cardinality $r+1$ so
$r+1\geq\beta_{1}^{(2)}(H_{n})$ and this is a contradiction. Hence, for almost
all $n$, $H_{n}$ has infinite index as required. It follows that
$\operatorname{Sub}_{[\infty]}(G)$ has no isolated points. ∎

######  Remark 4.2.

Kielak showed in [Kie20] that if $G$ is a finitely generated group which is
virtually residually finite rationally solvable (or RFRS) in the sense of
[Ago08] then $\beta_{1}^{(2)}(G)=0$ if and only if $G$ virtually fibers (i.e.
$G$ has a finite index subgroup which surjects onto $\mathbb{Z}$ with finitely
generated kernel). This allows us to sharpen our result in the following
sense:

If $G$ is finitely generated, subgroup separable and virtually RFRS then
$\operatorname{\mathcal{K}}(G)=\operatorname{Sub}_{[\infty]}(G)$ if and only
if $\beta_{1}^{(2)}(G)>0$.
One direction is given by Theorem 4.1 while the other follows from Corollary
3.6 since it implies that the kernel of a virtual fibering is an infinite
index subgroup which is not contained in the perfect kernel of $G$. In
particular, we characterise the property of having maximal perfect kernel in
terms of the first $\ell^{2}$-Betti number for a large class of subgroup
separable groups, including all those which virtually embed into right-angled
Artin groups.

The following corollary is a generalisation of a well-known fact about free
groups (see e.g. [CGLM20, Proposition 2.1]).

######  Corollary 4.3.

If $G$ is a virtually free group of rank $\geq 2$ then

$\mathcal{K}(G)=\begin{cases}\operatorname{Sub}_{[\infty]}(G)&\text{if
}G\text{ is finitely generated}\\\
\operatorname{Sub}(G)&\text{otherwise}.\end{cases}$

###### Proof.

If $\mathbf{F}_{r}$ is a free group of rank $r\geq 2$ then
$\beta_{1}^{(2)}(\mathbf{F}_{r})=r-1>0$. Moreover it was shown in [Hal49] that
$\mathbf{F}_{r}$ is subgroup separable. This implies that any group $G$
containing $\mathbf{F}_{r}$ as a finite index subgroup is subgroup separable
and has positive first $\ell^{2}$-Betti number so the result follows by
Theorem 4.1. ∎

######  Corollary 4.4.

If $G$ is virtually a finitely generated non-abelian limit group, then
$\operatorname{\mathcal{K}}(G)=\operatorname{Sub}_{[\infty]}(G)$.

###### Proof.

Limit groups are subgroup separable by [Wil08] and a finitely generated, non-
abelian limit group has positive first $\ell^{2}$-Betti number by [Pic06, p.
645, l. 2-3]. The result therefore follows from Theorem 4.1. ∎

In the special case of surface groups, the perfect kernel was also described
using a different proof in unpublished work by A. Carderi, the second author
and F. Le Maître.

######  Example 4.5 (Product of free groups over a cyclic subgroup).

Let $G=\mathbf{F}_{p}\ast_{\mathbb{Z}}\mathbf{F}_{q}$ for some finite $p,q\geq
2$. By [BBS84], free products of subgroup separable groups over cyclic
subgroups are subgroup separable. Moreover, using the Mayer-Vietoris formula
from [CG86, p. 204, §4], the first $\ell^{2}$-Betti number of $G$ satisfies
$\beta^{(2)}_{1}(G)=p+q-2>0$ so, by Theorem 4.1,
$\operatorname{\mathcal{K}}(G)=\operatorname{Sub}_{[\infty]}(G)$. For
information about the dynamics, see Corollary 5.23.

##  5 Actions on trees

We will refer throughout to the geometric realisation of an abstract graph as
a graph. More precisely, a graph is a one-dimensional cell complex $X$. We
denote by $V(X)$ its set of vertices (or 0-cells) and by $E(X)$ its set of
edges (or 1-cells). For each $e\in E(X)$ let $\varphi_{e}:\partial
e\rightarrow V(X)$ be the corresponding attaching map and define maps:

$o,t:E(X)\rightarrow V(X)$

where for each $e=[0,1]\in E(X)$ its endpoints are the vertices
$o(e)=\varphi_{e}(0)$ and $t(e)=\varphi_{e}(1)$. Note that $X$ is endowed with
a natural orientation where each edge $e$ is directed from $o(e)$ to $t(e)$. A
path $P$ in $X$ is a sequence of edges $(e_{n})_{n\in N}$, where
$N=\mathbb{N}=\mathbb{N}_{>0}$ or $N=\\{1,\dots,k\\}$ for some
$k\in\mathbb{N}$, such that $e_{n}=\gamma([n-1,n])$ for each $n\in N$ and for
some continuous map $\gamma:\cup_{n\in N}[n-1,n]\rightarrow X$. If $N$ has
cardinality $k\in\mathbb{N}$ then $k$ is the length of $P$ and if
$N=\mathbb{N}$ we say that $P$ is infinite. If $\gamma$ is injective then we
say that $P$ is simple.

If $X,Y$ are graphs and $\varphi:X\rightarrow Y$ is a bijection then $\varphi$
is an isometry if it preserves the cellular structure of $X$. It follows that
$\varphi$ preserves the length metrics $d_{X}$ and $d_{Y}$ on $X$ and $Y$
respectively. All actions on graphs are assumed to be by orientation
preserving isometries. A tree is a simply connected graph.

Recall that the action of a group $G$ on a tree $Y$ is minimal if $Y$ does not
contain any proper $G$-invariant subtree. The action is reducible if one of
the following holds: (1) there is a global fixed point; (2) $G$ fixes an end
of $Y$; or (3) there is a $G$-invariant line in $Y$. Otherwise, the action is
irreducible (aka of general type). If $x$ is a vertex or an edge of $Y$, we
denote its stabiliser under the action of a subgroup $H\leq G$ by
$\operatorname{Stab}_{H}(x)$.

The structure theorem of Bass-Serre Theory [Ser80, §5.4, Theorem 13] implies
that a group $G$ admits a minimal action on a tree with a single orbit of
edges if and only if it splits non-trivially as an amalgamated free product
$G=A\ast_{C}B$ or a HNN extension $G=A\ast_{C}$. The $G$-tree $Y$ associated
with such a splitting is called its Bass-Serre tree. In the former case $G$
acts on $Y$ with two orbits of vertices, $C$ is the stabiliser of an edge and
$A,B$ are the stabilisers of its endpoints and in the latter $G$ acts with a
single orbit of vertices, $C$ is the stabiliser of an edge and $A$ is the
stabiliser of one of its endpoints.

We will make use of the following standard characterisation of irreducibility.

######  Proposition 5.1.

Consider a minimal action $G\curvearrowright T$ on a tree that is transitive
on the set of edges. The action is irreducible if and only if, for some
(equivalently every) edge $e$ of $T$ with endpoints $v$ and $w$, the indices
of the stabilisers satisfy (up to relabelling $v$ and $w$):

  1. 1.

$[\operatorname{Stab}_{G}(v):\operatorname{Stab}_{G}(e)]\geq 2$ and
$[\operatorname{Stab}_{G}(w):\operatorname{Stab}_{G}(e)]\geq 2$, when
$G\backslash T$ has one vertex.

  2. 2.

$[\operatorname{Stab}_{G}(v):\operatorname{Stab}_{G}(e)]\geq 3$ and
$[\operatorname{Stab}_{G}(w):\operatorname{Stab}_{G}(e)]\geq 2$, when
$G\backslash T$ has two vertices.

###### Proof.

Recall that for an edge $e$ of $T$ and an endpoint $v$ of $e$, the set of
edges in the $G$-orbit of $e$ and adjacent to $v$ is indexed by the cosets
$\operatorname{Stab}_{G}(v)/\operatorname{Stab}_{G}(e)$.

Assume $G\backslash T$ has one edge and two vertices. If
$[\operatorname{Stab}_{G}(v):\operatorname{Stab}_{G}(e)]=[\operatorname{Stab}_{G}(w):\operatorname{Stab}_{G}(e)]=2$,
then $T$ is a line and the action is reducible. If
$[\operatorname{Stab}_{G}(v):\operatorname{Stab}_{G}(e)]=1$, then $v$ has
degree $1$ in $T$: this contradicts the minimality of the action.

Assume $G\backslash T$ has one edge and one vertex. If
$[\operatorname{Stab}_{G}(v):\operatorname{Stab}_{G}(e)]=1$, then up to
reversing the orientation of all the edges, every vertex of $T$ is the
startpoint of a single edge: for every $x\in V(T)$, there is a single edge
$f\in E(T)$ such that $o(f)=x$. There is a unique infinite simple path
$P=(e_{n})_{n\in\mathbb{N}}$ with startpoint $x$ that follows the orientations
of the edges. For any two vertices, the corresponding infinite simple positive
path must eventually agree. They thus define the same end of $T$. This
uniquely defined end is fixed by $G$ and the action on $T$ is reducible.

We have proved that irreducibility of the action implies the desired
conditions on the stabilisers.

Conversely, if $T$ admits an edge $e$ with an endpoint $v$ such that the index
of the stabiliser satisfies
$[\operatorname{Stab}_{G}(v):\operatorname{Stab}_{G}(e)]\geq 3$, it follows
that $T$ is not a line: $v$ has degree $\geq 3$. If $G\backslash T$ has only
one vertex and
$[\operatorname{Stab}_{G}(v):\operatorname{Stab}_{G}(e)]=[\operatorname{Stab}_{G}(w):\operatorname{Stab}_{G}(e)]=2$,
then every vertex has degree $\geq 4$, thus $T$ is not a line. There is no
global fixed point or $G$-invariant line in $T$ by minimality of the action.
Moreover, if $P=(e_{n})_{n\in\mathbb{N}}$ is an infinite simple path starting
at $v_{0}$ and
$g\in\operatorname{Stab}_{G}(v_{0})\smallsetminus\operatorname{Stab}_{G}(e_{1})$
then $gP$ is an infinite path in $T$ whose intersection with $P$ is
$\\{v_{0}\\}$. Therefore $P$ and $gP$ define distinct ends of $T$ so no end of
$T$ is fixed by $G$. ∎

Let $T$ be a tree and $\varphi:T\rightarrow T$ be a graph automorphism. Recall
that $\varphi$ is either elliptic (i.e. $\varphi$ fixes a point in $T$) or
hyperbolic (i.e. the translation length of $\varphi$ is realised in $T$ and is
strictly positive) [Ser80, Proposition 2.4]. In the later case, there is a
unique bi-infinite line $\mathrm{Axis}(\varphi)$ in $T$ on which $\varphi$
acts by translation.

######  Lemma 5.2.

Let $G\curvearrowright T$ be a minimal irreducible action of a countable
group. Let $e$ be an edge of $T$ and $Z$ be one of the two connected
components of $T\smallsetminus\\{e\\}$. Let $L$ be a finite segment of edges
$[f_{1},f_{2}]$. Then there are three hyperbolic elements
${\gamma}_{0},{\gamma}_{1},\gamma_{2}\in G$ such that
– $\mathrm{Axis}({\gamma}_{0})$ contains $e$;
– $\mathrm{Axis}({\gamma}_{1})$ is contained in $Z$;
– $\mathrm{Axis}({\gamma}_{0})\cap\mathrm{Axis}({\gamma}_{1})$ is non-empty
and bounded (the hyperbolic elements are “transverse”);
– the segment $\gamma_{2}(L)$ is contained in $Z$ and $\gamma_{2}\cdot f_{1}$
separates $\gamma_{2}\cdot f_{2}$ from $e$.

###### Proof.

By shrinking to a point each connected component of $T\smallsetminus G\cdot
e$, the complement in $T$ of the $G$-orbit of $e$, we obtain a tree
$T^{\prime}$ equipped with a $G$-action that is transitive on the set of edges
and for which the map $T\to T^{\prime}$ is $G$-equivariant. By minimality of
the action $G\curvearrowright T$, the convex hull of $G\cdot e$ is the whole
of $T$; thus $G\curvearrowright T^{\prime}$ is minimal and irreducible and
both actions have the same kernel. If ${\gamma}_{0},{\gamma}_{1}$ satisfy the
conditions of the lemma for $T^{\prime}$, then their action on $T$ also
satisfies these conditions.

Thus, one can start by assuming that $T$ has a single orbit of edges. There is
a hyperbolic element ${\gamma}_{0}$ whose axis contains $e$: by irreducibility
$G\curvearrowright T$ admits some hyperbolic element $h_{0}$, and for any edge
$f\in\mathrm{Axis}(h_{0})$, there is $h_{1}\in G$ such that $h_{1}\cdot f=e$;
then $h_{1}(\mathrm{Axis}(h_{0}))$ is the axis of
${\gamma}_{0}:=h_{1}h_{0}h_{1}^{-1}$ and it contains $e$.

Recall from [Ser80, I.6.4, p. 62] that two edges $f$ and $f^{\prime}$ are
coherent if they are oriented in the same way in the unique geodesic segment
which connects them. Let $a,b$ be edges of
$Z\smallsetminus\mathrm{Axis}({\gamma}_{0})$ that are adjacent to
$\mathrm{Axis}({\gamma}_{0})$ but at different vertices of
$\mathrm{Axis}({\gamma}_{0})$. If $a$ and $b$ are not coherent, one replaces
$b$ by an edge $b^{\prime}$ that is adjacent to $b$, that is separated from
$\mathrm{Axis}({\gamma}_{0})$ by $b$ and that is coherent with $a$ (every
vertex is adjacent to at least two incoherent edges – see Proposition 5.1).
Thus, one can take $a$ and $b$ to be coherent. Choose an element
${\gamma}_{1}\in G$ sending $a$ to $b$: then by [Ser80, Corollary, p. 63]
${\gamma}_{1}$ is hyperbolic and $[a,b]$ belongs to its axis. In particular
$\mathrm{Axis}({\gamma}_{1})$ has a non-trivial and bounded intersection with
$\mathrm{Axis}({\gamma}_{0})$. Moreover $[a,b]$ separates
$\mathrm{Axis}({\gamma_{1}})\smallsetminus[a,b]$ from $e$, in particular,
$\mathrm{Axis}({\gamma_{1}})\subset Z$.

There is an integer $r$ such that $\gamma_{0}^{r}(L)\subset Z$. If
$\gamma_{0}^{r}\cdot f_{1}$ separates $\gamma_{0}^{r}\cdot f_{2}$ from $e$,
set $\gamma_{2}\coloneqq\gamma_{0}^{r}$. Otherwise, the connected component
$Z_{2}$ of $T\smallsetminus\\{\gamma_{0}^{r}\cdot f_{1}\\}$ that does not
contain $\gamma_{0}^{r}\cdot f_{2}$ is entirely contained in $Z$. Choose
$\gamma^{\prime}_{1}\in G$ whose axis is entirely contained in $Z_{2}$. Then
the path from $\gamma^{\prime}_{1}\gamma_{0}^{r}\cdot f_{2}$ to $e$ traverses
$\gamma^{\prime}_{1}\gamma_{0}^{r}\cdot f_{1}$ and
$\gamma_{2}:=\gamma^{\prime}_{1}\gamma_{0}^{r}$ satisfies the condition. It is
easy to check that $\gamma_{2}$ is hyperbolic (the projection of $L$ onto
$\mathrm{Axis}({\gamma}_{0})$ belongs to the axis of $\gamma_{2}$). ∎

The kernel of an action $G\curvearrowright T$ is the normal subgroup
consisting of the elements which fix every vertex of $T$.

######  Proposition 5.3.

If $G\curvearrowright T$ is a minimal irreducible action on a tree with finite
kernel $N$, then
$\operatorname{Sub}(N)\subseteq\operatorname{\mathcal{K}}(G)$.

###### Proof.

By [CM87, Theorem 2.7] the irreducibility of the action implies that $G$ has a
free subgroup $\mathbf{F}_{2}$ of rank $2$. Therefore
$\operatorname{Sub}(N)\subseteq\operatorname{\mathcal{K}}(G)$ by Proposition
3.4. ∎

###  5.1 Main technical dendrological lemma

######  Lemma 5.4 (Main technical dendrological lemma).

Let $G\curvearrowright T$ be a minimal action on a tree $T$. Let $\Lambda\leq
G$ be a subgroup. Assume that $\Lambda$ admits a proper invariant subtree
$T_{\Lambda}\subsetneqq T$ for which there are sequences $(H_{n})_{n}$,
$({S_{n}})_{n}$ and $(e_{n})_{n}$ such that for every $n$

  1. 1.

$H_{n}$ is a subgroup of $G$;

  2. 2.

${S_{n}}$ is an $H_{n}$-invariant subtree of $T$;

  3. 3.

$e_{n}\in E(T)$ is an edge such that:

    1. (a)

$e_{n}$ is at distance greater than $n+1$ from ${T_{\Lambda}}$

    2. (b)

$e_{n}$ separates ${T_{\Lambda}}$ from ${S_{n}}$;

    3. (c)

$\operatorname{Stab}_{\Lambda}(e_{n})=\operatorname{Stab}_{H_{n}}(e_{n})$,
thus $\Lambda\cap H_{n}=\operatorname{Stab}_{\Lambda}(e_{n})$;

    4. (d)

$\operatorname{Stab}_{H_{n}}(e_{n})\not=H_{n}$;

    5. (e)

The $\Lambda$-orbit of $e_{n}$ does not separate ${T_{\Lambda}}$ from the
other $e_{k}$, $k\in\mathbb{N}\smallsetminus\\{n\\}$.

Let $\Lambda_{n}\coloneqq\langle\Lambda,H_{n}\rangle$. Let ${\mathcal{L}}_{T}$
be the set of subgroups of $G$ that satisfy assumptions 1-3. Then:

  1. (I)

Each $\Lambda_{n}$ splits as
$\Lambda_{n}=\Lambda*_{\operatorname{Stab}_{\Lambda}(e_{n})}H_{n}$ and
$\operatorname{Stab}_{\Lambda_{n}}(e_{n})=\operatorname{Stab}_{\Lambda}(e_{n})$.
More generally,
$\operatorname{Stab}_{\Lambda_{n}}(e_{k})=\operatorname{Stab}_{\Lambda}(e_{k})$
for all $k\in\mathbb{N}$ and $n\in\mathbb{N}$.

  2. (II)

The sequence $(\Lambda_{n})_{n}$ tends non trivially to $\Lambda$.

  3. (III)

Each $\Lambda_{n}$ belongs to ${\mathcal{L}}_{T}$, witnessed by a subsequence
of $(H_{n})_{n}$, $({S_{n}})_{n}$ and $(e_{n})_{n}$.

  4. (IV)

The set ${\mathcal{L}}_{T}$ has no isolated points for its induced topology.
In particular ${\mathcal{L}}_{T}\subseteq\operatorname{\mathcal{K}}(G)$.

  5. (V)

For every $\Lambda\in{\mathcal{L}}_{T}$, the quotient $\Lambda\backslash T$ is
infinite.

###### Proof.

Step (I). We fixed an integer $n$. We will first construct a tree $A_{n}$ on
which the subgroup $\Lambda_{n}\coloneqq\langle\Lambda,H_{n}\rangle$ of $G$
acts. Let $v_{n}$ be the endpoint of $e_{n}$ nearest to ${T_{\Lambda}}$ and
let $w_{n}$ be the other endpoint of $e_{n}$.

Denote by ${T^{*}_{n,\Lambda}}$ the connected component of
$T\smallsetminus\\{\Lambda\cdot e_{n}\\}$ that contains ${T_{\Lambda}}$ and by
${T^{*}_{H_{n}}}$ the connected component of $T\smallsetminus\\{H_{n}\cdot
e_{n}\\}$ that contains ${S_{n}}$.

Notice (by Assumption (3b)) that ${T^{*}_{n,\Lambda}}$ and ${T^{*}_{H_{n}}}$
are disjoint subtrees of $T$ that are $\Lambda$-invariant and
$H_{n}$-invariant respectively. Moreover, $e_{n}$ connects them. Also observe
for later use that by Assumption (3e), all the $e_{k}$,
$k\in\mathbb{N}\smallsetminus\\{n\\}$ are contained in ${T^{*}_{n,\Lambda}}$.

By an immediate induction based on the length of an element of
$\Lambda_{n}=\langle\Lambda,H_{n}\rangle$ as a product of elements from
$\Lambda$ and $H_{n}$, we obtain that the $\Lambda_{n}$-orbit of $e_{n}$
splits $T$ into connected components that are either a $\Lambda_{n}$-image of
${T^{*}_{n,\Lambda}}$ or a $\Lambda_{n}$-image of ${T^{*}_{H_{n}}}$.
Contracting each of these connected components to a point, we obtain a
$\Lambda_{n}$-equivariant map

$P_{n}:T\to A_{n}$

from $T$ to a tree $A_{n}$ equipped with the induced $\Lambda_{n}$-action,
with a single $\Lambda_{n}$-orbit of edges and two $\Lambda_{n}$-orbits of
vertices.

Let $\bar{v}_{n}$ be the vertex corresponding to the connected component
${T^{*}_{n,\Lambda}}$ and $\bar{w}_{n}$ be the vertex corresponding to the
connected component ${T^{*}_{H_{n}}}$. Let $\bar{e}_{n}$ be the image of
$e_{n}$; its endpoints are $\bar{v}_{n}$ and $\bar{w}_{n}$. It follows from
general Bass-Serre theory that $\Lambda_{n}$ splits as

$\Lambda_{n}=\operatorname{Stab}_{\Lambda_{n}}(\bar{v}_{n})*_{\operatorname{Stab}_{\Lambda_{n}}(\bar{e}_{n})}\operatorname{Stab}_{\Lambda_{n}}(\bar{w}_{n}).$
(6)

Our goal is now to identify this splitting with the splitting
$\Lambda_{n}=\Lambda*_{\operatorname{Stab}_{\Lambda}(e_{n})}H_{n}$ of (I).

The projection map $P_{n}:T\to A_{n}$ is injective on the $\Lambda_{n}$-orbit
of $e_{n}$, thus the stabilisers of $e_{n}$ and $\bar{e}_{n}$ are the same for
$\Lambda$ and $H_{n}$, respectively:

$\operatorname{Stab}_{\Lambda}(\bar{e}_{n})=\operatorname{Stab}_{\Lambda}(e_{n})\
\ \ \text{ and }\ \ \
\operatorname{Stab}_{H_{n}}(\bar{e}_{n})=\operatorname{Stab}_{H_{n}}(e_{n}).$

Thus, by Assumption (3c),
$\operatorname{Stab}_{\Lambda}(\bar{e}_{n})=\operatorname{Stab}_{H_{n}}(\bar{e}_{n})$.
By definition
$\operatorname{Stab}_{\Lambda_{n}}(\bar{e}_{n})\cap\Lambda=\operatorname{Stab}_{\Lambda}(\bar{e}_{n})$
and $\operatorname{Stab}_{\Lambda_{n}}(\bar{e}_{n})\cap
H_{n}=\operatorname{Stab}_{H_{n}}(\bar{e}_{n})$. It follows that:

$\operatorname{Stab}_{\Lambda_{n}}(\bar{e}_{n})\cap\Lambda=\operatorname{Stab}_{\Lambda_{n}}(\bar{e}_{n})\cap
H_{n}=\Lambda\cap H_{n}.$ (7)

Since ${T^{*}_{n,\Lambda}}$ is $\Lambda$-invariant and ${T^{*}_{H_{n}}}$ is
$H_{n}$-invariant in $T$, the stabilisers of their images
$P({T^{*}_{n,\Lambda}})=\bar{v}_{n}$ and $P({T^{*}_{H_{n}}})=\bar{w}_{n}$ for
the $\Lambda_{n}$-action on $A_{n}$ satisfy

$\Lambda\leq\operatorname{Stab}_{\Lambda_{n}}(\bar{v}_{n})\ \ \ \text{ and }\
\ \ H_{n}\leq\operatorname{Stab}_{\Lambda_{n}}(\bar{w}_{n}).$ (8)

We shall make use of the normal form theorem [LS77, Theorem 2.6, Section IV.2,
p. 187] for the amalgamated free product splitting (6). Since
$\langle\Lambda,H_{n}\rangle=\Lambda_{n}$, every non-trivial element
$g\in\operatorname{Stab}_{\Lambda_{n}}(\bar{v}_{n})$ can be written in the
form $g=c_{1}c_{2}\cdots c_{j}$ where

  * •

each $c_{i}$ is in one of the “factor” subgroups $\Lambda$ or $H_{n}$,

  * •

successive $c_{i},c_{i+1}$ come from different factors, and

  * •

if $j>1$ then no $c_{i}$ is in the intersection $\Lambda\cap H_{n}$.

Then (by (8) and (7)) $c_{1}c_{2}\cdots c_{j}$ is a reduced sequence for the
splitting (6). On the other hand, $g^{-1}c_{1}c_{2}\cdots c_{j}=1$ is not
reduced for the splitting (6). If $j>1$, then
$g^{-1}c_{1}\in\operatorname{Stab}_{\Lambda_{n}}(\bar{e}_{n})$, and
successively we get $g^{-1}c_{1}c_{2}\cdots
c_{j-1}\in\operatorname{Stab}_{\Lambda_{n}}(\bar{e}_{n})$. In the end, $c_{j}$
also belongs to $\operatorname{Stab}_{\Lambda_{n}}(\bar{e}_{n})$, and by (7),
$c_{j}\in\Lambda\cap H_{n}$, a contradiction. Thus $j=1$ and
$g=c_{1}\in\Lambda$ and
$\Lambda=\operatorname{Stab}_{\Lambda_{n}}(\bar{v}_{n})$. Similarly, the
symmetric argument with the other factor shows that
$H_{n}=\operatorname{Stab}_{\Lambda_{n}}(\bar{w}_{n})$. We have proved that

$\operatorname{Stab}_{\Lambda_{n}}(\bar{v}_{n})=\Lambda,\ \ \ \
\operatorname{Stab}_{\Lambda_{n}}(\bar{w}_{n})=H_{n}\ \ \ \ \text{ and }\ \ \
\
\operatorname{Stab}_{\Lambda_{n}}(e_{n})=\operatorname{Stab}_{\Lambda}(e_{n}).$
(9)

and the splitting (6) becomes

$\Lambda_{n}=\Lambda*_{\operatorname{Stab}_{\Lambda}(e_{n})}H_{n}.$ (10)

The “More generally” part of (I), i.e.
$\operatorname{Stab}_{\Lambda_{n}}(e_{k})=\operatorname{Stab}_{\Lambda}(e_{k})$
for all the other $k\in\mathbb{N}\smallsetminus\\{n\\}$, is proved later as
Equation (11).

Step (II). All the $\Lambda_{n}$ contain $\Lambda$, thus in order to show that
$(\Lambda_{n})_{n}$ tends to $\Lambda$, it is enough to prove that for every
$g\in G$, there is $n_{0}$ such that if $g\in\Lambda_{n}$ for some $n\geq
n_{0}$, then $g\in\Lambda$.

Let $g\in G$, let $v$ be a vertex $v$ in ${T_{\Lambda}}$ and let $n_{0}$ be
the distance in $T$ between $v$ and $g\cdot v$. Assume that $g\in\Lambda_{n}$
for some $n\geq n_{0}$. By Assumption (3a) the $n$-neighbourhood of
${T_{\Lambda}}$ is contained in ${T^{*}_{n,\Lambda}}$, thus both $v$ and
$g\cdot v$ belong to ${T^{*}_{n,\Lambda}}$. It follows that their images in
$A_{n}$ both equal $\bar{v}_{n}$. By equivariance of the quotient map $T\to
A_{n}$, the element $g$ fixes the vertex $\bar{v}_{n}$. Since
$\operatorname{Stab}_{\Lambda_{n}}(\bar{v}_{n})=\Lambda$ (by Equation (9)), it
follows that $g\in\Lambda$. We have proved that
$\Lambda_{n}\rightarrow\Lambda$ as $n\rightarrow\infty$.

By Assumption (3d), the injection
$\Lambda\leq\Lambda_{n}=\Lambda*_{\operatorname{Stab}_{\Lambda}(e_{n})}H_{n}$
is strict, so the convergence of the sequence $(\Lambda_{n})_{n}$ is non-
trivial.

Step (III). Let us show that $\Lambda_{n}\in{\mathcal{L}}_{T}$.

The invariant tree we propose for $T_{\Lambda_{n}}$ is the
$\Lambda_{n}$-saturation of the convex hull of ${T_{\Lambda}}\cup{S_{n}}$:

$T_{\Lambda_{n}}\coloneqq\Lambda_{n}\cdot\mathrm{Hull}({T_{\Lambda}}\cup{S_{n}}).$

As for the sequences of subgroups $(H^{\prime}_{k})_{k}$, subtrees
$(S^{\prime}_{k})_{k}$, and edges $(e^{\prime}_{k})_{k}$ associated to
$\Lambda_{n}$, we will take subsequences of those associated with $\Lambda$,
obtained by shifting the indices by $r_{n}$, the distance between
${T_{\Lambda}}$ and $v_{n}$, i.e. $r_{n}:=d(T_{\Lambda},v_{n})$,
$H^{\prime}_{k}:=H_{k+r_{n}}$, $S^{\prime}_{k}:=S_{k+r_{n}}$ and
$e^{\prime}_{k}:=e_{k+r_{n}}$.

As observed earlier, all the $e_{k}$, for
$k\in\mathbb{N}\smallsetminus\\{n\\}$ are contained in ${T^{*}_{n,\Lambda}}$
by Assumption (3e). If an element $g$ of $\Lambda_{n}$ were to send some
$e_{i}$ to $g\cdot e_{i}$ so that $g\cdot e_{i}$ separated some $e_{j}$ from
$T_{\Lambda_{n}}$ (for $i,j\not=n$), then $g\cdot e_{i}$ would also belong to
${T^{*}_{n,\Lambda}}$ and $g$ would fix the vertex $\bar{v}_{n}$ of $A_{n}$.
Thus $g$ would belong to $\Lambda$ (by Equation (9)). But Assumption (3e) for
$\Lambda$ and all $n$ prevents the $\Lambda$-orbit of $e_{i}$ from separating
${T_{\Lambda}}$ and $e_{j}$. Thus the sequence $(e^{\prime}_{k})_{k}$
satisfies Assumption (3e) for $\Lambda_{n}$, and in particular the
$\Lambda_{n}$-invariant tree $T_{\Lambda_{n}}$ is strictly contained in $T$.
It also follows that $e^{\prime}_{k}$ separates $T_{\Lambda_{n}}$ from
$S^{\prime}_{k}$, i.e. Assumption (3b) for $\Lambda_{n}$. Since
$T_{\Lambda_{n}}\cap{T^{*}_{n,\Lambda}}$ is contained in the
$r_{n}$-neighbourhood of ${T_{\Lambda}}$, this shift in the indices permits to
check Assumption (3a) for $\Lambda_{n}$.

An element $g$ of $\Lambda_{n}$ that fixes some $e_{k}$, for
$k\in\mathbb{N}\smallsetminus\\{n\\}$ (which again is an edge of
${T^{*}_{n,\Lambda}}$) belongs to the stabiliser of $\bar{v}_{n}$, thus to
$\Lambda$ by Equation (9). Thus for all $k\in\mathbb{N}$:

$\operatorname{Stab}_{\Lambda_{n}}(e_{k})=\operatorname{Stab}_{\Lambda}(e_{k}).$
(11)

Assumption (3c) for $\Lambda$ ensures that
$\operatorname{Stab}_{\Lambda}(e_{k})=\operatorname{Stab}_{H_{k}}(e_{k})$, so
$\operatorname{Stab}_{\Lambda_{n}}(e_{k})=\operatorname{Stab}_{H_{k}}(e_{k})$.
Thus the sequence $(e^{\prime}_{k})_{k}$ satisfies Assumption (3c) for
$\Lambda_{n}$. Assumption (3d) for $\Lambda_{n}$, i.e.
$\operatorname{Stab}_{H_{k}^{\prime}}(e^{\prime}_{k})\not=H_{k}^{\prime}$ is
immediate from the analogous condition
$\operatorname{Stab}_{H_{k+r_{n}}}(e_{k+r_{n}})\not=H_{k+r_{n}}$ for
$\Lambda$. We have proved that $\Lambda_{n}\in{\mathcal{L}}_{T}$.

Therefore, every point of ${\mathcal{L}}_{T}$ is a non-trivial limit of points
of ${\mathcal{L}}_{T}$. The proof of Item (IV) follows immediately.

For $\Lambda\in{\mathcal{L}}_{T}$ and any edge $e\in E(T)$, the distance from
$e$ to $T_{\Lambda}$ is an invariant of its $\Lambda$-orbit $\Lambda\cdot e$.
This is precisely the distance in $\Lambda\backslash T$ between
$\Lambda\backslash T_{\Lambda}$ and the image of $e$ in $\Lambda\backslash T$.
By Assumption(3a), the sequence of edges $(e_{n})$ thus delivers infinitely
many edges in $\Lambda\backslash T$. This proves Item (V), and concludes the
proof of the main technical dendrological Lemma 5.4. ∎

###  5.2 Main dendrological theorems

######  Theorem 5.5.

Let $G\curvearrowright T$ be a minimal and irreducible action with finite
kernel $N$ on a tree $T$. Let $\Lambda\leq G$ be a subgroup and suppose that
there exists an edge $e_{\Lambda}\in E(T)$ such that
$\operatorname{Stab}_{\Lambda}(e_{\Lambda})=\Lambda\cap N$ and one of the
connected component of $T\smallsetminus\\{e_{\Lambda}\\}$ contains a
$\Lambda$-invariant subtree $T_{\Lambda}\subseteq T$. Then $\Lambda$ belongs
to $\operatorname{\mathcal{K}}(G)$.
More precisely, the set ${\mathcal{N}}_{T}$ of subgroups that satisfy the
assumptions of this theorem for the action $G\curvearrowright T$ has no
isolated points with respect to its induced topology.

###### Proof.

We will show that $\Lambda$ satisfies the conditions of Lemma 5.4.

Let $w_{0}$ be the endpoint of $e_{\Lambda}$ that is contained in the
component $Z$ of $T\smallsetminus e_{\Lambda}$ which does not contain
$T_{\Lambda}$, and let $w_{1}$ be the other endpoint of $e_{\Lambda}$. Since
$e_{\Lambda}$ separates $Z$ from a $\Lambda$-invariant subtree, it follows
that if $u_{1},u_{2}$ are vertices of $Z$ and $\lambda\in\Lambda$ such that
$\lambda\cdot u_{1}=u_{2}$, then $\lambda$ fixes $w_{0}$ and $e_{\Lambda}$, so
$\lambda\in\Lambda\cap N$. In particular $\lambda$ belongs to the kernel of
the action, thus the quotient map $T\to\Lambda\backslash T$ is injective on
$Z$. Moreover, (taking $u_{1}=u_{2}$) for every vertex $v\in V(Z)$ and every
edge $e\in E(Z)$ the stabilisers satisfy:

$\operatorname{Stab}_{\Lambda}(v)=\Lambda\cap
N=\operatorname{Stab}_{\Lambda}(e).$ (12)

By Lemma 5.2, there is a hyperbolic element $g$ whose axis is contained in
$Z$. Thus the sequence of edges $(e_{n})_{n\in\mathbb{N}}\coloneqq(g^{n}\cdot
e_{\Lambda})_{n\in\mathbb{N}}$, satisfies two conditions of Lemma 5.4: (3a)
(distance to $T_{\Lambda}$) and (3e) (the $\Lambda$-orbit of the edge $e_{n}$
intersects $Z$ only in $e_{n}$, thus does not separate any other $e_{k}$ from
$T_{\Lambda}$).

By Lemma 5.2, there is a hyperbolic element $h_{n}$ for each $n\in\mathbb{N}$
whose axis is contained in the connected component of
$T\smallsetminus\\{e_{n}\\}$ that does not contain $T_{\Lambda}$. We define
the subtree $S_{n}$ as the axis of $h_{n}$. The sequence $(e_{n})_{n}$
satisfies (3b) of Lemma 5.4.

Since $N$ is finite and normal in $G$, there is a power $r_{n}$ of $h_{n}$
which commutes with $N$. Let $H_{n}$ be the subgroup generated by
$h_{n}^{r_{n}}$ and $\Lambda\cap N$:

$H_{n}\coloneqq\langle h_{n}^{r_{n}},\Lambda\cap N\rangle.$

Since $h_{n}^{r_{n}}$ is hyperbolic and commutes with $\Lambda\cap N$ and
since $\Lambda\cap N$ is in the kernel of the action, $g\in H_{n}$ fixes
$e_{n}$ (or any other edge) if and only if $g\in\Lambda\cap N$. Thus
$\operatorname{Stab}_{H_{n}}(e_{n})=\Lambda\cap
N=\operatorname{Stab}_{\Lambda}(e_{n})$. This shows condition (3c) of Lemma
5.4.

As for condition (3d), it is clear since $\operatorname{Stab}_{H_{n}}(e_{n})$
is contained in the kernel of the action $G\curvearrowright T$ while $H_{n}$
contains the hyperbolic element $h_{n}^{r_{n}}$. Thus $\Lambda$ satisfies the
conditions of Lemma 5.4.

By Conclusion (I) of Lemma 5.4, the group $\Lambda$ is the non-trivial limit
of

$\Lambda_{n}\coloneqq\langle\Lambda,H_{n}\rangle=\Lambda*_{(\Lambda\cap
N)}H_{n}.$

Since $\Lambda_{n}\cap N$ is contained in the edge stabiliser $\Lambda\cap N$
which is in turn contained in $N$, we have $\Lambda\cap N=\Lambda_{n}\cap N$.
By Conclusion (III) of Lemma 5.4, $\Lambda_{n}$ belongs to ${\mathcal{L}}_{T}$
for a subsequence of the $(e_{n})_{n}$ and a proper subtree
$T_{\Lambda_{n}}\subsetneq T$ that does not contain these edges. For any
$k\in\mathbb{N}$, we have
$\operatorname{Stab}_{\Lambda_{n}}(e_{k})=\operatorname{Stab}_{\Lambda}(e_{k})$
(by the “More generally” part of Conclusion (I) of Lemma 5.4) and, since all
the $e_{k}$ belong to $E(Z)$, Equation (12) gives
$\operatorname{Stab}_{\Lambda_{n}}(e_{k})=\Lambda\cap N=\Lambda_{n}\cap N$.
Thus each $\Lambda_{n}$ satisfies the assumptions of Theorem 5.5, thereby
completing the proof that ${\mathcal{N}}_{T}$ has no isolated points. This
concludes the proof of Theorem 5.5. ∎

If $G\curvearrowright T$ is a minimal and irreducible action on a tree, let us
denote by

$\operatorname{Sub}_{|\bullet\backslash
T|\infty}(G):=\\{\Lambda\in\operatorname{Sub}(G)\colon\left|\Lambda\backslash
T\right|=\infty\\}$

the set of subgroups whose action on $T$ has infinitely many orbits of edges.
Observe that $\operatorname{Sub}_{|\bullet\backslash T|\infty}(G)$ as well as
its closure $\overline{\operatorname{Sub}_{|\bullet\backslash T|\infty}(G)}$
are invariant under the action $G\curvearrowright\operatorname{Sub}(G)$.

######  Remark 5.6.

Bass-Serre Theory asserts that if $\Lambda\curvearrowright T$ is a _minimal_
action and if $\Lambda$ is finitely generated, then the quotient graph
$\Lambda\backslash T$ is compact. Indeed, by writing $\Lambda\backslash T$ as
an increasing union of compact connected subgraphs $\kappa_{n}$, we can write
$G$ as the increasing union of the subgroups $G_{n}$ corresponding to the
graphs of groups of $\kappa_{n}$. Since $G$ is finitely generated, this
sequence of subgroups is stationary: $G_{n}=G_{n_{0}}$ for $n$ greater than
some $n_{0}$. The pull-back of $\kappa_{n_{0}}$ is a $G$-invariant cocompact
subtree which coincides with $T$ by minimality.

######  Question 5.7.

What is the class of groups and minimal irreducible actions $G\curvearrowright
T$ such that $\Lambda\leq G$ has infinite index if and only if
$\Lambda\backslash T$ is infinite, i.e. for which

$\operatorname{Sub}_{|\bullet\backslash
T|\infty}(G)=\operatorname{Sub}_{[\infty]}(G)?$

Compare with Remarks 5.10 and 5.16, and with Proposition 5.25.

Thinking about this problem led us to the following question.

######  Question 5.8.

Let $H,A$ be infinite subgroups of the countable group $G$. Assume that $A$ is
amenable and that $\beta_{1}^{(2)}(G)>0$ and $\beta_{1}^{(2)}(H)<\infty$. If
$A\backslash G/H$ is finite, does this imply that $H$ has finite index in $G$?

######  Theorem 5.9 (A segment with finite stabiliser).

Assume $G\curvearrowright T$ is a minimal and irreducible action on a tree.
Assume that there are two edges $f_{1},f_{2}$ of $T$ such that
$\operatorname{Stab}_{G}(f_{1})\cap\operatorname{Stab}_{G}(f_{2})$ is finite.
Then the set $\operatorname{Sub}_{|\bullet\backslash T|\infty}(G)$ has no
isolated point, thus its closure
$\overline{\operatorname{Sub}_{|\bullet\backslash T|\infty}(G)}$ is a perfect
set. In particular:

$\overline{\operatorname{Sub}_{|\bullet\backslash
T|\infty}(G)}\subseteq\operatorname{\mathcal{K}}(G).$

######  Remark 5.10.

Observe that $G\backslash T$ may be infinite in the above theorem and in this
case

${\operatorname{Sub}_{|\bullet\backslash
T|\infty}(G)}=\operatorname{Sub}(G)=\operatorname{\mathcal{K}}(G).$

On the contrary, if $G\backslash T$ is finite, then
${\operatorname{Sub}_{|\bullet\backslash
T|\infty}(G)}\subseteq\operatorname{Sub}_{[\infty]}(G)$.

The subset $\overline{\operatorname{Sub}_{|\bullet\backslash T|\infty}(G)}$ is
invariant under the conjugacy $G$-action. We will see that it moreover admits
a natural finite partition into perfect $G$-invariant topologically transitive
pieces. Indeed, recall from Proposition 3.2 that for any finite normal
subgroup $N$ there is a finite clopen partition of $\operatorname{Sub}(G)$
indexed by the set $\mathrm{Conj}(N)^{G}$ of $G$-conjugacy classes of
subgroups of $N$. This partition induces a $G$-invariant partition of
$\overline{\operatorname{Sub}_{|\bullet\backslash T|\infty}(G)}$:

$\overline{\operatorname{Sub}_{|\bullet\backslash
T|\infty}(G)}=\bigsqcup_{C\in\mathrm{Conj}(N)^{G}}{\mathcal{F}}_{C}^{T}$

where, for each $C\in\mathrm{Conj}(N)^{G}$, the subset

${\mathcal{F}}^{T}_{C}:=\overline{\operatorname{Sub}_{|\bullet\backslash
T|\infty}(G)}\cap\\{\Lambda\in\operatorname{Sub}(G)\colon\Lambda\cap N\in
C\\}$ (13)

is a perfect set by Theorem 5.9.

######  Theorem 5.11 (Topological transitivity).

Suppose that $G\curvearrowright T$ is a minimal and irreducible action on a
tree. Assume that there are two edges $f_{1},f_{2}$ of $T$ such that
$\operatorname{Stab}_{G}(f_{1})\cap\operatorname{Stab}_{G}(f_{2})$ is finite,
thus the kernel $N_{0}$ of the action $G\curvearrowright T$ is finite. Then

  1. (i)

For each $C\in\mathrm{Conj}(N_{0})^{G}$ the action
$G\curvearrowright{\mathcal{F}}^{T}_{C}$ is topologically transitive, and

  2. (ii)

If $C\in\mathrm{Conj}(N_{0})^{G}$ has cardinality 1 then the action
$G\curvearrowright{\mathcal{F}}_{C}^{T}$ is highly topologically transitive.

In particular, if $N_{0}$ is trivial, then the action
$G\curvearrowright\overline{\operatorname{Sub}_{|\bullet\backslash
T|\infty}(G)}$ is highly topologically transitive.

######  Remark 5.12.

The condition “$C\in\mathrm{Conj}(N_{0})^{G}$ has cardinality $1$” means that
$M\in C$ is normalised by $G$; it is satisfied for instance when $N_{0}$ is
cyclic: a subgroup $M\leq N_{0}$ is uniquely defined by its cardinality. It is
also satisfied for some finite index subgroup $G^{\prime}\leq G$ that acts
trivially on $\operatorname{Sub}(N_{0})$ by conjugation. Then $G^{\prime}$
satisfies Theorem 5.11(ii), but for a finer partition.

######  Remark 5.13 (Obstruction to topological $2$-transitivity).

Note that the condition is necessary in part (ii) of Theorem 5.11. Indeed, if
$C$ contains at least two distinct elements $M,M^{\prime}$, then the action
$G\curvearrowright{\mathcal{F}}^{T}_{C}$ is not topologically $2$-transitive:
Consider the four non-empty open sets
$V_{1}=V_{2}=V_{3}={\mathcal{F}}^{T}_{C}\cap\\{\Lambda\in\operatorname{Sub}(G)\colon\Lambda\cap
N_{0}=M\\}$ and
$V_{4}={\mathcal{F}}^{T}_{C}\cap\\{\Lambda\in\operatorname{Sub}(G)\colon\Lambda\cap
N_{0}=M^{\prime}\\}$. For every element $\gamma\in G$ such that $\gamma
V_{1}\cap V_{3}\not=\emptyset$, then $\gamma M\gamma^{-1}=M$ thus $\gamma
V_{2}\cap V_{4}=\emptyset$.

######  Question 5.14.

Do the conclusions of the above two theorems still hold if we replace the
assumption about the edge stabilisers with the assumption that the kernel
$N_{0}$ of $G\curvearrowright T$ is finite and the action of $G/N_{0}$ on the
boundary of $T$ is topologically free?

###### Proof of Theorem 5.9.

Let us start by showing that, up to replacing $f_{2}$ with another edge, we
can assume that
$\operatorname{Stab}_{G}(f_{1})\cap\operatorname{Stab}_{G}(f_{2})=N_{0}$, the
kernel of $G\curvearrowright T$. In the connected component $Z_{2}$ of
$T\smallsetminus\\{f_{2}\\}$ that does not contain $f_{1}$, let us choose an
edge $f_{3}$ such that the intersection
$\operatorname{Stab}_{G}(f_{1})\cap\operatorname{Stab}_{G}(f_{3})$ has minimal
cardinality; this is possible by the finiteness of
$\operatorname{Stab}_{G}(f_{1})\cap\operatorname{Stab}_{G}(f_{2})$ and because
for every edge $f\in Z_{2}$, the edge $f_{2}$ belongs to the segment
$[f_{1},f]$. Let us call
$P:=\operatorname{Stab}_{G}(f_{1})\cap\operatorname{Stab}_{G}(f_{3})$. Let
$Z_{3}$ be the connected component of $T\smallsetminus\\{f_{3}\\}$ that does
not contain $f_{1}$. Every edge $f$ in $Z_{3}$ satisfies

$\operatorname{Stab}_{G}(f_{1})\cap\operatorname{Stab}_{G}(f)=P.$ (14)

By Lemma 5.2, there is an element $\gamma_{0}\in G$ whose axis contains
$f_{3}$: a “half of the axis” of $\gamma_{0}$ is in $Z_{3}$. Up to replacing
$\gamma_{0}$ by its inverse, one can assume that $\gamma_{0}^{-1}\cdot f_{3}$
belongs to $Z_{3}$.

For every edge $e$ of $T$, there is a positive integer $s$ such that
$e\in\gamma_{0}^{r}\cdot Z_{3}$ for every $r\geq s$. Indeed, first note that
$Z_{3}\subseteq\gamma_{0}^{r}Z_{3}$ for all $r\geq 1$. Now if $e\notin Z_{3}$,
let $y$ be the projection of $e$ on the axis of $\gamma_{0}$, i.e. the vertex
of $\mathrm{Axis}(\gamma_{0})$ that is the closest to $o(e)$. For every large
enough positive power $r$ of $\gamma_{0}$, the vertex $y$ lies between $f_{3}$
and $\gamma_{0}^{r}\cdot f_{3}$; and then $y$ and $e$ belong to
$\gamma_{0}^{r+1}\cdot Z_{3}$. In particular, for every $r$ larger than some
$s_{0}$, the segment $[f_{1},f_{3}]$ is contained in $\gamma_{0}^{r}\cdot
Z_{3}$. Every edge $e$ in $\gamma_{0}^{r}\cdot Z_{3}$ satisfies
$\operatorname{Stab}_{G}(\gamma_{0}^{r}\cdot
f_{1})\cap\operatorname{Stab}_{G}(e)=\gamma_{0}^{r}P\gamma_{0}^{-r}$, the
$\gamma_{0}^{r}$-translated version of (14). In particular, for the edges
$f_{1},f_{3}$ and $r\geq s_{0}$: $\operatorname{Stab}_{G}(\gamma_{0}^{r}\cdot
f_{1})\cap\operatorname{Stab}_{G}(f_{1})=\gamma_{0}^{r}P\gamma_{0}^{-r}=\operatorname{Stab}_{G}(\gamma_{0}^{r}\cdot
f_{1})\cap\operatorname{Stab}_{G}(f_{3})$. It follows that
$\gamma_{0}^{r}P\gamma_{0}^{-r}\leq\operatorname{Stab}_{G}(f_{1})\cap\operatorname{Stab}_{G}(f_{3})=P$.
But $P$ and $\gamma_{0}^{r}P\gamma_{0}^{-r}$ have the same finite cardinal,
thus $P=\gamma_{0}^{r}P\gamma_{0}^{-r}$. It follows that $P$ fixes every edge
of $T$ so that $P\leq N_{0}$. Since the kernel $N_{0}$ in contained in both
$\operatorname{Stab}_{G}(f_{1})$ and $\operatorname{Stab}_{G}(f_{3})$, we also
have $P\geq N_{0}$. We have proved that

$\operatorname{Stab}_{G}(f_{1})\cap\operatorname{Stab}_{G}(f_{3})=N_{0}.$ (15)

So, replace $f_{2}$ with $f_{3}$.

If $\Lambda\in\operatorname{Sub}_{|\bullet\backslash T|\infty}(G)$ is not
finitely generated, then $\Lambda$ is a limit of finitely generated subgroups
$\Lambda_{n}\leq\Lambda$ which satisfy a fortiori that $\Lambda_{n}\backslash
T$ is infinite. This is the case, in particular, if $G\backslash T$ is not
finite: by minimality, $G$ is not finitely generated.

It is thus enough to prove that $\Lambda\in\operatorname{\mathcal{K}}(G)$ when
$\Lambda\in\operatorname{Sub}_{|\bullet\backslash T|\infty}(G)$ is finitely
generated, by showing that $\Lambda$ satisfies the conditions of Theorem 5.5.
Since $\Lambda$ is finitely generated, it admits a cocompact (thus proper)
invariant subtree $T_{\Lambda}\subsetneq T$ (see Remark 5.6).

######  Claim 1.

Let $f$ be any edge of $T\smallsetminus T_{\Lambda}$ and let $Z_{f}$ be the
connected component of $T\smallsetminus\\{f\\}$ that does not contain
$T_{\Lambda}$. Let $L$ be the segment $[f_{1},f_{2}]$. One can assume, up to
replacing $L$ with some $G$-translate, that $L\subseteq Z_{f}$ and that
$f_{1}$ separates $f_{2}$ from $f$ and $T_{\Lambda}$.

###### Proof of Claim 1.

Since $N_{0}$ is normal, for any $\gamma\in G$ the pair $(\gamma\cdot
f_{1},\gamma\cdot f_{2})$ satisfies the same assumption as $(f_{1},f_{2})$.
Then Claim 1 is a direct application of the fourth item of Lemma 5.2
(existence of $\gamma_{2}$). ∎

######  Claim 2.

Suppose that $f,Z_{f}$ and $L$ are as in the Claim 1 and define the edge
$e_{\Lambda}:=f_{2}$. Then

$\operatorname{Stab}_{\Lambda}(e_{\Lambda})=\Lambda\cap N_{0}.$ (16)

###### Proof of Claim 2.

If $\lambda$ belongs to $\operatorname{Stab}_{\Lambda}(e_{\Lambda})$, then
(since $T_{\Lambda}$ is $\Lambda$-invariant) the whole path from $e_{\Lambda}$
to $T_{\Lambda}$ must be point-wise fixed by $\lambda$. This path uses the
subpath $L$ whose edges $f_{1}$ and $f_{2}$ satisfy
$\operatorname{Stab}_{\Lambda}(f_{1})\cap\operatorname{Stab}_{\Lambda}(f_{2})\leq\operatorname{Stab}_{G}(f_{1})\cap\operatorname{Stab}_{G}(f_{2})=N_{0}$.
Thus $\lambda\in\Lambda\cap N_{0}$, i.e.
$\operatorname{Stab}_{\Lambda}(e_{\Lambda})\leq\Lambda\cap
N_{0}\leq\operatorname{Stab}_{\Lambda}(e_{\Lambda})$ since $N_{0}$ is the
kernel of the action. Claim 2 is proved. ∎

We have proved that $\Lambda$ (when finitely generated) satisfies the
conditions of Theorem 5.5, i.e. $\Lambda\in{\mathcal{N}}_{T}$, in the
terminology of that theorem. It follows that
$\overline{\operatorname{Sub}_{|\bullet\backslash
T|\infty}(G)}\subseteq\overline{{\mathcal{N}}_{T}}$. Observe that for every
$\Lambda^{\prime}\in{\mathcal{N}}_{T}$, the quotient
$\Lambda^{\prime}\backslash T$ is infinite (one of the components of
$T\smallsetminus\\{e_{\Lambda^{\prime}}\\}$ is sent injectively in
$\Lambda^{\prime}\backslash T$), thus
${\mathcal{N}}_{T}\subseteq\operatorname{Sub}_{|\bullet\backslash
T|\infty}(G)$. Hence

$\overline{\operatorname{Sub}_{|\bullet\backslash
T|\infty}(G)}=\overline{{\mathcal{N}}_{T}}.$ (17)

Recall from Theorem 5.5 that ${\mathcal{N}}_{T}$ has no isolated points with
respect to its induced topology. This concludes the proof of Theorem 5.9. ∎

We now prove Theorem 5.11, keeping the same notation and numbering of claims.

###### Proof of Theorem 5.11.

Fix $C\in\mathrm{Conj}(N_{0})^{G}$.

In order to prove (i), we will have to show that if $V_{1},V_{2}$ are two
arbitrary non-empty open sets of ${\mathcal{F}}^{T}_{C}$, then there is
$\gamma\in G$ such that $\gamma V_{1}\gamma^{-1}\cap V_{2}\not=\emptyset$.

Finitely generated subgroups are dense in
$\operatorname{Sub}_{|\bullet\backslash T|\infty}(G)$ and
$\\{\Lambda\in\operatorname{Sub}(G)\colon\Lambda\cap N_{0}\in C\\}$ is open,
so there exist finitely generated $\Lambda_{i}\in V_{i}$ for $i=1,2$. By
Theorem 5.9, $\operatorname{Sub}_{|\bullet\backslash T|\infty}(G)$ has no
isolated points, so we can assume that
$\Lambda_{1},\Lambda_{2}\notin\operatorname{Sub}(N_{0})$. Moreover, we can
assume that $\Lambda_{1}\cap N_{0}=\Lambda_{2}\cap N_{0}=:M$. Indeed, if
$M_{i}:=\Lambda_{i}\cap N_{0}$ for $i=1,2$ then there exists $\gamma_{2}\in G$
such that $\gamma_{2}M_{2}\gamma_{2}^{-1}=M_{1}$, and we can replace $V_{2}$
by $V^{\prime}_{2}:=\gamma_{2}V_{2}\gamma_{2}^{-1}$: if there exists $\gamma$
such that $\gamma V_{1}\gamma^{-1}\cap V^{\prime}_{2}\not=\emptyset$ then
$\gamma_{2}^{-1}\gamma V_{1}\gamma^{-1}\gamma_{2}\cap V_{2}\neq\emptyset$.

To prove part (ii), we need to show that, if $C=\\{M\\}$ (i.e. $M$ is normal
in $G$) then, for all positive integers $r$ and for all non-empty open sets
$V_{1},\dots,V_{2r}\subseteq\mathcal{F}_{C}$, there exists $\gamma\in G$ such
that $\gamma V_{i}\cap V_{r+i}\neq\emptyset$ for each $i\in\\{1,\dots,r\\}$.
As above, we can pick $\Lambda_{i}\in V_{i}$ that is finitely generated and
not contained in $\operatorname{Sub}(N_{0})$. Moreover $\Lambda_{i}\cap
N_{0}=M$ for each $i$ by definition. Thus both part (i) and part (ii) follow
from the following statement:

Fix $C\in\mathrm{Conj}(N_{0})^{C}$ and $M\in C$ (we don’t assume that
$|C|=1$). Let $r\geq 1$ and $V_{1},\dots V_{2r}\subseteq\mathcal{F}_{C}$ be
open subsets such that, for each $i\in\\{1,\dots,2r\\}$, there exist finitely
generated subgroups $\Lambda_{i}\in V_{i}$ where $\Lambda_{i}\cap N_{0}=M$ and
$\Lambda_{i}\neq M$. Then there is an element $\gamma\in G$ such that $\gamma
V_{i}\cap V_{r+i}\neq\emptyset$ for each $i\in\\{1,\dots,r\\}$.

Let us prove that this statement indeed holds. For any subgroup $H\leq G$ let
$T_{H}$ denote a minimal $H$-invariant subtree of $T$. Recall that either
$T_{H}$ is a single vertex or $T_{H}$ is the union of the axes of the
hyperbolic elements of $H$.

######  Claim 3.

If $T_{H}$ contains a half-tree, i.e. a connected component $Z$ of
$T\smallsetminus\\{e\\}$ for some edge $e\in E(T)$, then $T_{H}=T$. If $H$ is
moreover finitely generated, then $H\backslash T$ is finite.

###### Proof of Claim 3.

Let $f$ be an edge in $Z$ and $h\in H$ a hyperbolic element whose axis
$\mathrm{Axis}(h)$ contains $f$. Then, up to replacing $h$ by $h^{-1}$, the
image $h\cdot e$ belongs to $Z$. Let $g^{\prime}\in H$ be a hyperbolic element
whose axis contains $h\cdot e$. Then the axis of the conjugate
$g=h^{-1}g^{\prime}h\in H$ contains $e$. Thus $e\in T_{H}$ and $T_{H}$
contains all the half-trees $g^{k}(Z)$ for $k\in\mathbb{Z}$. It follows that
$T=\cup_{k}\>g^{k}(Z)=T_{H}$. When $H$ is finitely generated, then (Remark
5.6) there is a finite connected subgraph $W\subseteq H\backslash T$ whose
associated graph of groups has fundamental group equal to $H$. If $H\backslash
T$ is infinite, then the pull-back of $W$ in $T$ is a proper $H$-invariant
subtree of $T$. This proves Claim 3. ∎

######  Claim 4.

Assume that $G\curvearrowright T$ is a minimal and irreducible action on a
tree and that $T_{i}:=T_{\Lambda_{i}}\subsetneq T$ are proper
$\Lambda_{i}$-invariant subtrees for finitely generated subgroups
$\Lambda_{i}\leq G$ for $i\in\\{1,\dots,2r\\}$. Then, there is an edge $e\in
E(T)$ such that $\cup_{i=1}^{2r}T_{i}$ is contained in one connected component
of $T\smallsetminus\\{e\\}$ and such that
$\operatorname{Stab}_{\Lambda_{i}}(e)=M$ for all $i$.

###### Proof of Claim 4.

If $U,V$ are subtrees of $T$ that do not contain any half-tree, then the
convex hull of $U\cup V$ does not contain any half-tree. To see this, pick an
edge $e_{0}\in E(T)$ and let $K$ be one of its half-trees. Since $U$ does not
contain $K$ nor any half-tree therein, there is an edge $e_{1}\in E(K)$ such
that $U$ is in the same half-tree of $T\smallsetminus\\{e_{1}\\}$ as $e_{0}$.
Let $K_{2}$ be the other half-tree of $T\smallsetminus\\{e_{1}\\}$ and note
that $K_{2}\subseteq K$. Since no half-tree is contained in $V$, there exists
an edge $e_{2}\in E(K_{2})$ such that $V$ is in the same half-tree of
$T\smallsetminus\\{e_{2}\\}$ as $e_{1}$ (and $e_{0}$). If $K_{3}\subseteq
T\smallsetminus\\{e_{2}\\}$ is the half-tree which does not contain $e_{1}$
then $K_{3}\subseteq K$ and $K_{3}$ has empty intersection with the convex
hull of $U\cup V$. Thus the half-tree $K$ is not entirely contained in the
convex hull of $U\cup V$.

Since minimal subtrees $T_{i}$ of $\Lambda_{i}$ do not contain any half-tree
by Claim 3, the same holds for their convex hull. It follows by induction that
$\cup_{i=1}^{2r}T_{i}$ is contained in a half-tree of $T\smallsetminus\\{f\\}$
for some $f\in E(T)$. This remains true if we replace $f$ by any edge in the
connected component of $T\smallsetminus\\{f\\}$ which does not contain
$\cup_{i=1}^{2r}T_{i}$.

By Claim 1, one can assume that the segment $L=[f_{1},f_{2}]$ is contained in
the connected component of $T\smallsetminus\\{f\\}$ that does not contain
$\cup_{i=1}^{2r}T_{i}$ and that $f_{1}$ separates $f_{2}$ from $f$, thus from
each $T_{i}$. By Claim 2, the edge $e=f_{2}$ satisfies
$\operatorname{Stab}_{\Lambda_{i}}(e)=\Lambda_{i}\cap N_{0}=M$ for each $i$
and this concludes the proof of Claim 4. ∎

We take an edge $e\in E(T)$ as in the Claim 4 and let $Z$ be the connected
component of $T\smallsetminus\\{e\\}$ that does not contain
$\cup_{i=1}^{2r}T_{i}$. By Lemma 5.2, there is a hyperbolic element $g\in G$
whose axis is contained in $Z$.

We now fix $i\in\\{1,\dots,r\\}$ and show that $\Lambda_{i}$ satisfies the
assumptions of Lemma 5.4 where the $\Lambda_{i}$-invariant proper subtree is
$T_{\Lambda_{i}}=T_{i}$ and the subgroups $H_{n}$ are of the form
$g^{q}\Lambda_{r+i}g^{-q}$. The sequences of subtrees $(S_{n})_{n}$ and of
edges $(e_{n})_{n}$ will be defined shortly.

Since the action by conjugation of $G$ on $\operatorname{Sub}(N_{0})$ has
finite orbits, there is a non-zero integer $k$ such that $g^{k}Mg^{-k}=M$. It
follows that $g^{kn}\Lambda_{r+i}g^{-kn}\cap N_{0}=g^{kn}(\Lambda_{r+i}\cap
N_{0})g^{-kn}=g^{kn}Mg^{-kn}=M$ for every integer $n$.

Let $e_{n}\coloneqq g^{kn}\cdot e$. Then the distance of $e_{n}$ from
$T_{\Lambda_{i}}$ is greater than $n+1$ times the translation length of $g$.
The subtree $S_{n}\coloneqq g^{kn}\cdot T_{r+i}$ is contained in $Z$ and
$e_{n}$ separates $S_{n}$ from $T_{i}$. The subtree $g^{kn}\cdot T_{r+i}$ is
an invariant subtree for the subgroup $H_{n}\coloneqq
g^{kn}\Lambda_{r+i}g^{-kn}$. We have defined $(H_{n})$, $(S_{n})$ and
$(e_{n})$ and checked Assumptions 1, 2, 3a and 3b of Lemma 5.4. The assumption
we made that $\Lambda_{r+i}$ is not a subgroup of $N_{0}$ implies that
$\operatorname{Stab}_{H_{n}}(e_{n})\not=H_{n}$, thus checking Assumption 3d.
If an element $\lambda\in\Lambda_{i}$ sends some $e_{n}$ to an edge
$\lambda\cdot e_{n}$ belonging to the path from some other $e_{k}$ to $T_{i}$
then $\lambda$ must fix the projection $p$ of $e_{k}$ in $T_{i}$ and
$\lambda\cdot e_{n}$ belongs to $Z$. It follows that $\lambda$ must fix the
edge $e$ and thus $\lambda\in\operatorname{Stab}_{\Lambda_{i}}(e)=M\leq
N_{0}$. So $\Lambda_{i}\cdot e_{n}$ does not separate $e_{k}$ from the tree
$T_{i}$. Therefore, $(e_{n})$ verifies condition 3e of Lemma 5.4. Concerning
the stabilisers:
$\operatorname{Stab}_{\Lambda_{i}}(e_{n})=g^{kn}\operatorname{Stab}_{\Lambda_{i}}(e)g^{-kn}=g^{kn}Mg^{-kn}=M$
while
$\operatorname{Stab}_{H_{n}}(e_{n})=\operatorname{Stab}_{g^{kn}\Lambda_{r+i}g^{-kn}}(g^{kn}\cdot
e)=\operatorname{Stab}_{\Lambda_{r+i}}(e)=M$. The Assumption 3c is satisfied.

Thus, by Lemma 5.4 (I) and (II),
$\Delta^{i}_{n}\coloneqq\langle\Lambda_{i},g^{kn}\Lambda_{r+i}\,g^{-kn}\rangle=\Lambda_{i}*_{M}g^{kn}\Lambda_{r+i}\,g^{-kn}$
and we have the non-trivial convergence:

$\Delta^{i}_{n}=\Lambda_{i}*_{M}g^{kn}\Lambda_{r+i}\,g^{-kn}\underset{n\to\infty}{\longrightarrow}\Lambda_{i}.$

Since $\\{\Lambda\leq G\colon\Lambda\cap N_{0}=M\\}$ is open, we have
$\Delta^{i}_{n}\cap N_{0}=\Lambda_{i}\cap N=M$ for all large enough $n$. By
(V) of Lemma 5.4, the quotient $\Delta^{i}_{n}\backslash T$ is infinite, i.e.
$\Delta^{i}_{n}\in\operatorname{Sub}_{|\bullet\backslash T|\infty}(G)$. Thus
$\Delta^{i}_{n}\in{\mathcal{F}}^{T}_{C}$. It follows that for $n$ larger than
some $n_{i}$, the subgroup $\Delta^{i}_{n}\in V_{i}$.

If remains to show that we have the non-trivial convergence:

$g^{-kn}\Delta^{i}_{n}\,g^{kn}=g^{-kn}\Lambda_{i}\,g^{kn}*_{M}\Lambda_{r+i}\underset{n\to\infty}{\longrightarrow}\Lambda_{r+i}.$

It is enough to observe that
$g^{-kn}\Delta^{i}_{n}\,g^{kn}=g^{-kn}\Lambda_{i}\,g^{kn}*_{M}\Lambda_{r+i}$
is obtained by the exact symmetric construction by exchanging the roles of
$\Lambda_{i},\Lambda_{r+i}$ and replacing $g$ by $g^{-1}$. So the symmetric
application of Lemma 5.4 (II) shows that
$\lim_{n\to\infty}g^{-kn}\Delta^{i}_{n}\,g^{kn}=\Lambda_{r+i}$. It follows as
above that for all $n$ larger than some $n_{r+i}$, the subgroup
$g^{-kn}\Delta^{i}_{n}\,g^{kn}\in V_{r+i}$.

Hence, for any $n\geq\max\\{n_{j}:j\in\\{1,\dots,2r\\}\\}$ and for each
$i\in\\{1,\dots,r\\}$, we have:

$\Delta^{i}_{n}\in V_{i}$ and $g^{-kn}\Delta^{i}_{n}\,g^{kn}\in V_{r+i}$, and
so $g^{-kn}V_{i}\,g^{kn}\cap V_{r+i}\neq\emptyset$.

This completes the proof of Theorem 5.11. ∎

###  5.3 Applications

Recall that the $G$-action $G\curvearrowright T$ on a tree is acylindrical if
there is $R>0$ such that the stabilisers of any path of length at least $R$ is
trivial.

Examples of $3$-acylindrical actions include the actions on the Bass-Serre
tree of non-trivial amalgamated free products $A\ast_{C}B$ where $C$ is
malnormal in $A$ (i.e. for all $g\in A\smallsetminus C$, $C\cap gCg^{-1}$ is
trivial).

######  Corollary 5.15 (Acylindrical actions).

Let $G\curvearrowright T$ be a minimal irreducible and acylindrical action of
a countable group on a tree. If $\Lambda\leq G$ is a subgroup such that the
quotient graph $\Lambda\backslash T$ is infinite, then
$\Lambda\in\operatorname{\mathcal{K}}(G)$. Moreover,
$G\curvearrowright\overline{\operatorname{Sub}_{|\bullet\backslash
T|\infty}(G)}$ is highly topologically transitive.

###### Proof.

The $G$-action on $T$ satisfies the assumptions of Theorem 5.9 and Theorem
5.11: the intersection of the stabilisers of any pair of edges at distance
greater than the acylindricity constant $R$ is trivial. ∎

######  Remark 5.16.

There exist acylindrical actions $G\curvearrowright T$ as in Corollary 5.15
where $G$ admits a finitely generated subgroup $\Lambda\leq G$ with infinite
index but where the quotient $\Lambda\backslash T$ is finite. Moreover, there
are examples where $\Lambda$ belongs to $\operatorname{\mathcal{K}}(G)$ and
others where it does not.

For example, let $C$ be any finitely generated infinite torsion-free group and
let $A,B$ be finitely generated non-trivial groups such that $C$ is malnormal
in $G_{1}=A\rtimes C$ and $G_{2}=B\rtimes C$ respectively. Such groups exist
by D. Osin’s Appendix in [HW14, Theorem 15].

Then let $G=G_{1}*_{C}G_{2}$ where the injections $C\hookrightarrow G_{i}$ are
given by the natural inclusion maps and consider the action of $G$ on the
Bass-Serre tree $T$ associated to this decomposition. It follows from the
malnormality of $C$ in $G_{1}$ and $G_{2}$ that the action $G\curvearrowright
T$ is $2$-acylindrical. Moreover, the normal subgroup $\Lambda=\langle
A,B\rangle\trianglelefteq G$ is finitely generated, has infinite index in $G$
since $G/\langle A,B\rangle\simeq C$, and acts transitively on the set of
edges of $T$, so $\Lambda\backslash T$ consists of a single edge.

Moreover, by Proposition 3.5, the finitely generated normal subgroup
$\Lambda=\langle A,B\rangle=A*B$ belongs to the perfect kernel
$\operatorname{\mathcal{K}}(G)$ if and only if the trivial group
$\\{\mathrm{id}\\}$ belongs to the perfect kernel
$\operatorname{\mathcal{K}}(C)$. If we choose for instance $C=\mathbb{Z}$ then
$\Lambda\not\in\operatorname{\mathcal{K}}(G)$ while if $C=\mathbf{F}_{2}$,
then $\Lambda\in\operatorname{\mathcal{K}}(G)$.

The results of the previous section imply that:
If $G$ is a finitely generated non-elementary free product111A free product
$G=A\ast B$ is non-elementary if $|A|,|B|\geq 2$ and at least one of $A$ and
$B$ has cardinality at least $3$. then
$\operatorname{\mathcal{K}}(G)=\operatorname{Sub}_{[\infty]}(G)$ and the
action of $G$ on its perfect kernel is highly topologically transitive. More
generally,

######  Corollary 5.17 (Infinitely many ends).

If $G$ is a finitely generated group with infinitely many ends, then the
perfect kernel of $G$ consists of its subgroups of infinite index:

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

Moreover, $G$ has a greatest finite normal subgroup $N_{0}$ and the set
$\mathrm{Conj}(N_{0})^{G}$ defines a finite clopen partition of
$\operatorname{\mathcal{K}}(G)$ into $G$-invariant and topologically
transitive perfect subsets:

$\operatorname{\mathcal{K}}(G)=\bigsqcup_{C\in\mathrm{Conj}(N_{0})^{G}}\mathcal{S}_{C}\text{,
\ \ where \
}\mathcal{S}_{C}:=\operatorname{Sub}_{[\infty]}(G)\cap\left\\{\Lambda\colon\Lambda\cap
N_{0}\in C\right\\}.$

If $N_{0}=\\{\mathrm{id}\\}$, resp. $|C|=1$, then
$G\curvearrowright\operatorname{\mathcal{K}}(G)$, resp.
$G\curvearrowright\mathcal{S}_{C}$, is highly topologically transitive.

###### Proof.

By Stallings’ theorem [Sta68, Sta71], $G$ admits a minimal and irreducible
action $G\curvearrowright T$ on a tree $T$ with finite edge stabilisers. If
$M$ is a finite normal subgroup of $G$, then it is contained in the kernel of
the action $G\curvearrowright T$. Conversely, this kernel is finite and
normal: this is $N_{0}$. If $\Lambda\in\operatorname{Sub}(G)$ has infinite
index, then the image in $\Lambda\backslash T$ of the $G$-orbit any edge $e$
of $T$ is in bijection with the double classes $\Lambda\backslash
G/\operatorname{Stab}_{G}(e)$, which is infinite since
$\operatorname{Stab}_{G}(e)$ is finite. Thus $\Lambda\backslash T$ is infinite
and $\mathcal{S}_{C}$ coincides with ${\mathcal{F}}^{T}_{C}$ (defined in
Equation 13). Since the edge stabilisers are finite, any two distinct edges
$f_{1},f_{2}$ of $T$ satisfy a fortiori that
$\operatorname{Stab}_{G}(f_{1})\cap\operatorname{Stab}_{G}(f_{2})$ is finite.
We conclude by applying Theorem 5.9 and Theorem 5.11. ∎

######  Example 5.18 ($\mathrm{SL}(2,\mathbb{Z})$).

This corollary applies for instance to $G=\mathrm{SL}(2,\mathbb{Z})$ for which
$N_{0}=\\{\pm\mathrm{id}\\}\simeq\mathbb{Z}/2\mathbb{Z}$ is the centre. The
perfect kernel decomposes into two invariant pieces

$\operatorname{\mathcal{K}}(G)=\operatorname{Sub}_{[\infty]}(\mathrm{SL}(2,\mathbb{Z}))=\left\\{\Lambda\colon
N_{0}\leq\Lambda\right\\}\sqcup\left\\{\Lambda\colon
N_{0}\not\leq\Lambda\right\\}$

on which the action is highly topologically transitive.

######  Corollary 5.19.

The action of the free group $\mathbf{F}_{\mathbb{N}}$ on countably many
generators on its perfect kernel
$\operatorname{\mathcal{K}}(\mathbf{F}_{\mathbb{N}})=\operatorname{Sub}(\mathbf{F}_{\mathbb{N}})$
is highly topologically transitive.

###### Proof.

By Proposition 3.8,
$\operatorname{\mathcal{K}}(\mathbf{F}_{\mathbb{N}})=\operatorname{Sub}(\mathbf{F}_{\mathbb{N}})$.
Every non-empty open subset $V$ of
$\operatorname{Sub}(\mathbf{F}_{\mathbb{N}})$ contains a finitely generated
subgroup, thus for every $2r$-tuple $(V_{1},V_{2},\cdots,V_{2r})$ of non-empty
open sets, there is a finite subset $J\Subset\mathbb{N}$ with $|J|\geq 2$ such
that $V_{i}\cap\operatorname{Sub}_{[\infty]}(\mathbf{F}_{J})\not=\emptyset$
for all $i$. By Corollary 5.17, there is $g\in\mathbf{F}_{J}$ such that
$g\cdot(V_{i}\cap\operatorname{Sub}_{[\infty]}(\mathbf{F}_{J}))\cap(V_{r+i}\cap\operatorname{Sub}_{[\infty]}(\mathbf{F}_{J}))\not=\emptyset$
for all $i$; in particular, $g\cdot V_{i}\cap V_{r+i}\not=\emptyset$. ∎

######  Example 5.20 (Weakly malnormal edge group).

A subgroup $C\leq G$ is said to be weakly malnormal if there exists $g\in G$
such that $gCg^{-1}\cap C$ is finite. Let $G$ be a free product with
amalgamation $G=A*_{C}B$ or a HNN-extension $G=A*_{C}$ with irreducible
associated Bass-Serre tree $T$. Assume that $C$ is weakly malnormal in $G$. It
follows that the action of $G$ on $T$ verifies the conditions of Theorem 5.9
(and of Theorem 5.11). Thus every subgroup $\Lambda$ such that
$\Lambda\backslash T$ is infinite belongs to $\operatorname{\mathcal{K}}(G)$.

For instance, if $A=\mathbf{F}_{p}$ is a non-abelian free group, then any
finitely generated subgroup $C$ with infinite-index is weakly malnormal in $A$
and thus in $G=A*_{C}B$ for any $B$. More generally, by [PT11, Theorem 5.12]
(with $\beta_{1}^{(2)}(.)$ the first $\ell^{2}$-Betti numbers): If $C\leq G$
is an infinite index subgroup with $\beta_{1}^{(2)}(C)<\infty$ and if
$\beta_{1}^{(2)}(G)>0$, then $C$ is weakly malnormal .

This leads to plenty of examples. Take $A\geq C$ and $B\geq C$ such that
$[A:C]\geq 2$ and $[B:C]\geq 3$: this will ensure the irreducibility of
$G\curvearrowright T$ by Proposition 5.1. Assume that
$\beta_{1}^{(2)}(C)<\infty$ and
$\beta_{1}^{(2)}(A)+\beta_{1}^{(2)}(B)-\beta_{1}^{(2)}(C)>0$: an application
of the Mayer-Vietoris formula from Cheeger-Gromov [CG86, p. 204, §4] gives
(when $C$ is infinite)
$\beta_{1}^{(2)}(A*_{C}B)\geq\beta_{1}^{(2)}(A)+\beta_{1}^{(2)}(B)-\beta_{1}^{(2)}(C)$.
Thus $C$ is weakly malnormal in $G=A*_{C}B$ and Theorems 5.9 and 5.11 apply.

###  5.4 A rich class of groups with maximal perfect kernel

In their study of the subgroups of direct products of limit groups, Bridson,
Howie, Miller, and Short [BHMS09] introduced a class $\mathcal{C}$ of finitely
presented groups defined in a hierarchical manner as the union of classes
$\mathcal{C}_{n}$ and proved a series of properties of this class, in
particular their Proposition 5.2. A close look at their proof reveals that
various properties that are of interest to us are verified by the larger class
$\mathcal{Q}$ defined below, with essentially the same arguments. Compare with
[BHMS09, Section 2.2]: $\mathcal{C}_{0}$ is the class of non-elementary free
products of finitely presented groups, and a group lies in $\mathcal{C}_{n}$
if and only if it is the fundamental group of a finite, acylindrical graph of
finitely presented groups, where all of the edge groups are cyclic, and at
least one of the vertex groups lies in $\mathcal{C}_{n-1}$.

######  Definition 5.21 (The classes $\mathcal{Q}$ and
$\mathcal{Q}^{\prime}$).

The class $\mathcal{Q}_{0}$ consists of finitely generated groups with
infinitely many ends. A group lies in $\mathcal{Q}_{n}$ if it is the
fundamental group of a finite graph of groups such that

  * •

every vertex group is finitely generated,

  * •

every edge group is finitely generated virtually abelian,

  * •

at least one vertex group lies in $\mathcal{Q}_{n-1}$ and

  * •

the associated Bass-Serre tree $T$ admits a pair $(f_{1},f_{2})$ of edges
whose stabilisers have finite intersection.

The class $\mathcal{Q}$ is the union of the classes $\mathcal{Q}_{n}$. The
class $\mathcal{Q}^{\prime}$ is the subclass of groups in $\mathcal{Q}$
without any non-trivial finite normal subgroup.

By Stallings’ Theorem a group in $\mathcal{Q}_{0}$ admits a minimal
irreducible action on a tree with a single orbit of edges and with finite edge
stabilisers. We call any such $G$-tree a “Stallings tree” of $G$. A $G$-tree
$T$ witnessing that $G\in\mathcal{Q}$ will be called “the” Bass-Serre tree of
$G$, although it is not uniquely defined.

Observe that the class $\mathcal{Q}$ does not contain any infinite virtually
abelian group as all its elements contain a non-abelian free subgroup.

######  Example 5.22.