2. Preliminaries

2·1. Quasi-2-transitive actions and finite bi-index

In this section, we record a few facts about quasi-2-transitive actions and their relation with subgroups of finite bi-index that we will need for the proof of Theorem A.

Recall from the introduction that an action of a group G on a set X is said to be quasi-2-transitive if the number of orbits of the diagonal action of G on $X^2$ is finite. Of course, any action on a finite set is trivially quasi-2-transitive, so in what follows, we will only concern ourselves with actions on infinite sets. It is an elementary and well-known fact that the existence of such an action is equivalent to the existence of an infinite index subgroup of finite bi-index. We include here a proof for completeness.

Definition 2·1. Let G be a group and let $H\leq G$ be a subgroup. We say that H has finite bi-index if the set $H\backslash G / H$ of double cosets of H is finite.

Proposition 2·2. A group G admits a quasi-2-transitive action on an infinite set if and only if there exists an infinite index subgroup $H\lt G$ of finite bi-index.

Proof. Let $H\lt G$ be a subgroup of infinite index and let us consider the action of G on the infinite set $G/H$ . We claim that this action is quasi-2-transitive if and only if H is of finite bi-index in G. Indeed, for $g_1,g_2,g_3,g_4\in G$ , the couples $(g_1H,g_2H)$ and $(g_3H,g_4H)$ are in the same orbit under the diagonal action of G on $G/H\times G/H$ if and only if $(H, g_1^{-1}g_2H)$ is in the same orbit as $(H, g_3^{-1}g_4H)$ , which is the case if and only if $g_3^{-1}g_4\in H g_1^{-1}g_2H$ . Therefore, the number of orbits is finite if and only if the set $H\backslash G / H$ of double cosets is finite.

To conclude the proof, we simply remark that if G admits a quasi-2-transitive action on an infinite set X, then there must exist some $x\in X$ whose orbit under the action of G is infinite. Then, by letting $H=\mathrm{St}_G(x)$ be the stabiliser of this point, we notice that H is of infinite index and that the action of G on $G/H$ is also quasi-2-transitive.

In light of the previous proposition, the following definition, due to de Cornulier [Reference de Cornulier11, Definition 3·9], will be very useful when investigating the existence of quasi-2-transitive actions on infinite sets.

Definition 2·3. Let G be a group. We say that G has property BF if every subgroup of finite bi-index has finite index.

Thus, a group admits a quasi-2-transitive action on an infinite set if and only if it does not have property BF.

In [Reference de Cornulier11], de Cornulier studied various consequences of property BF and different obstructions for a group to be of finite bi-index. Among these, we record here the fact that subgroups of finite bi-index, while not necessarily of finite index, must be contained in finitely many subgroups, a fact which we will make use of later on, in the proof of Lemma 4·1. We include a proof for completeness.

Proposition 2·4 ([11, Lemma 3·2]). Let G be a group and let H be a subgroup of finite bi-index. Then, H is almost maximal, meaning that there are only finitely many subgroups of G containing H.

Proof. Let $L\leq G$ be a subgroup containing H. Then, for every $g\in L$ , we have $HgH\subseteq L$ . As H is of finite bi-index in G, there are only finitely many, let us say n, different double cosets, from which we conclude that there can be at most $2^n$ different subgroups of G containing H.

If H is a subgroup of finite index of some group G, and if $K\leq G$ is any subgroup, then by elementary group theory, $H\cap K$ is a finite index subgroup of K. Under some slightly stronger assumptions on K, a similar and equally elementary fact holds for finite bi-index subgroup, which we record here for later use.

Proposition 2·5. Let G be a group and let $H\leq G$ be a subgroup of finite bi-index. If $K\leq G$ is such that $H\cap K$ is of finite index in H (which is the case, in particular, if K is of finite index in G), then $H\cap K$ is of finite bi-index in K.

Proof. Let n be the bi-index of H in G and let m be the index of $H\cap K$ in H. We claim that the bi-index of $H\cap K$ in K is at most $nm^2$ . Let us choose a transversal $T=\{t_1, \dots, t_n\}$ for the double cosets of H in G and left and right transversals $L=\{l_1,\dots, l_m\}$ and $R=\{r_1,\dots, r_m\}$ for the cosets of $H\cap K$ in H. Now, let $g\in K$ be an arbitrary element. Then, there exist $i\in \{1,\dots, n\}$ , $j,k\in \{1,\dots, m\}$ and $h_1,h_2\in H\cap K$ such that $g=h_1r_jt_il_kh_2$ . Thus, we conclude that

\[K = \bigcup_{i,j,k} (H\cap K) r_jt_il_k (H\cap K),\]

where the union is taken over all i,j,k such that $r_jt_il_k\in K$ . The result follows.

2·2. Branch groups

In this section, we recall the definition of branch groups and collect a few of their properties that will be useful in the sequel. We refer the reader to [Reference Bartholdi, Grigorchuk and Šuni2] for a more detailed introduction to branch groups.

2·2·1. Spherically homogeneous rooted trees

Recall that a tree is a connected graph that does not contain any cycles (or in other words, a graph such that any pair of vertices can be joined by a unique path without backtracking). A rooted tree is a tree with a distinguished vertex called the root. By using the combinatorial distance from the root, one can partition the vertex set of any rooted tree into spheres centred at the root, which are called levels. A rooted tree is said to be spherically homogeneous if all the vertices on the same level have the same degree.

It will be convenient in what follows to work with specific spherically homogeneous rooted trees, in which the vertex and edges sets are given concretely. To this end, let $(A_i)_{i\in \mathbb{N}}$ be a sequence of finite sets, that we will call alphabets, such that $|A_i|\geq 2$ for every $i\in \mathbb{N}$ . We will equip the set of all words in these alphabets (a notion that we will define just below) with a graph structure, turning it into a spherically homogeneous rooted tree. For every $k\in \mathbb{N}_{\gt0}$ , we define the set

\[(A_i)_{i\in \mathbb{N}}^{k} = A_0 \times A_{1} \times \dots \times A_{k-1}.\]

An element $v=(a_0, a_{1}, \dots, a_{k-1})$ of this set will be called a word of length k in the alphabet $(A_{i})_{i\in \mathbb{N}}$ . By convention, we set $(A_i)_{i\in\mathbb{N}}^0 = \{\epsilon\}$ , where $\epsilon$ is an element that we will call the empty word. We will denote by

\[(A_i)_{i\in \mathbb{N}}^{*} = \sqcup_{k=0}^{\infty} (A_i)_{i\in\mathbb{N}}^k\]

the set of all words in the alphabet $(A_i)_{i\in \mathbb{N}}$ . For $v\in (A_i)_{i\in \mathbb{N}}^*$ , we will call the length of v, denoted by $|v|$ , the unique $k\in \mathbb{N}$ such that $v\in (A_i)_{i\in \mathbb{N}}^k$ . In other words, if $|v|=k$ , then there exist $a_i\in A_i$ for all $0\leq i \lt k$ such that $v=(a_0, a_1, \dots, a_{k-1})$ .

For any given $k\in \mathbb{N}$ , there is a natural operation of concatenation between a word in $(A_i)_{i\in \mathbb{N}}^k$ and a word in $(A_{i+k})_{i\in \mathbb{N}}^*$ . Indeed, if $v=(a_0, \dots, a_{k-1})$ is a word of length k in the alphabet $(A_i)_{i\in \mathbb{N}}$ and $w=(b_0, \dots, b_{l-1})$ is a word of length l in the alphabet $(A_{i+k})_{i\in \mathbb{N}}$ , then we can define the word $vw=(a_0,\dots, a_{k-1}, b_0, \dots, b_{l-1})$ , which is a word of length $k+l$ in the alphabet $(A_i)_{i\in \mathbb{N}}$ . By convention, for any $v\in (A_i)_{i\in \mathbb{N}}^*$ , we have $\epsilon v = v\epsilon = v$ .

Let $v,w\in (A_i)_{i\in \mathbb{N}}^*$ be two words such that $|v|\leq |w|$ . We say that v is a prefix of w if there is a word $v'\in (A_{i+|v|})_{i\in \mathbb{N}}^*$ such that $vv'=w$ . Using prefixes, one can define a partial order on the set $(A_i)_{i\in \mathbb{N}}^*$ . Indeed, we will say that $v\leq w$ if v is a prefix of w. Notice that the empty word is a prefix of every word and is thus the least element of this partial order. If $v,w\in (A_i)_{i\in \mathbb{N}}^*$ are two words, we will say that v is an ancestor of w and that w is a descendant of v if $v\leq w$ . In the particular case where $v\leq w$ with $|v|+1=|w|$ , we will say that v is the parent of w and that w is a child of v.

Let $T=((A_i)_{i\in \mathbb{N}}^*, E, \epsilon)$ be the rooted graph whose vertex set is $(A_i)_{i\in \mathbb{N}}^*$ , whose root is $\epsilon$ and whose edge set is

\[E=\big\{ \{v,w\}\subset (A_i)_{i\in \mathbb{N}}^* \bigm\vert v\leq w, |w|=|v|+1 \big\}.\]

It is easy to deduce from the definition that the graph thus constructed is a rooted tree and that the length of a vertex in this tree (in the sense of the length of words defined above) corresponds to the combinatorial distance between the root and this vertex. Thus, for $n\in \mathbb{N}$ , the set $(A_i)_{i\in \mathbb{N}}^n$ of all vertices of length nth is the $n^\mathrm{th}$ level of the tree, and we will often prefer to denote this set by $\mathcal L_{n}$ . By an abuse of notation, we will also frequently write $v\in T$ instead of $v\in (A_i)_{i\in \mathbb{N}}^*$ to mean that v is a vertex of the rooted tree T. It is immediate from the construction of T that two vertices on the same level always have the same degree. Thus, T is a spherically homogeneous rooted tree.

It is easy to see that any spherically homogeneous rooted tree whose vertices have at least two children is isomorphic to a rooted tree of the form $T=((A_i)_{i\in \mathbb{N}}^*, E, \epsilon)$ as described above, so that no generality is lost when considering only trees of this form. In what follows, when we say rooted tree, we will always mean a spherically homogeneous rooted tree of the form $T=((A_i)_{i\in \mathbb{N}}^*, E, \epsilon)$ for some sequence of alphabets $(A_i)_{i\in \mathbb{N}}$ such that $|A_i|\geq 2$ for all $i\in \mathbb{N}$ , even though we will not generally mention this sequence of alphabets explicitly.

2·2·2. Rooted subtrees

Let $T=((A)_{i\in \mathbb{N}}^*, E, \epsilon)$ be a spherically homogeneous rooted tree, as described in the above section. For any vertex $v\in T$ , we can define a rooted subtree $T_v$ of T as the subgraph spanned by all the descendants of v, rooted at v. Thus, every vertex of $T_v$ is a word of the form vw for some $w\in (A_{i+|v|})_{i\in \mathbb{N}}^*$ . Let us denote by $T_{|v|} = ((A_{i+|v|})_{i\in\mathbb{N}}^*, E, \epsilon)$ the rooted tree corresponding to the sequence of alphabets $(A_{i+|v|})_{i\in \mathbb{N}}$ . It is clear that we have an isomorphism of rooted trees between $T_v$ and $T_{|v|}$ obtained simply by deleting the prefix v from any vertex of $T_v$ . Using this, we also have a canonical isomorphism between $T_v$ and $T_w$ for any $v,w\in T$ on the same level, obtained simply by replacing the prefix v with the prefix w. In what follows, we will frequently make use of these isomorphisms without mentioning them.

2·2·3. Automorphisms of rooted trees

Let $T=((A)_{i\in \mathbb{N}}^*, E, \epsilon)$ be a spherically homogeneous rooted tree, as described in section 2·2·1. An automorphism of T is simply an automorphism of graphs preserving the root. The group of all automorphisms of T will be denoted by $\mathrm{Aut}(T)$ .

Notice that an automorphism of T preserves the length of the vertices and the partial order. It follows that there is a well-defined map

\[\varphi_v\colon \mathrm{Aut}(T) \longrightarrow \mathrm{Aut}(T_{|v|})\]

such that for every $f\in \mathrm{Aut}(T)$ and $w\in T_{|v|}$ , we have

\[f(vw) = f(v)\varphi_v(f)(w).\]

Note that this map is in general not a group homomorphism. Indeed, for $g_1,g_2\in \mathrm{Aut}(T)$ and $v\in T$ , we have

\[\varphi_v(g_1g_2) = \varphi_{g_2v}(g_1)\varphi_v(g_2).\]

From the above formula, however, it follows that it does become a homomorphism when restricted to the stabiliser of the vertex v. From the definition, we also see easily that if $v\in T$ and $w\in T_{|v|}$ , then $\varphi_{vw} = \varphi_w \circ \varphi_v$ .

For $f\in \mathrm{Aut}(T)$ and $v\in T$ , the automorphism $\varphi_v(f)$ is often called the section of f at v. For this reason, we will call the map $\varphi_v$ the section map at v.

2·2·4. Subgroups of automorphisms of rooted trees

Let $G\leq \mathrm{Aut}(T)$ be a group of automorphisms of the rooted tree T. We can define many subgroups of G related to its action on T, which we do below.

For $v\in T$ , we will denote by $\mathrm{St}_G(v)$ the stabiliser of v in G. For $n\in \mathbb{N}$ , we will denote by $\mathrm{St}_G(n) = \bigcap_{v\in \mathcal L_{n}}\mathrm{St}_G(v)$ the subgroup of G fixing every vertex of the $n^\mathrm{th}$ level of T. We will call this subgroup the stabiliser of the $n^\mathrm{th}$ level, or the $n^\mathrm{th}$ -level stabiliser.

For $v\in T$ , the rigid stabiliser of v in G, denoted by $\mathrm{Rist}_G(v)$ , is the subgroup

\[\mathrm{Rist}_G(v) = \left\{g\in G \mid gw=w \quad \forall w\notin T_v \right\}.\]

In other words, the rigid stabiliser of a vertex v is the subgroup of all elements of G whose support is contained in the rooted subtree $T_v$ . In particular, notice that $\mathrm{Rist}_G(v) \leq \mathrm{St}_G(|v|) \leq \mathrm{St}_G(v)$ .

For $n\in \mathbb{N}$ , the rigid stabiliser of the $n^\mathrm{th}$ level, denoted by $\mathrm{Rist}_G(n)$ , is the subgroup of G generated by the rigid stabilisers of all vertices on the $n^\mathrm{th}$ level. Notice that, as elements in the rigid stabilisers of two different vertices on the same level have disjoint support, they commute, and thus we have

\[\mathrm{Rist}_G(n) \cong \prod_{v\in \mathcal L_{n}}\mathrm{Rist}_G(v).\]

2·2·5. Branch groups and their properties

We are now almost ready to define branch groups, which we do immediately after the following definition.

Definition 2·6. Let T be a spherically homogeneous tree, as defined in section 2·2·1, and let $G\leq \mathrm{Aut}(T)$ be a group of automorphisms of T. Then, G is said to act spherically transitively if the action of G on $\mathcal L_{n}$ is transitive for every $n\in \mathbb{N}$ .

Definition 2·7. Let T be a spherically homogeneous rooted tree as defined in section 2·2·1, and let $G\leq \mathrm{Aut}(T)$ be a group of automorphisms of T. Then, G is a branch group if:

1. (i) G acts spherically transitively on T;

2. (ii) for every $n\in \mathbb{N}$ , the subgroup $\mathrm{Rist}_G(n)$ is of finite index in G.

If, instead of (ii), the group G satisfies the weaker condition:

(ii^′) for every $n\in \mathbb{N}$ , the subgroup $\mathrm{Rist}_G(n)$ is non-trivial,

then G is said to be weakly branch.

In the rest of this section, we collect a few properties of branch groups that we will need later on. We begin with the following two lemmas, which are valid for any spherically transitive group. They show how spherical transitivity behaves when passing to sections and subgroups of finite index.

Lemma 2·8 . Let T be a spherically homogeneous rooted tree and let $G\leq \mathrm{Aut}(T)$ be a spherically transitive group of automorphisms of T. Then, for any $v\in T$ , the group $G_v = \varphi_v(\mathrm{St}_G(v))$ is a spherically transitive group of automorphisms of $T_{|v|}$ .

Proof. If G acts spherically transitively on T, then $\mathrm{St}_G(v)$ must act spherically transitively on $T_v$ , from which it follows that $G_v$ acts transitively on $T_{|v|}$ .

Lemma 2·9. Let T be a spherically homogeneous rooted tree and let $G\leq \mathrm{Aut}(T)$ be a spherically transitive group of automorphisms of T. Let $H\leq G$ be a subgroup of finite index of G. Then, there exists $N\in \mathbb{N}$ such that $\mathrm{St}_H(v)$ acts spherically transitively on $T_v$ for all $v\in \mathcal L_{N}$ .

Proof. For each $n\in \mathbb{N}$ , let l(n) denote the number of orbits of the action of H on $\mathcal L_{n}$ . Notice that if two vertices on a given level are not in the same orbit, then their descendants must also be in different orbits. It follows that the sequence $(l(n))_{n\in \mathbb{N}}$ is non-decreasing. On the other hand, since G acts transitively on $\mathcal L_{n}$ for all $n\in \mathbb{N}$ , the number of orbits in $\mathcal L_{n}$ under the action of H is bounded from above by the index of H in G. Therefore, the sequence $(l(n))_{n\in \mathbb{N}}$ must eventually stabilise.

Let $N\in \mathbb{N}$ be such that $l(n)=l(N)$ for all $n\geq N$ . For every $n\geq N$ , one can partition $\mathcal L_{n}$ into l(N) sets by saying that two vertices belong to the same set if their prefixes of length N are in the same H-orbit. These sets are obviously H-invariant, and since $l(n)=l(N)$ , it follows that H must act transitively on each of these sets. In particular, two vertices with the same prefix of length N must belong to the same H-orbit. It follows that $\mathrm{St}_H(v)$ acts spherically transitively on $T_v$ for all $v\in \mathcal L_{N}$ .

Note that one could formulate a similar version of the previous lemma in the more general context of groups with minimal actions by homeomorphisms on a topological space, but we will not require such generality here.

In branch or weakly branch groups, taking sections again yields a branch or weakly branch group, as the next lemma shows.

Lemma 2·10. Let G be a group of automorphisms of a spherically homogeneous rooted tree T and let $v\in T$ be any vertex. Then, for any $n\geq |v|$ , we have

\[\varphi_v(\mathrm{Rist}_G(n))\leq \mathrm{Rist}_{G_v}(n-|v|).\]

where $G_v=\varphi_v(\mathrm{St}_G(v))\leq \mathrm{Aut}(T_{|v|})$ . In particular, if G is a weakly branch group, then $G_v$ is also a weakly branch group, and if G is a branch group, then so is $G_v$ .

Proof. Let v and n be as above, and let $g\in \mathrm{Rist}_G(n)$ be an arbitrary element. By definition, we can write

\[g=\prod_{w\in \mathcal L_{n}}g_w\]

with $g_w\in \mathrm{Rist}_G(w)$ .

Since $n\geq |v|$ , we have $g_w\in \mathrm{St}_G(v)$ for all $w\in \mathcal L_{n}$ . Therefore, we have

\[\varphi_v(g) = \prod_{w\in \mathcal L_{n}}\varphi_v(g_w) = \prod_{vu\in \mathcal L_{n}}\varphi_v(g_{vu}),\]

where the first equality is due to the fact that $\varphi_v$ is a homomorphism when restricted to $\mathrm{St}_G(v)$ and the last equality follows from the fact that $\varphi_v(g_w)=1$ if $w\notin T_v$ . For every $vu\in \mathcal L_{n}$ , we have $\varphi_v(g_{vu}) \in \mathrm{Rist}_{G_v}(u)$ . Indeed, let $w\in T_{|v|}$ be such that $w\notin T_u$ . Then, $vw\notin T_{vu}$ , which implies that $g_{vu}vw = vw$ , since $g_{vu}\in \mathrm{Rist}_G(vu)$ , and thus $\varphi_v(g_{vu})w = w$ , which shows that $\varphi_v(g_{vu})\in \mathrm{Rist}_{G_v}(u)$ . We deduce that $\varphi_v(\mathrm{Rist}_G(n))\leq \mathrm{Rist}_{G_v}(n-|v|)$ .

For the second part of the lemma, let us first notice that if G acts spherically transitively on T, then $G_v$ acts spherically transitively on $T_{|v|}$ by Lemma 2·8. Thus, we only need to check condition (ii) or (ii^′) from Definition 2·7.

If G is a weakly branch group, then $\mathrm{Rist}_G(w)$ is non-trivial for every $w\in T$ , and it is not difficult to see that this implies that $\varphi_v(\mathrm{Rist}_G(n))$ is non-trivial for any $n\geq |v|$ . By the above, we conclude that $G_v$ is a weakly branch group. If G is a branch group, then $\mathrm{Rist}_G(n)$ is of finite index in $\mathrm{St}_G(v)$ for every $n\geq |v|$ , from which we conclude that $\varphi_v(\mathrm{Rist}_G(n))$ is of finite index in $G_v$ , and thus it follows from the above that $G_v$ is also a branch group.

Subgroups of groups of automorphisms of rooted trees have strong interactions with the elements of rigid stabilisers normalising them. In the context of branch groups, this phenomenon was first observed by Grigorchuk, who showed that a non-trivial normal subgroup of a branch group must always contain the derived subgroup of some rigid stabiliser [Reference Grigorchuk19]. This is a very useful result, since it means that in many respect, to understand normal subgroups of branch groups, it suffices to understand the derived subgroups of rigid stabilisers. In what follows, we will require a more powerful version of this fact than the one given by Grigorchuk, but the proof is essentially the same.

Lemma 2·11. Let G be a group of automorphisms of a rooted tree T, and let $H\leq G$ be a non-trivial subgroup of G. Let $v\in T$ be a vertex such that $Hv\ne \{v\}$ and let $R\leq \mathrm{Rist}_G(v)$ be a subgroup of $\mathrm{Rist}_G(v)$ normalising H. Then, $R'\leq H$ , where R’ denotes the derived subgroup (also known as the commutator subgroup) of R. In particular, if H is normal in G and if G acts spherically transitively on T, then $\mathrm{Rist}_G'(|v|)\leq H$ .

Proof. Let $h\in H$ be such that $hv\ne v$ . For every $r\in R$ , we have $hrh^{-1}\in \mathrm{Rist}_G(hv)$ , and this element must commute with every element of R, since $hv\ne v$ . Thus, using the fact that R normalises H, for every $r_1, r_2\in R$ we have

\begin{align*}H\ni [[r_1,h],r_2] &= [r_1hr_1^{-1}h^{-1}, r_2] \\&= [r_1,r_2].\end{align*}

We conclude that $R'\leq H$ .

If H is a normal subgroup of G, we can take $R=\mathrm{Rist}_G(v)$ , so that $\mathrm{Rist}_G'(v)\leq H$ . If G acts transitively on $\mathcal L_{|v|}$ , we conclude, using the fact that H is normal, that $\mathrm{Rist}_G'(w)\leq H$ for all $w\in \mathcal L_{|v|}$ , which implies that $\mathrm{Rist}_G'(|v|)\leq H$ .

As a direct consequence of the previous lemma, one sees that every proper quotient of a branch group is virtually abelian.

Lemma 2·12. Every proper quotient of a branch group G is virtually abelian.

Proof. Let $N\trianglelefteq G$ be a non-trivial normal subgroup. Then, by Lemma 2·11, there exists some $n\in \mathbb{N}$ such that $\mathrm{Rist}_G'(n)\leq N$ . It follows that $\mathrm{Rist}_G(n)N/N$ is an abelian subgroup of $G/N$ , and it must be of finite index in $G/N$ , since $\mathrm{Rist}_G(n)$ is of finite index in G.

Lemma 2·11 is more interesting if we know that the derived subgroup of a rigid stabiliser is non-trivial. This is always the case for weakly branch groups, as the following lemma shows.

Lemma 2·13. Let G be a weakly branch group acting on a rooted tree T. For any $v\in T$ and any $k\in \mathbb{N}$ , the subgroup $\mathrm{Rist}_G^{(k)}(v)$ is non-trivial, where $\mathrm{Rist}_G^{(0)}(v)=\mathrm{Rist}_G(v)$ and $\mathrm{Rist}_G^{(k+1)}(v) = [\mathrm{Rist}_G^{(k)}(v), \mathrm{Rist}_G^{(k)}(v)]$ .

Proof. We will proceed by induction on k. For $k=0$ , the result follows directly from the definition of a weakly branch group. Let us now suppose that for some $k\in \mathbb{N}$ , we have $\mathrm{Rist}_G^{(k)}(v)\ne 1$ for all $v\in T$ , and let us show that the same statement holds for $k+1$ .

Let us fix $v\in T$ . Since $\mathrm{Rist}_G^{(k)}(v)\ne 1$ , there exist $w\in T$ and $r\in \mathrm{Rist}_G^{(k)}(v)$ such that $rw\ne w$ . Since w is moved by an element of $\mathrm{Rist}_G^{(k)}(v)\leq \mathrm{Rist}_G(v)$ , we must have $v\leq w$ . Therefore, we have $\mathrm{Rist}_G(w)\leq \mathrm{Rist}_G(v)$ , which implies that $\mathrm{Rist}_G^{(k)}(w) \leq \mathrm{Rist}_G^{(k)}(v)$ .

By the induction hypothesis, we know that $\mathrm{Rist}_G^{(k)}(w)\ne 1$ , so there exists $h\in \mathrm{Rist}_G^{(k)}(w)$ with $h\ne 1$ . Since $\mathrm{Rist}_G^{(k)}(w)\leq \mathrm{Rist}_G^{(k)}(v)$ , we have $[r,h]\in \mathrm{Rist}_G^{(k+1)}(v)$ . However, notice that $[r,h]=(rhr^{-1})h^{-1}$ with $rhr^{-1}\in \mathrm{Rist}_G^{(k)}(rw)$ and $h^{-1}\in \mathrm{Rist}_G^{(k)}(w)$ both non-trivial. Using the fact that $\mathrm{Rist}_G^{(k)}(rw)\cap\mathrm{Rist}_G^{(k)}(w) = \{1\}$ , since $rw\ne w$ , we conclude that $[r,h]\ne 1$ . Thus, $\mathrm{Rist}_G^{(k+1)}(v)\ne 1$ , which concludes the proof.

Note that although non-trivial normal subgroups of branch groups must contain the derived subgroup of some rigid stabiliser, the same does not need to hold if one replaces the condition of being normal by the weaker condition of being normalised by a finite index subgroup. However, as an easy consequence of the previous lemmas, we obtain a condition under which this holds which, although technical, will prove very useful later on.

Lemma 2·14. Let G be a weakly branch group acting on a rooted tree T, let $H\leq G$ be a subgroup of finite index of G and let $K\trianglelefteq H$ be a normal subgroup of H such that $\mathrm{Rist}_K(v)\ne 1$ for all $v\in T$ . Then, there exists $n\in \mathbb{N}$ such that $\mathrm{Rist}_G'(n) \leq K$ .

Proof. Since H is of finite index in G, it contains a non-trivial normal subgroup of finite index $N\trianglelefteq G$ . By Lemma 2·11, there exists $m\in \mathbb{N}$ such that $\mathrm{Rist}_G'(m)\leq N \leq H$ . Using Lemma 2·9 and the fact that $\mathrm{Rist}_G'(m')\leq \mathrm{Rist}_G'(m)$ for all $m'\geq m$ , we may assume without loss of generality, replacing m by a bigger number if necessary, that $\mathrm{St}_H(v)$ acts spherically transitively on $T_v$ for all $v\in \mathcal L_{m}$ .

For every $v\in \mathcal L_{m}$ , we have $\mathrm{Rist}_K(v)\ne 1$ . Therefore, there exists $u_v\in T_v$ such that $\mathrm{Rist}_K(v)u_v\ne \{u_v\}$ . Notice that if $\mathrm{Rist}_K(v)u_v\ne \{u_v\}$ , then $\mathrm{Rist}_K(v)w\ne \{w\}$ for all $w\geq u_v$ . Therefore, we may assume that there exists $k\in \mathbb{N}$ such that $|u_v|=k$ for all $v\in \mathcal L_{m}$ .

Since $\mathrm{Rist}_G'(m)\leq H$ , it normalises K, so by Lemma 2·11, for every $v\in \mathcal L_{m}$ , we have $\mathrm{Rist}_G^{(2)}(u_v) \leq K$ . By the spherical transitivity of the action of H on each $T_v$ , we conclude that $\mathrm{Rist}_G^{(2)}(k)\leq K$ . Since $\mathrm{Rist}_G^{(2)}(k)$ is a normal subgroup of G, by applying Lemma 2·11, we obtain some $n\in \mathbb{N}$ such that $\mathrm{Rist}_G'(n)\leq \mathrm{Rist}_G^{(2)}(k)\leq K$ .

Recall that if G is a group, then a subgroup $K\leq G$ is subnormal if there exists subgroups $K_1,\dots K_k\leq G$ such that $K\trianglelefteq K_1 \trianglelefteq \dots \trianglelefteq K_k \trianglelefteq G$ . As the next lemma shows, subnormal subgroups of branch groups can only commute if their supports are disjoint (we call the support of a group H of automorphisms of some tree T the set of all vertices $v\in T$ such that $Hv\ne \{v\}$ ).

Lemma 2·15. Let G be a weakly branch group acting on a rooted tree T, let $H\leq G$ be a subgroup of finite index of G, and let $K_1,K_2\leq H$ be two subnormal subgroups of H. If $[K_1,K_2]=1$ , then $K_1$ and $K_2$ have disjoint support.

Proof. Since H is of finite index in G, it contains a non-trivial normal subgroup, and it thus follows from Lemma 2·11 that there exists $N\in \mathbb{N}$ such that $\mathrm{Rist}_G'(N)\leq H$ .

Let us fix $i\in \{1,2\}$ , and let $k_i\in \mathbb{N}$ be such that $K_i$ is a $k_i$ -subnormal subgroup of H. Let $v\in T$ be a vertex in the support of $K_i$ , meaning that $K_i v \ne \{v\}$ . Then, by inductively applying Lemma 2·11, we obtain that $\mathrm{Rist}_G^{(k_i+1)}(v)\leq K_i$ .

Therefore, if there existed $v\in T$ such that v were in the support of both $K_1$ and $K_2$ , we would have $\mathrm{Rist}_G^{(k_1+1)}(v)\leq K_1$ and $\mathrm{Rist}_G^{(k_2+1)}(v)\leq K_2$ , and thus $\mathrm{Rist}_G^{(k+1)}\leq K_i$ for $i=1,2$ , where $k=\max\{k_1,k_2\}$ . Since $[K_1,K_2]=1$ , this would imply that $\mathrm{Rist}_G^{(k+2)}(v)=1$ , which is impossible according to Lemma 2·13.

We conclude this section by observing that every proper quotient of a branch group has the property BF defined in Definition 2·3. This is a key fact that will allow us later on, in Lemma 4·1, to connect the theory of subgroups of finite bi-index with the theory of quasi-prodense subgroups of branch groups that we will develop in section 3, and thus to eventually prove that branch groups have property BF (Theorem 4·4).

Proposition 2·16. Every proper quotient of a branch group G has property BF.

Proof. Let $N\trianglelefteq G$ be a non-trivial normal subgroup of G. Let $H\leq G/N$ be a subgroup of finite bi-index. We need to show that H has finite index in $G/N$ .

By Lemma 2·12, $G/N$ is a virtually abelian group. Therefore, there exists a subgroup $K\leq G/N$ of finite index in $G/N$ such that K is abelian. According to Proposition 2·5, the subgroup $H\cap K$ is of finite bi-index in K. Since K is abelian, a subgroup of finite bi-index must obviously be a subgroup of finite index. Thus, we have that $H\cap K$ is of finite index in K, which is itself of finite index in $G/N$ . It follows that H is of finite index in $G/N$ .

3. Quasi-prodense subgroups of branch groups

In this section, we define the notion of a quasi-prodense subgroup, which is one of the key ingredients in the proof of our main result. Indeed, as we will show in Lemma 4·1 below, any subgroup of finite index of a branch group is quasi-prodense. Thus, in light of Proposition 2·2, to show that a branch group does not admit a quasi-2-transitive action on an infinite set, it will suffice to show that every quasi-prodense subgroup of finite bi-index is also of finite index.

To this end, we obtain results about the structure of quasi-prodense subgroups of branch groups. Our main result here is Theorem 3·4, which states that in weakly branch groups, quasi-prodense subgroups of infinite index always have at least one section which is also a quasi-prodense subgroup of infinite index. While rather technical, this theorem is really useful, as it allows us to relate the structure of a quasi-prodense subgroup of infinite index with the structure of the rooted tree on which it acts.

The results in this section are mostly generalisations of results on prodense subgroups of branch groups obtained by the author in [Reference Francoeur14]. Unfortunately, although the main ideas are mostly the same, the results in [Reference Francoeur14] are not sufficiently strong for our purposes and cannot be used to deduce what we require. We therefore present here complete proofs of all results, following closely the outline of the proofs in [Reference Francoeur14] but with a few key modifications necessary to deal with the added complications.

We begin by defining quasi-prodense subgroups.

Definition 3·1. A subgroup H of a group G is said to be quasi-prodense if its pro-normal closure

\[\overline{H}=\bigcap_{\substack{N\trianglelefteq G\\ N\ne 1}}HN\]

is of finite index in G.

In other words, a subgroup H of a group G is quasi-prodense if there exists a finite index subgroup $\overline{H}$ such that H and $\overline{H}$ cannot be distinguished in any proper quotient of G. This notion is very similar to the notion of finite proindex subgroup defined by de Cornulier in [Reference de Cornulier11]. A subgroup H of a group G is said to be of finite proindex if its closure in the profinite topology is of finite index, and it follows from the definition that the closure of H in the profinite topology is the intersection of all the subgroups of the form HN for $N\trianglelefteq G$ a normal subgroup of finite index. The difference with our definition is that here, we consider the intersection of the subgroups HN for all non-trivial normal subgroups of G. In groups where the intersection of any two non-trivial normal subgroup is again non-trivial (which is the case for branch and weakly branch groups by Lemmas 2·11 and 2·13), this so-called pro-normal closure of a subgroup is indeed the closure of this subgroup in a finer topology, known as the pro-normal topology, which is defined by declaring as a basis the set of all cosets of non-trivial normal subgroups (see for example [Reference Glasner, Souto and Storm17]). In this case, what we call a quasi-prodense subgroup is simply a subgroup which is dense (with respect to the pro-normal topology) in a subgroup of finite index (namely, its pro-normal closure). In groups that do not satisfy this property, the notions of pro-normal closure and quasi-prodense subgroups as we have defined them need not arise from an actual topology on the group, but they still make sense.

As the next lemma shows, in branch and weakly branch groups, sections of quasi-prodense subgroups are quasi-prodense in sections.

Proof. Let us fix $v\in T$ . We have

\[\overline{H_v} = \bigcap_{\substack{N\trianglelefteq G_v\\ N\ne 1}}H_vN \leq \bigcap_{n\in \mathbb{N}}H_v\mathrm{Rist}'_{G_v}(n).\]

On the other hand, since $G_v$ is a weakly branch group by Lemma 2·10, for every non-trivial normal subgroup $N\trianglelefteq G$ , there exists some $n\in \mathbb{N}$ such that $\mathrm{Rist}'_{G_v}(n)\leq N$ by Lemma 2·11. Therefore, we have

\[\overline{H_v} = \bigcap_{n\in \mathbb{N}}H_v\mathrm{Rist}'_{G_v}(n).\]

Since H is quasi-prodense by assumption, we have by definition that $\overline{H}$ is of finite index in G, which implies that $(\overline{H})_v=\varphi_v(\mathrm{St}_{\overline{H}}(v))$ is of finite index in $G_v$ . Let $h\in \mathrm{St}_{\overline{H}}(v)$ be an arbitrary element. Then, for any $n\in \mathbb{N}$ , there exist $h_n\in H$ and $r_n\in \mathrm{Rist}'_G(n)$ such that $h=h_nr_n$ . For every $n\in \mathbb{N}$ greater that $|v|$ , we have $r_n\in \mathrm{Rist}'_G(n) \leq \mathrm{St}_G(v)$ and thus $h_n\in \mathrm{St}_H(v)$ . Therefore, we have

\[\varphi_v(h)=\varphi_v(h_n)\varphi_v(r_n)\in H_v\mathrm{Rist}'_{G_v}(n-|v|)\]

by Lemma 2·10. This implies that $\varphi_v(h)\in \bigcap_{n\in \mathbb{N}}H_v\mathrm{Rist}'_{G_v}(n)$ and thus that $(\overline{H})_v\leq \overline{H_v}$ . As $(\overline{H})_v$ is of finite index in $G_v$ , we conclude that $H_v$ is quasi-prodense in $G_v$ .

As we have just seen, sections of quasi-prodense subgroups are quasi-prodense. However, a priori, sections of quasi-prodense subgroups of infinite index need not be again of infinite index. Our goal here is to show that there must in fact always exist such a section of infinite index, which we do in Theorem 3·4 below. Before we can do that, however, we need the following lemma, which allows us to understand what happens when sections of quasi-prodense subgroups are of finite index.

Proof. We begin by fixing some notation that we will use throughout the proof. If $U\subset T$ is a set of vertices of T, we define the subgroup

\[\mathrm{Rist}_H(U) = \left\{h\in H \hspace{0.3em}\Big\vert\hspace{0.3em} hw=w \quad \forall w\notin \bigcup_{u\in U} T_u\right\}\cap \bigcap_{u\in U}\mathrm{St}_H(u).\]

In particular, we have $\mathrm{Rist}_H(\{v\}) = \mathrm{Rist}_H(v)$ for all $v\in T$ , and $\mathrm{Rist}_H(\mathcal L_{n}) = \mathrm{St}_H(n)$ for all $n\in \mathbb{N}$ . Take notice that for convenience, we have given here a slightly different definition of $\mathrm{Rist}_H(U)$ than the one given in [Reference Francoeur14].

Now, let us suppose for a contradiction that there exists a vertex $v\in T$ different from the root such that $\mathrm{Rist}_H(v)=1$ , and let us set $n=|v|$ . Let $U\subseteq \mathcal L_{n}$ be a subset of minimal cardinality such that $\varphi_v(\mathrm{Rist}_H(U)) \ne 1$ . Such as subset exists, since $\varphi_v(\mathrm{Rist}_H(\mathcal L_{n}))=\varphi_v(\mathrm{St}_H(n))$ , and as $\varphi_v(\mathrm{St}_H(n))$ is of finite index in $\varphi_v(\mathrm{St}_H(v))=H_v$ , which is itself of finite index in the weakly branch group $G_v$ by assumption, we know that $\varphi_v(\mathrm{St}_H(n))\ne 1$ . Notice that we must have $v\in U$ , otherwise any element of $\mathrm{Rist}_H(U)$ would act trivially on $T_v$ , which would imply that $\varphi_v(\mathrm{Rist}_H(U))=1$ . Since $\mathrm{Rist}_H(v)=1$ , it follows that $|U|\geq 2$ . Let us pick some $u\in U\setminus \{v\}$ .

Writing $K_1=\mathrm{Rist}_H(U)$ and $K_2=\mathrm{Rist}_H(\mathcal L_{n}\setminus \{u\})$ , we have

\[[K_1,K_2] \leq \mathrm{Rist}_H(U\setminus\{u\}).\]

Indeed, if $w\in \mathcal L_{n}$ is a vertex not belonging to $U\setminus \{u\} = U\cap (\mathcal L_{n}\setminus \{u\})$ , then $K_1T_w = K_2T_w = T_w$ , since both $K_1$ and $K_2$ are in $\mathrm{St}_H(n)$ , and one of $K_1$ or $K_2$ must act trivially on $T_w$ , since w is either not in U or not in $\mathcal L_{n}\setminus \{u\}$ , from which it follows that $[K_1,K_2]$ must also act trivially on $T_w$ . By the minimality of the cardinality of U, we must have $\varphi_v(\mathrm{Rist}_H(U\setminus\{u\}))=1$ . It follows that

\[[\varphi_v(K_1), \varphi_v(K_2)] = \varphi_v([K_1,K_2])\leq \varphi_v(\mathrm{Rist}_H(U\setminus\{u\})) = 1,\]

and thus that $\varphi_v(K_1)$ and $\varphi_v(K_2)$ commute.

Both $K_1$ and $K_2$ are normal subgroups of $\mathrm{St}_H(n)$ , which is itself a normal subgroup of finite index of $\mathrm{St}_H(v)$ . Thus, $\varphi_v(K_1)$ and $\varphi_v(K_2)$ are two commuting subnormal subgroups of $H_v$ , which is by assumption a finite index subgroup of the weakly branch group $G_v$ . It follows from Lemma 2·15 that $\varphi_v(K_1)$ and $\varphi_v(K_2)$ must have disjoint support. Since $\varphi_v(K_1)\ne 1$ by the definition of $K_1$ , there must exist some $w\in T_{|v|}$ such that $\varphi_v(K_1)w \ne \{w\}$ , and it follows that we must have $\varphi_v(K_2)w = \{w\}$ . Furthermore, since every descendant of w will also be moved by $\varphi_v(K_1)$ , they must all be fixed by $\varphi_v(K_2)$ . In other words, we have $\varphi_v(K_2)\leq \mathrm{St}_{H_v}(w)$ and $\varphi_w(\varphi_v(K_2))=1$ . It follows that $K_2\leq \mathrm{St}_H(vw)$ , with $\varphi_{vw}(K_2)=1$ .

Let us write $m=|w|$ . The map

\begin{align*}\alpha\colon \varphi_u(\mathrm{St}_H(n+m)) & \rightarrow \varphi_{vw}(\mathrm{St}_H(n+m)) \\\varphi_u(g) & \mapsto \varphi_{vw}(g)\end{align*}

is a well-defined homomorphism. Indeed, using the fact that both $\varphi_u$ and $\varphi_{vw}$ are homomorphisms when restricted to $\mathrm{St}_H(n+m)$ , if $g_1,g_2\in \mathrm{St}_H(n+m)$ are such that $\varphi_u(g_1)=\varphi_u(g_2)$ , then $g_1g_2^{-1}\in K_2$ , which implies that $\varphi_{vw}(g_1g_2^{-1})=1$ and thus $\varphi_{vw}(g_1)=\varphi_{vw}(g_2)$ . We will show that the existence of such a homomorphism is in contradiction with the fact that H is quasi-prodense.

Let $r\in \mathrm{Rist}_{\overline{H}}(vw)$ be a non-trivial element. Such an element must exist, since $\overline{H}$ is a subgroup of finite index of the weakly branch group G. Notice however that $r\notin H$ , since $\mathrm{Rist}_H(vw) \leq \mathrm{Rist}_H(v)=1$ . As r belongs to the rigid stabiliser of vw and is non-trivial, we must have $\varphi_{vw}(r)\ne 1$ , and therefore there must exist $m_1\in \mathbb{N}$ such that $\varphi_{vw}(r) \notin \mathrm{St}_{G_{vw}}(m_1)$ .

Let us consider the subgroup $L=\alpha^{-1}(\varphi_{vw}(\mathrm{St}_{H}(n+m+m_1))) \geq \varphi_u(\mathrm{St}_{H}(n+m+m_1))$ . As $\mathrm{St}_H(n+m+m_1)$ is of finite index in $\mathrm{St}_H(u)$ , the subgroup L is of finite index in $H_u$ , which is itself of finite index in $G_u$ . Thus, by Lemma 2·11, there exists some $m_2\in \mathbb{N}$ such that $\mathrm{Rist}'_{G_u}(m_2)\leq L$ .

Now, let $m_3=\max\{m_1,m_2-m\}$ . By definition, we have $\overline{H}\leq H \mathrm{Rist}'_G(n+m+m_3)$ . This implies that $r=hg$ for some $h\in H$ and some $g\in \mathrm{Rist}'_G(n+m+m_3)$ . Since we have $r,g\in \mathrm{St}_G(n+m)$ , we must have $h\in \mathrm{St}_H(n+m)$ . Therefore, applying $\varphi_u$ to both sides of this equality, we find $\varphi_u(r)=\varphi_u(h)\varphi_u(g)$ . As $r\in \mathrm{Rist}_H(vw)$ , we must have $\varphi_u(r)=1$ . Therefore, $\varphi_u(h) = \varphi_u(g^{-1})$ . Since $g\in \mathrm{Rist}'_G(n+m+m_3)$ , we conclude, using Lemma 2·10, that

\[\varphi_u(h)\in \varphi_u(\mathrm{Rist}'_G(n+m+m_3))\leq \mathrm{Rist}'_{G_u}(m+m_3)\leq \mathrm{Rist}'_{G_u}(m_2) \leq L.\]

Using the fact that $\varphi_{vw}(h)=\alpha(\varphi_u(h))$ , we see that

\[\varphi_{vw}(h)\in \alpha(L) = \varphi_{vw}(\mathrm{St}_H(n+m+m_1)) \leq \mathrm{St}_{H_{vw}}(m_1).\]

Since we also have $g\in \mathrm{Rist}'_G(n+m+m_3)\leq \mathrm{St}_G(n+m+m_1)$ , we also get $\varphi_{vw}(g)\in \mathrm{St}_{G_{vw}}(m_1)$ .

Applying $\varphi_{vw}$ to both sides of the equality $r=hg$ , we obtain $\varphi_{vw}(r)=\varphi_{vw}(h)\varphi_{vw}(g)$ , and since $\varphi_{vw}(h),\varphi_{vw}(g)\in \mathrm{St}_{G_{vw}}(m_1)$ , we conclude that $\varphi_{vw}(r)\in \mathrm{St}_{G_{vw}}(m_1)$ . However, this is a contradiction, since we established above that $\varphi_{vw}(r) \notin \mathrm{St}_{G_{vw}}(m_1)$ . We conclude that our initial assumption that $\mathrm{Rist}_H(v)=1$ was false.

Proof. To prove the result, it suffices to show that there exists a vertex $v\in T$ such that $H_v$ is of infinite index in $G_v$ . Indeed, if that is the case, then we may assume without loss of generality that $v\in \mathcal L_{1}$ , since $H_u$ being of infinite index in $G_u$ for some $u\in T$ implies that $H_w$ is of infinite index in $G_w$ for all $w\leq u$ . It then follows from Lemma 3·2 that the result is true for $n=1$ . By induction, we then conclude that the result is true for all $n\in \mathbb{N}$ .

For the sake of contradiction, let us suppose that $H_v$ is of finite index in $G_v$ for all $v\in T$ except the root. Then, by Lemma 3·3, we have $\mathrm{Rist}_H(v)\ne 1$ for all $v\in T$ . Thus, as $\mathrm{Rist}_H(v)$ is a normal subgroup of $\mathrm{St}_H(v)$ , we see that $\varphi_v(\mathrm{Rist}_H(v))$ is a non-trivial normal subgroup of $H_v$ , which is itself a subgroup of finite index of the weakly branch group $G_v$ . Therefore, it follows from Lemma 2·14 that there exists some $m_v\in \mathbb{N}$ such that $\mathrm{Rist}'_{G_v}(m_v)\leq \varphi_v(\mathrm{Rist}_{H}(v))$ , which implies, using Lemma 2·10, that $\varphi_v(\mathrm{Rist}'_G(m_v+|v|))\leq \varphi_v(\mathrm{Rist}_H(v))$ .

Let us set $m=\max\{m_v \mid v\in \mathcal L_{1}\}$ . Then, by the above, we have $\varphi_v(\mathrm{Rist}'_G(m+1))\leq \varphi_v(\mathrm{Rist}_H(v))$ for all $v\in \mathcal L_{1}$ . Since $\varphi_v(\mathrm{Rist}'_G(m+1)) = \varphi_v(\mathrm{Rist}'_G(m+1)\cap \mathrm{Rist}_G(v))$ , using the fact that $\varphi_v$ is injective on $\mathrm{Rist}_G(v)$ , we find that $\mathrm{Rist}'_G(m+1)\cap \mathrm{Rist}_G(v)\leq H$ . As this is true for every $v\in \mathcal L_{1}$ , we conclude that $\mathrm{Rist}'_G(m+1)\leq H$ .

However, this contradicts the fact that H is a quasi-prodense subgroup of infinite index. Indeed, we have $\overline{H}\leq H\ \mathrm{Rist}'_G(m+1) =H$ , with $\overline{H}$ of finite index and H of infinite index, which is absurd. Thus, we conclude that there must exist some vertex $v\in T$ different from the root such that $H_v$ is of infinite index. This concludes the proof.

Using Theorem 3·4, one can show that for a quasi-prodense subgroup of infinite index of a weakly branch group, the number of vertices on a given level at which the section is of infinite index is unbounded. This will be crucial to the proof of Theorem A in the next section, since we will establish a relation between the bi-index of the subgroup and the number of vertices where the section is of infinite index.

Proposition 3·5. Let G be a weakly branch group acting on a rooted tree T, and let $H\lt G$ be a quasi-prodense subgroup of infinite index. Then, for every $M\in \mathbb{N}$ , there exists $n\in \mathbb{N}$ such that

\[|\{v\in\mathcal L_{n} \mid [G_v:H_v]=\infty\}| \gt M,\]

where $H_v=\varphi_v(\mathrm{St}_H(v))$ and $G_v=\varphi_v(\mathrm{St}_G(v))$ .

Proof. By the definition of a quasi-prodense subgroup, the subgroup

\[\overline{H}=\bigcap_{\substack{N\trianglelefteq G\\ N\ne 1}}HN\]

must be of finite index in G. Therefore, by Lemma 2·9, there exists $m\in \mathbb{N}$ such that $\mathrm{St}_{\overline{H}}(v)$ acts spherically transitively on $T_v$ for every $v\in \mathcal L_{m}$ . It follows that for every $M\in \mathbb{N}$ , one can find $n\in \mathbb{N}$ such that

\[|\overline{H}w|\gt M\]

for every $w\in \mathcal L_{n}$ .

Notice that for every $u\in T$ , we must have $Hu=\overline{H}u$ . Indeed, it is clear that $Hu\subseteq \overline{H}u$ , and to show the reverse inclusion, one simply has to observe that $\overline{H}\leq H\mathrm{St}_G(|u|)$ , so $\overline{H}u \subseteq H\mathrm{St}_G(|u|)u = Hu$ .

Therefore, for every $M\in \mathbb{N}$ , there exists $n\in \mathbb{N}$ such that $|Hw|\gt M$ for every $w\in \mathcal L_{n}$ . By Theorem 3·4, there exists $v\in \mathcal L_{n}$ such that $H_v$ is a quasi-prodense subgroup of infinite index of $G_v$ . This implies that $[G_u:H_u]=\infty$ for every $u\in Hv$ , since in this case, $H_u$ and $G_u$ are conjugates of $H_v$ and $G_v$ . As $|Hv|\gt M$ , the result follows.

4. Quasi-2-transitive actions of branch groups

In this section, we finally prove our main result, namely that a branch group cannot have a quasi-2-transitive action on an infinite set. By Proposition 2·2, this is equivalent to showing that branch groups possess property BF, as defined in Definition 2·3. In other words, we will show that every subgroup of infinite index of a branch group must have infinite bi-index.

The first step to establish this is to show that it suffices to restrict our attention to quasi-prodense subgroups.

Proof. Let $H\leq G$ be a subgroup of finite bi-index and let us consider its pro-normal closure

\[\overline{H}=\bigcap_{\substack{N\trianglelefteq G\\ N\ne 1}}HN.\]

We want to show that $\overline{H}$ is of finite index in G.

Notice first that for every non-trivial normal subgroup $1\neq N\trianglelefteq G$ , the subgroup HN must also be of finite bi-index. Consequently, $HN/N$ is a subgroup of finite bi-index in $G/N$ . As every proper quotient of G possesses property BF by assumption, $HN/N$ must be of finite index in $G/N$ , and thus HN must be of finite index in G.

Furthermore, as H is a subgroup of finite bi-index, it is almost maximal by Proposition 2·4, meaning that it is contained in only finitely many subgroups. We conclude that $\overline{H}$ is an intersection of finitely many subgroups of finite index, and it thus of finite index.

By Proposition 2·16, every proper quotient of a branch group has property BF. Thus, in light of Lemma 4·1, it suffices to show that a quasi-prodense subgroup of infinite index of a branch group must also have infinite bi-index. To do this, we will establish a relation between the bi-index of a quasi-prodense subgroup of infinite index and the number of sections of infinite index of this subgroup, and then apply Proposition 3·5, which states that this number is unbounded.

In order to achieve this, we first establish in the next lemma a family of invariants of bi-cosets.

Lemma 4·2. Let G be a group acting on a spherically homogeneous rooted tree T, and let $H\leq G$ be a subgroup of G. For every $n\in \mathbb{N}$ , let us define a map

\begin{align*}\nu_n \colon \mathrm{St}_G(n) &\longrightarrow \mathbb{N} \\g &\longmapsto |\{v\in \mathcal L_{n} \mid \varphi_v(g)\in H_v\}|,\end{align*}

where $H_v = \varphi_v(\mathrm{St}_H(v))$ . Then, $\nu_n$ is an invariant of bi-cosets, in the sense that for every $g_1,g_2\in \mathrm{St}_G(n)$ with $Hg_1H=Hg_2H$ , we have $\nu_n(g_1)=\nu_n(g_2)$ .

Proof. Let us fix $n\in \mathbb{N}$ , and let $g_1,g_2\in \mathrm{St}_G(n)$ be such that $Hg_1H=Hg_2H$ . Then, there exist $h_1,h_2 \in H$ such that $g_2=h_1g_1h_2$ . Notice that since both $g_1$ and $g_2$ are in the stabiliser of the $n^\mathrm{th}$ level, for every $v\in \mathcal L_{n}$ we must have $h_1h_2v=v$ .

For every $v\in \mathcal L_{n}$ , we have

\[\varphi_v(g_2)=\varphi_{g_1h_2v}(h_1)\varphi_{h_2v}(g_1)\varphi_v(h_2)=\varphi_{h_2v}(h_1)\varphi_{h_2v}(g_1)\varphi_v(h_2),\]

where the last equality follows from the fact that $g_1\in \mathrm{St}_G(n)$ . We want to show that $\varphi_v(g_2)$ is in $H_v$ if $\varphi_{h_2v}(g_1)$ is in $H_{h_2v}$ . Suppose that $\varphi_{h_2v}(g_1)\in H_{h_2v}$ . Then, by definition, there exists $h_3\in \mathrm{St}_H(h_2v)$ such that $\varphi_{h_2v}(h_3)=\varphi_{h_2v}(g_1)$ . Thus, we have

\[\varphi_v(g_2)=\varphi_{h_2v}(h_1)\varphi_{h_2v}(h_3)\varphi_v(h_2) = \varphi_{h_3h_2v}(h_1)\varphi_{h_2v}(h_3)\varphi_v(h_2) = \varphi_v(h_1h_3h_2),\]

and since $h_1h_3h_2 \in \mathrm{St}_H(v)$ due to the fact that $h_3h_2v=h_2v$ and $h_1h_2v=v$ , we conclude that $\varphi_v(g_2) \in H_v$ .

We have thus shown that for each $v\in \mathcal L_{n}$ , we have $\varphi_v(g_2)\in H_v$ if $\varphi_{h_2v}(g_1)\in H_{h_2v}$ . Since $h_2$ is a bijection on $\mathcal L_{n}$ , we conclude that $\nu_n(g_2)\geq\nu_n(g_1)$ . By symmetry, we must have $\nu_n(g_1) = \nu_n(g_2)$ .

We are now in position to combine the invariants of Lemma 4·2 with Proposition 3·5 to prove that quasi-prodense subgroups of infinite index of branch groups are of infinite bi-index.

Proof. We will show that for any $M\in \mathbb{N}$ , the number of bi-cosets of H must be larger than M.

Let us fix an arbitrary $M\in \mathbb{N}$ . By Proposition 3·5, there exists $n\in \mathbb{N}$ such that $ |\mathcal{V}_n|\gt M$ , where

\[\mathcal{V}_n=\{v\in\mathcal L_{n} \mid [G_v:H_v]=\infty\}\]

with $G_v=\varphi_v(\mathrm{St}_G(v))$ and $H_v = \varphi_v(\mathrm{St}_H(v))$ .

Since G is a branch group, $\mathrm{Rist}_G(n)$ is of finite index in G. It follows that $\varphi_v(\mathrm{Rist}_G(n))$ is of finite index in $G_v$ for all $v\in \mathcal L_{n}$ , and thus in particular for all $v\in \mathcal{V}_n$ . It then follows that for every $v\in \mathcal{V}_n$ , one can find an element $g_v\in \mathrm{Rist}_G(v)$ such that $\varphi_v(g_v) \notin H_v$ , using the fact that the index of $H_v$ in $G_v$ is infinite. Notice that since $g_v\in \mathrm{Rist}_G(v)$ , we have $\varphi_w(g_v)=1\in H_w$ for all $w\in \mathcal L_{n}$ with $w\ne v$ .

Let us write $\mathcal{V}_n=\{v_1, v_2, \dots, v_{|\mathcal{V}_n|}\}$ . For every $i\in \{1,2,\dots, |\mathcal{V}_n|\}$ , we define an element $g_i\in \mathrm{Rist}_G(n)$ as follows:

\[g_i = \prod_{k=1}^{i}g_{v_k}.\]

By construction, for $w\in \mathcal L_{n}$ , we have $\varphi_w(g_i)\in H_w$ if and only if $w\notin \{v_1, \dots, v_i\}$ . Therefore, for every $i\in \{1,2, \dots, |\mathcal{V}_n|\}$ , we have

\[\nu_n(g_i) = |\mathcal L_{n}|-i,\]

where $\nu_n(g_i)$ is the map defined in Lemma 4·2. As i ranges from 1 to $|\mathcal{V}_n|$ , we conclude that the map $\nu_n$ takes at least $|\mathcal{V}_n|\gt M$ different values. Since two elements in the same bi-coset must have the same image under $\nu_n$ according to Lemma 4·2, there must exist more than M different bi-cosets. This concludes the proof.

Thanks to the previous lemma, we are finally ready to show that branch groups have property BF, which will complete the proof of Theorem A.

Proof. If $H\lt G$ is a subgroup of finite bi-index, then it is a quasi-prodense subgroup of G by Lemma 4·1 and Proposition 2·16. Therefore, by Lemma 4·3, H must be of finite index in G. Thus, G has property BF. It then follows directly from Proposition 2·2 that G cannot act quasi-2-transitively on an infinite set.

Note that our proof of Theorem 4·4 does not immediately extend to weakly branch groups. One major obstacle is that there is a priori no reason to believe that every proper quotient of a weakly branch group has property BF. However, if one considers weakly branch groups whose proper quotients have property BF, then there is only one point in the proof of Lemma 4·3 where we need the fact that the group is branch instead of weakly branch. It therefore seems reasonable to hope that the result could extend to those groups.

Question 4·5. If G is a weakly branch group whose proper quotients all have property BF, must G also have property BF?