On strongly just infinite profinite branch groups

Definition 1.2 (see Section 2.4)

A profinite branch group is a closed spherically transitive subgroup of the automorphism group of a rooted locally finite tree such that every rigid level stabilizer is an open subgroup.

Proposition 1.3 (see Theorem 3.2)

Suppose that G⩽Aut⁡(Tα) is a profinite branch group. Then the following are equivalent:

1. G is strongly just infinite.

2. For every vertex v∈Tα, the derived subgroup of ristG⁢(v) is open in ristG⁢(v).

3. For every level n⩾1, the derived subgroup of ristG⁢(n) is open.

Theorem 1.4 (see Theorem 4.8)

Let G be a profinite branch group. Then the following are equivalent:

1. G is strongly just infinite.

2. Every commensurated subgroup of G is either finite or open.

3. G has the normal countable index property.

4. G has the countable index property.

5. G has the weak Steinhaus property.

6. G has uncountable cofinality.

7. G is Cayley-bounded.

8. G has property (FA).

9. G has the Bergman property.

Infinite products of non-abelian finite simple groups are known to satisfy a similar characterization; see [27, 28].

Theorem 1.6 (see Theorem 5.10)

If G is a strongly just infinite profinite branch group which locally has derangements and has uniform commutator widths, then G enjoys the equivalent properties of Theorem 1.4, the invariant automatic continuity property, and the locally compact automatic continuity property.

Theorem 1.7 (see Theorem 8.4)

Suppose that F⩽Sd is non-trivial, perfect, and two transitive. Suppose further the point stabilizers of F are also perfect. The Burger–Mozes universal simple group U⁢(F)+ then enjoys the countable index property, the invariant automatic continuity property, and the locally compact automatic continuity property.

Theorem 1.8 (see Theorem 8.6)

Suppose that F⩽Sd is non-trivial, perfect, and two transitive. Suppose further that the point stabilizers of F are also perfect. Then every commensurated subgroup of U⁢(F)+ is either finite, compact and open, or equal to U⁢(F)+.

If A is an uncountable abelian group, then A has an infinitely generated countable quotient.

Proof.

Let Tor⁢(A) be the torsion subgroup of A and form A~:=A/Tor⁢(A). Suppose first that A~ is uncountable, so A~ is an uncountable torsion free abelian group. The extension of scalars A~⊗ℤℚ is an uncountable ℚ-vector space, and there is a canonical injection A~↪A~⊗ℤℚ, since A~ is torsion free. We may find a set {ai⊗1}i∈ℕ with ai∈A~ linearly independent vectors in A~⊗ℤℚ. We then have a projection

The composition A→A~→A~⊗ℤℚ→V has a countably infinite image that is infinitely generated, verifying the lemma in this case.

If A/Tor⁢(A) is countable, then it suffices to find a countable quotient of Tor⁢(A) that is infinitely generated; we thus assume A=Tor⁢(A). By [12, Theorem 8.4], we have a decomposition A=⊕p⁢ primeAp where Ap are abelian p-groups, and since A is uncountable, there is a prime p such that Ap is uncountable. We may thus also assume A is an uncountable p-group.

Appealing to [12, Theorem 32.3], there exists B⩽A such that B is a direct sum of cyclic p-groups and A/B is divisible. Suppose first A/B is non-trivial. Divisible abelian groups are direct sums of copies of ℚ and Prüfer p-groups via [12, Theorem 23.1], and both of these are countably generated. Projecting onto one of these summands, we obtain a countable quotient of A which is infinity generated. If A/B is trivial, then A is an uncountable direct sum of cyclic p-groups. Projecting onto a countable direct sum gives the desired countable quotient which is infinitely generated. The lemma is thus verified. ∎

Lemma 2.2 (Hartley [14, Lemma 1])

If G is a profinite group such that D⁢(G) is open in G, then cw⁢(G)<∞.

A topological group G is called strongly just infinite if every non-trivial normal subgroup is open with finite index.

Let G be a group.

1. G has the Bergman property if every G-action by isometries on a metric space has bounded orbits.

2. G has uncountable cofinality if there is no increasing chain (Gn)n∈ℕ of proper subgroups of G such that ⋃n∈ℕGn=G. Otherwise, G has countable cofinality.

3. G is Cayley bounded if for any symmetric generating set U containing 1, there is some n⩾1 such that Un=G. Equivalently, every Cayley graph for G has finite diameter.

A sequence (An)n∈ℕ of subsets of a group G is called a Bergman sequence if it is increasing, each of its elements is symmetric, 1∈A0, and the union ⋃n∈ℕAn is equal to G.

Theorem 2.6 (Cornulier, cf. [9, Proposition 2.7])

Let G be a group. The following assertions are equivalent.

1. G has the Bergman property.

2. If (An)n∈ℕ is a Bergman sequence such that An⁢An⊆An+1 for all n∈ℕ, then there exists k∈ℕ such that Ak=G.

3. If (An)n∈ℕ is a Bergman sequence, then there exist two integers k,n∈ℕ such that Ank=G.

Proposition 2.7 (Cornulier [9, Proposition 2.4])

A group G has the Bergman property if and only if it has uncountable cofinality and is Cayley bounded

Theorem 2.8 (Khelif [17, Théorème 10])

If Γ is a countable infinite subgroup of a compact group G, then Γ is not Cayley bounded.

Proof (Khelif).

First note that we may assume G is metrizable: By the Peter–Weyl Theorem, for each γ∈Γ there is a finite dimensional unitary representation πγ:G→𝒰⁢(ℂnγ) such that πγ⁢(γ)≠1, so Γ embeds into the compact metrizable group ∏γ∈Γ𝒰⁢(ℂnγ). Furthermore, by taking the closure of Γ in G, we may assume that Γ is dense in G. Fix a compatible right-invariant metric d on G.

Enumerate Γ={γn∣n∈ℕ} and for all n∈ℕ let Γn:=〈γ0,…,γn〉. Consider the sequence of continuous functions fn:G→[0,+∞[ given by

This sequence of functions decreases pointwise to zero, so by Dini’s theorem, they converge uniformly to zero. We can thus find a sequence (un)n∈ℕ such that un∈Γn and d⁢(un,γn+1) tends to zero.

Set S:={1,γ0}∪{γn+1⁢un-1∣n∈ℕ} and let U:=S∪S-1. The set U is a symmetric generating set for Γ which contains 1, and since γn+1⁢un-1→1, the set U is also a compact subset of G. If Γ is Cayley bounded, we can find k∈ℕ such that Γ=Uk. In particular, Γ is then compact, but by the Baire category theorem, there is no countable infinite compact group. We thus deduce that Γ is not Cayley bounded. ∎

Corollary 2.9 (Khelif [17, Corollaire 11])

If G is an infinite solvable-by-finite group, then G is not Cayley bounded.

Proof (Khelif).

Let H be a finite index normal subgroup of G such that H is solvable and consider the derived series H(n+1):=D⁢(H(n)), where H(0):=H. Let n be the smallest integer such that H/H(n) is infinite. Since H(n-1)/H(n) is abelian, the quotient group G/H(n) is infinite and abelian-by-finite. It thus suffices to show that no infinite abelian-by-finite group is Cayley bounded, so we assume G has a finite index normal subgroup N which is abelian.

By Lemma 2.1, the group N has a subgroup B⩽N with countable index. Letting x1,…,xk be coset representatives for N in G, the subgroup

is a normal subgroup of G such that G/L is a countably infinite abelian-by-finite group. Passing to G/L, we may also assume G is countable.

Since N is abelian, Pontryagin duality gives a morphism ψ:N→K with countable infinite image where K is a compact group. Taking coset representatives x1,…,xk for N in G, define ρ:N→Kk by

The action of G on N by conjugation induces an action of G on ρ⁢(N) by defining

For each g∈G and h∈N, it follows that there is a σ∈𝔖⁢(k) such that

Each element g∈G thus acts on ρ⁢(N) by permuting coordinates. Such an action extends to an action on ρ⁢(N)¯ by continuous automorphisms.

Under this action, the group ρ⁢(N)¯⋊G is a locally compact group. A straightforward calculation shows the subset {(ρ⁢(h),h-1)∣h∈N} is a discrete normal subgroup of ρ⁢(N)¯⋊G. The quotient is furthermore a compact group into which G embeds. The desired conclusion now follows via Theorem 2.8. ∎

A topological group is Polish if the underlying topology is separable and admits a compatible, complete metric. A Polish group is called non-archimedean if the topology admits a basis at 1 of open subgroups.

Let G be a Polish group.

1. The group G has the normal countable index property if every countable index normal subgroup of G is open.

2. The group G has the countable index property if every countable index subgroup of G is open.

Let G be a Polish group.

1. The group G has the normal countable index property if and only if every homomorphism ψ:G→H with H a countable discrete group is continuous.

2. The group G has the countable index property if and only if every homomorphism ψ:G→H with H a non-archimedean Polish group is continuous.

Proof.

The first claim is immediate, so we prove the second. Suppose first G has the countable index property and ψ:G→H is a homomorphism with H a non-archimedean Polish group. Taking O⩽H an open subgroup of H, the index of O is countable, since H is second countable. Hence, ψ-1⁢(O) is a countable index subgroup of G. That G has the countable index property now ensures ψ-1⁢(O) is open, and we deduce that ψ is continuous.

Conversely, suppose G is such that every homomorphism ψ:G→H is continuous where H is a non-archimedean Polish group. Take O⩽G a countable index subgroup. The action of G on the left cosets G/O induces a homomorphism σ:G→𝔖⁢(G/O). The group 𝔖⁢(G/O) with the topology of pointwise convergence is a non-archimedean Polish group, hence σ is continuous. Furthermore, the collection of permutations in 𝔖⁢(G/O) which fix the coset O, denoted Σ, form an open subgroup of 𝔖⁢(G/O). We thus see that σ-1⁢(Σ)=O is open in G, whereby G has the countable index property. ∎

A Polish group G has the weak Steinhaus property if for any symmetric σ-syndetic set A, there exists n∈ℕ such that the set An contains a neighborhood of 1.

Lemma 2.14 ([29, Lemma 4])

Suppose that G is a group and H is a subgroup of G. If A is a symmetric σ-syndetic set containing 1 for G, then H∩A2 is a symmetric σ-syndetic set containing 1 for H.

Suppose that G is a group and g0,…,gn∈G. If A is a symmetric σ-syndetic set containing 1, then ⋂i=0ngi⁢A2n⁢gi-1 is symmetric σ-syndetic set containing 1.

Proof.

The fact that ⋂i=0ngi⁢A2n⁢gi-1 is again symmetric and contains 1 is straightforward. To see that this set is also σ-syndetic, one argues via the obvious induction. This induction comes down to the following claim, which we prove here: If A,B⊆G are two σ-syndetic symmetric sets containing 1, then the set B2∩A2 is also σ-syndetic.

Let A and B be two σ-syndetic symmetric sets containing 1, fix a sequence (hk)k∈ℕ such that G=⋃k∈ℕhk⁢B and set

For each k∈K, fix ak∈hk⁢B∩A and say ak=hk⁢bk with bk∈B.

For a∈A, there are k∈K and b∈B such that a=hk⁢b, hence a=ak⁢bk-1⁢b. We also have a=ak⁢ak-1⁢a, so a∈ak⁢(B2∩A2). It now follows that

Since the set A is σ-syndetic, we conclude that B2∩A2 is σ-syndetic, completing the proof. ∎

A profinite group G is said to be a profinite branch group if there is a tree Tα for some α∈ℕ⩾2ℕ such that the following hold:

1. G is isomorphic to a closed subgroup of Aut⁡(Tα).

2. G acts transitively on each level of Tα.

3. For each level n, the index |G:ristG(n)| is finite.

Suppose that G⩽Aut⁡(Tα) is a profinite branch group and v∈Tα. Then each element g∈CG⁢(ristG⁢(v)) fixes pointwise Tαv. In particular, the center of ristG⁢(v) is trivial for all v∈Tα.

Proof.

Fix v∈Tα and suppose for contradiction that there is a w⩾v such that g.w≠w. The subgroup ristG⁢(w) is non-trivial since G is infinite, so we may find y∈ristG⁢(w) and u⩾w with y.u≠u. The element gy then sends u to g.(y.u), but the element yg sends u to g.u. Hence, g⁢y≠y⁢g contradicting the hypothesis that g centralizes ristG⁢(v). ∎

Suppose G⩽Aut⁡(Tα) is a profinite branch group. If v∈Tα is at level n, then

Let X be a set and let G⩽S⁢(X) be a permutation group with τ∈G. If σ1,σ2∈G are such that τ⁢(supp⁡(σ1)) is disjoint from supp⁡(σ1)∪supp⁡(σ2), then the commutator [σ1,σ2] is the product of four conjugates of τ±1 by elements of G.

Proof.

Whenever two permutations have disjoint support, they commute. Moreover,

so by our hypothesis, τ⁢σ1⁢τ-1 commutes with both σ1 and σ2. It follows τ⁢σ1-1⁢τ-1 also commutes with both σ1 and σ2.

Setting σ1¯:=[σ1,τ]=σ1⁢(τ⁢σ1-1⁢τ-1), the fact that τ⁢σ1-1⁢τ-1 commutes with both σ1 and σ2 yields that [σ1,σ2]=[σ¯1,σ2]. The permutation σ¯1=(σ1⁢τ⁢σ1-1)⁢τ-1 is the product of two conjugates of τ±1 by elements of the group G. Hence, [σ¯1,σ2] is the product of four conjugates of τ±1 by elements of G, verifying the lemma. ∎

Suppose that G⩽Aut⁡(Tα) is a profinite branch group. Then the following are equivalent:

1. G is strongly just infinite.

2. For all v∈Tα, the abstract derived subgroup D⁢(ristG⁢(v)) is open in ristG⁢(v).

3. For all n⩾1, the abstract derived subgroup D⁢(ristG⁢(n)) is open in G.

Proof.

For (1) ⇒ (2), we prove the contrapositive. Suppose for some v∈Tα the abstract commutator D⁢(ristG⁢(v)) is not open in ristG⁢(v); Lemma 2.17 ensures D⁢(ristG⁢(v)) is also non-trivial. Letting n be the level of v, the subgroup

is a characteristic subgroup of ristG⁢(n) which is non-trivial and not open. Since ristG⁢(n) is a normal subgroup of G, we deduce that G has a non-trivial normal subgroup which is not open, hence G is not strongly just infinite.

The implication (2) ⇒ (3) is immediate.

For (3) ⇒ (1), let H be a non-trivial normal subgroup of G and let τ∈H∖{1}. There exists a vertex v such that τ⁢(v)≠v; let n be its level. Taking two elements σ1,σ2 in ristG⁢(v), their support is a subset of Tαv, and since τ⁢(Tαv)=Tατ.v is disjoint from Tαv, we apply the commutator trick and deduce that the commutator [σ1,σ2] is the product of four conjugates of τ±1.

The commutator group D⁢(ristG⁢(v)) is thus a subgroup of H. Since H is normal and G acts spherically transitively on Tα, it follows that the open subgroup

is a subgroup of H, hence H is open. ∎

For G⩽Aut⁡(Tα), we call a subset A of Gfull above the vertex v∈Tα if every element of ristG⁢(v) coincides with an element of A restricted to Tαv.

Suppose that G⩽Aut⁡(Tα) is a strongly just infinite profinite branch group and that A⊆G is full above v. If A∩ristG⁢(v)⊈{1}, then there is a w⩾v such that D⁢(ristG⁢(w))⩽A10⁢k with k:=cw⁢(ristG⁢(w)).

Proof.

Take x∈A∩ristG⁢(v)∖{1} and find w⩾v such that x.w≠w. We now consider g,h∈ristG⁢(w). Since A is full above v, there are g~,h~∈A with the same action on Tαv as g and h, respectively. The element x is supported on ristG⁢(v), so h~⁢x-1⁢h~-1=h⁢x-1⁢h-1. In particular, [h~,x]=[h,x]. The element [h,x] is again supported on Tαv, so we have that [g~,[h,x]]=[g,[h,x]].

Now x⁢h-1⁢x-1 commutes with both h and g since supp⁡(x⁢h-1⁢x-1)⊆Tαx.w, so we further have that [g,h]=[g,[h,x]]. Therefore,

and we deduce that [g,h]∈A10.

The set A10 thus contains every commutator of ristG⁢(w). In view of Lemma 2.2, Theorem 3.2 implies that k:=cw⁢(ristG⁢(w)) is finite, so D⁢(ristG⁢(w))⩽A10⁢k. ∎

Let G⩽Aut⁡(Tα) be a closed subgroup and let (An)n∈N be a countable family of subsets of G such that G=⋃n∈NAn. Then for any vertex w∈Tα, there exists a vertex v⩾w and n∈N such that An is full above v.

Proof.

Let (wn)n∈ℕ enumerate the vertices of the rightmost branch of Tαw. For each n, let vn be a child of wn different from wn+1. Let us prove by contradiction that there is some n∈ℕ such that An is full above vn; this implies the lemma.

If not, for each n∈ℕ there is a tree automorphism gn∈G supported on Tαvn such that its restriction to Tαvn does not extend to an element of An. The products ∏i=0ngi converge to a limit g, and g extends gi for all i. Since G is closed in Aut⁡(Tα), we have that g∈G, but for all n∈ℕ, the restriction of g to Tαvn does not extend to an element of An. We conclude g∉An for any n, contradicting the assertion ⋃n∈ℕAn=G. ∎

Let G⩽Aut⁡(Tα) be a closed subgroup and let A be a symmetric σ-syndetic subset of G. Then for any w∈Tα, there exists v⩾w such that A2 is full above v.

Proof.

We may find a sequence (gn)n∈ℕ of elements of G such that ⋃n∈ℕgn⁢A is equal to G. Lemma 3.5 provides n∈ℕ and a vertex v⩾w such that gn⁢A is full above v. Since 1∈ristG⁢(v), there exists a∈gn⁢A such that a↾Tαv is the identity. It now follows that A⁢gn-1⁢gn⁢A=A2 is full above v. ∎

If G is a profinite group with uncountable cofinality, then the derived group D⁢(O) is open for every open normal O⁢⊴⁢G.

Proof.

We prove the contrapositive. Suppose that O⁢⊴⁢G is open but that D⁢(O) is not. The subgroup D⁢(O) is then meagre in O, hence the quotient group O/D⁢(O) is uncountable. Appealing to Lemma 2.1, we may find D⁢(O)⩽A⁢⊴⁢O such that O/A is countable and infinitely generated.

Taking g0,…,gn left coset representatives for O in G, the subgroup

is normal in G, and G/A~ is countable. The group G/A~ must be infinitely generated since O/A~ is a finite index subgroup. Since infinitely generated countable groups plainly have countable cofinality, we conclude that G has countable cofinality. ∎

Let G⩽Aut⁡(Tα) be a profinite branch group. Then the following are equivalent:

1. G is strongly just infinite.

2. G has the Bergman property.

3. G has uncountable cofinality.

4. G has property (FA).

5. G is Cayley bounded.

Proof.

For (1) ⇒ (2), let (An)n⩾0 be a Bergman sequence. By Lemma 3.5, there are n0⩾0 and v∈Tα such that An0 is full above v. Appealing to Lemma 3.4, we may find w⩾v on some level l such that

for k:=cw⁢(ristG⁢(w)).

Since G acts transitively on the levels, there are g1,…,gn∈G such that

and the fact that (An)n⩾0 is a Bergman sequence ensures there is an n1⩾n0 for which g1,…,gn∈An1. It now follows that

We infer D⁢(ristG⁢(l))⩽An110⁢n⁢k+2⁢n.

The set An110⁢n⁢k+2⁢n contains the open subgroup D⁢(ristG⁢(l)) of G, which has finite index. Letting h1,…,hm be left coset representatives for this subgroup, there is an n2⩾n1 such that h1,…,hm∈An2. We deduce that An210⁢n⁢k+2⁢n+1=G, hence G has the Bergman property.

The implications (2) ⇒ (3) and (2) ⇒ (5) are given by Proposition 2.7. H. Bass’ work [3] establishes the equivalence (3) ⇔ (4). It thus remains to show (3) ⇒ (1) and (5) ⇒ (1). The former is an easy exercise: The contrapositive follows from Lemma 4.1 and Theorem 3.2.

To show (5) ⇒ (1), we prove the contrapositive. Suppose G is not strongly just infinite. In view of Theorem 3.2, there is a level k such that D⁢(ristG⁢(k)) is not open in G. The quotient G/D⁢(ristG⁢(k)) is then an infinite abelian-by-finite group. Applying Corollary 2.9, G/D⁢(ristG⁢(k)) is not Cayley bounded, whereby G is not Cayley bounded. ∎

Suppose that G is a profinite branch group. Then the following are equivalent:

1. G is strongly just infinite.

2. G has the weak Steinhaus property.

3. G has the countable index property.

4. G has the normal countable index property.

Proof.

The implications (2) ⇒ (3) and (3) ⇒ (4) are immediate. The contrapositive of (4) ⇒ (1) follows from Lemma 2.1 and Theorem 3.2.

For (1) ⇒ (2), suppose G is strongly just infinite and let A⊆G be a σ-syndetic symmetric set. The set A¯ is then also σ-syndetic. Applying the Baire category theorem, some left translate of A¯ has non-empty interior, so A is dense in some open set V. Since V-1⁢V is a neighborhood of the identity and A is symmetric, we deduce that A2 is dense in a neighborhood of the identity. There is thus a level n∈ℕ such that A2 is dense in the pointwise stabilizer of Vn, denoted stG⁢(n). Fixing a symmetric set of right coset representatives g0,…,gl for stG⁢(n) in G, put B:=⋂i=0lgi-1⁢A2l⁢gi. Via Lemma 2.15, B is again a symmetric σ-syndetic set.

For v∈Vn, we apply Lemma 3.6 to find w⩾v such that B2 is full above w. By Lemma 2.14, the set B4∩ristG⁢(w) is σ-syndetic in ristG⁢(w), and since ristG⁢(w) is uncountable, we have B4∩ristG⁢(w)⊈{1}. Lemma 3.4 now implies we may find s⩾w such that D⁢(ristG⁢(s))⊆(B4)10⁢k=B40⁢k.

Let m be the level of s. The group G acts transitively on Vm, so for all t∈Vm, there exists z∈G such that z.s=t. We may write z=x⁢gi for some x∈stG⁢(n) and gi one of the previously fixed right coset representatives. Since A2 is dense in stG⁢(n), there is an h∈A2 such that h⁢gi.s=x⁢gi.s=t. We now have that h⁢gi⁢D⁢(ristG⁢(s))⁢gi-1⁢h-1=D⁢(ristG⁢(t)). Moreover,

We conclude that A40⁢k⁢2l+4 contains D⁢(ristG⁢(t)) for all t∈Vm. The open subgroup D⁢(ristG⁢(m)) is thus contained in A(40⁢k⁢2l+4)⁢|Vm|, whereby G enjoys the weak Steinhaus property. ∎

For n⩾2, the profinite group PSLn⁢(Zp) fails the countable index property but is strongly just infinite.

Proof.

The profinite group PSLn⁢(ℤp) is strongly just infinite via the main theorem of the appendix of [23].

On the other hand, via [16, Theorem 1], there is an injective homomorphism ξ:GLn⁢(ℂ)→𝔖∞. Considered as abstract fields, the algebraic closure of ℚp is isomorphic to ℂ, hence we may see SLn⁢(ℚp)⩽GLn⁢(ℂ). For each α∈Aut⁡(ℂ), the map ϕα given by applying α to the entries of a matrix is an automorphism of GLn⁢(ℂ). We therefore obtain maps ξ∘ϕα:SLn⁢(ℚp)→𝔖∞ for each automorphism α∈Aut⁡(ℂ).

For α and β in Aut⁡(ℂ), the maps ξ∘ϕα and ξ∘ϕβ are equal if and only if

The maps ϕα and ϕβ agree on SLn⁢(ℚp) if and only if ϕα-1∘β is the identity on SLn⁢(ℚp). The group SLn⁢(ℚp) contains elementary matrices Ei,j⁢(a) for i≠j and a∈ℚp, where Ei,j⁢(a) has ones on the diagonal, a in the (i,j)-entry, and zeros elsewhere. We conclude α-1∘β⁢(a)=a, so α-1∘β fixes ℚp pointwise. Therefore, ξ∘ϕα and ξ∘ϕβ are equal if and only if α-1∘β∈Aut⁡(ℂ/ℚp).

It is well known that |Aut⁡(ℂ)|=2𝔠. On the other hand, Aut⁡(ℂ/ℚp) is a second countable profinite group and thus has size 𝔠. We conclude there are 2𝔠 distinct left cosets of Aut⁡(ℂ/ℚp) in Aut⁡(ℂ). In view of the previous paragraph, there must be 2𝔠 distinct homomorphisms ξ∘ϕα:SLn⁢(ℚp)→𝔖∞. Since there can be at most continuum many continuous homomorphisms, we conclude that SLn⁢(ℚp) fails the countable index property, and as SLn⁢(ℤp) is an open subgroup of SLn⁢(ℚp), the group SLn⁢(ℤp) also fails the countable index property. It now follows PSLn⁢(ℤp) fails the countable index property. ∎

Theorem 4.5 (Bergman–Lenstra [4, Theorem 6])

Let G be a group with subgroups H and K. Then the following are equivalent:

1. supk∈K|H:H∩kHk-1|<∞.

2. There is an N normalized by K such that N∼cH.

If a Polish group G has the Bergman property, then every commensurated subgroup is commensurate to a normal subgroup.

Proof.

Suppose that C⩽G is commensurated. For each n⩾1, set