Nontrivial abelianization of torsion-free pro-$p$-group which contains a dense free subgroup is infinite?

Let $$G$$ be a topologically finitely generated pro-$$p$$-group. Assume that $$G$$ is torsion-free, it contains a dense free subgroup of infinite rank and the (topological) abelianization $$G^{\text{ab}}$$ of $$G$$ is not trivial. My question is that:

is $$G^{\text{ab}}$$ an infinite group?

Note that the derived group of $$G$$ is closed in $$G$$ in my setting, and hence the topological abelianization is the same as abstract abelianization. My motivation for asking this question comes from the so-called "topological Tits alternative", see A Topological Tits Alternative. For example, let $${\rm GL}^1_n(\mathbb{Z}_p)$$ denote the first congruence subgroup, i.e. the kernel of the homomorphism $${\rm GL}_n(\mathbb{Z}_p)\to {\rm GL}_n(\mathbb{F}_p)$$ where $$\mathbb{Z}_p$$ is the ring of integers of $$p$$-adic numbers. Now suppose that $$G$$ is a closed subgroup of $${\rm GL}^1_n(\mathbb{Z}_p)$$ which does not contain an open solvable subgroup and $$G^{\text{ab}}\neq 1$$. Then $$G$$ will satisfy the above assumptions by the topological Tits alternative.

No.

Let $$H$$ be a torsion-free pro-$$p$$-subgroup of finite index in $$\mathrm{SL}_3(\mathbf{Z}_p)$$. Since $$\mathrm{SL}_3(\mathbf{Z})$$ has the property that every finite index subgroup has finite abelianization, the same holds, topologically, for $$\mathrm{SL}_3(\mathbf{Z}_p)$$. Then $$H$$ has a dense free subgroup $$F$$. So the closure $$G$$ of $$[F,F]$$ has finite index in $$H$$, and has the dense subgroup $$[F,F]$$ that is free of infinite rank.

So $$G$$ has a finite, nontrivial abelianization (every nontrivial pro-$$p$$-group has a nontrivial abelianization).

• thanks! great answer. Oct 26, 2022 at 21:41