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.