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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.

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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).

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  • $\begingroup$ thanks! great answer. $\endgroup$
    – stupid boy
    Oct 26, 2022 at 21:41

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