The congruence subgroup problem for the free metabelian group on $n\ge 4$ generators

Abstract

The congruence subgroup problem for a finitely generated group $\Gamma$ asks whether the map $\widehat{\mathrm{Aut}(\Gamma )}\to \mathrm{Aut}(\hat{\Gamma })$ is injective, or more generally, what is its kernel $C(\Gamma )$? Here $\hat{X}$ denotes the profinite completion of $X$. It is well known that for finitely generated free abelian groups $C({\mathbb{Z}}^n)=\{1\}$ for every $n\ge 3$, but $C({\mathbb{Z}}^2)={\hat{F}}_{\omega }$, where ${\hat{F}}_{\omega }$ is the free profinite group on countably many generators.

Considering${\Phi }_n$, the free metabelian group on $n$ generators, it was also proven that $C({\Phi }_2)={\hat{F}}_{\omega }$ and $C({\Phi }_3)\supseteq {\hat{F}}_{\omega }$. In this paper we prove that $C({\Phi }_n)$ for $n\ge 4$ is abelian. So, while the dichotomy in the abelian case is between $n=2$ and $n\ge 3$, in the metabelian case it is between $n=2,3$ and $n\ge 4$.