The congruence subgroup problem for the free metabelian group on $n\geq4$ generators

Abstract

The congruence subgroup problem for a finitely generated group $\Gamma$ asks whether the map $\widehat{\mathrm{Aut}(\Gamma)}\to \mathrm{Aut}(\widehat{\Gamma})$ is injective, or more generally, what is its kernel $C(\Gamma)$? Here $\widehat{X}$ denotes the profinite completion of $X$. It is well known that for finitely generated free abelian groups $C(\mathbb{Z}^{n})={ 1}$ for every $n\geq3$, but $C(\mathbb{Z}^{2})=\widehat{F}{\omega}$, where $\widehat{F}{\omega}$ is the free profinite group on countably many generators. Considering$\Phi\_{n}$, the free metabelian group on $n$ generators, it was also proven that $C(\Phi\_{2})=\widehat{F}{\omega}$ and $C(\Phi{3})\supseteq\widehat{F}{\omega}$. In this paper we prove that $C(\Phi{n})$ for $n\geq4$ is abelian. So, while the dichotomy in the abelian case is between $n=2$ and $n\geq3$, in the metabelian case it is between $n=2,3$ and $n\geq4$.