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 $\widehat{Aut\left(\Gamma \right)}\to Aut(\hat{\Gamma })$ is injective, or more generally, what is its kernel $C\left(\Gamma \right)$? Here $\hat{X}$ denotes the profinite completion of $X$. It is well known that for finitely generated free abelian groups $C\left({\mathbb{Z}}^n\right)=\left\{1\right\}$ for every $n\ge 3$, but $C\left({\mathbb{Z}}^2\right)={\hat{F}}_{\omega }$, 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 $C\left({\Phi }_2\right)={\hat{F}}_{\omega }$ and $C\left({\Phi }_3\right)\supseteq {\hat{F}}_{\omega }$. In this paper we prove that $C\left({\Phi }_n\right)$ 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 $n\ge 4$.

arXiv comment30 pages