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

### David El-Chai Ben-Ezra

The Hebrew University of Jerusalem, Israel

## 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 $n\geq4$.

## Cite this article

David El-Chai Ben-Ezra, The congruence subgroup problem for the free metabelian group on $n\geq4$ generators. Groups Geom. Dyn. 14 (2020), no. 3, pp. 729–764

DOI 10.4171/GGD/561