# The congruence subgroup problem for the free metabelian group on $n≥4$ generators

### David El-Chai Ben-Ezra

The Hebrew University of Jerusalem, Israel

## Abstract

The congruence subgroup problem for a finitely generated group $Γ$ asks whether the map $Aut(Γ) →Aut(Γ)$ is injective, or more generally, what is its kernel $C(Γ)$? Here $X$ denotes the profinite completion of $X$. It is well known that for finitely generated free abelian groups $C(Z_{n})={1}$ for every $n≥3$, but $C(Z_{2})=F_{ω}$, where $F_{ω}$ is the free profinite group on countably many generators.

Considering$Φ_{n}$, the free metabelian group on $n$ generators, it was also proven that $C(Φ_{2})=F_{ω}$ and $C(Φ_{3})⊇F_{ω}$. In this paper we prove that $C(Φ_{n})$ for $n≥4$ is abelian. So, while the dichotomy in the abelian case is between $n=2$ and $n≥3$, in the metabelian case it is between $n=2,3$ and $n≥4$.

## Cite this article

David El-Chai Ben-Ezra, The congruence subgroup problem for the free metabelian group on $n≥4$ generators. Groups Geom. Dyn. 14 (2020), no. 3, pp. 729–764

DOI 10.4171/GGD/561