# The Congruence Subgroup Problem for the Free Metabelian group on $n\geq4$ generators

## TL;DR

This paper investigates the congruence subgroup problem for free metabelian groups, showing that for four or more generators, the kernel of the associated map is abelian, revealing a new structural property.

## Contribution

It proves that the kernel $C(\

## Key findings

- For $n \\geq 4$, $C(\\\Phi_{n})$ is abelian.
- The dichotomy in the kernel's structure shifts at $n=4$ in the metabelian case.
- Extends understanding of the congruence subgroup problem for free metabelian groups.

## Abstract

The congruence subgroup problem for a finitely generated group $\Gamma$ asks whether the map $\hat{Aut\left(\Gamma\right)}\to Aut(\hat{\Gamma})$ 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 $n\geq3$, 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 $n\geq4$ 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 $n\geq4$.