Abstract

The congruence subgroup problem for a finitely generated group $\Gamma$ and $G\le Aut\left(\Gamma \right)$ asks whether the map $Ĝ\to Aut\left(\widehat{\Gamma }\right)$ is injective, or more generally, what its kernel $C\left(G,\Gamma \right)$ is. Here $\widehat{X}$ denotes the profinite completion of $X$. In this paper we investigate $C\left(IA\left({\Phi }_n\right),{\Phi }_n\right)$, where ${\Phi }_n$ is a free metabelian group on $n\ge 4$ generators, and $IA\left({\Phi }_n\right)=ker\left(Aut\left({\Phi }_n\right)\to {GL}_n\left(\mathbb{Z}\right)\right)$.

We show that in this case $C\left(IA\left({\Phi }_n\right),{\Phi }_n\right)$ is abelian, but not trivial, and not even finitely generated. This behavior is very different from what happens for a free metabelian group on $n=2$ or $3$ generators, or for finitely generated nilpotent groups.

Keywords

Mathematical Subject Classification 2010

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