The IA-congruence kernel of high rank free metabelian groups

Abstract

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

We show that in this case $C\left(IA\left({\mathrm{\Phi }}_n\right),{\mathrm{\Phi }}_n\right)$ 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.