The congruence subgroup problem for a finitely generated group $\Gamma$ and $G\leq Aut(\Gamma)$ asks whether the map $\hat{ G}\to Aut(\hat{ \Gamma})$ is injective, or more generally, what is its kernel $C\left(G,\Gamma\right)$? Here $\hat{ X}$ denotes the profinite completion of $X$. In this paper we investigate $C\left(IA(\Phi_{ n}),\Phi_{ n}\right)$, where $\Phi_{ n}$ is a free metabelian group on $n\geq4$ generators, and $IA(\Phi_{ n})=\ker(Aut(\Phi_{ n})\to GL_{ n}(\mathbb{ Z}))$. We show that in this case $C(IA(\Phi_{ n}),\Phi_{ n})$ is abelian, but not trivial, and not even finitely generated. This behavior is very different from what happens for free metabelian group on $n=2,3$ generators, or for finitely generated nilpotent groups.
Categories: math.KT math.GR