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The congruence subgroup problem for a finitely generated group $\Gamma$ askswhether the map $\hat{Aut\left(\Gamma\right)}\to Aut(\hat{\Gamma})$ isinjective, 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 forfinitely 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 $\hat{F}_{\omega}$ isthe free profinite group on countably many generators.  Considering $\Phi_{n}$, the free metabelian group on $n$ generators, it wasalso 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 inthe abelian case is between $n=2$ and $n\geq3$, in the metabelian case it isbetween $n=2,3$ and $n\geq4$.

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