  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 $n\geq4$.
