The congruence subgroup problem for the free metabelian group on n ≥ 4 generators

The congruence subgroup problem for a finitely generated group Gamma asks whether the map <(Aut(Gamma))over cap> -> Aut((Gamma) over cap) is injective, or more generally, what is its kernel C(Gamma)? Here (X ) over cap denotes the profinite completion of X. It is well known that for finitely generated free abelian groups C(Z(n)) = {1} for every n >= 3, but C(Z(2)) (F) over cap omega = where (F) over cap 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(Phi(2)) = (F) over cap omega, and C(Phi(3)) superset of (F) over cap omega. In this paper we prove that C(Phi(n)) for n >= 4 is abelian. So, while the dichotomy in the abelian case is between n = 2 and n >= 3, in the metabelian case it is between n = 2, 3 and n >= 4.