The congruence subgroup problem for the free metabelian group on $n\geq4$ generators

Abstract

The congruence subgroup problem for a finitely generated group \Gamma asks whether the map \widehat{\mathrm{Aut}(\Gamma)}\to \mathrm{Aut}(\widehat{\Gamma}) is injective, or more generally, what is its kernel C(\Gamma) ? Here \widehat{X} denotes the profinite completion of X . It is well known that for finitely generated free abelian groups C(\mathbb{Z}^{n})=\{ 1\} for every n\geq3 , but C(\mathbb{Z}^{2})=\widehat{F}_{\omega} , where \widehat{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(\Phi_{2})=\widehat{F}_{\omega} and C(\Phi_{3})\supseteq\widehat{F}_{\omega} . In this paper we prove that C(\Phi_{n}) 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 .