The congruence subgroup problem for a finitely generated group asks whether the map is injective, or more generally, what is its kernel ? Here denotes the profinite completion of . It is well known that for finitely generated free abelian groups for every , but , where is the free profinite group on countably many generators.
Considering, the free metabelian group on generators, it was also proven that and . In this paper we prove that for is abelian. So, while the dichotomy in the abelian case is between and , in the metabelian case it is between and .
Cite this article
David El-Chai Ben-Ezra, The congruence subgroup problem for the free metabelian group on generators. Groups Geom. Dyn. 14 (2020), no. 3, pp. 729–764DOI 10.4171/GGD/561