The Congruence Subgroup Problem for low rank Free and Free Metabelian groups

Abstract

The congruence subgroup problem for a finitely generated group asks whether Aut() Aut() is injective, or more generally, what is its kernel C()? Here X denotes the profinite completion of X. In this paper we first give two new short proofs of two known results (for =F2 and 2) and a new result for =3: 1. C(F2)=\ e\ when F2 is the free group on two generators. 2. C(2)=Fω when n is the free metabelian group on n generators, and Fω is the free profinite group on 0 generators. 3. C(3) contains Fω. Results 2. and 3. should be contrasted with an upcoming result of the first author showing that C(n) is abelian for n≥4.

0

Discussion (0)

Sign in to join the discussion.

Loading comments…