The Congruence Subgroup Problem for finitely generated Nilpotent Groups

Abstract

The congruence subgroup problem for a finitely generated group $\Gamma$ and $G\le Aut(\Gamma )$ asks whether the map $\hat{G}\to Aut(\hat{\Gamma })$ is injective, or more generally, what is its kernel $C\left(G,\Gamma \right)$? Here $\hat{X}$ denotes the profinite completion of $X$. In the case $G=Aut(\Gamma )$ we denote $C\left(\Gamma \right)=C\left(Aut(\Gamma ),\Gamma \right)$. Let $\Gamma$ be a finitely generated group, $\bar{\Gamma }=\Gamma /[\Gamma ,\Gamma ]$, and ${\Gamma }^{\ast }=\bar{\Gamma }/tor(\bar{\Gamma })\cong {\mathbb{Z}}^{(d)}$. Denote $Au{t}^{\ast }(\Gamma )=\text{Im}(Aut(\Gamma )\to Aut({\Gamma }^{\ast }))\le G{L}_d(\mathbb{Z})$. In this paper we show that when $\Gamma$ is nilpotent, there is a canonical isomorphism $C\left(\Gamma \right)\simeq C(Au{t}^{\ast }(\Gamma ),{\Gamma }^{\ast })$. In other words, $C\left(\Gamma \right)$ is completely determined by the solution to the classical congruence subgroup problem for the arithmetic group $Au{t}^{\ast }(\Gamma )$. In particular, in the case where $\Gamma ={\Psi }_{n,c}$ is a finitely generated free nilpotent group of class $c$ on $n$ elements, we get that $C({\Psi }_{n,c})=C({\mathbb{Z}}^{(n)})=\{e\}$ whenever $n\ge 3$, and $C({\Psi }_{2,c})=C({\mathbb{Z}}^{(2)})={\hat{F}}_{\omega }$ = the free profinite group on countable number of generators.

18 pages