The Congruence Subgroup Problem for finitely generated Nilpotent Groups (2005.03263v1)

Abstract: The congruence subgroup problem for a finitely generated group $\Gamma$ and $G\leq Aut(\Gamma)$ 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 $\Gamma^{*}=\bar{\Gamma}/tor(\bar{\Gamma})\cong\mathbb{Z}^{(d)}$. Denote $Aut^{*}(\Gamma)=\textrm{Im}(Aut(\Gamma)\to Aut(\Gamma^{*}))\leq GL_{d}(\mathbb{Z})$. In this paper we show that when $\Gamma$ is nilpotent, there is a canonical isomorphism $C\left(\Gamma\right)\simeq C(Aut^{}(\Gamma),\Gamma^{})$. In other words, $C\left(\Gamma\right)$ is completely determined by the solution to the classical congruence subgroup problem for the arithmetic group $Aut^{*}(\Gamma)$. 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 $n\geq3$, and $C(\Psi_{2,c})=C(\mathbb{Z}^{(2)})=\hat{F}_{\omega}$ = the free profinite group on countable number of generators.