The IA-congruence kernel of high rank free Metabelian groups

TL;DR

This paper studies the kernel of the IA-congruence subgroup for high-rank free metabelian groups, revealing it is abelian, non-trivial, and infinitely generated, contrasting with lower-rank or nilpotent cases.

Contribution

It demonstrates that for free metabelian groups with at least four generators, the IA-congruence kernel is abelian, non-trivial, and infinitely generated, a novel finding in this area.

Findings

Kernel is abelian for n≥4

Kernel is non-trivial and not finitely generated

Behavior differs from lower-rank or nilpotent groups

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 this paper we investigate $C\left(IA(\Phi_{n}),\Phi_{n}\right)$, where $\Phi_{n}$ is a free metabelian group on $n\geq4$ generators, and $IA(\Phi_{n})=\ker(Aut(\Phi_{n})\to GL_{n}(\mathbb{Z}))$. We show that in this case $C(IA(\Phi_{n}),\Phi_{n})$ is abelian, but not trivial, and not even finitely generated. This behavior is very different from what happens for free metabelian group on $n=2,3$ generators, or for finitely generated nilpotent groups.