JournalsggdVol. 14, No. 3pp. 729–764

The congruence subgroup problem for the free metabelian group on n4n\geq4 generators

  • David El-Chai Ben-Ezra

    The Hebrew University of Jerusalem, Israel
The congruence subgroup problem for the free metabelian group on $n\geq4$ generators cover
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Abstract

The congruence subgroup problem for a finitely generated group Γ\Gamma asks whether the map Aut(Γ)^Aut(Γ^)\widehat{\mathrm{Aut}(\Gamma)}\to \mathrm{Aut}(\widehat{\Gamma}) is injective, or more generally, what is its kernel C(Γ)C(\Gamma)? Here X^\widehat{X} denotes the profinite completion of XX. It is well known that for finitely generated free abelian groups C(Zn)={1}C(\mathbb{Z}^{n})=\{ 1\} for every n3n\geq3, but C(Z2)=F^ωC(\mathbb{Z}^{2})=\widehat{F}_{\omega}, where F^ω\widehat{F}_{\omega} is the free profinite group on countably many generators.

ConsideringΦn\Phi_{n}, the free metabelian group on nn generators, it was also proven that C(Φ2)=F^ωC(\Phi_{2})=\widehat{F}_{\omega} and C(Φ3)F^ωC(\Phi_{3})\supseteq\widehat{F}_{\omega}. In this paper we prove that C(Φn)C(\Phi_{n}) for n4n\geq4 is abelian. So, while the dichotomy in the abelian case is between n=2n=2 and n3n\geq3, in the metabelian case it is between n=2,3n=2,3 and n4n\geq4.

Cite this article

David El-Chai Ben-Ezra, The congruence subgroup problem for the free metabelian group on n4n\geq4 generators. Groups Geom. Dyn. 14 (2020), no. 3, pp. 729–764

DOI 10.4171/GGD/561