On the profinite rigidity of free and surface groups
Ismael Morales
Abstract
Open-access reader
Ismael Morales
Abstract
Open-access reader
Let $S$ be either a free group or the fundamental group of a closed hyperbolic surface. We show that if $G$ is a finitely generated residually-$p$ group with the same pro-$p$ completion as $S$, then two-generated subgroups of $G$ are free. This generalises (and gives a new proof of) the analogous result of Baumslag for parafree groups. Our argument relies on the following new ingredient: if $G$ is a residually-(torsion-free nilpotent) group and $H\leq G$ is a virtually polycyclic subgroup, then $H$ is nilpotent and the pro-$p$ topology of $G$ induces on $H$ its full pro-$p$ topology. Then we study applications to profinite rigidity. Remeslennikov conjectured that a finitely generated residually finite $G$ with profinite completion $\widehat G\cong \widehat S$ is necessarily $G\cong S$. We confirm this when $G$ belongs to a class of groups $\mathcal{H}_{ab}$ that has a finite abelian hierarchy starting with finitely generated residually free groups. This strengthens a previous result of Wilton that relies on the hyperbolicity assumption. Lastly, we prove that the group $S\times {\mathbb Z}^n$ is profinitely rigid within finitely generated residually free groups.
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Let $S$ be either a free group or the fundamental group of a closed hyperbolic surface. We show that if $G$ is a finitely generated residually-$p$ group with the same pro-$p$ completion as $S$, then two-generated subgroups of $G$ are free. This generalises (and gives a new proof of) the analogous result of Baumslag for parafree groups. Our argument relies on the following new ingredient: if $G$ is a residually-(torsion-free nilpotent) group and $H\leq G$ is a virtually polycyclic subgroup, then $H$ is nilpotent and the pro-$p$ topology of $G$ induces on $H$ its full pro-$p$ topology. Then we study applications to profinite rigidity. Remeslennikov conjectured that a finitely generated residually finite $G$ with profinite completion $\widehat G\cong \widehat S$ is necessarily $G\cong S$. We confirm this when $G$ belongs to a class of groups $\mathcal{H}_{ab}$ that has a finite abelian hierarchy starting with finitely generated residually free groups. This strengthens a previous result of Wilton that relies on the hyperbolicity assumption. Lastly, we prove that the group $S\times {\mathbb Z}^n$ is profinitely rigid within finitely generated residually free groups.
Key concepts: Mathematics, Profinite group, Finitely-generated abelian group, Stallings theorem about ends of groups, Nilpotent, Finitely generated group, Abelian group, Group (periodic table)