Roots of outer automorphisms of free groups and centralizers of abelian subgroups of $\mathrm{Out}(F_N)$

TL;DR

This paper proves that a certain subgroup of the outer automorphism group of a free group is an R-group, and uses this to analyze centralizers and normalizers of abelian subgroups, providing new structural insights.

Contribution

It establishes that the subgroup IA_N(Z/3Z) is an R-group, and applies this to describe normalizers and centralizers of abelian subgroups in Out(F_N).

Findings

IA_N(Z/3Z) is an R-group.

Normalizers of abelian subgroups equal their centralizers.

Alternative proof that certain elements have virtually abelian centralizers.

Abstract

Let $N \geq 2$ and let $\mathrm{Out}(F_N)$ be the outer automorphism group of a nonabelian free group of rank $N$. Let $\mathrm{IA}_N(\mathbb{Z}/3\mathbb{Z})$ be the finite index subgroup of $\mathrm{Out}(F_N)$ which is the kernel of the natural action of $\mathrm{Out}(F_N)$ on $H_1(F_N,\mathbb{Z}/3\mathbb{Z})$. We show that $\mathrm{IA}_N(\mathbb{Z}/3\mathbb{Z})$ is an $R$-group, that is, for every $\phi,\psi \in \mathrm{IA}_N(\mathbb{Z}/3\mathbb{Z})$, if there exists $k \in \mathbb{N}^*$ such that $\phi^k=\psi^k$, then $\phi=\psi$. This answers a question of Handel and Mosher. We then use the fact that $\mathrm{IA}_N(\mathbb{Z}/3\mathbb{Z})$ is an $R$-group in order to prove that the normalizer in $\mathrm{IA}_N(\mathbb{Z}/3\mathbb{Z})$ of every abelian subgroup of $\mathrm{IA}_N(\mathbb{Z}/3\mathbb{Z})$ is equal to its centralizer. We finally give an alternative proof of a result, due to Feighn and Handel, that the centralizer of an element of $\mathrm{Out}(F_N)$ which has only finitely many periodic orbits of conjugacy classes of maximal cyclic subgroups of $F_N$ is virtually abelian.