Roots of Outer Automorphisms of Free Groups and Centralizers of Abelian Subgroups of Out(FN)

Abstract

Let $\mathit{N}\ge 2$ and let $\mathrm{Out}({\mathit{F}}_{\mathit{N}})$ be the outer automorphism group of a non-abelian free group of rank N. Let ${\mathrm{IA}}_{\mathit{N}}(\mathbb{Z}/3\mathbb{Z})$ be the finite index subgroup of $\mathrm{Out}({\mathit{F}}_{\mathit{N}})$, which is the kernel of the natural action of $\mathrm{Out}({\mathit{F}}_{\mathit{N}})$ on ${\mathit{H}}_1({\mathit{F}}_{\mathit{N}},\mathbb{Z}/3\mathbb{Z})$. We show that ${\mathrm{IA}}_{\mathit{N}}(\mathbb{Z}/3\mathbb{Z})$ is an R-group, that is, for every $\mathit{\varphi },\mathit{\psi }\in {\mathrm{IA}}_{\mathit{N}}(\mathbb{Z}/3\mathbb{Z})$, if there exists $\mathit{k}\ge 1$ such that ${\mathit{\varphi }}^{\mathit{k}}={\mathit{\psi }}^{\mathit{k}}$, then $\mathit{\varphi }=\mathit{\psi }$. This answers a question of Handel and Mosher. We then use the fact that ${\mathrm{IA}}_{\mathit{N}}(\mathbb{Z}/3\mathbb{Z})$ is an R-group in order to prove that the normalizer in ${\mathrm{IA}}_{\mathit{N}}(\mathbb{Z}/3\mathbb{Z})$ of every abelian subgroup of ${\mathrm{IA}}_{\mathit{N}}(\mathbb{Z}/3\mathbb{Z})$ 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 $\mathrm{Out}({\mathit{F}}_{\mathit{N}})$, which has only finitely many periodic orbits of conjugacy classes of maximal cyclic subgroups of ${\mathit{F}}_{\mathit{N}}$, is virtually abelian.