{
  "id": "2212.07674",
  "version": "v1",
  "published": "2022-12-15T09:18:06.000Z",
  "updated": "2022-12-15T09:18:06.000Z",
  "title": "Roots of outer automorphisms of free groups and centralizers of abelian subgroups of $\\mathrm{Out}(F_N)$",
  "authors": [
    "Yassine Guerch"
  ],
  "comment": "18 pages",
  "categories": [
    "math.GR",
    "math.GT"
  ],
  "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.",
  "revisions": [
    {
      "version": "v1",
      "updated": "2022-12-15T09:18:06.000Z"
    }
  ],
  "analyses": {
    "subjects": [
      "20E05",
      "20E08",
      "20E36",
      "20F65"
    ],
    "keywords": [
      "abelian subgroup",
      "centralizer",
      "outer automorphism group",
      "finite index subgroup",
      "nonabelian free group"
    ],
    "note": {
      "typesetting": "TeX",
      "pages": 18,
      "language": "en",
      "license": "arXiv",
      "status": "editable"
    }
  }
}