6533b7d7fe1ef96bd1269167

RESEARCH PRODUCT

Roots in the mapping class groups

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The purpose of this paper is the study of the roots in the mapping class groups. Let $\Sigma$ be a compact oriented surface, possibly with boundary, let $\PP$ be a finite set of punctures in the interior of $\Sigma$, and let $\MM (\Sigma, \PP)$ denote the mapping class group of $(\Sigma, \PP)$. We prove that, if $\Sigma$ is of genus 0, then each $f \in \MM (\Sigma)$ has at most one $m$-root for all $m \ge 1$. We prove that, if $\Sigma$ is of genus 1 and has non-empty boundary, then each $f \in \MM (\Sigma)$ has at most one $m$-root up to conjugation for all $m \ge 1$. We prove that, however, if $\Sigma$ is of genus $\ge 2$, then there exist $f,g \in \MM (\Sigma, \PP)$ such that $f^2=g^2$, $f$ is not conjugate to $g$, and none of the conjugates of $f$ commutes with $g$. Afterwards, we focus our study on the roots of the pseudo-Anosov elements. We prove that, if $\partial \Sigma \neq \emptyset$, then each pseudo-Anosov element $f \in \MM(\Sigma, \PP)$ has at most one $m$-root for all $m \ge 1$. We prove that, however, if $\partial \Sigma = \emptyset$ and the genus of $\Sigma$ is $\ge 2$, then there exist two pseudo-Anosov elements $f,g \in \MM (\Sigma)$ (explicitely constructed) such that $f^m=g^m$ for some $m\ge 2$, $f$ is not conjugate to $g$, and none of the conjugates of $f$ commutes with $g$. Furthermore, if the genus of $\Sigma$ is $\equiv 0 (\mod 4)$, then we can take $m=2$. Finally, we show that, if $\Gamma$ is a pure subgroup of $\MM (\Sigma, \PP)$ and $f \in \Gamma$, then $f$ has at most one $m$-root in $\Gamma$ for all $m \ge 1$. Note that there are finite index pure subgroups in $\MM (\Sigma, \PP)$.