 Solution

Let the lengths in the disjoint cycle decomposition of $\tau$ be ${\ell }_1,\dots ,{\ell }_s$, with fixed points contributing length one. Disjoint cycles commute, and ${\tau }^m=e$ if and only if every cycle raised to $m$ is the identity, equivalently ${\ell }_j$ divides $m$ for every $j$. Hence the order of a group element isIn particular, the identity has order one.