Suppose that $S_n$, the symmetric group of order $n!$ is given and for given $m\in \mathbb N$ fixed, we are to find the number of solutions to $\theta^m=e, \theta\in S_n$.
Can someone tell me or give some idea how to proceed ? what condition shall I impose on $n, m$ ? etc
Some particular cases can be helpful: for $m = 2,$ we can give a general formula for number of solutions $s(n)$ provided here:
$$s(n) = \sum\limits_{k=0}^{\lfloor \frac{n}{2} \rfloor} \dfrac{n!}{2^k k! (n - 2k)!}.$$
Moreover, some information about existence of solutions for the mentioned equation can be found here. For example, by using Corollary 1 in section 2, we conclude that if $\prod\limits_{i=1}^k p_i^{a_i}$ is the prime factorization of $m,$ then $\theta^m = e$ has at least one solution in $S_n$ if and only if $\sum\limits_{i=1}^k p_i^{a_i} \le m.$