How many permutations of order $8$ in $S_{10}$?

Question

Find the number of permutations of order $8$ in $S_{10}$. If $\sigma$ is one of them, find the number of subgroups in the subgroup generated by $\sigma$.

Answer 1

An element of $S_{10}$ has order $8$ iff its cycle type is $(8,1,1)$ or $(8,2)$.

There are are ${10\choose 8}\cdot7!$ of type $(8,1,1)$. There are ${10\choose 8}\cdot 7!$ of type $(8,2)$.

Together we get $900\cdot7!$

If $\sigma$ is one of them, $\langle\sigma \rangle $ is cyclic of order $8$. Therefore there are two proper subgroups, of orders $2$ and $4$, other than $\{e\}$.

Answer 2

Hint:

The order of a permutation decomposed as a product of disjoint cycles is the l.c.m. of the orders of the cycles, i.e. the l.c.m. of the lengths of the cycles.