Find the number of abelian subgroups with order $15$ in symmetric group of degree $8$.
I was trying to find the number of elements of order $15$ in $S_8$ but I think this is not true. Also I have read this discussion but have not solved it yet.
Please give me hints or solution. Thanks in advance.
I think it's easier to reason this way.
We know that every abelian subgroup of $S_8$ of order $15$ is cyclic and is generated by an element of type $(12345)(678)$.
In the group of permutations $S_5$ on symbols $\{1,2,3,4,5\}$ there are exactly $24$ cycles of length $5$. Hence, there are exactly $24/4=6$ cyclic subgroups of order $5$.
The set of $8$ symbols can be split into two parts of $3$ and $5$ elements in each parts in $\binom{8}{3}$ ways.
It follows that $S_8$ has $$ 6\cdot\binom{8}{3} $$ of subgroups of the order of $15$.