show elements of order 15 in $S_8$ satisfy a relation

Question

Let $G$ be a group. Define a relation $\sim$ on $G$ by $a \sim b$ if there exists $g \in G$ such that $a = gbg^{-1}$. Prove that all elements of order 15 in $S_8$ are related by $\sim$.

I noticed that in order to have order $15$ in $S_8$. We need a $5$ cycle and a $3$ cycle. Then, I am stuck. Can anyone give me some ideas about what to do? Thank you in advance.

Answer 1

Conjugate permutations have the same cycle types. Since $15$ cannot be split into any other cycle types than $5$ and $3$, all of them must be conjugate. For a proof that conjugate permutations must have the same cycle type look in any abstract algebra textbook.

The idea is that when we conjugate a permutation, we are simply relabeling the permutation but the "essential structure" (ie. the cycle types) do not change. Instead of cycling (1 2 3) in a conjugate permutation we may cycle (2 3 4). But the permutation is essentially the same if we associate 1 with 2, 2 with 3, and 3 with 4. So, all elements of order 15 must be of the 3-cycle,5-cycle type (by the unique prime factors of 15).

Edit: Proof

Answer 2

The conjugacy class of a permutation is defined by the multiset of lengths of independent cycles. To prove it, note that $\pi\, (1\,2\ldots\,k)\,\pi^{-1}=(\pi(1)\,\pi(2)\ldots\,\pi(k))$.