As a part of my proof that $\mathbb{A}_n$ is simple, I have to prove this statement.
Let $\pi \in \mathbb{S}_n$ contain only two-cycles, at least three of them. Then it is possible to obtain a permutation with its appropriate conjugation, which has two three-cycles.
It seems to be pretty clear for me, but how am I supposed to find $g \in \mathbb{S}_n$ such that $\pi (g \pi g^{-1})$ contains two three-cycle? Should I use brute force, or is there some more convenient way?