I know there are questions that may look similiar at first glance, but they ask about different aspects.
In a book I read there is a certain result proven($[\sigma]$ denotes the conjugacy class of $\sigma$):
Let $n \geq 2$, and let $\sigma \in A_n$. Then $[\sigma]_{A_n} = [\sigma]_{S_n}$ or the size of $[\sigma]_{A_n}$ is half the size of $[\sigma]_{S_n}$, according to whether the centralizer $Z_{S_n}(\sigma)$ is not or is contained in $A_n$.
Then it says that in case if $Z_{S_n}(\sigma)$ lies in $A_n$(that is, $|[\sigma]_{S_n}| = 2|[\sigma]_{A_n}|$ ) then $[\sigma]_{S_n}$ actually splits into two conjugacy classes in $A_n$(one of which is, obviously, $[\sigma]_{A_n}$).
But how come $[\sigma]_{S_n} \setminus [\sigma]_{A_n}$ is even a conjugacy class in $A_n$?
What I know is that $g \in [\sigma]_{S_n} \setminus [\sigma]_{A_n}$ iff $g = \tau \sigma \tau^{-1}$ for some odd permutation such that for all $\rho \in A_n \ \ \ \rho^{-1} \tau \notin Z_{S_n}( \sigma ) \Leftrightarrow \rho^{-1} \tau \sigma \tau^{-1} \rho \neq \sigma \Leftrightarrow \tau \sigma \tau^{-1} \neq \rho \sigma \rho^{-1}$.
It's obvious that $g \in A_n$. But why all such $g$'s are conjugates to some even permutation?