Browsing the web I came across this:
The conjugacy class of an element $g\in A_{n}$:
- splits if the cycle decomposition of $g\in A_{n}$ comprises cycles of distinct odd length. Note that the fixed points are here treated as cycles of length $1$, so it cannot have more than one fixed point; and
- does not split if the cycle decomposition of $g$ contains an even cycle or contains two cycles of the same length.
Anybody with a proof?
Best Answer
Note the following: (1) The conjugacy class in $S_n$ of an element $\sigma \in A_n$ splits, iff there is no element $\tau \in S_n\setminus A_n$ commuting with $\sigma$. For if there is one, for each $\tau' \in S_n \setminus A_n$ we have $$ \tau'\sigma{\tau'}^{-1} = \tau'\sigma\tau\tau^{-1}\tau'{}^{-1} = (\tau'\tau)\sigma(\tau'\tau)^{-1} $$ and $\tau\tau' \in A_n$.On the other hand, if $\tau\sigma\tau^{-1}$ and $\sigma$ with $\tau \in S_n\setminus A_n$ are conjugated in $A_n$, then for some $\tau' \in A_n$, we have $\tau\sigma\tau^{-1} = \tau'\sigma\tau'^{-1}$, giving $$ \tau'{}^{-1}\tau \sigma = \sigma\tau'{}^{-1}\tau $$ and hence $\tau'{}^{-1}\tau \in S_n\setminus A_n$ commutes with $\sigma$.
Now suppose, $\sigma$ has a cycle $c_i$ of even length. A cycle of even length is an element of $S_n \setminus A_n$, and as $\sigma$ commutes with its cycles, we are done by the above. If $\sigma$ has to cycles $(a_1\ldots a_\ell)$ and $(b_1 \ldots b_\ell)$ of the same odd length $\ell$, then $(a_1b_1) \ldots (a_\ell b_\ell)$ is a product of $\ell$ permutations (hence odd, so an element of $S_n \setminus A_n$) commuting with $\sigma$.
Now suppose $\sigma = c_1 \cdots c_s$ is a product of odd cycles $c_i$ of distinct length $d_i$. Let $\tau \in S_n$ be a permutation commuting with $\sigma$. Then $\tau$ must fix each of the $c_i$, that is, $\tau$ must be of the form $\tau = c_1^{a_1} \cdots c_s^{a_s}$ for some $a_i \in \mathbb Z$. But as the $c_i$ are even permutations (as cycles of odd length), we have $\tau \in A_n$. So no $\tau \in S_n \setminus A_n$ commutes with $\sigma$ and we are done.