Splitting of conjugacy class in $A_n$

Question

During reading, I have encountered this, in several places:

The following are equivalent for a permutation $\sigma \in A_n$:
1) the $S_n$-conjugacy class of $\sigma$ splits into two $A_n$-classes
2) there is no odd permutation which commutes with $\sigma$
3) $\sigma$ has no cycles of even length, and all its cycles have distinct lengths

I don't understand the meaning of the first bullet, what is the meaning of Conjugacy class splitting in two? what does it look like?

thanks

Answer 1

In $S_n$, conjugacy classes are determined by cycle type. For example, the conjugacy class $C$ of $(123)$ in $S_4$ consists of all 3-cycles. The $S_4$-conjugacy class $C$ is contained in $A_4$, where it becomes the union of two $A_4$-conjugacy classes. These are $$ \{ (123), (421), (243), (341) \} $$ and $$ \{ (132), (412), (234), (314) \} . $$ Notice that $(123)$ and $(132)$ are conjugate in $S_4$ via any of $(12)$, $(13)$ and $(23)$, but none of these lie in $A_4$.

Answer 2

I wrote this as comment, but it probably looks better as answer.

In general, given a group $G$ and a subgroup $H\leq G$, for $x\in G$, we have $x^G\cap H=\bigcup_{i=1}^n x_i^H$, for some $x_1,...,x_n\in H$ and $n\leq |G:H|$. Notice that this intersection may be empty, in which case $x$ is called derangement (a derangement always exists).

In your case, since $A_n$ is a $2$-index subgroup of $S_n$ you have $\sigma^{S_n}\cap A_n=\sigma_1^{A_n}\cap \sigma_2^{A_n}$. In this sense we say that the $S_n$-conjugacy class of $\sigma$ splits into $2$ $A_n$-conjugacy classes. (Notice that your conditions 2 and 3 tell you precisely when you have a union of two distinct classes or not.)