I'm trying to understand a proof about the splitting criterion for the conjugacy classes of $A_n$, but I think I'm getting bogged down in earlier terminology. Here's a setup.
Definitions: Let $G$ be a group and $N \unlhd G$.
- A conjugacy class in $G$ is a set $\{ghg^{-1} \mid g \in G \}$ for fixed $h \in G$.
- A $G$-conjugacy class in $N$ is a subset of $N$ of the form $\{ghg^{-1} \mid g \in G\}$ for fixed $h \in N$.
The proposition I'm trying to prove is the following.
Proposition: For any $\sigma \in A_n$ the $S_n$ conjugacy class of $\sigma$ equals the $A_n$ conjugacy class of $\sigma$ if and only if there is an odd permutation $\tau \in S_n$ such that $\tau \sigma = \sigma \tau$. If this does not hold then the $S_n$ conjugacy class of $\sigma$ is the disjoint union of two $A_n$ conjugacy classes; the conjugacy class of $\sigma$ and the conjugacy class of $\tau\sigma\tau^{-1}$ for any $\tau \in S_n$.
The proof begins as follows. Let $C = \{g\sigma g^{-1} \mid g \in S_n \}$. I.e $C$ is the conjugacy class of $\sigma$. Then let $A_n$ act on $C$ by conjugation, and let $C_1,\dots,C_m$ be the distinct orbits of this action; i.e the distinct $A_n$ conjugacy classes contained in $C$.
This last line doesn't make sense to me, but maybe it's because the definition of an $G$-conjugacy class didn't make sense. Why would the orbits be $A_n$ conjugacy classes? Because in my mind the action is $\tau \ast a = \tau a \tau^{-1}$ for $\tau \in A_n$ and $a \in C$, but then an orbit is just $\text{Orb}(a) = \{\tau(g \sigma g^{-1}) \tau^{-1} \mid (g \sigma g^{-1}) \in C \}$. But why is this an $A_n$ conjugacy class? To be an $A_n$ conjugacy class wouldn't we need I get that $\tau$ is an element of $A_n$ so we're conjugating by elements of the right set, but wouldn't we need $g\sigma g^{-1} \in A_n$ by the definition? Sorry I'm not sure if this makes much sense, I think I'm misunderstanding the definitions somewhere. If I need to elaborate anywhere let me know, and thanks in advance for the help.
