I've been working with this problem for my own exercise, as I haven't touched on Group theory for quite a while. But I am too incompetent to give it a proof. I've searched for answers but surprisingly, while it seems like a common question, none had a fully comprehensible answer.
Q. Let $\sigma \in A_n$. Suppose that there is some odd $\tau \in S_n$ that commutes $\sigma \tau = \tau \sigma$. Show that then, $\text{Cl}_{S_n}(\sigma)=\text{Cl}_{A_n}(\sigma)$. Where $\text{Cl}_G$ denotes the conjugacy class of some $G$.
My attempt was to use the fact that the orbits of a normal subgroup are equal in order when the entire group acts transitively, sine conjugacy is a transitive action, and $\sigma \in A_n$ would split $\text{Cl}_{S_n}(\sigma)$ to $|S_n:A_n\text{Cl}_{S_n}(\sigma)|$ classes under the action of $A_n$.
...But that's all I got to, maybe it's only an idea rather than saying it's any part of the proof. I've just tried giving it a kick start from whatever that pops up in my mind that seem relevant.
Please ignore that attempt if it seems like it's going nowhere that has t do with proving the question...
Can someone please give me proof to the above? It would be great if I can study a valid proof of this and perhaps ask in the comments if I am not clear about anything...Thank you so much
Okay, so clearly $CL_{A_n}(\sigma)\subseteq CL_{S_n}(\sigma)$. Now suppose $x=g\sigma g^{-1}\in CL_{S_n}(\sigma)$. If $g\in A_n$ then $x\in CL_{A_n}(\sigma)$ and we are done. Otherwise, $g$ is odd, which means it can be written $\alpha\tau$ for $\alpha\in A_n$. But then $g\sigma g^{-1}= \alpha\tau\sigma\tau^{-1}\alpha^{-1}=\alpha\sigma\alpha^{-1}$ because $\tau$ commutes with $\sigma$. So again $x\in CL_{A_n}(\sigma)$ and we are done.