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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

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  • $\begingroup$ Hint: use the fact that every element of $S_n$ is either even or is $\tau\alpha$ for $\alpha$ even. $\endgroup$ – Cheerful Parsnip Dec 1 '15 at 18:20
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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.

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    $\begingroup$ Hello, thanks, sorry i didn't see the comment earlier but this was a very simple yet concise proof. Thank you $\endgroup$ – Melba1993 Dec 1 '15 at 19:00

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