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Let $\sigma \in A_n$ such that the disjoint cycle decomposition of $\sigma$ contains no even cycles or cycles of the same length. Although there is one conjugacy class of $\sigma$ in $S_n$, there are two conjugacy classes of $\sigma$ in $A_n$. We can choose any permutation in the original $S_n$ conjugacy class to represent one of the two $A_n$ classes. But how can we find another permutation to represent the other $A_n$ class? These answers suggest the following to be true:

Let $\sigma \in A_n$. If the conjugacy class of $\sigma$ splits in $A_n$, then $\tau \sigma \tau^{-1}$ is not conjugate (in $A_n$) to $\sigma$ for all $\tau \in S_n \setminus A_n$.

What is a proof of this fact?

From here we know that $\tau \sigma \tau^{-1} \neq \sigma$ for all odd $\tau$. But proving $\tau \sigma \tau^{-1} $ is not conjugate (in $A_n$) to $\sigma$ seems to be a stronger claim.

jskattt797
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    Not really. Since $\rho\tau$ is odd for all $\rho \in A_n$, we have $\rho\tau\sigma\tau^{-1}\rho^{-1} \ne \sigma$. – Derek Holt Nov 24 '20 at 08:13

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