Let $n>2$ and $\tau\in A_n$ and let $C=\{g\tau g^{-1}|g\in S_n\}$ be the conjugacy class of $\tau$ in $S_n$.
Suppose that there is no odd element $x\in S_n$ such that $\tau x=x\tau$.
Prove that $C$ is a union of two disjoint conjugacy classes in $A_n$.
I managed to observe that if $\sigma_1\in S_n$ an even element and $\sigma_2\in S_n$ an odd element, then:
$\sigma_1\tau\sigma_1^{-1}\neq\sigma_2\tau\sigma_2^{-1}$ otherwise we get that the odd element $x=\sigma_2^{-1}\sigma_1$ commutes with $\tau$.
So, I got one conjugacy class to be the conjugacy class of $\tau$ in $A_n$, say $Q$. It is now clear that it contains all the elements in $C$ of the form $g\tau g^{-1}$ for $g$ even. But I couldn't prove that $C\setminus Q$ can be generated by a conjugacy class in $A_n$ and moreover the sizes equivalence $|Q|=|C\setminus Q|$.
Best Answer
Let $g_0\in S_n$ be a fixed odd permutation (for example, a transposition).
Then $C$ is the disjoint union of the conjugacy classes in $A_n$ of $\tau$ and $g_0\tau g_o^{-1}.$
First, let $g\tau g^{-1}\in C$. Then either $g\in A_n$, and $g\tau g^{-1}$ lies in the $A_n$-conjugacy class of $\tau$, or $g$ is odd and in this case, $g=g'g_0$ with $g'\in A_n$ and $g\tau g^{-1}$ lies in the $A_n$-conjugacy class of $g_0\tau g_0^{-1}.$ (Indeed $g'=g g_0^{-1}$ is the product of two odd permutations, hence is even).
It remains to see that the two $A_n$-conjugacy classes are disjoint: assume that $g\tau g^{-1}= h (g_0 \tau g_0^{-1})h^{-1}$ with $g,h$ even. Then $g^{-1}hh_0$ commutes with $\tau$, but is odd, contradicting the assumption on $\tau.$