split conjugacy class in $S_n$ into two equal size conjugacy classes in $A_n$ 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 $|Q|=|C\setminus Q|$.
 A: 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 $\tau.$
A: $\newcommand{\Size}[1]{\left\lvert #1 \right\rvert}$Consider the more general situation where the finite group $G$ acts on the finite set $\Omega$, and $H$ is a subgroup of $G$ of index $2$.
Let us compare the orbit of $\alpha \in \Omega$ under $G$ and under $H$, using orbit-stabiliser. I write $\alpha^{G}, \alpha^{H}$ for the orbits, $G_{\alpha}, H_{\alpha}$ for the stabiliser.
$$
\Size{\alpha^{H}}
=
\Size{H : H_{\alpha}}
=
\Size{H : H \cap G_{\alpha}}
=
\Size{H G_{\alpha} : G_{\alpha}}
=
\frac{\Size{H G_{\alpha}}}{\Size{G}} \Size{G : G_{\alpha}}
=
\frac{\Size{H G_{\alpha}}}{\Size{G}} \Size{\alpha^{G}}.
$$
Now note that $H G_{\alpha}$ is a subgroup of $G$, as $H$ is normal in $G$, and that $H \le H G_{\alpha} \le G$,
so that
$$
\frac{\Size{H G_{\alpha}}}{\Size{G}}
=
\begin{cases}
1/2 & \text{if $G_{\alpha} \le H$}\\
1 & \text{if $G_{\alpha} \nleq H$}\\
\end{cases}
$$
Apply for $G = S_{n}$ and $H = A_{n}$ acting by conjugacy on $H$.
