Show that all elements in the conjugacy class of $\sigma$ in $S_n$ are conjugate in $A_n$ if and only if $\sigma$ commutes with an odd permutation. Let $\sigma\in A_n$. Show that all elements in the conjugacy class of $\sigma$ in $S_n$ (i.e. all elements of the same cycle type as $\sigma$) are conjugate in $A_n$ if and only if $\sigma$ commutes with an odd permutation. 
Hint: Use the previous proven fact: Assume $H$ is normal subgroup of $G$, $\mathcal{K}$ is a conjugacy class of $G$ contained in $H$ and $x\in\mathcal{K}$. Prove that $\mathcal{K}$ is a union of $k$ conjugacy classes of equal size in $H$, where $k=|G:HC_G(x)|$. Deduce that a conjugacy class in $S_n$ which consists of even permutations is either a single conjugacy class under the action of $A_n$ or is a union of two classes of the same size in $A_n$. 
Proof ($\Rightarrow$): Let $\sigma\in A_n$. Assume that all elements in the conjugacy class of $\sigma$ in $S_n$ (i.e. all elements of the same cycle type as $\sigma$) are conjugate in $A_n$. We  need to show that $\sigma$ commutes with an odd permutation. By our assumption we have for all $x\in \sigma S_n\sigma^{-1}$, $\sigma\in \sigma A_n \sigma^{-1}$
Proof ($\Leftarrow$): Let $\sigma\in A_n$. Assume that $\sigma$ commutes with an odd permutation. We need to show all the elements in the conjugacy class of $\sigma$ in $S_n$ (i.e. all elements of the same cycle type as $\sigma$) are conjugate in $A_n$.
Let $\tau$ be an odd permutation in $S_n$. We know that $\sigma$ commutes with $\tau$. i.e. $$\sigma\tau=\tau\sigma \iff \tau\sigma\tau^{-1}=\sigma.$$ Let $\rho$ be an element of the conjugacy class of $\sigma$ in $S_n$
I have tried to do something for each direction but I don't see how the hint is going to help me. Can I get any tips on finishing each direction?
 A: It's not very hard to prove without using the hint:


*

*$ \Rightarrow \,:$ Suppose a permutation in the conjugacy class in $S_n$ of $\sigma$ is also in the conjugacy class of $\sigma$ in $A_n$, i.e. a permutation $ \tau \sigma \tau^{-1} $, $\tau \in S_n\smallsetminus A_n$ can be written as $\alpha \sigma \alpha^{-1} $, $\alpha \in A_n$. Then
$$\tau \sigma \tau^{-1} = \alpha \sigma \alpha^{-1} \iff (\alpha^{-1}\tau )\sigma= \sigma (\alpha^{-1}\tau ),$$
multiplying both sides by $\alpha^{-1}$ on the left and by $\tau$ on the right. This means $\sigma$ commutes with $\beta=\alpha^{-1}\tau $, which is an odd permutation.

*$ \Leftarrow \::$ If $\sigma$ commutes with an odd $\beta$ and $\tau \sigma \tau^{-1} $ is the conjugate of $\sigma$ by an odd permutation, then
\begin{alignat}{2}
\tau \sigma \tau^{-1}&= \tau (\beta \sigma \beta^{-1})\tau^{-1}&&\qquad\text{since $\beta$ and $\sigma$ commute}\\
&=(\tau \beta) \sigma (\beta^{-1}\tau^{-1})&&=(\tau \beta) \sigma (\tau \beta)^{-1}
\end{alignat}
As $\beta$ is odd, we've finished since $\tau \beta$ is even

A: If no odd permutation commutes with $\sigma$, then $H=C_{S_n}(\sigma)\leq A_n$, and so half of the cosets of $H$ lie in $A_n$ and the other half in $S_n\backslash A_n$. So the orbit of $A_n$ containing $\sigma$ contains just half of the conjugacy class $\mathcal{K}$, and $\mathcal{K}$ splits into two classes of equal size.
On the other hand, suppose that $H$ contains an odd permutation. Then $C_{A_n}(\sigma)=C_{S_n}(\sigma)\cap A_n$ is a subgroup of index 2 in $H$. Now the size of the $A_n$-conjugacy class is the index of the stabilizer, which is 
$$|A_n:C_{A_n}(\sigma)|=\dfrac{|S_n:C_{A_n}(\sigma)|}{2}=|S_n:C_{S_n}(\sigma)|=|\mathcal{K}|.$$
So $\mathcal{K}$ is a conjugacy class in $A_n$.
