# 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?

• Another solution using the given exercise. ⇐: Suppose that $\sigma$ commutes with an odd permutation. We show that its conjugacy class in $S_n$ cannot split in $A_n$ and we are done, since there are only two options. If that happens, by the orbit-stabilizer theorem we get: $|S_n:C_{S_n}(\sigma)| = 2|A_n:C_{A_n}(\sigma)|$ and then by opening the definition of these indexes, we get $|C_{S_n}(\sigma)|=|C_{A_n}(\sigma)|$ and therefore $C_{A_n}(\sigma)=C_{S_n}(\sigma)$, but this condradicts the fact that $\sigma$ commutes with and odd element. – math.h Apr 1 '18 at 1:27

• $\Rightarrow \,:$ 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
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 $A_n$.