Conjugacy class of an element in $A_n$

Question

I need help to answer the following problem:

Let $n\ge 3$, $\sigma\in A_n$, $cl_{S_n}(\sigma)$ be the conjugacy class of $\sigma$ in $S_n$ and $cl_{A_n}(\sigma)$ be the conjugacy class of $\sigma$ in $A_n$.

Show that if $C_{S_n}(\sigma)\not\subset A_n$ then $cl_{A_n}(\sigma)=cl_{S_n}(\sigma)$

Thanks

Answer 1

We know that $|cl_{S_n}(\sigma)| = [S_n:C_{S_n}(\sigma)]$ and $|cl_{A_n}(\sigma)| = [A_n:C_{A_n}(\sigma)]$. Now we have that $C_{A_n}(\sigma)$ is a proper subgroup of $C_{S_n}(\sigma)$, as $C_{S_n}(\sigma) \not \subset A_n$. So we have that $[C_{S_n}(\sigma):C_{A_n}(\sigma)] \ge 2$. Hence

$$\frac{|cl_{S_n}(\sigma)|}{|cl_{A_n}(\sigma)|} = \frac{[S_n:C_{S_n}(\sigma)]}{[A_n:C_{A_n}(\sigma)]} = \frac{[S_n:A_n]}{[C_{S_n}(\sigma):C_{A_n}(\sigma)]} \le \frac{2}{2} = 1$$

But also trivially $cl_{A_n}(\sigma) \subset cl_{S_n}(\sigma)$, so we must have that $|cl_{S_n}(\sigma)| = |cl_{A_n}(\sigma)|$. Hence $cl_{S_n}(\sigma) = cl_{A_n}(\sigma)$.