Let $\sigma \in A_n$ s.t $\sigma$ has a 2-cycle. Prove the conjugacy class of $\sigma$ is the same in $A_n, S_n$.

Question

Question:

Let $\sigma \in A_n$ s.t $\sigma$ has a 2-cycle. Prove the conjugacy class of $\sigma$ in $A_n$ is the same as its conjugacy class in $S_n$.

So I tried going down this path but I'm not sure from where I can derive contradiction. Even if there are other ways to solve this I'd love help completing this soulution path.

Say $c$ is a congujacy class in $S_n$ we know it has two options :

1. split in $A_n$.
2. remain intact.

Therfore, assume with contradiction that $\sigma$ doesn't have the same congjucay class in $A_n$. This implies $\exists \tau \in S_n$ s.t $\tau \notin A_n$ s.t $\tau \sigma \tau{^-1} = \sigma$.

Now consider $\sigma = c_1 ... c_k$ where $c_i$ represents some cycle and assume without loss of generality $|c_1| = 2$.

Then we have $\tau \sigma \tau{^-1} = \tau (c_1) ... \tau (c_k)$.

Now I want to derive the contradiction from this last statement by somehow showing $\tau \in A_n$ due to its relationship with $c_1$ but I'm clueless.

Thanks in Advance for all Help!

Answer 1

Let $\sigma = \sigma' c$, where $c$ is a 2-cycle. Note that if $\rho \in S_n \setminus A_n$, then $\tau := \rho c \in A_n$. Now you can show that $$\rho \sigma \rho^{-1} = \tau\sigma\tau^{-1}.$$ What does this mean for the conjugacy classes in $A_n$ and $S_n$?