TITLE: $(123)$ and $(132)$ are not in the same conjugacy class in $A_4$
QUESTION [7 upvotes]: Could you tell me how to show that $(123)$ and $(132)$ are not in the same conjugacy class in $A_4$? 
I know that all 3-cycles can't be in the same class, because the order of each class must divide $|A_4| = 12$ and there are eight 3-cycles in $A_4$.
If $\sigma$ is a 3-cycle ﬁxing $n$ and $τ \in A_4$ then $\tau \circ \sigma \circ \tau^{-1}$
is a 3-cycle
ﬁxing $τ(n)$. (And there are 3 such permutations). If we conjugate $(123)$ by $(132)$ we will get $(123)$ and vice versa. So it would appear that it only makes sense to conjugate $(123)$ by something with a $4$. If we conjugated $(123)$ by $(124)$, we would need $(124)(123)(124)^{-1}$ to send 4 to 4, but it doesn't. It turns out that such $\tau$ cannot have anything "common" with $(123)$ - what I mean by this is that there can't be $(12...), (23...), (31...)$ in $\tau$. So the only cycles that remain are $(134), (142), (243)$ - and they are indeed ion the same class as $(123)$.
As you see, this isn't the best way to solve this problem.
I also know that if $g \in A_n$ commutes with an odd permutation, then all permutations with the same cycle type as $g$ are in one conjugacy class, but I have no idea how to show that in $A_4$ no odd permutation commutes with $(123)$.
Could you make it more mature?
Thank you.

REPLY [2 votes]: For an easy argument, you can use that the size of the conjugacy class $C_G(g)$ of $g\in G$ is $|G|/|Z_G(g)|$ where $Z_G(g)=\{\,h\in G\mid gh=hg\,\}$ is the centraliser of$~g$ (this is a special case of the orbit-stabiliser theorem). The centraliser contains at least the powers of $g=(123)$, which here gives $|Z_{A_4}(g)|\geq3$ and therefore $|C(g)|\leq12/3=4$. So as soon as we have $4$ conjugates of $g$ there cannot be more of them; one easily sees that $\{(123),(134),(142),(243)\}\subseteq C_{A_4}(g)$, and hence this must be all.