$(123)$ and $(132)$ are not in the same conjugacy class in $A_4$ 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.
 A: One way to see this is to work in $S_4$ instead. There $(1\,2\,3)$ and $(1\,3\,2)$ are conjugates, of course, since $(2\,3)^{-1}(1\,2\,3)(2\,3)=(1\,3\,2)$.
If we have a $\tau$ such that  $\tau^{-1}(1\,2\,3)\tau=(1\,3\,2)$, it must factor as $\tau=\rho(2\,3)$ for some $\rho$, and we have $$(2\,3)^{-1}\rho^{-1}(1\,2\,3)\rho(2\,3)=(1\,3\,2)=(2\,3)^{-1}(1\,2\,3)(2\,3)$$
Canceling the $(2\,3)$s we find $\rho^{-1}(1\,2\,3)\rho=(1\,2\,3)$, and it's then easy to see that $\rho(1)$ determines $\rho$ and there are only three possibilities, which are all even permutations. So $\tau$ must be odd and so not in $A_4$.

This may or may not be easier to follow than Tobias' suggestion to brute-force the possible $\tau$s directly. Personally I think it's easier to think about possible stabilizers of a cycle than to keep things straight while trying to conjugate one cycle into another.
A: 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.
