Conjugation in $A_n$ Definition: Let $G$ be a group and $a,b\in G$. We say that $b$ is a conjugate of $a$ if $b=cac^{-1}$ for $c\in G$. 
It's easy to verify that conjugation is an equivalence class in $G$.
Also, we know the following fact: $\sigma, \tau\in S_n$ are conjugate in $S_n$ $\Leftrightarrow$ $\sigma, \tau$ have the same cycle decomposition.
Let's consider the normal subgroup $A_n$ instead of $S_n$. We know that $A_n$ is generated by $3$-cycles. 
Theorem: If $n\geq 5$ we can prove that $3$-cycles are conjugate in $A_n$.
Let's consider $n<5$. Firstly, let's deal with $A_3$ and take $(123)$ and $(132)$. Even though $(123)$ and $(132)$ are $3$-cycles, they are not conjugare since $(123)\neq\theta(132)\theta^{-1}$ for any $\theta \in A_3=\{e, (123), (321)\}$. Thus, the theorem fails for $n=3$.
Let's take the group $A_4$. In one lecture (namely, Keith Conrad's lecture) I have read the following: Let's take $(123)$ and $(132)$. Although they are $3$-cycles, $(123)=(32)(132)(32)$ hence they are not conjugate. 
How does it follow that they are not conjugate? What if for some $\theta\in A_4$ we get $(123)=\theta (132)\theta^{-1}$? 
Would be grateful if somebody could explain my question, please.
 A: Consider the action of $A_4$ on itself and the associated conjugacy classes. 
Let $(u,v,w) \in A_4$. Then, $(u,v,w)^{-1} = (w,v,u)$.
Suppose $\exists \tau \in A_4: (\tau(u),\tau(v),\tau(w)) = \tau\circ (u,v,w)\circ \tau^{-1} = (w,v,u) = (u,w,v)$.
Then, $\tau(u) = u$ and $\tau(v) = w$ and $\tau(w) = v$, that is, $\tau \in \{(1,2)\circ (3,4), (1,3)\circ (2,4),(1,4)\circ (2,3)\}$
which is a contradiction to $\tau(u) = u$.
Therefore, $(u,v,w)$ and $(u,v,w)^{-1}$ are in different conjugacy classes in $A_4$.
EDIT: $A_4$ has the following conjugacy classes


*

*$\{\iota\}$

*$\{(1,2)\circ (3,4), (1,3)\circ (2,4),(1,4)\circ (2,3)\}$

*$\{(1,2,3),(2,4,3),(1,3,4),(1,4,2)\}$

*$\{(1,3,2),(2,3,4),(1,4,3),(1,2,4)\}$


What I was trying to argue above is that the inverse $(w,v,u)$ of any $3$-cycle $(u,v,w)$ in $A_4$ lies in another conjugacy class than $(u,v,w)$.
So, I suppose (wrongly) that $(w,v,u)$ and $(u,v,w)$ lie in the same conjugacy class, that is, there is a permutation $\tau$ in $A_4$ which maps $(w,v,u)$ to $(u,v,w)$. Iny symbols: $\tau\circ (u,v,w)\circ \tau^{-1} = (w,v,u)$.
We know $(\tau(u),\tau(v),\tau(w)) = \tau\circ (u,v,w)\circ \tau^{-1}$. If you look really closely at $(\tau(u),\tau(v),\tau(w)) = \tau\circ (u,v,w)\circ \tau^{-1} = (w,v,u)$ you will see that $\tau$ has a fixed point, that is, either  $\tau(u) = u$ or $\tau(v) = v$ or $\tau(w) = w$. I chose $\tau(u) = u$. 
Moreover, $\tau$ contains a transposition: $\tau(v) = w$ and $\tau(w) = v$ by my choice $\tau(u) = u$. The only permutations in $A_4$ which contain a transposition are $\{(1,2)\circ (3,4), (1,3)\circ (2,4),(1,4)\circ (2,3)\}$. But none of these permutations has a fixed point. Contradiction! 
Does my argument make more sense to you now?
