Conjugacy class $A_4$ I want to find all conjugacy classes of $A_4$. So basically what I did, I took all elements of $A_4$ and calculated their conjugates. I had no problems with $$\{e\}, \{(123),(134),(142),(243)\}, \{(132),(143),(124),(234)\}$$but I don't understand why the rest of $A_4$ elements $$(12)(34),(13)(24),(23)(14)
$$is in one conjugacy class? Because, for example, computing the cojugates of (12)(34) gives me:
$$(12)(34)(12)(34)(12)(34)=(12)(34)$$
$$(13)(24)(12)(34)(13)(24)=(12)(34)$$
$$(14)(23)(12)(34)(14)(23)=(12)(34)$$
(here $(ab)(cd)^{-1}=(ab)(cd)$). My result is that $(12)(34)$ is the only member of the conjugacy class generated by it. Same for $(13)(24)$ and $(14)(23)$. I get three different conjugacy classes. Where am I wrong?
Hint-
For $\pi \in A_n$, its conjugacy class in $S_n$ remains as a single conjugacy class in $A_n$ or it breaks into two conjugacy classes in $A_n$ of equal size. The conjugacy breaks up if and only if the lengths in the cycle type of $\pi$ are distinct odd numbers.

NB-1.See Keith Conard's notes on Conjugacy class for details

2.Follow this link for details
The conjugacy classes of a subgroup of the symmetric group are all of the permutations with the same structure.

if we look at $S_5,$ a representative from each congjugacy class would be:

$()\\
(12)\\
(123)\\
(1234)\\
(12345)\\
(12)(34)\\
(12)(345)$

I think of linear algebra.  $P^{-1}AP$ gives a matrix similar to $A.$  It keeps the eigenvalues unchanged, and for the right choice of $P$ it diagonalizes the matrix.

With the symmetric group, we get a conjugate, but something remains unchanged by conjugation, and that is the structure of the permutation.

The symmetric group is a set of functions with group action of composition.  Whenever we have the a concatenation of functions of the form, $g{-1}hg$  We can think that $g$ translates from the domain of $g$ to the domain of $h, h$ does whatever $h$ does, and $g^{1}$ takes you back to the domain of $g.$

The congugacy class of $(12)(34)$ in $A_4$

$()(12)(34)() = (12)(34)\\
(132)(12)(34)(123) = (13)(24)\\
(123)(12)(34)(132) = (14)(23)$

and if conjugate $(12)(34)$ with any of the other 3 cycles we will get the same members.