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?
[Math] Conjugacy class $A_4$
abstract-algebrapermutationsrepresentation-theory
Best Answer
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