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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?

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3 Answers

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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

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So, you didn't conjugate by all the other elements of $A_4$.

But, as pointed out in the comments and the other answer, $(12)(34),(13)(24)$ and $(23)(14)$ are in the same conjugacy class, because of the cycle type (cycle lengths not distinct odd numbers).

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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.

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