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?

• The conjugacy class takes $ghg^{-1}$ for every $g \in G$, not just for the $g$'s in the conjugacy class. For example $(123)(12)(34)(132) \ne (12)(34)$ – Mathmo123 Dec 29 '14 at 0:10
• Have you seen the theorem relating the conjugacy classes of $S(n)$ to the cycle types of its elements? – Mathmo123 Dec 29 '14 at 0:11
• @Mathmo123 Right, I somehow forgot we are in $A_4$ :) – user2345215 Dec 29 '14 at 0:16
• @Andrew exactly. But are you familiar with the theorem I mentioned above? Computing these things manually really isn't the best way!! – Mathmo123 Dec 29 '14 at 0:16
• @Andrew ah ok. All it says is that two elements are conjugate in $S(n)$ if and only if they have the same cycle type. (Where cycle type is the set of sizes of the cycles of a permutation in disjoint cycle notation). This holds for $S(n)$ not $A(n)$, but it will make computation easier - if two things are conjugate in $A(n)$, then they must also be conjugate in $S(n)$, so have the same cycle type, but the converse is false in general. – Mathmo123 Dec 29 '14 at 0:21

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

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).

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.