Compute all the conjugacy classes of $Alt(4)$
The order of $Alt(4)$ is 12, so do I really have to compute 144 conjugations and see what the classes are?. I was reading online and we can use the tetrahedron to fulfill this task. Does anyone know how to use it?
First, note that in $S_4$ (or any $S_n$), two elements are conjugate if and only if they have the same cycle structure.
Since $A_4$ is a subgroup of $S_4$, two elements which are conjugate in $A_4$ are also conjugate in $S_4$, so they have the same cycle structure. However, the converse is not true: two elements with the same cycle structure are conjugate in $S_4$, but they need not be conjugate in $A_4$.
Start by listing the elements of $A_4$. We have $$1$$ $$(123), (132), (124), (142), (134), (143), (234), (243)$$ $$(12)(34), (13)(24), (14)(23)$$ The size of the conjugacy class containing an element $\alpha$ is equal to $|A_4 : G_{\alpha}|$, which is the index of the stabilizer (centralizer) of $\alpha$: $$G_{\alpha} = \{g \in A_4 : g\alpha g^{-1} = \alpha\}$$ Note that $|A_4 : G_{\alpha}| = |A_4|/|G_{\alpha}| = 12/|G_{\alpha}|$ so the size of each conjugacy class must be a divisor of $12$.
Note that $A_4$ has subgroups of index $1$, $3$, $4$, $6$, and $12$, so these are the only possible sizes of conjugacy classes.
Obviously $1$ is in its own conjugacy class.
There are $3$ elements of cycle type $(12)(34)$. Therefore the size of the conjugacy class containing $(12)(34)$ must be either $1$ or $3$. Note that if we conjugate $\alpha = (12)(34)$ using $g = (123)$ we get $g(12)(34)g^{-1} = (123)(12)(34)(132) = (14)(23)$, hence the size of the conjugacy class containing $\alpha$ cannot be $1$, and so it must be $3$. This means that the elements of type $(12)(34)$ are in a single conjugacy class, $\{(12)(34), (13)(24), (14)(23)\}$.
There are $8$ elements of cycle type $(123)$. Since $8$ is not a divisor of $12$, the size of the conjugacy class containing $(123)$ cannot be $8$, so there must be at least two conjugacy classes for this cycle type.
Also, all conjugacy classes for a given cycle type must have the same size. This is because if $\alpha$ and $\beta$ have the same cycle type, then they are conjugate in $S_4$, say $\beta = g\alpha g^{-1}$, and so if $G_{\alpha}$ is the stabilizer of $\alpha$ in $A_4$, then $g^{-1}G_{\alpha}g$ is the stabilizer of $\beta$, and it has the same order as $G_{\alpha}$. (Note that $g^{-1}G_{\alpha}g$ is a subgroup of $A_4$ even if $g \not\in A_4$, since $G_{\alpha}$ is a subgroup of $A_4$, and $A_4$ is normal in $S_4$.)
There are $8$ elements of cycle type $(123)$, so the conjugacy class containing $\alpha = (123)$ must have size $1$, $3$, $4$, or $6$. But of these, only $1$ and $4$ are divisors of $8$, so they are the only possibilities, since the $8$ elements of this cycle type must be partitioned into conjugacy classes of equal size.
If we conjugate $\alpha$ by $g = (124)$, we get $g(123)g^{-1} = (124)(123)(142) = (243)$, so the size of the conjugacy class containing $(123)$ is not $1$, therefore it is $4$.
How do we determine which elements are conjugate to $(123)?$ Any such element must be of the form $g(123)g^{-1}$, where $g \in A_4$ and therefore $g$ can be written as the product of an even number of transpositions. Conjugation by a transposition swaps two digits in the full cycle representation $(123)(4)$, so conjugation by an even number of transpositions causes an even number of swaps.
Since $(142)(3)$, $(134)(2)$, and $(243)(1)$ can each be obtained from $(123)(4)$ by performing two swaps, the conjugacy class containing $(123)$ is $\{(123), (142), (134), (243)\}$.
Similarly, the conjugacy class containing $(132)$ is $\{(132), (124), (234), (143)\}$.
Note that we could have shortened the argument considerably by using this result about splitting of conjugacy classes in the alternating group. This tells us that each conjugacy class in $S_4$ is either a single conjugacy class in $A_4$, or it is split evenly into two classes. We can use this argument as follows.
In $S_4$, two elements are conjugate if and only if they have the same cycle type. So the class containing $(12)(34)$ has size $3$. Since this can't be split evenly into two classes, the same conjugacy class must exist in $A_4$.
In $S_4$, the class containing $(123)$ has size $8$. Since this is impossible in $A_4$ as $8$ is not a divisor of $12$, this class must be split in half by $A_4$ to form two classes of size $4$.