The split of conjugate classes of $A_4$

Question

We know that the class equation of $A_4$, the alternating group of $S_4$, is $1+3+4+4$.

The order of $cl((1))$ is $1$, and order of $cl((12)(34))$ is $3$. And we split the rest of the $8$ elements into $cl((123))$ and $cl((132))$, and both of these have order $4$.

My question is, since $2$ and $6$ can still divide the order of the group $A_4$, why cannot these $8$ elements be divided into two conjugate classes with each orders of $2$ and $6$?

Answer 1

Let $H$ be a normal subgroup of $G$. If $C \subset H$ is an $H$-conjugacy class, then $gCg^{-1} \subset H$ is also an $H$-conjugacy class, so they will have the same order. To see this, it suffies to show that: $$\text{If} \ c \in C, \ \text{so} \ gcg^{-1} \in gCg^{-1}, \text{then} \ h(gcg^{-1})h^{-1} \in gCg^{-1}.$$

But we have

$$h(gcg^{-1})h^{-1} = g g^{-1} (h(gcg^{-1})h^{-1}) g g^{-1} = g (g^{-1} h g) c (g^{-1} h^{-1} g) g^{-1}= g \gamma c \gamma^{-1} g^{-1},$$ where $\gamma = g^{-1} h g \in H$. Since $\gamma c \gamma^{-1} \in C$, we are done.

Thus $G$ acts on the conjugacy classes of $H$ by conjugation. Certainly the union of $gCg^{-1}$ for a fixed $C$ over all $g \in G$ is just the $G$-conjugacy class of any $c \in C$. But clearly $C$ and $gCg^{-1}$ have the same order. Hence a conjgugacy class of $G$ inside $H$ breaks up into $H$-conjugacy classes all of the same order. Since $G$ permutes them and $H$ fixes them, the $G$ action factors through $G/H$, and thus the number of pieces has to divide $G/H$.

If $H = A_n$ and $G=S_n$ then $G/H$ has order $2$, so any $S_n$-conjugacy class inside $A_n$ breaks up into either $1$ or $2$ $A_n$-conjugacy classes of the same order.

Answer 2

You talk about it as if we arbitrarily decide how to split the elements into conjugacy classes. The conjugacy classes are defined by which elements are conjugate to each other. Note that the $8$ permutations of type $(123)(4)$, with one fixed point and one $3$-cycle, are all conjugate in $S_4$, but they are not all conjugate in $A_4$, where there is no odd permutation that would relabel $(123)$ as $(132)$, and hence $(123)(4)$ cannot be turned into $(132)(4)$ by conjugation. (Recall that conjugating a permutation is equivalent to relabeling its cycle notation with the conjugating permutation.) Thus these $8$ elements split into two conjugacy classes, according to whether you need an odd or even permutation to relabel their cycle notation as $(123)(4)$.