Here are two motivating problems. I will begin with $S_4$.
Problem 1. Why is there no proper non-trivial normal subgroup of order 2 in $S_4$?
Using the class equation, we know that
$$|S_4| = |Z(S_4)| + \sum_{i = 1}^N [S_4 : C_{S_4}(g_i)]$$ where $N$ is the number of non-singular conjugacy classes in $S_4$ (the singular conjugacy classes are counted in $|Z(S_4)|$) and $g_i$ are representative elements for each of the $N$ conjugacy classes. There's a trick we can use to calculate the conjugacy classes of $S_4$: the fact that the conjugacy classes of $S_4$ correspond to the "shape" of elements when each element is written in cycle notation. These are representative elements of the conjugacy classes.
$$E = \{(12), (123), (1234), (12)(34)\}$$
And these are how the orders of the conjugacy classes are calculated for all $e \in E$. Recall that $[S_4 : C_{S_4}(g_i)]$ is equal to the size of the conjugacy class that contains $g_i$.
$[S_4 : C_{S_4}((12))] = {4\choose 2} = 6$
$[S_4 : C_{S_4}((123))] = {4\choose3}2 = 8$
$[S_4 : C_{S_4}((1234))] = 4!/\langle \text{symmetry of 4-cycle} \rangle = 24/4 = 6$
$[S_4 : C_{S_4}((12)(34))] = {4\choose 2}/\langle \text{symmetry from the fact that disjoint cycles commute} \rangle = 6/2 = 3$
So the class equation is expanded as thus:
$$|S_4| = 1 + 6 + 8 + 6 + 3.$$ Because a normal subgroup is any subgroup $H$ such that $gHg^{-1} = H$ for any element $g \in S_4$, any normal group can only be made from whole conjugacy classes, not a part, which means they can only be addition subsets of the class equation. In conjunction with Lagrange's theorem (the fact that the order of the subgroups of $S_4$ must divide $24$), $2$ is not in the intersection of possible sums of the class equation with possible divisors of 24.
Problem 2. Why is there no proper nontrivial normal subgroup of order 6 in $A_4$.
The problem is that I can't use the fact that elements of the same equivalence class have the same shape when written in cycle notation, because $A_4$ only has even permutations. So it could be possible that for $x, y \in S_4$, $x = g y g^{-1}$ only for an odd permutation $g$, which would make them conjugate in $S_4$ but non-conjugate in $A_4$. How would I count the conjugacy classes in $A_4$?