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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 $A_4$?

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  • $\begingroup$ How do you get $3$ (or $6$, or $8$) as a possible answer for "Problem 1"? $\endgroup$ – Lord Shark the Unknown Feb 19 at 3:46
  • $\begingroup$ @LordSharktheUnknown Err... my question was ill-posed. I rewrote it to better match the spirit I wanted to get at. $\endgroup$ – BalancedTryteOperators Feb 19 at 3:51
  • $\begingroup$ I don't see what the new Problem 1 is getting at. Anyway, $A_4$ has conjugacy classes of sizes $1$, $3$, $4$ and $4$, so from this information alone, cannot have a normal subgroup of order $6$. $\endgroup$ – Lord Shark the Unknown Feb 19 at 3:54
  • $\begingroup$ @LordSharktheUnknown how do you know that $A_4$ has conjugacy classes of sizes 1, 3, 4, and 4? $\endgroup$ – BalancedTryteOperators Feb 19 at 3:58
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    $\begingroup$ Should that "Question 2" be about subgroups of $A_4$ instead, as the following text implies? $\endgroup$ – jmerry Feb 19 at 5:48
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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 ...

Well, actually, we can use that fact. We just can't stop there. We can find the sets of permutations with the same cycle shapes, and then we'll have to check to see which of these split further into pairs of conjugacy classes.

$1+1+1+1$: The identity. A single conjugacy class of one element, which obviously doesn't split.
$2+2$: Pairs of disjoint transpositions. There are three of these, and they can't split. Why? Because any split would have to be into two subsets of equal size. Conjugation by a transposition would be a bijection between the two classes.
$3+1$: $3$-cycles. There are eight of these, and this set does split into two conjugacy classes of size $4$ each. As it turns out, a $3$-cycle and its inverse aren't conjugate in $A_4$; any even permutation which leaves the fixed element alone commutes with the cycle.
Alternately, we can prove this splits with an orbit-stabilizer count. $A_4$ acts on itself by conjugation, and the orbits are the conjugacy classes. The stabilizer of a $3$-cycle consists of all permutations that fix the fourth element not in that cycle, which are the powers of that $3$-cycle. That's a subgroup of order $3$, leaving room for four elements in the orbit. There's no way to get an eight-element orbit, because $\frac{12}{8}$ isn't an integer; the size of any conjugacy class must always divide the order of the group.

The other shapes are all odd permutations.

So then, the class equation is $$|A_4|=1+3+4+4$$ No way to get $6$ out of that. Every subgroup contains the identity, so the only possible orders of normal subgroups are $1$, $1+3=4$, and $1+3+4+4=12$. All three of those possibilities are in fact normal subgroups in this case.

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  • $\begingroup$ I'm sorry, but I didn't really understand how you determined which conjugate classes split and which ones didn't. $\endgroup$ – BalancedTryteOperators Feb 19 at 4:44
  • $\begingroup$ The critical part is that if a class splits, it has to split into two equal parts (one for each coset of $A_4$ in $S_4$). That means that a class of odd size can't split. As for the big class - it doesn't even matter for the problem you're looking at. The possible normal subgroups are the same either way. $\endgroup$ – jmerry Feb 19 at 4:56
  • $\begingroup$ Still, I can explain the big class splitting. In order to conjugate a cycle $(abc)$ to its inverse $(acb)$, we'd have to conjugate by a permutation that fixed the fourth element $d$. The only such permutations in $A_4$ are the powers $(abc)$, $(acb)$, and $e$ (the identity) of that cycle. Those all commute with $(abc)$, so conjugating by them just gives back $(abc)$. Alternately, an orbit-stabilizer count will do it. Actually, I'll edit that into the main answer - it's probably better than what I had. $\endgroup$ – jmerry Feb 19 at 5:02
  • $\begingroup$ "The critical part is that if a class splits, it has to split into two equal parts (one for each coset of $A_4$ in $S_4$)." How did cosets enter the picture? $\endgroup$ – BalancedTryteOperators Feb 19 at 7:14
  • $\begingroup$ If you conjugate the elements in an orbit by something in $A_4$, you get the same orbit back, permuted. If you conjugate by something in the other coset $S_4\setminus A_4$, you get the other orbit in some order. $\endgroup$ – jmerry Feb 19 at 7:20
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It's relatively easy to come up with the class equation of $A_4$: there's a theorem that permutations with the same cycle structure are conjugate in $S_n$ (and vice-versa) ; and there's a theorem that of those one splits in $A_n$ iff its cycle type is a product of cycles of distinct odd length.

I'll let you work out (based on the cycle types), that the class equation is $1+3+4+4$. Note that the eight $3$ cycles split into two classes of order $4$.

Once we have the class equation, we get that there can't be a normal subgroup of order $6$.

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