Need to prove that the A4 group is Normal sub-group of S4

Question

I already proved that N, wich is a sub-group of S4 (4-permutations), which is all the permutations, which look's like: $(a,b)(c,d)$ (which are defintly are in A4 (even permutations of S4)) are a normal sub-group.

Now I want to show that S4 and N are the only non-trivial normal sub-groups, what lead me to prove that A4 is a normal sub-group.

So as I understand A4 is the union of N and more eight 3-permutations.

Now, my question is if those 3-permutations are a normal sub-group (and they are, because there are all the options of 3-permutations therefore from conjugacy we can conclude that it is normal) and N is a normal sub-group, is there union, which is A4, is also a normal sub-group?

Or did I a mistake in my previous conclusions ?

P.S - for a better understanding $$N=\left\{id,\left(12\right)\left(34\right),\left(13\right)\left(24\right),\left(14\right)\left(23\right)\right\}$$

And I need to show eather that they are the only normal sub-groups, how do I do that?

Answer 1

$A_n$ is the kernel of the signature morphism $\;\varepsilon\colon S_n\longrightarrow \{-1,1\}$.

Answer 2

A normal subgroup is a union of conjugacy classes, including the identity. The classes of $G=S_4$ are: $\{1\}$, $(1,2)(3,4)^G$ (length 3), $(1,2)^G$ (length 6), $(1,2,3)^G$ (length 8), and $(1,2,3,4)^G$ (length 6).

The only ways of adding up 1 and a submultiset of $\{3,6,6,8\}$ to get a divisor of $24$ are $1$, $1+3=4$, $1+3+8=12$, and $1+3+6+6+8=24$, and all of these combinations really do give normal subgroups of $S_4$, namely $1$, $N$, $A_4$, and $S_4$.