normal subgroups of the symmetric groups

Theorem 1.

For $n\mathrm{\ge }\mathrm{5}$, $A_n$ is the only proper nontrivial normal subgroup of $S_n$.

Proof.

This is essentially a corollary of the simplicity of the alternating groups $A_n$ for $n\ge 5$. Let $N⊴S_n$ be normal. Clearly $N\cap A_n⊴A_n$. But $A_n$ is simple, so $N\cap A_n=A_n$ or $N\cap A_n=\{e\}$. In the first case, either $N=A_n$, or else $N$ also contains an odd (http://planetmath.org/SignatureOfAPermutation) permutation, in which case $N=S_n$. In the second case, either $N=\{e\}$ or else $N$ consists solely of one or more odd permutations in addition to $\{e\}$. But if $N$ contains two distinct odd permutations, $\sigma$ and $\tau$, then either ${\sigma }^2\ne e$ or $\sigma \tau \ne e$, and both ${\sigma }^2$ and $\sigma \tau$ are even (http://planetmath.org/SignatureOfAPermutation), contradicting the assumption that $N$ contains only odd nontrivial permutations. Thus $N$ must be of order $2$, consisting of a single odd permutation of order 2 together with the identity.

It is easy to see, however, that such a subgroup cannot be normal. An odd permutation of order $2$, $\sigma$, has as its cycle decomposition one or more (an odd number, in fact, though this does not matter here) of disjoint transpositions. Suppose wlog that $(12)$ is one of these transpositions. Then $\tau =(13)\sigma (13)=(13)(12)(\mathrm{\dots })(13)$ takes $2$ to $3$ and thus is neither $\sigma$ nor $e$. So this group is not normal.∎

If $n=1$, $S_1$ is the trivial group, so it has no nontrivial [normal] subgroups.

If $n=2$, $S_2=C_2$, the unique group on $2$ elements, so it has no nontrivial [normal] subgroups.

If $n=3$, $S_3$ has one nontrivial proper normal subgroup, namely the group generated by $(123)$.

$S_4$ is the most interesting case for $n\le 5$. The arguments in the theorem above do not apply since $A_4$ is not simple. Recall that a normal subgroup must be a union of conjugacy classes of elements, and that conjugate elements in $S_n$ have the same cycle type. If we examine the sizes of the various conjugacy classes of $S_4$, we get

Cycle Type	Size
4	6
3,1	8
2,2	3
2,1,1	6
1,1,1,1	1

A subgroup of $S_4$ must be of order $1,2,3,4,6,8$, or $12$ (the factors of $|S_4|=24$). Since each subgroup must contain $\{e\}$, it is easy to see that the only possible nontrivial normal subgroups have orders $4$ and $12$. The order $4$ subgroup is $H=\{e,(12)(34),(13)(24),(14)(23)\}$, while the order $12$ subgroup is $A_4$. $A_4$ is obviously normal, being of index $2$, and one can easily check that $H\cong V_4$ is also normal in $S_4$. So these are the only two nontrivial proper normal subgroups of $S_4$.