normal subgroups of the symmetric groups

Theorem 1.

For $n\geq 5$ , $A_{n}$ 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 $n\geq 5$ . Let $N\unlhd S_{n}$ be normal. Clearly $N\cap A_{n}\unlhd A_{n}$ . 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 $\sigma^{2}\neq e$ or $\sigma\tau\neq e$ , 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 $(1~{}2)$ is one of these transpositions. Then $\tau=(1~{}3)\sigma(1~{}3)=(1~{}3)(1~{}2)(\ldots)(1~{}3)$ 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 $(1~{}2~{}3)$ .

$S_{4}$ is the most interesting case for $n\leq 5$ . 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 $\lvert S_{4}\rvert=24$ ). 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 $H=\{e,(1~{}2)(3~{}4),(1~{}3)(2~{}4),(1~{}4)(2~{}3)\}$ , 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 $S_{4}$ .

Title	normal subgroups of the symmetric groups
Canonical name	NormalSubgroupsOfTheSymmetricGroups
Date of creation	2013-03-22 17:31:38
Last modified on	2013-03-22 17:31:38
Owner	rm50 (10146)
Last modified by	rm50 (10146)
Numerical id	8
Author	rm50 (10146)
Entry type	Theorem
Classification	msc 20B35
Classification	msc 20E07
Classification	msc 20B30