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   NormalSubgroupsOfTheSymmetricGroups 2013-03-22 17:31:38 2013-03-22 17:31:38 rm50 (10146) rm50 (10146) 8 rm50 (10146) Theorem msc 20B35 msc 20E07 msc 20B30