# 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 $N\u22b4{S}_{n}$ be normal. Clearly $N\cap {A}_{n}\u22b4{A}_{n}$. 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 ${S}_{4}$.

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 |