# alternating group is a normal subgroup of the symmetric group

###### Theorem 1.

The alternating group^{} ${A}_{n}$ is a normal subgroup^{} of the symmetric group^{} ${S}_{n}$

###### Proof.

Define the epimorphism^{} $f:{S}_{n}\to {\mathbb{Z}}_{2}$ by
$:\sigma \mapsto 0$ if $\sigma $ is an even permutation^{} and
$:\sigma \mapsto 1$ if $\sigma $ is an odd permutation. Hence,
${A}_{n}$ is the kernel of $f$ and so it is a normal subgroup of the
domain ${S}_{n}$. Furthermore ${S}_{n}/{A}_{n}\cong {\mathbb{Z}}_{2}$ by
the first isomorphism theorem^{}. So by Lagrange’s theorem

$$|{S}_{n}|=|{A}_{n}||{S}_{n}/{A}_{n}|.$$ |

Therefore, $|{A}_{n}|=n!/2$. That is, there are $n!/2$ many elements in ${A}_{n}$ ∎

Remark. What we have shown in the theorem is that, in fact, ${A}_{n}$ has index $2$ in ${S}_{n}$. In general, if a subgroup^{} $H$ of $G$ has index $2$, then $H$ is normal in $G$. (Since $[G:H]=2$, there is an element $g\in G-H$, so that $gH\cap H=\mathrm{\varnothing}$ and thus $gH=Hg$).

Title | alternating group is a normal subgroup of the symmetric group |
---|---|

Canonical name | AlternatingGroupIsANormalSubgroupOfTheSymmetricGroup |

Date of creation | 2013-03-22 13:42:32 |

Last modified on | 2013-03-22 13:42:32 |

Owner | CWoo (3771) |

Last modified by | CWoo (3771) |

Numerical id | 8 |

Author | CWoo (3771) |

Entry type | Theorem |

Classification | msc 20-00 |