alternating group is a normal subgroup of the symmetric group

Proof.

Define the epimorphism        $f:S_{n}\rightarrow\mathbb{Z}_{2}$ 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 $gH\cap H=\varnothing$ and thus $gH=Hg$).

Title alternating group is a normal subgroup of the symmetric group AlternatingGroupIsANormalSubgroupOfTheSymmetricGroup 2013-03-22 13:42:32 2013-03-22 13:42:32 CWoo (3771) CWoo (3771) 8 CWoo (3771) Theorem msc 20-00