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 thedomain $S_n$. Furthermore $S_n/A_n\cong {\mathbb{Z}}_2$ bythe 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 $n!/2$ manyelements 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 $gH=Hg$).