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 $$ \vert S_{n} \vert=\vert A_{n} \vert\vert S_{n}/A_{n}\vert. $$ Therefore, $\vert A_{n}\vert=n!/2$ . 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 $gH=Hg$ ).