The Index of the Alternating group in the Symmetric group.

We know that the order of $S_{n}$ is $n!$ . The Alternating group $A_{n}$ is defined to be the kernel of the sign homomorphism $\varphi: S_{n} \rightarrow \{\pm 1\}$ . Then the First Isomorphism Theorem has that $S_{n}/A_{n} \cong \{\pm 1\}$ . Then by the property of isomorphic groups, they share the same order. Thus $[S_{n} : A_{n}] = 2$ .