Q:

If the left regular representation of a finite group G contains an odd permutation, then G has a subgroup of index 2

I got to the point where I showed: $$[\pi(G):A_G \cap \pi(G)]=2$$
I have a trouble convincing myself of the final step. So if $\pi:G \rightarrow S_G$ is faithful (i.e. $\pi$ is injective), then $G$ can be identified with $\pi(G)$ and thus the preimage of $A_G \cap \pi(G)$ gives the subgroup of index as required.
Is there any other (possibly, more "formal") way to argue this?
Thanks! 

A:

Let $f = \operatorname{sgn}\circ \pi$ be the composite map $f:G\to S_n \to C_2$ to the cyclic group of order $2$.  There are two cases:
Case I: $f(G)$ is trivial.  Then $\pi(G) \subset A_n$.
Case II: $f(G)=C_2$.  Then, by the first isomorphism theorem, $\ker{f}$ has index $2$ in $G$.

