I have what seems to me a very simple proof that $A_{n}$ is the only subgroup of $S_{n}$ of index 2.  Since I've seen other people prove it with what feel like really complicated methods (), I'm wondering if I've overlooked something.  Proof: Let $H$ be any subgroup of index 2 in $S_{n}$.  Then $H\cap A_{n}$ is a normal subgroup of $A_{n}$ and since $A_{n}$ is simple, $H$ is either the trivial or the improper subgroup of $A_{n}$.  If it's the trivial subgroup it doesn't have index 2 in $S_{n}$, and otherwise the theorem is proved.
How to prove that the only subgroup of the symmetric group $S_n$ of order $n!/2$ is $A_n$?  Why isn't there other possibility?  Thanks :)
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