For the $n=3$ and $n \geq 5$, $A_n$ is simple, and suppose for a contradiction $H\leq S_n$ has index 2 and $H\neq A_n$, then since $H\cap A_n \unlhd A_n$ and $A_n$ is simple, $H \cap A_n$ must be trivial.   Since $H$ has order equal to that of $A_n$ and $|A_n|&gt;3$, there must exist at least two non-identity elements in $H$. Let $h_1 \neq h_2 \in H$ be non-identity elements of $H$. Then $h_1$ and $h_2$ must be odd permutations, hence $h_1 h_2 \neq h_1^2$. But the product of two odd permutations is even, so $h_1h_2 \in A_n$, so $h_1h_2 \in H\cap A_n = 1$, so $h_1h_2=h_1^2=1$, which is a contradiction.   Therefore $A_n$ is the only subgroup of $S_n$ with index 2 for $n=3$ and $n\geq 5$.  Also note that $A_1$ and $A_2$ are trivial, so we only have to prove the case for $A_4$. But then I'm stuck.
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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