Show $A_n$ has no subgroups of index 2 
I want to prove $A_n$ has no subgroups of index 2. 

I know that if there exists such a subgroup $H$ then $\vert H \vert = \frac{n!}{4}$ and that $\vert \frac{A_n}{H} \vert = 2$ but am stuck there. I have tried using the proof that $A_4$ has no subgroup of order 6 to get some ideas but am still stuck. Sorry I don't have much else to add at this point. Thanks a bunch.
 A: This elementary argument works for all $n$ at once and does not need simplicity of the alternating group: Let $N$ be of index $2$ in $A_n$ and let $x$ be a $3$-cycle. We have $xN=x^4N=(xN)^4=N$, where the latter holds since the factor group is of order $2$. We derive that $N$ contains all $3$-cycles. But it is a well known (and easy to show) fact that the alternating group is generated by $3$-cycles.
Of course we implictly used that a subgroup of index $2$ is always normal.
A: As has been stated in the comments, if you know that $A_n$ is simple for $n\geq 5$, and subgroups of index 2 are normal, you are done with that case. 
For the $n=4$ case, you do need to do a bit more work. Your idea to show that $A_n$ has no subgroup of order 6 is correct. Well, a subgroup of order $6$ in $A_4$ is either cyclic, or isomorphic to $S_3$. Since $A_4$ has no element of order 6, we can exclude the case of a cyclic subgroup and any subgroup of $A_4$ of order 6 must be isomorphic to $S_3$. 
To see that this can't happen, suppose
$$\phi:S_3\to A_4$$
is an injective homomorphism and use the fact that $S_3$ is generated by the simple transpositions $(12),(23)$. The image of these elements must be elements of order 2 in $A_4$ which are disjoint 2-cycles. Any two of these generate a subgroup of order 4 in $A_4$. Hence, no such homomorphism exists.
A: Hint: What do we know about a subgroup of $G$ whose index is the smallest prime dividing $|G|$?
