For $n \ge 3$, every subgroup of $A_n$ with index $n$, is isomorphic to $A_{n-1}$ For $n \ge 3$, every subgroup of $A_n$ with index $n$, is isomorphic to $A_{n-1}$.
Any idea to solving it?
 A: Let $H \le G$ with $|G:H|=n$ and $G \cong A_n$. The action of $G$ by multiplication  on the cosets of $H$ gives a homomorphism $\phi: G \to S_n$. For $n \ge 5$ the simplicity of $G$ implies that $\phi$ is injective. So $|{\rm Im}(\phi)| = n!/2$ and hence ${\rm Im}(\phi) = A_n$. Now $\phi(H)$ is a point stabilizer in $A_n$ which is $A_{n-1}$. So the restriction of $\phi$ to $H$ gives an isomorphism $H \cong A_{n-1}$.
You need to do the cases $n=3,4$ separately.
A: Note that this question proves that a subgroup of $S_n$ of index $n$ is isomorphic to $S_{n-1}$.
Let $H$ be a subgroup of $A_n$ of index $n$. Then $H \rtimes \Bbb{Z}_2$ is a subgroup of $S_n$ of index $n$. By the result cited in the above question, $J = H \rtimes \Bbb{Z}_2$ is isomorphic to $S_{n-1}$. But $H$ is a normal subgroup of $J$, so it can only be isomorphic to $\{e\},A_{n-1}$ or $S_{n-1}$.
On the other hand, the cardinality of $H$ is $\frac{(n-1)!}{2}$ which is the cardinality of $A_{n-1}$. Thus the only possibility is that $H \simeq A_{n-1}$.
