Subgroup of $A_n$ isomorphic to $A_{n-1}$ Let $G$ be a subgroup of $A_n$, for $n\geq 5$, if $|G|=\frac{(n-1)!}{2}$, then prove that $G$ is isomorphic to $A_{n-1}$.
I think that I have to use the simplicity of $A_n$, but I can not see how.
 A: Given $G$ is subgroup of index $n$ in $A_n$. $A_n$ acts on the $n$ left cosets cosets of $G$ in $A_n$ by left multiplication. This gives a homomorphism from $A_n$ to permutation on $n$ letters:
$$ (x\in A_n) \mapsto \begin{pmatrix} \cdots & tG & \cdots\\ \cdots & xtG & \cdots \end{pmatrix}.$$
The kernel of homomorphism should be trivial since $A_n$ is simple. Thus, the map is injective. Where elements of $G$ go under this injection? They go to permutations on $n-1$ letters, since if $x\in G$ then 
$$ x\mapsto \begin{pmatrix} G & \cdots & tG & \cdots\\ xG &\cdots & xtG & \cdots \end{pmatrix}=\begin{pmatrix} G & \cdots & tG & \cdots\\ G &\cdots & xtG & \cdots \end{pmatrix}.$$
Thus, $G$ under the $1-1$ homomorphism goes into $A_{n-1}$; since both have same orders, they are isomorphic.
A: I haven't figured this one out yet, but a suggestion: $G$ consists of certain even permutations of the numbers $1, ... , n$.  If you can show that one of the numbers $1 \leq k \leq n$ does not occur in any of the permutations of $G$, then you are done.  By pigeonhole, $G$ must then consist of all even permutations of the numbers $\{1, ... , n\} \setminus \{k \}$, and therefore $G \cong A_{n-1}$.
