According to wikipedia
For n > 1, the group $A_n$ is the commutator subgroup of the symmetric group $S_n$ with index 2 and has therefore $\frac{n!}{ 2}$ elements.
I have a question about what the index is. Usually it is the number of left cosets, in this case I assume it includes something about a mapping from $S_n$ to {$\pm 1$} instead of the number of left cosets?
I can see that clearly {$\pm1$} has index 2, but $A_n$ is a subgroup that has elements of $S_n$, so i don't see why that would have index 2. I am also not sure what the "commutator" subgroup is, which might be where the confusion is.
Once we have index 2, I understand that by Lagrange's theorem, the order of $A_n$ is $\dfrac{n!}{ 2}$.
As $A_n\le S_n$ is the subgroup consisting of even permutations, and exactly half the elements of $S_n$ are even, the index is two.
Recall that a permutation is called even if it can be written as a product of an even number of transpositions. Odd otherwise.
Define $\operatorname {sgn}:S_n\to \Bbb Z_2$ by $\operatorname {sgn}(\pi)=1$ if $\pi$ is odd, and $\operatorname {sgn}(\pi)=0$ if $\pi$ is even. It can be shown that $\operatorname {sgn}$ is a surjective homomorphism (one key is that the parity of a permutation is an invariant).
Now $A_n=\operatorname {ker}(\operatorname {sgn})$.
Thus by the first isomorphism theorem, $S_n/A_n\cong \Bbb Z_2$.
The claim about the index follows.