How to prove that index of second smallest index subgroup in $A_n$ is $n \choose 2$? How to prove that 

in $A_n$ (Alternating group), the subgroup of second smallest index has index $n \choose 2$ if $n\ge 9$ ?

I know how to prove it for the smallest index, but for the second smallest I don't know how to prove it.
 A: Although some fiddling will be necessary to get a precise result, one can understand problems of this type without using powerful modern results in group theory.
For large enough $n$, there are a lot of primes strictly between $n/2$ and $n-2$.  Any subgroup whose index is at most $n^2$ will have order divisible by at least one of these primes, call it $p$, and thus contain a $p$-cycle.  C. Jordan showed long ago that a proper subgroup of $A_n$ containing a $p$-cycle with $p<n-2$ cannot be primitive (see for example Theorem 3.3E in Dixon and Mortimer's ``Permutation Groups").  The indices of the intransitive and imprimitive maximal subgroups of $A_n$are straightforward to compute. 
A: This should follow from the O'Nan-Scott theorem, as described in the accepted answer to this question: maximal subgroups of finite simple groups
A completely explicit classification is given by Liebeck, Praeger, Saxl, using the methodology suggested above.
Liebeck, Martin W.; Praeger, Cheryl E.; Saxl, Jan, A classification of the maximal subgroups of the finite alternating and symmetric groups, J. Algebra 111, 365-383 (1987). ZBL0632.20011.
