How to prove that 



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> in $A\_n$ (Alternating group), the subgroup of second smallest index has index $n \choose 2$ if $n\ge 9$ ?
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I know how to prove it for the smallest index, but for the second smallest I don't know how to prove it.



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


