Prove that does not have a proper subgroup of index less than for all .
We begin with a lemma.
Lemma: If , then . Proof: For the base case, note that . For the inductive step, suppose . Now , so that .
Let , and let be a proper subgroup. Note that acts on the left cosets of by left multiplication; let be the permutation representation induced by this action, and let be the kernel of this action. Note that . Since is simple and is normal, we have . Thus is injective.
If , we have . By the lemma this is a contradiction. Hence no proper subgroup has order less than .