For n at least 5, the index of a subgroup of Alt(n) is at least n

Prove that A_n does not have a proper subgroup of index less than n for all n \geq 5.

We begin with a lemma.

Lemma: If n \geq 3, then n!/2 > (n-1)!. Proof: For the base case, note that 3!/2 = 3 > 2 = 2!. For the inductive step, suppose n!/2 > (n-1)!. Now n+1 > n, so that (n+1)!/2 > n!. \square

Let n \geq 5, and let H < A_n be a proper subgroup. Note that A_n acts on the left cosets of H by left multiplication; let \varphi : G \rightarrow S_{[A_n : H]} be the permutation representation induced by this action, and let K be the kernel of this action. Note that K \leq \mathsf{stab}(H) = H < A_n. Since A_n is simple and K is normal, we have K = 1. Thus \varphi is injective.

If [A_n : H] < n, we have A_n \leq S_{n-1}. By the lemma this is a contradiction. Hence no proper subgroup has order less than n.

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