Order Limit of Proper Subgroups of the Alternating Group (4.6.1)

Dummit and Foote Abstract Algebra, section 4.6, exercise 1:

Prove that there are no proper subgroups of A_n of index < n for n > 4.

Proof: Reword it to mean no proper subgroup of index ≤ n-1 exists. Let H be such a proper subgroup and let m = | A_n : H | ≤ n-1. Now let A_n act on H by left multiplication. We have the kernel of this action is the largest normal subgroup in H, and since A_n is simple for n > 4, we have the kernel is the identity and A_n maps injectively into S_m. But | A_n | = n!/2 and | S_m | ≤ (n-1)!, and we have that n!/2 ≤ (n-1)! ⇒ n ≤ 2, a contradiction of premises.

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