How can it be proved that any subgroup of $A_5$ has order at most 12? This is [Herstein, Problem 2.10.15], which also gives the hint that I can assume the result of the previous problem that $A_5$ has no normal subgroups $N \ne (e),A_5$.
This problem appears in an earlier section of the text than the Sylow theorems. There is a proof given at Subgroups of $A_5$ have order at most $12$?, but it uses the Sylow theorems, and I wonder if a more elementary proof is available.
So far, I can prove the following: For $n \ge 3$, the subgroup generated by the 3-cycles is $A_n$; if a normal subgroup of $A_n$ contains even a single 3-cycle it must be all of $A_n$; $A_5$ has no normal subgroups $N \ne (e),A_5$. I showed the latter by repeatedly conjugating a given nontrivial element in $A_5$ by 3-cycles to eventually obtain elements whose product is a 3-cycle.