Prove that A5 is a group of order 60 having no subgroup of order 30.
Have you covered the fact that $\displaystyle A_5$ is simple yet? As if so, you can apply the fact that a subgroup of order 30 would have index 2, and so is normal, a contradiction.
Otherwise, I am sorry but I can't think of a neat way of doing it. I keep wanting to prove that $\displaystyle A_5$ is simple...