Suppose that G is a group with where is a prime and . Prove that is not simple.
There are always Sylow p-subgroups. In this case, it follows immediately from Sylow's theorems that there is a unique one. Unique Sylow subgroups are normal.
But as you say, you haven't covered that yet, so I'm sure you're expected to make a more direct argument.
Let be such that . It is trivial that there is a homomorphism by having . Moreover, one can prove that . Now, since is prime and we must have that or . Suppose that then is a subgroup of of order and so but since is prime and this is impossible. Thus, and so so that isn't simple.