Proper subgroup is cyclic if $|G|=p^2$

Question

Let G be a group with |G| = p^2, where p is prime. Show that every proper subgroup of G is cyclic.

Proof: Let H be a proper subgroup of G where |G| = p^2 and p is prime. Then |H| divides |G| by Lagrange's Theorem. So |H| = 1, p, or p^2. Since H is proper, then |H| does not equal |G|. So |H| = 1 or p.

I do not know if I am correct or where to go from this point if I am correct.

Answer 1

This is correct so far. Now you need to show that every group of prime order or of order 1 is cyclic. The order 1 case is trivial. For the prime order case, assume it is not cyclic. Then there exists a nontrivial proper subgroup of our subgroup of order p. But Lagrange's theorem says that the order of this subgroup divides p, which means it can't be nontrivial and proper.

Answer 2

As you say, either $|H|=1$ (and there is nothing to prove) or $|H|=p$. In this latter case pick any $h\in H,\;\;h\neq1$ and consider the cyclic subgroup generated by $h$, namely $\langle h\rangle$.

Clearly $\{1\}\neq\langle h\rangle\le H$, from which it follows that $H$ needs to be $\langle h\rangle$, which is cyclic.