Math Help - Abstract Algebra: Suppose that G is a group with |G|=p3. Prove that |Z(G)| != p2.

1. Abstract Algebra: Suppose that G is a group with |G|=p3. Prove that |Z(G)| != p2.

Suppose that $G$ is a group with $|G|=p^3$ where $p$ is prime. Prove that $|Z(G)|\ne p^2$.

2. If so, then |G/Z(G)|=p, and so G/Z(G) is cyclic. But G/Z(G) cyclic implies G is abelian, a contradiction.