Prove if there is a group whose order is $p^2$,and it is non-abelian, then it is cyclic

Question

Suppose the non-abelian group G whose order is $p^2$ where p is a prime number, prove it is a cyclic group.

My work:

there is a $\tau\not=e$ in the group and the order of $\tau$ is either $p$ or $p^2$, if the latter, the problem can be solved. so if it exists a element whose order is $p^2$, the group G is a cyclic group. But I can't prove it. Can you prove it or give some hint so I can continue do it? Thank you!

Answer 1

Well this is odd. It is well-known that a group of order $p^2$ is abelian. A proof would go like this: $p$-groups have nontrivial center $\implies G/Z(G)$ is cyclic $\implies G$ is abelian.

Furthermore your setup is incorrect. A cyclic group is always abelian.