$O(G)=p^2 $, $p$ prime, $|Z(G)|>1$; prove $G$ abelian We know that $Z(G)<G,\;$  then $O(Z(G)) \mid  O(G). $
If $\;O(Z(G))= p^2, $ then $\;Z(G)=G$ and we are done.
Now, if $O(Z(G))= p,\,$ how can I prove that $G$ is abelian ?
Is it by proving that $G/Z(G)$ is cyclic? And if so, then $G$ is abelian.
If yes how to prove that that  $G/Z(G)$ is cyclic?

Prove the following (easy):
Lemma: For any  group $\;G\;$, the quotient group $\;G/Z(G)\;$ cannot be cyclic non-trivial.
From the above it follows that $\;G/Z(G)\;$ cyclic $\;\implies G\;$ is abelian.
