TITLE: Given a finite non abelian group G, prove the order of the center is less than or equal to 1/4 of the order of the group.
QUESTION [5 upvotes]: I am a bit confused on where to start with this one. The question given was

Let $G$ be a non abelian finite group. Prove $|Z(G)| \leq \frac {1}{4} |G|$.

Am I supposed to use some property of centers or am I just blanking on something really basic

REPLY [6 votes]: Hint: If $G/Z(G)$ is cyclic, then $G$ is abelian.