TITLE: $G$ is a nonabelian finite group, then $|Z(G)|\leq \frac{1}{4}|G|$
QUESTION [7 upvotes]: If $G$ is a nonabelian finite group, then I have to show that $|Z(G)|\leq\frac{1}{4}|G|$, where $|Z(G)|$ denotes the center of group.  

I have got this question through random search on net. I am finding it difficult to solve. Moreover, I am not sure whether is it true or not. 
Thanks for help.

REPLY [10 votes]: Hint: If $G/Z(G)$ is cyclic, then $G$ is abelian.