Show that the center of a group of order 60 cannot have order 4?

Question

This seems to be related to Sylow's theorems, but I have no idea how?

Answer 1

Assume that $|Z(G)|=4$. $Z(G)$ is normal, so the group $G/Z(G)$ has order 15. But every group of order 15 is cyclic, and that means that $G$ is abelian - contradiction. (Because if $G$ is abelian, then $G=Z(G)$)