IF a group G of order 8. Then, it is impossible that |Z(G)| = 4

Question

I have no idea what equation or proposition I can use to prove that. I know Z(G)≤G.Is that help?

Answer 1

Use that $G$ is abelian if $G/Z(G)$ is cyclic.

Answer 2

HINT: Suppose that $H$ is a subgroup of $G$ of order $4$ contained in the centre of $G$. Show that if $x,y\in G\setminus H$, then $xy=yx$, and conclude that $G$ is Abelian. Note that there must be $h\in H$ such that $y=hx$; why?