This question already has an answer here:
Is it true that when $G/C(G)$ is cyclic, then $G$ is abelian?
The original question is
Prove that if $G$ is a finite non-abelian group, then $4|C(G)|\le G$.
Proof from the answer book: suppose that $1<G/C(G)<4$, then $G/C(G)$ is of order 2 or 3 and therefore cyclic. Hence $G$ would be abelian, a contradiction.
As is said in the title, I have no idea how the implication in the last sentence goes...