# Question on Abelian groups can anyone help?

Prove that G/Z(G) is not cyclic if G is a non abelian group

I assume going by contrapositive is the easiest way to prove this right?

Suppose it is cyclic, say $$G/Z(G)=\langle gZ(G) \rangle$$. Let $$a \in G$$. Then $$a \cdot Z(G)= [gZ(g)]^n=g^nZ(G)$$ so $$a=g^nz$$ where $$z \in Z(G)$$. Consequently $$a \in C(g)$$, the centralizer of $$g$$, since $$g^n,z \in C(g)$$. Since "$$a$$" is arbitrary, it follows that every element of $$G$$ commutes with $$g$$, so $$g \in Z(G)$$ and so $$gZ(G)=Z(G)$$. Thus $$Z(G)=G$$ and so $$G$$ is Abelian