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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?

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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

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