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This question already has an answer here:

How to prove that Z(G) is not a maximal subgroup of G, where G is an arbitrary group?

Thanks in advance.

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Hint: Consider some $x\in G$ with $x\not\in Z(G)$. What can you say about $C_G(x)$? – Tobias Kildetoft Mar 18 at 19:32
@TobiasKildetoft Please stick that in an answer. – rschwieb Mar 18 at 19:37

marked as duplicate by Tara B, Henry T. Horton, Cameron Buie, rschwieb, Thomas Mar 18 at 20:25

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The hint given in the comments is probably the easiest way to do this. Here's an idea for a different solution.

Hint 1: If a maximal subgroup $M$ is normal in $G$, then $G/M$ is cyclic.

Hint 2: If $G/Z(G)$ is cyclic, then $G$ is abelian.

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