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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 '13 at 19:32
    
@TobiasKildetoft Please stick that in an answer. –  rschwieb Mar 18 '13 at 19:37

2 Answers 2

Hints: Take care of the case where $G$ is abelian first (that should be very easy). Now, for nonabelian $G$, clearly $Z(G)$ is a proper normal subgroup. If it were a maximal subgroup, what can you say about the quotient $G/Z(G)$? (Use the correspondence theorem to find out what are the subgroups of $G/Z(G)$.) What does that say about $G/Z(G)$? By now, you should have figured out that $G/Z(G)$ is cyclic. What does that say about $G$?

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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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