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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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marked as duplicate by Tara B, Henry T. Horton, Cameron Buie, rschwieb, Thomas Mar 18 '13 at 20:25

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

2  
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

1 Answer 1

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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