I read the proof of this theorem but the question i have is If $G/Z(G)$ IS Cyclic then $G$ is abelian therefore $G = Z(G)$ which implies that $G/Z(G) = \{e\}$. Am i right? For example If $o(G/Z(G)) =11$ then $G/Z(G)$ is cyclic therefore $G$ is abelian which implies that $o(G/Z(G)) = 1$ a contradiction. Please explain or correct me if i am wrong.
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1$G/Z(G)$, if non-trivial, cannot be cyclic. (Although we usually do not regard the trivial group as cyclic.) – Groups May 04 '19 at 14:21
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@HongyiHuang Why is the trivial group not cyclic, see here? – Dietrich Burde May 04 '19 at 14:43
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@DietrichBurde Yes, but I just neglect it. Thank you! – Groups May 04 '19 at 14:46
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The question is if G/Z(G) is cyclic then can it only be trivial? – Sam Christopher May 04 '19 at 15:16
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Yes; I don’t like that phrasing of the problem precisely because it leads to the sort of “wait-a-minute” moment you’re having. A better phrasing for the problem is the following: “Let $G$ be a group. Prove that if $G/N$ is cyclic with $N\leq Z(G)$, then $G$ is abelian.” – Arturo Magidin May 04 '19 at 18:00
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@HongyiHuang: “Although we usually do not regard the trivial group as cyclic”... ehr, yes, we do. – Arturo Magidin May 04 '19 at 18:00