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Theorem: Prove that if $G/Z(G)$ is cyclic then $G$ is abelian.

My confusion is that if theorem is true i.e. $G/Z(G)$ is cyclic then $G$ is abelian, then $Z(G)=G$ implies $|G/Z(G)|$ always equal to 1. Am I thinking correctly?

If I am then let $|G|=pq$ where $p,q$ are distinct primes. Then if $|Z(G)|=p$ then $|G/Z(G)|=q$ implies $|G/Z(G)|$ is cyclic. Hence $G$ is abelian. Is this true?

If this is true then $|Z(G)|=pq$. What is going on?

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    $\begingroup$ Yes. Therefore, $G/Z(G)\cong\left\{e\right\}$ is the trivial group, and in particular it is cyclic. $\endgroup$
    – Guy
    Oct 4, 2016 at 18:57
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    $\begingroup$ Yes. It is a bit like showing "If pigs can fly, then ..." - except that there actually exists a tiny flying pig $\endgroup$
    – Hagen von Eitzen
    Oct 4, 2016 at 18:58
  • $\begingroup$ You are right, and what is going on (as you have basically deduced, assuming the theorem) is that for a group $G$ of order $pq$, with $p,q$ primes, it is impossible to have $\vert Z(G)\vert = p$. $\endgroup$
    – Steve Kass
    Oct 4, 2016 at 19:16

2 Answers 2

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Yes. In other words, the factor group $G / Z(G)$ cannot be simultaneously cyclic and nontrivial.

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The theorem in the question is usually stated in the same form, but it is better to say it as

If $G$ is non-abelian then $G/Z(G)$ should not be cyclic.

With this statement, you may immediately solve your question. Let $|G|=pq$. If $|Z(G)|=p$, then $Z(G)<G$, i.e. $G$ should be non-abelian. Then $|G/Z(G)|=q$ i.e. $G/Z(G)$ is cyclic, contradiction.

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  • $\begingroup$ Your new "theorem" is false. Take any centerless group $G$. $\endgroup$
    – Nex
    Oct 5, 2016 at 6:00

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