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Gallian, in his Contemporary Abstract Algebra, first proves this theorem:

Theorem: Let $G$ be a group and let $Z(G)$ be the center of $G$. If $G/Z(G)$ is cyclic, then $G$ is Abelian.

Proof: Let $G/Z(G)=\langle gZ(G)\rangle$ and let $a\in G$. Then there exists an integer $i$ such that $aZ(G) = (gZ(G))^i= g^ i Z(G)$. Thus, $a= g^ i z$ for some $z$ in $Z(G)$. Since both $g^ i$ and $z$ belong to $C(g)$, so does $a$. Because $a$ is an arbitrary element of $G$ this means that every element of $G$ commutes with $g$ so $g \in Z(G)$. Thus, $gZ(G) = Z(G)$ is the only element of $G/Z(G)$.

Then the author goes on to say that: our proof shows that a better result is possible: If $G/H$ is cyclic, where $H$ is a subgroup of $Z(G)$, then $G$ is Abelian.

So, I follow the same proof to prove this new statement. Everything goes fine for the first three sentences, when I replace $Z(G)$ with $H$ in above proof, but, I can't see how to deduce here $g\in Z(G)$ that $G$ indeed abelian.

Community
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Silent
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1 Answers1

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If $G/H$ is cyclic then also $G/Z(G)$ is it because $$ aH=bH\Rightarrow b^{-1}a\in H\leq Z(G)\Rightarrow aZ(G)=bZ(G) $$ then $G/H=\langle gH\rangle\Rightarrow G/Z(G)=\langle gZ(G)\rangle$

Jihlbert
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  • Can you explain this in more detail? I'm a beginner in group theory. – John Mars Oct 02 '20 at 02:29
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    @JohnMars Suppose that $G/H=\langle aH\rangle$ (in other words $aH$ generates $G/H$) where $H\leq Z(G)$. Then for every $g\in G$ exists $n\in\mathbb{N}_0$ such that $gH=(aH)^n=a^nH$ so $g^{-1}a^n\in H$ and so $g^{-1}a^n\in Z(G)$. This means also that $gZ(G)=[aZ(G)]^n$ so $aZ(G)$ is a generator of $G/Z(G)$. – Jihlbert Oct 04 '20 at 17:59
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    So $G/H$ cyclic $\Rightarrow G/Z(G)$ cyclic $\Rightarrow G$ Abelian. – Jihlbert Oct 04 '20 at 18:00