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Why $G/Z(G)$ cyclic can imply that $G$ is a commutative group, where $Z(G)$ is the center of $G$? I have a proof by Schreier's theorem as follow. But I think a more brief proof definitely exists. My proof are as follow. Also I hope you can help check if this is fake. (To avoid the ambiguity, I state the Schreier theorem here: If $G$ is abelian, then for any extension of group $G$ by the group $Q$, denoted as the group $E$, there exist a homomorphism $\phi: Q\rightarrow \mathrm{Aut}G$ and a mapping $s:Q\times Q\rightarrow G$ which satisfies $s_{a,1}=s_{1,a}=1$ and $s_{a,b}s_{ab,c}=a(s_{b,c})s_{a,bc}$, such that $E$ is isomorphic to the set $G\times Q$ with the operation: $(x,a)(y,b)=(x\cdot a(x)s_{a,b}, ab)$, where $a(x)$ is the automorphism $\phi$ maps $a$ into. )
If we denote $G/Z(G)$ by $C=\langle a\rangle$,then by Schreier's theorem, there exist a mapping $s:C\times C\rightarrow Z(G)$ and a homomorphism $\phi: C\rightarrow \mathrm{Aut}(Z(G))$, which satisfies $s_{a,1}=s_{1,a}=1$ and $s_{a,b}s_{ab,c}=a(s_{b,c})s_{a,bc}$, where a(x) is the automorphism which $\phi$ maps $a$ to. Then G is isomorphic to $Z(G)\times C$ with the binary operation $(x,a)(y,b)=(x\cdot a(x)s_{a,b}, ab)$. By $(x, 1)$ is in the center of $G$ we learn $a^k(x)=x, \forall x$. Hence $s_{a,b}s_{ab,c}=s_{b,c}s_{a,bc}$. By plugging in $b=a^k, c=a$ and induction we learn that $s_{a^k,a}=s_{a,a^k}$ for all positive integer k. (Note that $G$ is commutative). If we plug into $b=a^{-k}, c=a$ we learn $s_{a^k,a}=s_{a,a^k}$ for k=<0 But by calculation we learn the above implies that $(x, a)$ is in the center of $G$, from which we can infer $a=1$ and $C=\{1\}$. So G is commutative.

Asigan
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  • Basically, the reason why $G/Z(G)$ cyclic implies that $G$ is abelian is that, if $G/Z(G)$ is cyclic, and $Z(G)\cdot x$ is a generator for the quotient, then $G=\langle Z(G),x\rangle$. The centre commutes with everything, so all generators commute. If it is generated by (say) two elements, then $G=\langle Z(G),x,y\rangle$. You know that $Z(G)$ commutes with $x$ and with $y$, but there's no reason that $x$ and $y$ commute. – David A. Craven Oct 01 '21 at 15:11
  • The general statement is that if $A$ is an abelian subgroup such that $G=Z(G)\cdot A$ then $G$ is abelian. We simply take $A=\langle x\rangle$. – David A. Craven Oct 01 '21 at 15:12
  • This question is very popular and has several duplicates, see for example here. It is worth to compare the answers to yours (which mostly are what David said above). – Dietrich Burde Oct 01 '21 at 15:13
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    By the way, while I appreciate my groups talking to me, I guess it should be commutative instead of communicative groups ;) – Severin Schraven Oct 01 '21 at 16:31
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    Schreier proved several theorems. What "Schreier's Theorem" are you using? – Arturo Magidin Oct 01 '21 at 16:40
  • See Gallian's, "Contemporary Abstract Algebra (Eighth Edition)", Theorem 9.3. – Shaun Oct 01 '21 at 20:04
  • @ArturoMagidin I mean this one: If $G$ is abelian, then for any extension of group $G$ by the group $Q$, denoted as the group $E$, there exist a homomorphism $\phi: Q\rightarrow \mathrm{Aut}G$ and a mapping $s:Q\times Q\rightarrow G$, which satisfies $s_{a,1}=s_{1,a}=1$ and $s_{a,b}s_{ab,c}=a(s_{b,c})s_{a,bc}$ such that $E$ is isomorphic to the set $G\times Q$ with the operation: $(x,a)(y,b)=(x\cdot a(x)s_{a,b}, ab)$, where $a(x)$ is the automorphism $\phi$ maps $a$ into. I also put it in my original post. – Asigan Oct 02 '21 at 12:49

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Suppose $G/Z=\langle x\rangle$. Pick any $a,b$ then if one of them in the center then $ab=ba$ if both of them not in the center then they can be written as $a=c_1x^n, b=c_2x^m, c_1,c_2 \in Z(G)$ then $$ab = (c_1x^nc_2x^m) = ba$$ because $c_1,c_2$ commutes

Arturo Magidin
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IrbidMath
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