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If the factor group with respect to the center of $G$ is cyclic, then $(aZ(G))^n=gZ(G)$ for some $n$ and any $g$, where both $a$ and $g$ are from $G$ (and $a^n$ is, too).

Because of the definition of the operation on the factor group it should be correct that $a^nZ(G)=gZ(G)$.

How come $G$ is not cyclic, too?

Shaun
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355durch113
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    Pick an abelian group $G$ which is not cyclic. What is $G/Z(G)$? – hardmath Apr 19 '16 at 18:22
  • I'll go a step further - start with the simplest case, $C_2\times C_2$, and see what goes wrong. – Steven Stadnicki Apr 19 '16 at 18:31
  • Why would $a^nZ(G) = gZ(G)$ imply that $G$ is cyclic? It just means that $a^n$ and $g$ are in the same coset of $Z(G)$. –  Apr 19 '16 at 18:31
  • @hardmath Ok I'll try: If abelian, G=Z(G). Left coset of some element g is all the combinations gz where z in Z(G) (here z in G). So the set of all cosets is basically all the combinations of elements from G, which is G, but that can't be it. – 355durch113 Apr 19 '16 at 18:42
  • My point was that if $G$ is abelian, $G/Z(G)$ is the trivial group (and hence cyclic). But $G$ can be abelian without being cyclic, as @StevenStadnicki's minimal example shows. – hardmath Apr 19 '16 at 18:46
  • I understand the example. How do you obtain the trivial group? – 355durch113 Apr 19 '16 at 18:53
  • @StevenStadnicki I see, there's no generator. – 355durch113 Apr 19 '16 at 18:59
  • We have $G/Z$ cyclic $\implies$ $G/Z=${$e$} because of $G=Z$. So, $G/Z$ is never a non-trivial cyclic group. In particular, the index of $Z$ cannot be a cyclic number greater than $1$, especially it cannot be a prime. – Peter Apr 19 '16 at 20:00
  • @Peter Is there a simple way to get $G/Z={e}$? – 355durch113 Apr 19 '16 at 20:20
  • Note that $G$ must be abelian, so $G=Z$. – Peter Apr 19 '16 at 20:39
  • @Peter I get that, but when I try to compute G/Z via the definitions of cosets G/H I get the entire G instead of {e}. – 355durch113 Apr 19 '16 at 21:23
  • @TanzWalzer No, you get the one-element group $G/G = {G}$. –  Apr 19 '16 at 23:07
  • @Bungo Now I understand, thank you. – 355durch113 Apr 20 '16 at 07:21

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This question has been ressurrected, so we might as well.

The fact that if $G/N$ is cyclic and $N\subseteq Z(G)$ then $G$ is abelian (and hence $G/Z(G)$ is in fact trivial) has been asked and answered before (many times) and I have expressed my dislike for that phrasing several times, too.

Now the question here is: why is it we can't conclude that $G$ itself was in fact cyclic? The argument given is okay as far as it goes: there exists $g\in G$ such that for every $a\in G$ there exists $n\in\mathbb{N}$ such that $aN = g^nN$ (I'm using my formulation of a central subgroup $N$). That means that $a^{-1}g^n\in N$, not that $a=g^n$, so we cannot conclude that $G=\langle g\rangle$.

Of course, this is a much more general phenomenon. It is just a reflection that, in general, if $H$ is a subgroup of $G$, and $x,y\in G$ are such that $xH=yH$, we cannot conclude $x=y$ (except in the literally trivial case of $H=\{e\}$). We can only conclude that there exists $h\in H$ with $x=yh$.

Arturo Magidin
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