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I am trying understand this proof http://www.proofwiki.org/wiki/Quotient_of_Group_by_Center_Cyclic_implies_Abelian

but I am confused what its trying to prove.

Wouldn't $G/Z(G)$ group have just one element $G$ if $G$ is abelian because $Z(G)$ is all of $G$ for abelian groups. So is $Z/Z(G)$ a cyclic group with just one element? I am not quite sure what that theorem is saying.

Surya
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    One can use the same argument to show that if $G/N$ is cyclic for a subgroup $N$ with $N\leq Z(G)$ then $G$ is abelian (this might make the statement seem less strange). – Tobias Kildetoft Oct 31 '13 at 14:47
  • A nice way, imo, to say this is: for any group $;G;$ , the factor group $;G/Z(G);$ is never cyclic non-trivial. – DonAntonio Oct 31 '13 at 15:38

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The theorem is saying that if $G/Z(G)$ is cyclic then $G$ is abelian so $G/Z(G)$ has one element. Putting these together: if $G/Z(G)$ is cyclic then it has one element.

The hypothesis here is not that $G$ is abelian. That is a consequence of the assertion that $G/Z(G)$ is cyclic. The interesting result here is that $G/Z(G)$ can never be a nontrivial cyclic group, like $\mathbb{Z}/12\mathbb{Z}$ or $\mathbb{Z}/2013\mathbb{Z}$.

The title of the theorem is not "factor group of an abelian group by its center is cyclic". It is "if the factor by the center is cyclic, then the group was abelian (so the factor was trivial)."

Elchanan Solomon
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  • Thank you Isaac. I am trying to understand how this theorem is useful. If I already know $Z(G)$ i know right away if $G$ is abelian or not. Why would one go through the trouble of constructing factor group. – Surya Oct 31 '13 at 15:00
  • I guess you answered my question 'The interesting result here is that G/Z(G) can never be a nontrivial cyclic group, like Z/12Z or Z/2013Z' – Surya Oct 31 '13 at 15:05