0

Let $G/Z(G)$ be the quotient group with center $Z(G)$. If $G/Z(G)$ is cyclic, then $G$ is abelian.

I was thinking that since $G/Z$ is cyclic, this implies the induced projection isomorphic quotient map $\phi : G/Z \to G$ would give the result. So such a $\phi$ is given by say $$\phi(gZ(G)) = gz$$ for $z \in Z(G).$

I feel like I made an assumption that is incorrect. I looked up the canoical homomorphism and it appears it usually is a map $\pi : G \to G/Z$.

quid
  • 42,135
jacob smith
  • 713
  • It can be defined both ways....one can be more useful than the other one depending on what you are doing. – ReverseFlowControl Aug 20 '16 at 13:33
  • @ReverseFlow, are you talking about my maps $\pi$ and $\phi$? – jacob smith Aug 20 '16 at 13:33
  • Quotient maps in general. – ReverseFlowControl Aug 20 '16 at 13:34
  • You don't need quotient maps for this....just use what you are given...$G/Z(G)$ is cyclic, start with the generator and go from there. – ReverseFlowControl Aug 20 '16 at 13:41
  • @ReverseFlow, I know. All I must do is that let $gZ(G)$ be the generator, then compare $g^kZ(G)g^nZ(G) = g^{k +n}Z(G).$ – jacob smith Aug 20 '16 at 13:45
  • @ReverseFlow, but i just want to know if I can end it with my answer. – jacob smith Aug 20 '16 at 13:47
  • No. I do not see how your written statements imply abelian...plus, the point of the proof is to show that $G$ is abelian not to imply it. – ReverseFlowControl Aug 20 '16 at 13:55
  • @ReverseFlow, cyclic groups are abelian. The isomorphism preserves such group properties is what I am saying. – jacob smith Aug 20 '16 at 14:05
  • 1
    Sorry but "the induced projection isomorphic quotient map" is just gibberish in my opinion. Try to write down an actual map precisely and you see where problems will emerge. Other than that this question was asked frequently. I'll come up with a dupe very soon. – quid Aug 20 '16 at 14:22
  • Finally on your last comment the groups $G/Z(G)$ and $G$ are in general not isomorphic. For finite $G$ the groups $G/H$ and $G$ are iso if and only if $H$ is trivial. Recall that $|G/H| = |G|/|H|$. – quid Aug 20 '16 at 14:30

0 Answers0