I have problems doing this exercise:

Let $G$ be a group and $Z(G)$ the center of $G$.

Prove: If $G/Z(G)$ is cyclic, then $G$ is an abelian group.

My attempt:

I think that I have to show $G/Z(G) = \{G\}$, since from this it follows $Z(G) = G$, and so $G$ is abelian. Maybe one way of proving that $G/Z(G) = \{G\}$ is by showing $|G/Z(G)|=1$.

$G/Z(G) \neq \emptyset$ should be clear. Now let's take $g_1Z(G), g_2Z(G) \in G/Z(G)$ aribtrary, and show that they are equal.

Since $G/Z(G)$ is cyclic, I know that there is a $k \in \mathbb{N}$ so that $g_1Z(G) = (g_2Z(G))^k = g_2^kZ(G)$. Since $Z(G)$ is the center, I think I should use that the elements of $Z(G)$ commutate with the elements of $G$. But I have no idea, how I can continue now.

Tank you for your help.

Regards, S. M. Roch

  • $\begingroup$ See math.stackexchange.com/questions/78690/… and links there. $\endgroup$ – quid Aug 16 '16 at 15:52
  • $\begingroup$ The links should provide you correct arguments. The argument in the final paragraph is close to part of what you need. It is also true that if $G$ is abelian $G/Z(G)$ is in fact trivial, yet to show this directly is harder than necessary. Instead show that any two elements commute. $\endgroup$ – quid Aug 16 '16 at 18:16
  • $\begingroup$ Thank you all for providing these helpful links! $\endgroup$ – S. M. Roch Aug 16 '16 at 18:25