Abelian group whose center is proper subgroup

I've seen a statement that if $G/C(G)$ is cyclic, then $G$ is abelian, and I could prove this. And then I was confused because if G is abelian, then it should be the same with its center $C(G)$. So the quotient group just becomes a trivial group.

Is there an abelian group whose center is not the same with itself? Orelse the statement means nothing i guess.

• You seem to know how the center is defined, so what do you think? – Edward Evans Apr 10 '17 at 16:02
• The theorem may be rephrased as "For any finite group $G$, either $G/C(G)$ is non-cyclic or it is trivial". Remember that the trivial group is cyclic. – Arthur Apr 10 '17 at 16:04
• The statement means a lot: it means that the quotient of any group by its center (which, btw, is always abelian) cannot be cyclic non-trivial. Now, the title of your question is rather trivial. – DonAntonio Apr 10 '17 at 16:04
• @Arthur's comment might be better for clearing up the confusion. But the fact that the center of abelian groups isthe whole group is trivial, you may want to back up if you don't see it clearly. – Jorge Fernández Hidalgo Apr 10 '17 at 16:09