It’s known that if $G/Z(G)$ is cyclic, then $G$ is abelian. This implies that $Z(G)=G$, which in turn implies that $G/Z(G)$ is trivial. Does this mean that if the quotient group $G/Z(G)$ is cyclic then it’s order is $1$?
More specifically, if $|G/Z(G)|=p$ for a prime $p$, then does it imply that $p=1$?
Question regarding the quotient of a group with it’s centre.
abelian-groupsfinite-groupsgroup-theory
Best Answer
Question. If the quotient group $G/Z(G)$ is given to be cyclic and of order $n$, would it imply that $n=1$?
Answer. Yes. As we know from the OP, if $G/Z(G)$ is cyclic then $G$ is abelian. Therefore, if $G/Z(G)$ is cyclic $G=Z(G)$, and so $G/Z(G)$ is trivial, i.e. $|G/Z(G)|=1$.
As an example application: if $|G|=35=5\times7$ then the order of the center is $1$ or $35$. To see this, suppose otherwise. Then $|Z(G)|$ is $5$ or $7$, so $|G/Z(G)|$ is $7$ or $5$, and so prime, and so cyclic. Hence, $G$ is abelian, so $G=Z(G)$, so $|Z(G)|=35$, a contradiction.