I'm trying to do Exercise 2.3.9 from textbook Groups, Matrices, and Vector Spaces - A Group Theoretic Approach to Linear Algebra by James B. Carrell.
Prove that if $G$ is a finite group such that $G / Z(G)$ is cyclic then $G$ is abelian. (Recall that $Z(G)$ is the center of $G$.)
Could you please verify if my attempt is fine or contains errors? In my proof, I do not use the finiteness of $G$. Is this assumption redundant?
Thank you so much for your help!
My attempt:
We have $G / Z(G) = \langle x Z(G) \rangle$ for some $x \in G$. Then any $y,z \in G$ can be written as $y = x^m p, z = x^n q$ for some $p,q \in Z(G)$ and $m,n \in \mathbb Z$.
As a result, $yz = (x^m p) (x^n q) = (x^m p) (q x^n) = x^m (pq) x^n =(pq) x^mx^n$. Similarly, $zy = (qp)x^n x^m$. Because $p,q \in Z(G)$, we have $pq \in Z(G)$ and thus $pq=qp$. This completes the proof.
