If $\vert G\vert=pq$ with $p$ and $q$ prime numbers, then either $G$ is abelian or $Z(G) = \{ e\}$. What are the options for the order of $G/Z(G)$?
So I know that if $\vert Z(G)\vert = 1$, then $Z(G) = {e}$. And to prove that $G$ is abelian when $Z(G)$ is not just the identity element, I know that if $\vert G/Z(G)\vert \in \{p,q\}$, it would imply that $G/Z(G)$ is cyclic because $p$ and $q$ are prime numbers, and $G$ would be abelian (I've already proven that bit); but I am not quite sure on how to determine the possible orders of the quotient group, so that I could use the previous argument.
And we have not yet learned about centralizer and class equation and those things so I can't use them.
Any help or hints are very much appreciated, thank you.
By Lagrange's Theorem, since $Z(G)$ is a subgroup of $G$ (always), we have that $|Z(G)|\in \{1, p, q, pq\}$.
If $|Z(G)|=p$, then $|G/Z(G)|=\frac{pq}{p}=q$, a prime. Since all groups of prime order are cyclic, this means that $G/Z(G)$ is cyclic. But then, $G$ is abelian (have you proven this?).
Similarly, if $|Z(G)|=q$, then $|G/Z(G)|=p$, so again $G$ is abelian.
Why does it matter that $G$ is abelian? Well, if $G$ is abelian, then all of its elements commute with everything, so $G=Z(G)$. That means that $|Z(G)|=|G|=pq$, a contradiction. Thus $|Z(G)|\neq p, q$.
This leaves us with two options: $|Z(G)|=1$, $|Z(G)|=pq$.
If $|Z(G)|=1$ then $Z(G)=\{e\}$ (all subgroups contain the identity).
If $|Z(G)|=pq$, then $|Z(G)|=pq=|G|$ so $Z(G)=G$ (they have the same elements).
Thus either $G/Z(G)=G/\{e\}\cong G$ or $G/Z(G)=G/G\cong \{e\}$.
Hence, $|G/Z(G)|\in \{1, pq\}$.