# Show that if $\vert G\vert = pq$, then either $G$ is abelian or $Z(G) = \{ e\}$.

The center of a group $G$ is defined as $Z(G):=\{ z\in G : gz = zg, \; \forall g \in G\}$.

The goal is to show that if $\vert G\vert = pq$, where $p$ and $q$ are not necessarily distinct primes then either $G$ is abelian or $Z(G) = \{ e\}$.

I want to suppose that $Z(G) \neq \{ e\}$ and then use the fact that $G/Z(G)$ is cyclic to imply that $G$ is abelian, which is something I have already proven. But how do I show that $G/Z(G)$ is cyclic when I am not certain what exactly $Z(G)$ looks like. I only know that it has at least one non-identity element in it, which will be of order $p$ WLOG, (the case where it is of order $pq$ is trivial).

Any help is appreciated. Thank you.

• You're almost there! If $Z(G)\neq e$ then what is it's order? What must the order of $G/Z(G)$ be? Sep 25, 2017 at 7:21
• So $Z(G)$ can only have order $pq$ or $p$, the former resulting in $G/Z(G)$ having order 1 which is trivially cyclic, and the latter case leaving $G/Z(G)$ with order $q$, which must also be cyclic? Right? Sep 25, 2017 at 7:41
• Yes, because we're given that $q$ is prime and groups of prime order are always cyclic (which isn't too difficult to prove, use Lagrange's theorem). Sep 25, 2017 at 7:43

Hint: assume $Z(G) \neq \{1\}$. Then look at $|G:Z(G)| \in \{1,p,q\}$
You already suppose that $Z(G)\neq 1$. Then the order of the quotient group $G/Z(G)$ is one of 1,p,q.
You can follow this question to see that all group of prime order is cyclic. So, the group $G/Z(G)$ is a cyclic group.
This argument admittedly doesn't use the result $$G/Z(G) \space\text{cyclic}\Rightarrow G\space\text{abelian}$$.
If $$q=p$$, it is well known that $$G$$ of order $$p^2$$ is abelian. Next, let's take $$p,q$$ distinct primes. By contradiction, let's assume $$|Z(G)|=q$$; therefore, the noncentral elements $$g\in G$$ have centralizer of order $$q$$, whence $$\frac{|G|}{|C_G(g)|}=\frac{pq}{q}=p$$ for every noncentral $$g\in G$$, and finally (take the noncentral part of the Class Equation) $$q(p-1)=kp$$, where $$k$$ is the number of noncentral conjugacy classes of $$G$$: contradiction, since $$p\nmid q(p-1)$$. Thus, $$|Z(G)|\ne q$$. The same argument works by swapping $$p$$ and $$q$$, whence $$|Z(G)|\ne p$$ as well. Therefore, either $$|Z(G)|=pq$$ and $$G$$ is abelian, or $$Z(G)=\{e\}$$.