How to prove the group $G$ is abelian? 
Question: Assume $G$ is a group of order $pq$, where $p$ and $q$ are primes (not necessarily distinct) with $p\leqslant q$. If $p$ does not divide $q-1$, then $G$ is Abelian.

I know that if the order of $Z(G)$ is not equal to $1$, then I can prove $G$ is Abelian. However, suppose $|Z(G)|=1$, how can I know $G$ is Abelian too?
 A: Remember if $|G|=p$, with $p$ then $G$ is cyclic. Use Sylow Theorems to show that $G\cong H\times K$ where $H,K\le G$ and $|H|=p$, $|K|=q$ and $p\nmid q-1$.
A: Assume first that $\vert G \vert = p^2$. Since $Z(G) < G$, we have that $\vert Z(G) \vert \mid \vert G \vert$. One can show that the center of a group $G$ whose order is a prime power is non-trivial - it's not too hard, but too long to state here. You might want to look that up in an Algebra book. 
If $\vert Z(G) \vert = p$, the quotient $G / Z(G)$ is of prime order by Lagrange's theorem, hence cyclic and generated by - say - $g \in G$. Let $a, b \in G$ and $a \in g^n Z(G)$, $b \in g^m Z(G)$. Then $ab = g^n z g^m z' = g^n g^m z z' = g^{m + n} z z' = g^m z' g^n z = ba$ by choosing $z$, $z'$ to be elements in the centralizer. We've just shown that $G$ is abelian, implying $Z(G) = G$, contradiction. 
Hence $\vert Z(G) \vert = p^2$, proving the first statement. 
A sketch of a solution of the second statement may look like this: 
If now $\vert G \vert = pq$ for $p, q$ distinct primes and $p \nmid q - 1$,
we get by Sylow's Theorem's that there are $$n_p = 1 \mod p$$ $p$-subgroups and $$n_q = 1 \mod q$$ $q$-subgroups, and also $n_p \mid q$, $n_q \mid p$. But by what we just said $n_p$ is of the form $1 + kp$ and as $p \mid q - 1$, we have $k = 0$. As we assumed $p \leq q$ and $n_q \mid p$, we get $n_q = 1$ also. 
The group $H_p \cdot H_q < G$ will now be a subgroup of order $pq$ for $H_p$ a $p$-subgroup and $H_q$ a $q$-subgroup, hence $G = H_p \cdot H_q \simeq H_p \times H_q$. Both subgroups are cyclic, so is their product (they are of relatively prime order).
