Problem: Prove the following: Every Abelian group $G$ with $|G| = pq$ where $p$ and $q$ are two distinct primes, is cyclic.
Attempt: According to the first Sylow theorem, exists a Sylow $p$-subgroup of $G$. Let $n_p(G)$ be amount of Sylow $p$-subgroups of $G$. Then by the third Sylow theorem we have $n_p(G) \equiv 1 \mod p$ and $n_p(G)$ divides $q$. From this it follows that $n_p(G) = 1$ or $n_p(G) = q$.
Now I wanted to treat the two cases separately.
Case 1: Let $n_p(G) = 1$. Let $P$ be the unique Sylow $p$-subgroup. Then $P$ is a normal subgroup of $G$ because $P$ is conjugated to itself. Hence $$ [G: P] = \frac{ |G|}{|P|} = \frac{pq}{ |P|} = q. $$ This is because $|P| = p$, since a finite group is a $p$-group if and only if its order is a power of $p$. The above then shows that $G/P$ is cyclic since $q$ is prime.
Now I wanted to construct an isomorphism of some kind, which relates $G$ to $G/P$ and then conclude that since $G/P$ is cyclic, so is $G$. But I can't find the right map.
Case2: For this case I didn't have much inspiration. Suppose $n_p(G) = q$. I know there are then $q$ amount of Sylow $p$-subgroups of $G$, which are conjugated to each other. How to continue?
Help is appreciated!
The suggestion in the comment is perfectly good. To gain some perspective, I would like to point out the Fundamental Theorem of Abelian groups for you:
https://proofwiki.org/wiki/Fundamental_Theorem_of_Finite_Abelian_Groups
So your group is isomorphic to $(\mathbb{Z}_p, +) \times (\mathbb{Z}_q, +)$.
There is another general theorem you should know (and usually all these important facts are explained in any introductory course before Sylow's theorems): the direct product of two nontrivial groups $G,H$ is cyclic if and only if the groups $G,H$ are finite cyclic groups with coprime order.