TITLE: Structure of groups of order $pq$, where $p,q$ are distinct primes.
QUESTION [21 upvotes]: I don't know about the Sylow Theorems.
But I have been wondering about a proof of the fact that a group or order $pq$ where $p$ and $q$ are distinct primes must be cyclic. I can't quite work out the details, but here is the general idea. I would like help with filling in details. I assume that it is already known that $G$ has subgroup(s) of order $p$ and subgroup(s) of order $q$.
If $G$ is a group of order $pq$ ($p\neq q$), then I know that $G$ has a subgroup $H$ of order $p$ and a subgroup $K$ of order $q$. Then $H\simeq \mathbb{Z}_p$ and $K\simeq \mathbb{Z}_q$. 
But then $H\oplus K \simeq \mathbb{Z}_{pq}$, so I would think that $H\oplus K \simeq G$. I guess one could do an internal direct product instead of an external direct product, but I don't know that $H$ and $K$ are normal subgroups.
I am asking for help completing this argument.
Edit: I see from the comments below that I might need to assume that the smaller prime does not divide the larger prime minus $1$. Or maybe it is enough to assume that the primes are greater than or equal to $3$ (Still distinct).

REPLY [11 votes]: Let $\lvert G \rvert = pq$ for primes $p, q$ such that $q < p$ and $q \not \mid p-1$.
Let $n_p$ and $n_q$ be the number of Sylow $p$-subgroups and Sylow $q$-subgroups, respectively. 
By the Third Sylow Theorem, $n_p \mid q$ and $n_p \equiv 1 \pmod p$ which implies $n_p = 1$ since $q < p$. 
Similarly, $n_q \mid p$ and $n_q \equiv 1 \pmod q$ implies $n_q = 1$ since $p \not\equiv 1 \pmod q$. 
Let $P$ be the unique Sylow $p$-subgroup and $Q$ be the unique Sylow $q$-subgroup.
Since $p$ and $q$ are prime, $P$ and $Q$ are cyclic: $P \cong \mathbb Z / p \mathbb Z$ and $Q \cong \mathbb Z / q \mathbb Z$.
Restatement of 1. from David Hill's answer: $P \cap Q \le P, Q$ so by Lagrange's theorem we have  $|P\cap Q|$ divides both $p$ and $q$, and we must have $|P\cap Q|=1$. It follows that $$|PQ|=\frac{|P||Q|}{|P\cap Q|}=pq=|G|$$
Hence, $PQ=G$. Since $P$ and $Q$ are unique, by consequence of the Third Sylow Theorem, $P,Q \lhd G$. Then the internal and external direct product are isomorphic, so $G \cong P \times Q$. 
An analogue to the Chinese Remainder Theorem for groups shows that if $\operatorname{gcd}(m,n)=1$ then  $\mathbb Z / m \mathbb Z \times \mathbb Z / n \mathbb Z \cong \mathbb Z / mn \mathbb Z$. (prove with First Isomorphism Theorem and the standard Chinese Remainder Theorem) So by this theorem, $G \cong P \times Q \cong \mathbb Z / pq \mathbb Z$ and hence cyclic.