If I have a group $G$ with order $pq$ with $p,q$ primes, $p\neq q$. Then I want to try to apply Sylow's third theorem.
I want to argue that there is only 2 proper subgroups of $G$.
Since $p$ divides $|G|$ then let $n_p$ be the number of subgroups of $G$ with order $p$. Then by sylow's third theorem we have $$n_p =_p 1 \quad $$ $$n_p | \frac{pq}{p}=q$$
Since $q$ is prime, so we can only have $n_p = 1$ or $n_p=q$. But $n_p=q$ doesn't satisfy the first, and so $n_p=1$. A similar argument can be said on $n_q$
So there is 1 subgroup of order $p$, and 1 subgroup of order $q$. Does that work?