If G is a group of order $pq$, where $p$ and $q$ are primes. How do I prove that any nontrivial subgroup of $G$ must be cyclic? This makes sense since $p$ and $q$ are primes but I'm not sure how I can prove that the subgroup will be cyclic? 
Will the generator of the subgroup be $pq$?
 A: This is just an application of Lagrange theorem and the definition of primes.
I give a completely fleshed out answer in light of the comments. Note that all groups are assumed finite.
From Lagrange theorem we know that if $H\leq G$ (is a subgroup of) then $|H|$ divides the order of $G$.
Now if we suppose that $H\leq G$ and $H$ is non-trivial (not equal to $\{e\}$ or $G$) then we have that $|H| | |G|$ so that $|H| | pq$
But by the definition of prime this gives that either $|H| | p$ or $|H| | q$ and as $H$ was assumed to be non-trivial this then gives that either $|H|=p$ or $|H|=q$
We now need to show that if a group has order $p$ then it is cyclic and we are done.
So let $H$ be a group of prime order and pick $g\in H$ such that $g\neq e$ then again from Lagrange theorem we have that $o(g)|H|$ and so $o(g)|p$ but as $g$ was assume not to be the identity we have that $o(g)>1$ and so $o(g)=p$
This then gives that $\langle g \rangle $ (cyclic subgroup generated by $g$) has order $p$ and as $\langle g \rangle \leq H$ we have that they are equal and so $H$ is cyclic.
A: A non trivial subgroup of $G$ has order $p$ or $q$, hence it's cyclic beacuse $p$ and $q$ are prime numbers.
