Group of order $pq$ with $p
If $p<q$ are primes and $G$ is a group of order $pq$, then $G$ has exactly $q-1$ elements of order $q$.
My attempt was:
Use Sylow theorem to show that $G$ has only one subgroup of order $q$, so it's cyclic and then generated by let's call $x$. Now the order of $\langle x\rangle$ is $q$, so it has $q$ elements but the unit is in it and $order(e)=1$ so it has a total of $q-1$ elements, thus G has only $q-1$ elements.
Did i go wrong by assuming $G$ has only that many of elements of order $q$ by looking only at the fact that $p$ and $q$ are primes?
 A: Summarizing the comments: your idea is fine but you should explain why every nonidentity element of $\langle x \rangle$ has order $q$, and you should also explain why $G$ has no other elements of $q$ besides the ones in $\langle x \rangle$.
Here is an argument which addresses these questions.
First, as you noted, we can use the Sylow theorems to deduce that $G$ has a unique subgroup of order $q$. This is because the number of $q$-Sylow subgroups, call it $n_q$, must divide $p$ and must equal to $1$ modulo $q$. This forces $n_q = 1$.
Let's call the unique $q$-Sylow subgroup $Q$. If $x \in Q$ then the order of $x$ must divide $q$ (Lagrange's theorem). Since $q$ is prime, this means that $x$ has order $1$ or $q$. Therefore every nonidentity element of $Q$ has order $q$. This means that $Q$ contains $q-1$ elements of order $q$.
Can $G$ contain any other elements of order $q$? If $x \in G$ has order $q$, then $\langle x \rangle$ is a subgroup of order $q$. But $Q$ is the unique subgroup of order $q$, so $\langle x \rangle = Q$, and in particular, $x \in Q$, so $x$ is one of the $q-1$ elements we already counted.
We conclude that $G$ has exactly $q-1$ elements of order $q$.
