Find the number of elements of orders $p$ and $q$ in a nonabelian group $G$ of order $pq$, with $p<q$ (both prime).
Approach: Let the number of elements of order $t$ with $n(t)$.
Let $n(p) = m$ and $n(q)=pq-1-m$. By Sylow's theorem, the number of subgroups of order $p$ is either $1$ or $q$. It cannot be $1$, so it must be $q$. ($p$ must be of the form $qk+1$ in order for $G$ to be nonabelian ).
Now, each of the sylow-p subgroups being of order $p$, it has $\phi(p)=p-1$ generators in it. There are $q$ of them. So, $n(p)=q(p-1)=m$, and $n(q)=pq-1-q(p-1)=q-1$.
Is this correct?
All you really need is that if there were an element of order $pq$ the group would be abelian, and that there is only one $q$-Sylow subgroup. That means you have $1$ element of order $1$, $q-1$ elements of order $q$, and all the rest must have order $p$.