$G$ is a group of order $pq$ and $P_q$ and $P_p$ are Sylow subgroups... If $(G,*)$ is a group of order $pq$, then it is clear that there are Sylow subgroups $P_q$ and $P_p$ of order $q$ and $p$ in $G$. If $q>p$ then $P_q$ is normal. 
I want to find a decomposition for all $g \in G$, there exists $(q',p') \in P_q \times P_p$ such that $g = q'*p'$, but I have been stuck for about 8 hours. How can I find such decomposition?

Since $p$ and $q$ are different primes, Bézout's Lemma guarantees the existence of $m,n \in \mathbb{Z}$, with $1=mp+nq$. Hence $g=(g^p)^m \cdot (g^q)^n$. Now prove that $g^p \in P_q$ and $g^q \in P_p$. Hence powers of these elements are in the same respective subgroups.
