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?
On
A group $G$ of order $pq$ is either abelian, or has trivial center. In the second case, it cannot be nilpotent, hence $G$ cannot be the direct product of its Sylow subgroups $G=P_q\times P_p$. Moreover, if we have $G=P_q\times P_p$, then $P_q$ and $P_p$ are cyclic, hence $G$ is abelian. But this is not true in general. Take the semidirect product $G= C_p\rtimes_\varphi C_q$ for $q\mid p-1$. This is a non-abelian, non-nilpotent group.
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.