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**clic-clac** If you've never used the fact that $\displaystyle N_1$ and $\displaystyle N_2$ were normal subgroups:

$\displaystyle S_3$ is a group of order $\displaystyle 6,$ it contains an element of order $\displaystyle 2$ and another one of order $\displaystyle 3,$ which of course generate subgroups of orders $\displaystyle 2$ and $\displaystyle 3.$

But $\displaystyle S_3$ is not cyclic...

The hypothesis "$\displaystyle N_1,N_2$ normal" allows you to prove $\displaystyle \text{order}(ab)=pq,$ for instance, consider the commutator $\displaystyle [a,b]=aba^{-1}b^{-1}$ and prove it is the identity element, that will mean $\displaystyle a,b$ commute and then you will be able to conclude $\displaystyle \text{order}(ab)=pq.$