Showing that a group of order $pq$ is cyclic if it has normal subgroups of order $p$ and $q$

Look first at Lagrange's theorem, that tells you that $N \cap M = \{ e \}$.

This implies that if $a \in N$, $b \in M$, then $a b = b a$, because $$ (ba)^{-1} a b = a^{-1} b^{-1} a b = (a^{-1} b^{-1} a) b \in M, \qquad a^{-1} b^{-1} a b = a^{-1} (b^{-1} a b) \in N, $$ since $N, M \triangleleft G$,

If you take in particular $a \ne 1$ (so that $a$ has order $p$) and $b \ne 1$ (so that $b$ has order $q$), then $ab$ will have order $pq$.

Since $M$ is normal we can form $G/M$. Now $G/M$ has order $p$ and thus is cyclic, in particular abelian. It follows that $[G,G] \subseteq M$. Now similarly we find also that $[G,G] \subseteq N$ and so $$[G,G] \subseteq M \cap N = \{e\}.$$ Now recall that $[G,G]$ is such that $G/[G,G]$ is abelian. Since $[G,G] =\{e\}$ we conclude that

$$G/[G,G] \cong G/\{e\} \cong G$$

is abelian too. The structure theorem for finite abelian groups now gives that $G \cong \Bbb{Z}/p\Bbb{Z} \times \Bbb{Z}/q\Bbb{Z}$ or $G \cong \Bbb{Z}/pq\Bbb{Z}$. However the former is isomorphic to $\Bbb{Z}/pq\Bbb{Z}$ by CRT and so in any case we see that $G$ is cyclic of order $pq$.