Let $a$ be a primitive root modulo $mn$. Show that $a$ is also primitive root modulo $m$ and $n$.

Showing $(a,mn)=1\Longrightarrow (a,m)=(a,n)=1$ is not a problem. The problem is showing $a^{\varphi (m)} \equiv 1\pmod{m}$ where $\varphi (m)$ is the least natural number for which the congruence holds.

Initial thought:

Let $A :=\{x|a^x\equiv 1\pmod{m}\}\subset\mathbb{N}$, obviously $A\neq\emptyset$ so it contains the smallest element $r$. Must show that $r=\varphi (m)$. Supposing for a contradiction that $r<\varphi (m)$ I would need it somehow to contradict the fact that
$$a^{\varphi (mn)}\equiv 1\pmod{mn} $$
where $\varphi (mn)$ is the least for which said congruence holds.

How should one go about finding a contradiction?