Let $a, b$ be two integers coprime with $n$. We denote $\omega(a)$ and $\omega(b)$ as the respective orders of $a$ and $b$ modulo $n$. We assume that $\omega(a)$ and $\omega(b)$ are coprime. What is the order of $a b$ modulo $n$?
It is clear that $(a b)^{\omega(a) \omega(b)}=1$. We deduce that $\omega(a b)$ divides $\omega(a) \omega(b)$.

Now let $k \geq 1$ be such that $(a b)^{k}=1$ and let us show that $k \geq \omega(a) \omega(b)$. By raising to the power $\omega(a)$, we get $b^{k \omega(a)}=1$. This implies that $\omega(b)$ divides $k \omega(a)$, and according to the hypothesis, this implies that $\omega(b)$ divides $k$. We similarly show that $\omega(a)$ divides $k$. According to the hypothesis, we then deduce that $\omega(a) \omega(b)$ divides $k$, hence the result.