Originally Posted by
Bruno J. Yes, and that's what the problem essentially is about!
Suppose $\displaystyle (ab)^k=1=a^kb^k$ (where the last equality is justified because $\displaystyle a,b$ commute). Use the division algorithm to write $\displaystyle k=q_1m+r_1=q_2n+r_2$, with $\displaystyle 0\leq r_1 <m$ and $\displaystyle 0\leq r_2 <n$. Then $\displaystyle a^{r_1}b^{r_2}=1$. Now raise this expression to the power $\displaystyle m$, eliminating $\displaystyle a$ from the left side; you obtain $\displaystyle b^{r_2m}=1$. Now we know $\displaystyle b^{m}$ has the same order as $\displaystyle b$ (prove it, using $\displaystyle (m,n)=1$), and therefore we must have $\displaystyle n\mid r_2$, which implies $\displaystyle r_2=0$. Similarily we must have $\displaystyle r_1=0$. Therefore $\displaystyle k$ is divisible both by $\displaystyle m,n$ and therefore divisible by $\displaystyle mn$.