If $a$ and $b$ commute and $\text{gcd}\left(\text{ord}(a),\text{ord}(b)\right)=1$, then $\text{ord}(ab)=\text{ord}(a)\text{ord}(b)$.
Hint/roadmap:
If $|a|$ and $|b|$ are coprime and $a$ and $b$ commute, then $|ab|=|a||b|$. In particular, this holds for all pairs of elements in abelian groups.
This follows from these facts:
if $ab=ba$, then $(ab)^x=a^xb^x$, $\hspace{17pt}$(note: this is the step which fails without commutativity)
if $g$ is an element in any group $G$ and $g^x=\operatorname{id}_G$, then $|g|$ divides $x$,
if $|a|$ and $|b|$ are coprime, the smallest number divisible by both $|a|$ and $|b|$ (the least common multiple, one might say) is $|a||b|$.
From these facts, one may show that the smallest number $x$ for which $(ab)^x$ is the identity is $|a||b|$, hence $|ab|=|a||b|$.
Hint $\rm\ \ (ab)^k\! = 1\,\Rightarrow\, a^k\! = b^{-k}\! = c \in \langle a\rangle\cap\langle b\rangle\Rightarrow ord\,c\mid m,n\,\Rightarrow\, ord\,c\mid(m,n)=1\,\Rightarrow\, c = 1.\,$ Therefore $\rm\ a^k\! = 1 = b^k\Rightarrow\: m,n\mid k\:\Rightarrow\:\ell = lcm(m,n)\mid k.\ $ Conversely $\rm\:m,n\mid \ell \:\Rightarrow\:(ab)^\ell\! = 1.$