Let $a$ and $b$ commute. If $m$ and $n$ are relatively prime, then ord($ab$) = $mn$. This is exercise $10.E.4$ from Pinter:

Let $a$ and $b$ be elements of a group $G$.
Let ord($a$) = $m$ and ord($b$) = $n$.
Prove:
Let $a$ and $b$ commute.
If $m$ and $n$ are relatively prime, then ord($ab$) = $mn$.
(HINT: Use $10.E.2$.)

Here is $10.E.2$, which Pinter suggests that we use:

If $m$ and $n$ are relatively prime, then no power of $a$ can be equal to any power of $b$ (except for $e$).

I'm also going to use $10.E.1$:

If $a$ and $b$ commute, then ord($ab$) is a divisor of lcm($m$,$n$).

As well as $B.T6.i$ (Theorem $6.i$ from Appendix $B$. A fact from basic number theory.):

If   gcd($m$,$n$) = 1   then   lcm($m$,$n$) = $mn$

Let's begin.
We are given that $m$ and $n$ are relatively prime which means that:
$$\text{gcd}(m,n) = 1$$
By $B.T6.i$:
$$  \text{lcm}(m,n) = mn \tag{1}  $$
By $10.E.1$:
$$ \text{ord}(ab)\ |\ \text{lcm}(m,n) $$
Substituting $(1)$:
$$ \text{ord}(ab)\ |\ mn $$
Which means that there is an integer $x$ such that:
$$ \text{ord}(ab) x = mn $$
This is so close! For the theorem to be true, we'd have to show that $x = 1$.
However, the original exercise statement says to use $10.E.2$.


*

*Is that helpful in showing that $x = 1$?

*Or is there some other completely different approach whereby $10.E.2$ is used?



UPDATE: Some comments regarding the answer below.
The proof uses the following fact:

If $a|c$ and $b|c$ then $lcm(a,b)|c$.

For this exercise, I wanted to only use theorems that had been presented in the book up to that point. And, I didn't seem to recall seeing this theorem. (If anyone spots this in Pinter, please comment below with the location.)
However, I did notice that the following similar fact is in Appendix B (REVIEW OF THE INTEGERS) as exercise B.9:

If $a|c$ and $b|c$ and $gcd(a,b) = 1$ then $ab|c$.

So yeah, it looks like the approach shown below is definitely a valid way to go if you want to stick to what's presented in the book.
 A: $$\boxed{\textit{You can show $mn\ \big|\ \text{ord}(ab)$ using $10.E.2$:}}$$
Since $a$ and $b$ commute we can distribute common powers of $a$ and $b$:
$$
e = (ab)^{\text{ord}(ab)} = a^{\text{ord}(ab)}b^{\text{ord}(ab)}
$$ This implies $a^{\text{ord}(ab)} = b^{-\text{ord}(ab)}$ or, in other words, some power of $a$ equals some power of $b$. But $10.E.2$ says that since $m = \text{ord}(a)$ and $n = \text{ord}(b)$ are relatively prime, no power of $a$ can equal any power of $b$ unless those powers of $a$ and $b$ both evaluate to $e$.
So we must have $a^{\text{ord}(ab)} = e = a^m$ and $b^{-\text{ord}(ab)} = e = b^n$.
But then, $m\ \big|\ \text{ord}(ab)$. This is because $m$ is the order of $a$ i.e. the least positive integer power of $a$ that evaluates to $e$; so if any other positive integer power of $a$ evaluates to $e$ (like $\text{ord}(ab)$ here), then $m$ must divide that integer. For a similar reason, $n\ \big|\ \text{ord}(ab)$.
Therefore since both $m$ and $n$ divide $\text{ord}(ab)$, their least common multiple $\text{lcm}(m,n)$ must divide $\text{ord}(ab)$. But in our case $\text{lcm}(m,n) = mn$ so we finally have $mn\ \big|\ \text{ord}(ab)$.
Along with the result you already established $\text{ord}(ab)\ \big|\ mn$, this suggests $\text{ord}(ab) = mn$.
P.S. If you don't believe the bolded statements above, use Euclidean Division. E.g. to see $m\ \big|\ \text{ord}(ab)$, use Euclidean Division on the integer pair $\big(m, \text{ord}(ab)\big)$ to get $\text{ord}(ab) = km + r$ for some $k, r \in \Bbb Z$ with $0 \leq r < m$. Do you see why $r$ must be $0$?
