I am trying to prove this following result.
Let $G$ be an abelian group. Let $a,b$ be elements with orders $m,n$, respectively, where $m$ and $n$ are relatively prime. Prove that $|ab| = mn$.
Here is my attempt. There is one final step that I cannot figure out.
Let $G$ be an abelian group with identity $e$, and let $a,b \in G$ with the property that $|a| = m$ and $|b| = n$. We have: \begin{align*} (ab)^{mn} & = a^{mn} b^{mn} & & \text{since $G$ is abelian} \\ & = (a^n)^m (b^m)^n \\ & = e^m e^n \\ & = ee \\ & = e \end{align*} Define $t := |ab|$. I claim that $t \mid mn$. By the division algorithm, there exist unique $k \in \mathbb{Z}$ and $r$ with the property that $0 \leq r < t$ such that $$mn = kt + r.$$ We therefore have \begin{align*} (ab)^{mn} & = (ab)^{kt + r} \\ & = (ab)^{kt} (ab)^r \\ & = ((ab)^t)^k (ab)^r \\ & = e^k (ab)^r \\ & = e(ab)^r \\ & = (ab)^r \end{align*} Since $(ab)^{mn} = e$, and $(ab)^{mn} = (ab)^r$, we deduce $(ab)^r = e$ by transitivity of equality. But, $r$ is non-negative by definition and is strictly less than $t$, the order of $ab$. If $r$ is positive, this contradicts the definition of order, so we must have $r = 0$. Hence, $mn = kt$. That is, $t \mid mn$, or $|ab| \mid mn$.
Now, since $t$ is the order of $ab$, we have $$(ab)^t = a^t b^t = e,$$ but we have $$e = e^n = (a^t b^t)^n = (a^t)^n (b^t)^n = a^{tn} b^{tn} = a^{tn} (b^n)^t = a^{tn} e^t = a^{tn} e = a^{tn}.$$ Since $a^{tn} = e$, by the earlier result, the order of $a$ divides the $tn$. That is, $m \mid tn$. Since $m$ and $n$ are relatively prime, $m \mid t$. By an exactly analogous argument, we deduce $n \mid t$. Indeed, $$e = e^m = (a^t b^t)^m = (a^m)^t b^{tm} = e^t b^{tm} = eb^{tm} = b^{tm},$$ which again implies that the order of $b$ divides $tm$, that is, $n \mid tm$, and since $m$ and $n$ are relatively prime, $n \mid t$.
The solution I am looking at goes one step further and argues that $n \mid t$ and $m \mid t$, so $mn \mid t$. I cannot figure out why this is the case. If in fact it were the case, the proof would be done since we earlier found that $t \mid mn$, so together with this fact, we'd have $t = mn$.
$$(ab)^{|a| |b|}=a^{|a||b|}b^{|a||b|}=1$$ Therefore, $|ab|$ divides $|a||b|$. Conversely, $$1=(ab)^{|ab|}=a^{|ab|}b^{|ab|}$$
implies that $$1=1^{|b|}=a^{|b||ab|}b^{|b||ab|}=a^{|b||ab|}$$ Therefore, $|a|$ divides $|b||ab|$. Since $|a|$ and $|b|$ are coprime, you may write $$1=|a|x+|b|y$$ for suitable integers $x$ and $y$. Then $$|ab|=|a||ab|x+|b||ab|y$$ Since $|a|$ divides the right-hand-side, it also must divide $|ab|$. With a similar argument, you get that also $|b|$ divides $|ab|$. Hence $|a||b|=\operatorname{lcm}(|a|,|b|)$ divides $|ab|$.