$|a|=m, |b|=n,k=\text{gcd}(m,n)$, Then $|ab|=\frac{mn}{k^2}i$ for some $i$ such that $i|k$.
Proof.
Let $c=|ab|$.
$(ab)^{mn/k}=(a^m)^{n/k}(b^n)^{m/k}=e$, $c|\frac{mn}{k}$.
$e=(ab)^{cm}=(a^m)^{c}b^{cm}=b^{cm}$, $n|cm$, $\frac{n}{k}|c\frac{m}{k}$, $\text{gcd}(\frac{n}{k},\frac{m}{k})=1 \rightarrow \frac{n}{k}|c$. By the same logic, $\frac{m}{k}|c$, $\frac{nm}{k^2}|c$.
Conclusion.
$\frac{nm}{k^2}|c|\frac{mn}{k}$, meaning $|ab|=\frac{mn}{k^2}i$ for some $i$ such that $i|k$.
Corollary.
Consider the case $k=1$, (so $\text{gcd}(m,n)=1$). Then, $|ab|=|a||b|$.
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