Why is this elementary group theory problem infamously hard? I was working on an exercise in Artin about orders of products of elements in abelian groups, and found the following question about my problem. There, I saw people talking about how it was one of the most infamous exercises in basic abstract algebra, but I found it quite trivial (relatively)! The problem is:

Given $\vert a\vert = m$ and $\vert b\vert = n$ in an abelian group, find $\vert ab\vert$.

My solution was this (fairly routine and simple) one:

Note that in an abelian group we have $$(ab)^{mn} = ((ab)^m)^n = (a^mb^m)^n = (eb^m)^n = (b^m)^n = b^{mn} = (b^n)^m = e^m = e$$ where commutativity has been excessively used. Hence, we find $m\vert(\vert ab\vert)$ and $n\vert(\vert ab\vert)$. Since $\vert ab\vert$ is the minimal exponent with these properties, we find by definition $\vert ab\vert = \operatorname{lcm}(m,n)$. $\square$

Yet, in the linked question, people had ridiculously long proofs and even said that the exercise is hated and feared in certain books! Is my trivial solution somehow incorrect?
 A: Here are some major falsehoods implicit in your argument:


*

*The relation $g^{mn}=e$ does not imply either of $m$ or $n$ divide $|g|$; even if $(ab)^{mn}=e$ and $m=|a|,n=|b|$ in an abelian group it still does not imply $m$ or $n$ divide $|ab|$.

*If integers $m$ and $n$ divide another integer $|ab|$, it doesn't imply $|ab|={\rm lcm}(m,n)$.


(You say "by definition" we get $|ab|={\rm lcm}(m,n)$, but in fact the defition of lcm via universal property is correctly given by $m,n\mid k\Leftrightarrow {\rm lcm}(m,n)\color{Red}\mid k$, not ${\rm lcm}(m,n)=k$.)
Here are some true facts:


*

*In a group, $g^k=e$ does not imply $k=|g|$ but it does imply $k$ is a multiple of $|g|$. It should be obvious that $k=|g|$ isn't necessarily true, because setting $k$ to be any multiple of $|g|$ will force the relation $g^k=e$ to be true. And conversely if $g^k=e$ is true, we know $g^r=e$ is true where $r$ is the residue of $k$ modulo $|g|$, which contradicts minimality of $|g|$ unless $r=0$ i.e. $|g|$ divides $k$.

*In an abelian group, $\ell:={\rm lcm}(m,n)$ is a multiple of $|ab|$ but not necessarily equal. To see divisibility, note that $a^\ell=e$ since $\ell$ is a multiple of $|a|$ and $b^\ell=e$ since $\ell$ is a multiple of $|b|$, and therefore $(ab)^\ell=a^\ell b^\ell=e$ implies $\ell$ is a multiple of $|ab|$ (see previous point). They are not necessarily equal, though: suppose $a\ne e$ and $b=a^{-1}$, then $|ab|=0$ and $\ell=|a|>0$. Note that the question you link to says there is an element of order $\ell$, not that $ab$ has order $\ell$.


As for why it's considered infamously hard (relative to how easy it is to state), I think Herstein's note quoted in Dubuque's answer in the linked question illustrates that well enough.
