Why is this elementary group theory problem infamously hard?

Question

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?

Where did you get the «hence» from? The equality $(ab)^{mn}=e$ only implies that $|ab|$ divides $mn$. — Mariano Suárez-Álvarez, Dec 30 '14 at 18:47
You've only shown that $|ab|$ divides $nm$ (not the other way around), which is not sufficient. — anomaly, Dec 30 '14 at 18:48
Still it doesn't seem "ridiculouosly hard". By the Structure Theorem applied to the group generated by $a$ and $b$ (and the fact that lcm is associative) we can reduce the problem to the case of a cyclic group of prime power order, where it is easy. — hmakholm left over Monica, Dec 30 '14 at 19:15
@Henning Iirc "hard" means hard using the results presented up to that point in Herstein's textbook. — Bill Dubuque, Dec 30 '14 at 19:22
@BillDubuque: Umm.. but the OP found the problem in Artin's book? — hmakholm left over Monica, Dec 30 '14 at 19:25
@Henning It's a famous hard problem from Herstein's textbook - see the remark in my answer that the OP links to. — Bill Dubuque, Dec 30 '14 at 19:27
@theage Please read the comments on my answer in the linked thread. — Bill Dubuque, Dec 30 '14 at 19:30
(Hmm, it's (more or less) exercise 2.2.20 in Artin and the Structure Theorem is only in chapter 12 ...) — hmakholm left over Monica, Dec 30 '14 at 19:37

score 2 · Answer 1 · answered Dec 31 '14 at 00:01

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.

Why is this elementary group theory problem infamously hard?

1 Answers1