In the book of Algebra by Hungerford, at page 36, question 2, it is asked that
For the case where $(m,n) = 1$, I have done the following:
Observe that $$(ab)^{mn} = a^{mn} b^{mn} = e.$$
Now, let $0\leq r \leq mn$ and assume that $(ab)^r = e$. Then $$a^r b^r = e => a^r = b^{-r} => e=(a^r)^m = (b^{-r})^m => b^{rm} = e,$$ hence $n$ divides $rm$.
A similar process gives $a^{rn} = e$, hence $m$ divides $rn$.
Since $gcm(m,n)=1$, it must be true that $n | r$ and $m|r$.
So, $$ nq = r \quad \&\quad mp = r$$for some $p,q \in \mathbb{N}$.
Also, since $nq = mp$, we have $m | q$, $n|p$, so for some $k,l \in \mathbb{N}$,
$$nq\cdot mp = r^2 = nm \cdot nm \cdot kl = r^2,$$ hence $$(nm)^2 \leq r^2 => nm \leq r$$, but we have chosen $$r \leq nm,$$
hence $$r = nm.$$
QED.
Question:
First of all, is there any flaw in the proof ?
Secondly, I think the proof is too long and it can be made shorter by some means, but I have come up with this proof in almost 2.5 hours, so I cannot see how can we make this proof, or any other proof of the statement, shorter and elegant. Hence, is there any more clever and elegant proofs ?
Thirdly, after showing that $|ab|$ divides $mn$, how can we show the other way around. Is there any general method for that direction of the proof. If I don't remember wrong, I had a general method for showing that direction, such as looking the order of some powers of $|ab|$, but I cannot remember.
Your proof looks fine. Though I think you could prove that lcm($m,n$)|$r$ in a bit shorter way which is also easier to understand. Let $l$=lcm($m,n$). We can divide $r$ by $l$ with remainder. We get $r=ql+s$ when $0\leq s<l$. But note that $r$ and $ql$ are both common multiples of $m$ and $n$, so it is easy to see that $s=r-ql$ is also a common multiple. But $l$ by definition is the smallest positive common multiple of $m$ and $n$ and hence $s=0$, so $l|r$.