# Order of products of elements in a finite Abelian group

We want to show that if $a,b\in G$ where $G$ is a finite Abelian group, we have $\operatorname{LCM}(|a|,|b|) = |ab|$ given that $ab \neq e$.

How I approached this question was by saying let $\operatorname{LCM}(|a|,|b|) = L$. Then if we show that $L$ divides $|ab|$ and $|ab|$ divides $L$ then $|ab| = L$. Showing that $|ab|$ divides $L$ was fine. But then I am having trouble with the second part that is showing $L$ divides $|ab|$. For this, so far I have:

Consider $(ab)^{|ab|} = e = a^{|ab|}b^{|ab|}$ since we know $a \neq b^{-1}$ by assumption then we can conclude that $|a|$ divides $|ab|$ and $|b|$ divides $|ab|$. We know that $L = \frac{|a||b|}{\operatorname{gcd}(|a|,|b|)}$ from here can we conclude that $L$ divides $|ab|$?

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This first part was fine, I understand that |ab| divides L, I wanted to know if I approached the second part of trying to show that L divides |ab| was correct. –  user77404 Jun 19 '13 at 1:21
This is very likely to be a duplicate, though I can't find one at the moment. –  Zev Chonoles Jun 19 '13 at 1:24
How exactly did you get that |a| and |b| divide |ab|? –  anon Jun 19 '13 at 1:46
@Zev I think this might be a duplicate It's close, but it talks about the existence of an element of order lcm(|a|, |b|), in an abelian group with elements $a$, $b$, not specifically that $|ab| = \operatorname{lcm}(|a|, |b|)$ –  amWhy Jun 19 '13 at 1:46
oops, that's not quite right, it is in fact a duplicate! –  amWhy Jun 19 '13 at 1:53

But then I am having trouble with the second part that is showing |ab| divides L.

Recall that if $g\in G$ and $g^n=e$, then this implies implies $|g|$ divides $n.$ So $(ab)^L = e \implies |ab|$ divides $L.$

Edit: If you are trying to show that $L$ divides $|ab|$, then with respect to your last question, the work you have, knowing that $$L = \frac{|a||b|}{\gcd(|a|,|b|)}$$ tells you $L$ divides $|a||b|$. If you can show it follows that $$L = \frac{|ab|}{\gcd(|a||b|)}$$ then you have shown that $L$ indeed divides $|ab|$.

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No, in general |a||b| does not divide |ab|. –  anon Jun 19 '13 at 1:36
We don't know that $|a||b|$ divides $|ab|$. –  amWhy Jun 19 '13 at 1:39
You've made some good progress. I did find a "duplicate sort" of question, take a look at this post for a nudge to completion! Recall, What you want is to conclude lcm(|a|,|b|) = |ab|, and for this part of the problem, that $L$ divides |ab|, so don't take on more than you need to! –  amWhy Jun 19 '13 at 1:54
@amWhy , i think there is intuitive proof as follows , we can show easily that if $g\in G$ and if $g^n= 1$ then $|g|$ \ $n$ , i think every one will accept that ! " it's easy to prove it also".now , let $L.C.M(|g|,|h|)=L$ first , it's easy to prove that $(gh)^L=1$ using the fact that $(gh)^k=g^k * h^k$ as $G$ is abelian so $|gh|$ \ $L$ , now , if $L$ is not the order of $gh$ then there is some $r <L$ and $(gh)^r =1$ so $r$ must be a multiple of $|g|$ and $|h|$ so $r\geqslant L$ contradiction since $L<r$ so $|gh|=L=L.C.M(|g|,|h|)$ ,it needs some details to be done,but i think it works –  Maths Lover Jun 19 '13 at 9:50
@MathsLover I really like your idea, I think I will go with that thank you. Another approach I discovered is to consider cases. We can show that if |a|,|b| are relatively prime (gcd(|a|,|b|)=1) then lcm(|a|,|b|)=|a||b| = |ab| this is not to hard to show. If |a| = |b|, then also lcm(|a|,|b|)=|a|=|b| = |ab|. However, I wasn't sure about the case where gcd(a,b) = k > 1. But your approach fixes the last part of my proof, thanks. –  user77404 Jun 19 '13 at 15:47
If you are having difficulty proving, you might chack a few cases to see whether it is true. In the cyclic group of order $6$ with generator $g$, and $a=g$, $b=g^2$ one has $|a|=6$, $|b|=3$ and $|ab|=|g^3|=2\neq L=6=\operatorname{lcm}(6,3)$. So it fails. This also shows that your argument that $|a|$ and $|b|$ both divide $|ab|$ is wrong. Indeed while $a,b$ aren't inverses of each other, $a^2$ and $b^2$ are.