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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 $|ab|$?

amWhy
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user77404
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2 Answers2

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The stated result is wrong

In the (additively written) cyclic group $\Bbb Z/6\Bbb Z$ of order$~6$, the elements $a=1$ and $b=2$ respectively have orders $|a|=6$ and $|b|=3$. Then $L=\operatorname{lcm}(6,3)=6$, but that is not the order of $a+b=3$, since that element has order$~2$. So the statement fails.

This also shows that your argument that $|a|$ and $|b|$ both divide $|ab|$ cannot work. You seem to think that since $a,b$ are not inverses of each other, $a^m$ and $b^m$ are not inverses of each other either, with $m=|ab|$. But in the example $m=2$, and here it is the case that $a^2$ and $b^2$ are inverses of each other (additively written these elements are $2a=2$ and $2b=4$). When you say "by assumption then we can conclude that $|a|$ divides $|ab|$" it is not clear what assumption you are referring to, but your conclusion is wrong.

Marc van Leeuwen
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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 $|ab|$.

amWhy
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  • Sorry, just edited the question, I meant I am having trouble with L divides |ab| – user77404 Jun 19 '13 at 01:26
  • So are you trying to say that so far since we know that L divides |a||b| and we know that |a||b| divides |ab| then that implies L must divide |ab|? – user77404 Jun 19 '13 at 01:35
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    No, in general |a||b| does not divide |ab|. – anon Jun 19 '13 at 01:36
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    We don't know that $|a||b|$ divides $|ab|$. – amWhy Jun 19 '13 at 01:39
  • That's a good point, any ideas of I could try show that |a||b| divides |ab|? Or somehow show in this case |a||b| = |ab|. – user77404 Jun 19 '13 at 01:48
  • @user77404 You will never be able to show that |a||b| divides |ab| because it is not generally even true. – anon Jun 19 '13 at 01:50
  • but in some groups it could be possible that |a||b| = |ab| I know this is obviously not true in general. I feel really close to the answer, however some detail seems to be missing. If this approach doesn't seem to work, is there other approaches for a question like this? – user77404 Jun 19 '13 at 01:51
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    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 01:54
  • @amWhy: excellent discourse and advice =1 – Amzoti Jun 19 '13 at 01:59
  • Hey amWhy, I have read Arturo's answer on the post you have linked on comment. However, I apologize if its obvious to see from his post, that how in this case from that we can justify why L divides |ab| (I can't see that yet). – user77404 Jun 19 '13 at 02:23
  • Maybe try another direction...did his argument make sense to you? The thing is, we may not have, for abelian groups, that $|a||b|$ divides |ab|, and so we're not going to get $L$ divides |ab| taking that argument...but you're very close... – amWhy Jun 19 '13 at 02:33
  • I don't fully understand the argument, because I am not familiar with the notation of $ord_p()$, in his proof in the first part he also shows that |ab| divides L, but then in the second part of the proof, he uses a different argument to conclude L = |ab| could I also fix my second argument to show that? – user77404 Jun 19 '13 at 02:39
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    @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 – FNH Jun 19 '13 at 09:50
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    @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
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    Exactly, @user77404. When relatively prime, your idea works! You're doing great work with this proof (hence my earlier +1). Is this making sense now? – amWhy Jun 19 '13 at 15:52
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    @amWhy, thanks for all the help, combining what I have with the suggestion of Maths Lover completes the proof. I will post a good copy of the work we have here so it helps others who come across this or a similar problem. – user77404 Jun 19 '13 at 16:10
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    You're welcome! It's a pleasure to work with dedicated students! – amWhy Jun 19 '13 at 16:22
  • @user77404 , i think my idea came to me since this moment when i was reading dummit and foote chapter 1 , section 3 and they asked me to prove that the order of the product of two cycles , namely , $m$-cycle and $n$-cycle is $L.C.M(n,m)$ , in that moment i said to myself , this is so clear as 1 + 1 = 2 ! how can i prove this ?! and it remains So clear in my mind till i say to myself, i have to make it into a proof because saying " this is so clear for me " is not a mathematical reasoning since it may be not "so clear" for others!. – FNH Jun 19 '13 at 17:59
  • @amWhy , it's more than a pleasure for me to see and read your Great posts and comments and answers " i started watching your answers on Groups theory which you have posted as it is full of new , important and great information for me , for me , you - and of course many members in the site - make international work task - function - which needs to be honored by the whole of the world . Regards . – FNH Jun 19 '13 at 18:09
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    This is a non-answer; compare with the one I gave. To begin with the opening "quotation" of the question inverses the direction OP was asking about; the question says "Showing that $|ab|$ divides $L$ was fine." Also I think the answer should maybe indicate somewhere that the result OP is trying to prove is wrong. – Marc van Leeuwen Jun 29 '14 at 03:55