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Let $a,b\in G$, a finite abelian group and $|a|=r, |b|=s$ with $\gcd(r,s)=1$. Prove that $|ab|=rs$.

My attempt: Let $|ab|=n$. Since $G$ is ableian, $(ab)^n=a^nb^n=1$. Thus $r\mid n$ and $s\mid n$. Together with $\gcd(r,s)=1$, it follows that $rs\mid n$. This is where I'm stuck; need to show that $rs=n$. Any hints on how to proceed?

Edit: I've come up with a solution that is a somewhat different approach to what has been provided in the hints. Here it goes:

Since $G$ is abelian, $n\mid{\rm lcm}(r,s)$. But since $\gcd(r,s)=1$, ${\rm lcm}(r,s)=rs$ by an elementary result in number theory. Thus $n\mid rs$. Together with $rs\mid n$, we have that $n=rs$, which is what we want to prove.

tmaj
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    Have you thought about the orders of $(ab)^r$ and $(ab)^s$? – user1729 Jul 12 '21 at 16:10
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    You've shown that $n=\vert ab \vert \geq rs.$ To show equality, all you need to show is that $(ab)^{rs}=a^{rs}b^{rs}=1$. I'm thinking you can figure that part out without too much trouble. – Robert Shore Jul 12 '21 at 16:11
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    @lulu it's not a duplicate of that question. One answer says "anything is possible (in the non-abelian case)", and the second answer contains a statement of a more general fact for abelian groups, but with only a sketchy proof. – user1729 Jul 12 '21 at 16:12
  • @lulu Is there a way to solve it using my approach directly? – tmaj Jul 12 '21 at 16:14
  • @user1729 Yes, you are right. Those answers really don't address the question at all. I will delete the close vote while I track down a better duplicate...I'm sure the question has been answered before. – lulu Jul 12 '21 at 16:15
  • @lulu I agree, it is definitely a duplicate! If you find a target then comment here and I can use gold badge powers to close it. (Will try to find one myself too.) – user1729 Jul 12 '21 at 16:17
  • @user1729 Here is a duplicate, which is itself a duplicate. But both it and the original appear to contain useful arguments. – lulu Jul 12 '21 at 16:18
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    @user1729 I get the answer following your hint. It was quite obvious once I got it, thank you. – tmaj Jul 12 '21 at 16:19
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    @lulu The answer there is clearer than the other two, at least is my eyes, and it's not a "check my proof" question, so I closed it as a duplicate of that one. – user1729 Jul 12 '21 at 16:25
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    I am struggling to understand why this question was reopened. It is clearly a duplicate. – user1729 Jul 14 '21 at 12:09

1 Answers1

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You've shown that $n=\vert ab \vert \geq rs.$ To show equality, all you need to show is that $(ab)^{rs}=a^{rs}b^{rs}=(a^r)^s(b^s)^r=1$. I'm thinking you can figure that part out without too much trouble.

Robert Shore
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