Let $a,b\in G$ (a group) such that $ab=ba$, $|a| = m$ and $|b| = n$. If $(m, n) = 1$, then $|ab|= mn$

Asked Mar 06 '21 at 20:05

Active Mar 06 '21 at 20:19

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Let $a,b\in G$ (a group) such that $ab=ba$, $|a| = m$ and $|b| = n$. If $(m, n) = 1$, then $|ab|= mn$.

Here I understand that $G$ is abelian and that this computes to $a^{mn} = b^{mn} = 1,$ and $m,n$ are relatively prime. But I really don't know about a step-by-step process of proving this.

edited Mar 06 '21 at 20:19

Shaun

47,747

asked Mar 06 '21 at 20:05

KKR

4

I think you mean $|ab|=mn.$ – Thomas Andrews Mar 06 '21 at 20:07
And no, it is not true that $m=n.$ – Thomas Andrews Mar 06 '21 at 20:07
@ThomasAndrews I thought so too... but this is what the question says.. and it is confusing me. – KKR Mar 06 '21 at 20:08
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Probably a mistake, because as stated, the problem doesn’t make any sense. Specifically, $ab=|mn|$ is meaningless. – Thomas Andrews Mar 06 '21 at 20:11
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@ThomasAndrews Oh thank you for that! Just realised that (m,n)=1 was referring to the gcd, and that m,n are relatively prime. – KKR Mar 06 '21 at 20:11

Let $a,b\in G$ (a group) such that $ab=ba$, $|a| = m$ and $|b| = n$. If $(m, n) = 1$, then $|ab|= mn$

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