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If $G$ is a group with $xy=yx$ for $x,y\in G$, and $x^m=y^n=e$, then: (1) $o(xy)|mn$, where $o(xy)$ is the order of $xy$, and (2) if $(m,n)=1$ then $o(xy)=mn$.

Well, I've already proved (1), but I'm having trouble with (2), I was thinking that maybe, if we already know (1), $o(xy)|mn$ then we just have to show that $mn|o(xy)$, that way we get equality. So we get: $mn|o(xy)$ iff $o(xy)=mn*l$ for some $l\in\Bbb Z$. On the other hand, we know that $(xy)^{mn}=e$ then it's obvious that for any $l\in\Bbb Z$, $(xy)^{mn*l}=e^l=e$, we get what we wanted, but if there's an infinite number of $l's$, what does this means? I feel that what I'm doing is wrong since I didn't used the principal hypothesis: $(m,n)=1$, but this is the only way so far I've been able to do something.

Abodi
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  • Note that if $(xy)^m=x^my^m=1$ then $x^m=y^{-m}$ so $x^m\in\langle y\rangle$ and $y^{-m}\in \langle x\rangle$. What can you say about $\langle x\rangle \cap \langle y\rangle$ when $(o(y),o(x))=1$? – Pedro Feb 02 '14 at 04:05
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    You should burn into your brain forever that the condition $x^m = e$ does not mean $x$ has order $m$. It only means the order of $x$ divides $m$. Since $e^m = e$ and $e^n = e$, as far as the statement of the problem is concerned (as written at the time I write this) both $x$ and $y$ could be $e$. Therefore the statement in part (2) is false. You could use $x = y = e$ and $m$ and $n$ any relatively prime numbers at all. – KCd Feb 02 '14 at 04:21
  • @PedroTamaroff ok, I know that, since $,$ are subgroups then $ \cap $ is a subgroup, moreover $ \cap \leq , $, but... I've already checked the link of that other question, and I've seen what you wrote there, however I still don't understand, (but I'm trying!). If we have $ \cap $ then its order is smaller than $$ and $$, but I don't see more... – Abodi Feb 02 '14 at 04:26
  • It is not only smaller, it divides both the order of $x$ and the order of $y$... but those numbers are coprime, so the order of $\langle x\rangle\cap\langle y\rangle$ is...? – Pedro Feb 02 '14 at 04:32
  • @KCd oh! I see my mistake, thank you. "You should burn into your brain forever..." and it'll be done. – Abodi Feb 02 '14 at 04:32
  • @PedroTamaroff (sorry this takes me so long, I wish I could go faster). So $o( \cap )|o()$ and $o( \cap )|o()$? how do you know that? In that case doesn't it follows that $o( \cap )|(o(),o())$? – Abodi Feb 02 '14 at 04:45

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