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My attempt: Let $O(a)=m, O(b)=n$, then $mx+ny=1$

Let $O(ab)=p$, then using commutative property, $(ab)^p=a^pb^p=e$, which is the identity.

Then $a^p=e, b^p=e$, hence, $m | p$ and $n | p$.

So, $p=mk_1=nk_2$. I'm stuck here. I understand I need to prove $p=mn$, please help.

user26857
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Diya
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    Why, from $a^pb^p=e$, do you deduce that $a^p=b^p=e$? Can't you have (for example) $a^p=b^{-p}$? If $mk_1=nk_2$ then $m$ divides $nk_2$ but since $m$ and $n$ are coprime then $m$ divides $k_2$.. – rafforaffo Jul 14 '15 at 13:38
  • this question already have answer here

    http://math.stackexchange.com/questions/78544/if-orda-m-ordb-n-then-does-there-exist-c-such-that-ord-c-lcmm-n

     – Chiranjeev_Kumar Jul 14 '15 at 14:08

1 Answers1

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You're close!

Unfortunately your proof isn't correct because $a^pb^p=e\Rightarrow a^p=b^{-p}$ but it may be still not $e$. But the result is true: it is in fact $e$ because if you consider the two cyclc subgroups of $G$: $<a>$ and $<b>$ and then you consider $H=<a>\cap <b>$, then according to Lagrange's Theorem $|H||m$ and $|H||n$ because $H$ is a subgroup of both $<a>$ and $<b>$ and $|<a>|=O(a)$. Thus $|H||\gcd(m,n)$ which is $e$. Thus $H=1$. Hence: $$a^p=b^{-p}\Rightarrow a^p\in\, <a>\cap <b>\Rightarrow a^p=e$$.

This proves that $m|p$ and $n|p$. Since $\gcd(m,n)=1$ therefore $mn|p$.

Now $(ab)^{mn}=a^{mn}b^{mn}=(a^m)^n(b^n)^m=e$. Thus $p|mn$.

Scientifica
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