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Let $H$ be an abelian group and $a,b\in H$ with $\mathrm{ord}(a)<\infty$ and $\mathrm{ord}(b)<\infty$.

My question is why $\mathrm{ord}(ab)|(\mathrm{ord}(a)\cdot\mathrm{ord}(b))$ and why there are in a non-abelian group elements with $\mathrm{ord}(ab)\not|(\mathrm{ord}(a)\cdot\mathrm{ord}(b)$ ?

Thank you very much!

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    For arbitrary groups, I believe the order of $ab$ is completely independent of the order of $a$ and $b$. If you take the group $\langle a,b\vert a^n,b^m,(ab)^k\rangle$ with $2\leq n,m,k\leq \infty$, $a$ will be of order $n$, $b$ of order $m$ and $ab$ of order $k$2012-10-30

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Suppose $a$ has order $n$ and $b$ has order $m$; that is, $a^n=e$ and $b^m=e$. Then, since $H$ is Abelian, we have $(ab)^{nm}=a^{nm}b^{nm}=(a^n)^m(b^m)^n=e$. Now, the order of $ab$ may be smaller, but it must divide $nm=ord(a)\cdot ord(b)$.

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    Good, and more direct than what I had in mind.2012-10-30
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If $o(a)=n , \ o(b)=m$ and your group is abelian then $(ab)^{nm}=abab\ldots ab=a^{nm}b^{nm}=(a^n)^m(b^m)^n=1$ so $o(ab)|o(a)o(b)$.

For non abelian groups this is not correct. For example consider the permutations $a=(1 \ 2) , \ b=(2 \ 3)$. Then $o(a)=o(b)=2$ but $o(ab)=o((1 \ 3 \ 2))=3$.