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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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Hint: $(ab)^n = a^n b^n$ for abelian groups. However, groups satisfying said property for all $a,b$ needn't be abelian. – Lord_Farin Oct 30 '12 at 17:59
The only answer to your second question is to give an example, and the symmetric group on three things does this nicely, since the product of two distinct elements of order two is of order three. – Lubin Oct 30 '12 at 18:11
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 $k$ – tomasz Oct 30 '12 at 18:37
up vote 4 down vote accepted

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. – Lubin Oct 30 '12 at 18:07

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 $o(ab)=o((1 \ 3 \ 2))=3$.

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