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I've read somewhere that if $x,y$ commute, and $gcd(|x|,|y|) = 1$, then $|x*y|$ is the product of their individual order, but I don't even know why the criterion of commutativity is needed there. Also, is there a formula to find the order of $x*y$ in general? Do you need to have any restriction on the size of $G$, by any chance (as in, does it matter whether its finite or not, in regards to this question)?

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    consider the following example: you can take $G$ to be a group generated by two elements $a$ and $b$, modulo the relations $a^2=b^2=0$. Then the order of $a$ and $b$ is $2$ but $ab$ has infinite order.2012-12-10
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    You need commutativity to expand $(x\ast y)^n$. Then you can involve the order of $x$ and $y$.2012-12-10
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    adding also the relation $(ab)^n=0$ in the example of my previous comment, for some positive integer $n$, you can construct a group where $ab$ as the order you prefer but the order of $a$ and $b$ is $2$. Clearly, if you also have the relation $aba^{-1}b^{-1}=0$, that is, $a$ and $b$ commute, you cannot construct examples as the one above.2012-12-10
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    (also, I guess that in you question you want that the gcd of **the orders** of $x$ and $y$ is $1$)2012-12-10
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    yes, I'll correct that2012-12-10
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    To answer your last question: If $G$ has infinite order then what does Lagrange's theorem mean? It tells you that $|x|$ divides $\infty = |G|$. But $\infty$ is not a number so this is non-sensical. Hence one needs that $G$ has finite order.2012-12-10
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    @Simone So, I think you're saying that you can construct a group containing $ab$ with whatever order I please, where the order of $a$ and $b$ is $2$. Why? Also, that still differs from my question, which asks about the order of $ab$ as it is in $G$. Or is it not always possible to tell its order? I'd assume that there should be some way to do so, at least in the case where $G$ is finite, no?2012-12-10
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    well... I wanted just to show you that you have no chance to bound the order of $ab$ just knowing the order of $a$ and $b$. @MattN. The statement "if $a$ and $b$ commute, then $|ab|\leq |a|\cdot |b|$ with equality if $|a|,|b|<\infty$ and $gcd(|a|,|b|)=1$" is perfectly correct in infinite groups. (Also because the finiteness of the orders of $a$ and $b$ and their commutativity imply that the group they generate is a finite subgroup of the original group)2012-12-10
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    @Simone I can accept that there is not chance to bound the order in general, but surely one can do so in the case that $G$ is finite. What would be the order of $ab$ then? Or does this question need to specify more things about the relation of $a,b$?2012-12-10
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    well, the order of $ab$ is certainly smaller than or equal to the order of the group $\langle a,b\rangle$ generated by $a$ and $b$, which divides the order of $G$. Anyway I do not think that in general you will be able to find upper bounds that are much better than $|ab|\leq |G|$.2012-12-10

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Consider the dihedral group of order $2n$. It contains 2 reflections (which have order 2) whose product is a rotation of order $n$, which is half the order of the group. Half the order of the group is the highest possible order of an element in a noncyclic group, so no better bounds are possible in general.