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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. – Simone Dec 10 '12 at 7:52
You need commutativity to expand $(x\ast y)^n$. Then you can involve the order of $x$ and $y$. – Hans Giebenrath Dec 10 '12 at 7:53
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. – Simone Dec 10 '12 at 7:57
(also, I guess that in you question you want that the gcd of the orders of $x$ and $y$ is $1$) – Simone Dec 10 '12 at 8:00
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 $|ab|\leq |G|$. – Simone Dec 10 '12 at 8:21
up vote 1 down vote accepted

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.

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