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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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    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.