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Given the order of $x=36$ in a group, how do I compute the order of $x^{-8},x^{27} $.

Also, a similar question, for $x, y \in G$, if order of $x=2$ and order of $y=3$, what can we say about order of $xy$?

The only facts I am aware of are order of $xy$=$yx$ and for any $x,y\in G$ order of $x=y^{-1}xy$

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    $\begingroup$ Yet another mysterious negative vote! $\endgroup$
    – Derek Holt
    Mar 16, 2013 at 12:50

3 Answers 3

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For the first question, the order of $x^n$ is the smallest $k$ such that $(x^n)^k = e$, so you have to find the smalles $k$ such that $nk$ is a multiple of 36. You can easily prove that $k = \frac{36}{\gcd(36,n)}$.

For the second question, if you only have $|x| = 2$ and $|y| = 3$, you can't say anything about $|xy|$: just think about the group $\left<x,y\mid x^2,y^3\right>$. If else you have $xy=yx$ as a relation, then notice that $(xy)^n=x^ny^n$, so you have to find the smallest $n$ which is a multiple of both $2$ and $3$, ie $n=6$.

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For cyclic groups, there's a known result that says:

$$o(x^k)=\frac{o(x)}{\gcd(o(x),k)}$$

I will leave it to you to prove it but it's really simple, if you take a couple of cyclic small groups you will see a clear pattern

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for your first part: if $n=|x|$

$$|x^k|=\frac{n}{\gcd(k,n)}$$

in the 2nd part $|xy|$ can be $6 $ if the group is cyclic . If the group is abelian then $|xy|$ is also $6$

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