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Has anyone an idea how to prove this one??

If $G$ is a finite group and $a$ is an element of $G$ with $\mathrm{ord}(a) = r$., then $\mathrm{ord}(a^k) = \dfrac{r}{\gcd(r,k)}$.

Thank you in advance!

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Have you tried anything? – user61527 Jan 4 '14 at 21:40
Note that since you are looking for the order of $a^k$, you are essentially working in the cyclic subgroup of $G$ generated by $a$, which here is isomorphic to integers modulo $r$. So look at what happens in integers modulo $r$, this greatly helps for visualisation (you can even draw a perfectly accurate picture!). – fkraiem Jan 4 '14 at 21:51

Hint: Let $d:=\gcd(r,k)$. Then we can write $r=bd$ and $k=cd$, where $\gcd(b,c)=1$.

We are then asked to prove that $\text{ord}(a^k)=b$. Now, $$ (a^k)^b=a^{bcd}=(a^{cd})^b=(a^r)^d=1, $$ so that the order of $a^k$ must divide $b$. Now, what's left is to show that there is no smaller number $\tilde{b}$ so that $(a^k)^{\tilde{b}}=1$. Can you see how to do that? Try looking for a contradiction.

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