Let $g\in G$ such that $o(g)=n<\infty$ (of a finite order). Show that $o(g^r)=\frac{n}{(n,r)}, 0<r<n$

Question

Let $g\in G$ such that $o(g)=n<\infty$ (of a finite order). Show that $o(g^r)=\frac{n}{(n,r)}, 0<r<n$ where $(n,r)$ is gcd$(n,r)$.

First of all, it is quite easy to verify that $(g^r)^{\dfrac{n}{(n,r)}}=e$.

Then, let $s>0$ such that $g^{rs}=e$. To show that $o(g^r)=s$, we have to show that $s$ is the smallest integer such that $g^{rs}=e$. Clearly, $n$ must divide $rs$. My problem is how to show that $s=\frac{n}{(n,r)}$. I was told to use the prime number decomposition, but I don't really see how to get the desired result. If someone could help, I would really appreciate it. Thank you in advance.

Answer 1

The order of $g^{r}$ is the least $m$ such that $rm=kn$ for some integer $k$. Therefore, $m$ is of the form $\frac{n}{r/k}$ ($*$), and $m=o(g^{r})$ ("the least") when the denominator in ($*$) is the greatest divisor of $r$ which is also a divisor of $n$, namely: $$o(g^r)=\frac{n}{\operatorname{gcd}(n,r)}$$

Answer 2

Let $d=(n,r)$. Since $n|rs$ we have $\frac{n}{d}|\frac{r}{d}s$. Now, one can check that $\frac{n}{d},\frac{r}{d}$ are coprime, and so $\frac{n}{d}|s$. Hence $s$ must be at least $\frac{n}{d}$.