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

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

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: $$o(g^r)=\frac{n}{\operatorname{gcd}(n,r)}$$
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 $$\frac{n}{d}$$.