Prove that the steps needed to move from $w_i$ with steps of length $i$ to $w_0$ divides $n$ On the circle with radius $1$, consider these points:  
$w_i=(\cos{\frac{2\pi i}{n}},\sin{\frac{2\pi i}{n}}) \space\space i = 0,\dots,n-1$  
Prove that If we start moving from $w_i$ with the objective of reaching $w_0$ with steps of length $i$, the steps we take will divide $n$.  
Note 1: By steps of length $i$, I mean passing the part of the circle's perimeter, which is equal to the distance of $w_o$ and $w_i$.  
Note 2: This question should be related to this theorem:
If $G$ is a group and $O(g)=n$ and $t$ is a positive integer, then we have:
$O(g^t)=\frac{n}{gcd(n,t)}$  
But i can't understand how! Someone told me that considering $O(w_1^i)=\frac{n}{gcd(n,i)}$ and $O(w_1)=n$ solves the problem. I don't know what it means!  
So, I have a part of the answer (maybe all of it) but i can't understand it.  Any idea? 
 A: What you are trying to prove is actually incorrect, which we can see from an easy example with $n=3$ and $i=1$: We start at $w_1$, go to $w_2$ in the first step and arrive at $w_3 = w_0$ with the second step. Obviously, $2$ does not divide $3$. What is true, however, is that the number of steps plus one (which is equal to the number of steps required to go around the circle) divides n, which we can even prove without your theorem:
We consider the group $G$ of all $w_i$, with $w_i + w_j = w_{i+j}$ (i.e. going from $w_i$ the distance from $w_0$ to $w_j$). Obviously, $w_0$ is the neutral element.
The number of steps is then the number $k$, so that
\begin{align}
w_i + w_i^k &= w_i\\
\Leftrightarrow w_i^{k} &= w_0\\
\Leftrightarrow k &= O(w_i)
\end{align}
We also see that all the points $w_i^{l}$, with $l$ ranging from $0$ to $k-1$ make up a subgroup $H$ of $G$ (The most difficult to prove part here is the closure, which comes from $w_i^a + w_i^b = w_i^{a+b}$) The order of that group is $k$. From Lagrange's theorem, we find that the order $H$, which is equal to $k$, has to divide the order of $G$, which is equal to $n$, which is exactly what we wanted to prove.
If you want to solve it with your theorem: The theorem says that
$$O(g^t) = \frac{O(g)}{gcd(O(g), t)}$$
The length from $w_0$ all the way around the circle back to $w_0 = w_n$ is $n$ times the length from $w_0$ to $w_1$. Therefore, $w_1^n = w_0$ (with $n$ being the smallest $k$, so that $w_1^k = w_0$), which, by definition, means that $O(w_1) = n$. Hence we have from your theorem:
$$O(w_1^t) = \frac{n}{gcd(n, t)}$$
With the same reasoning we find that $w_i = w_1^i$. Setting $t = i$ we then have
\begin{align}
k = O(w_1^i) &= \frac{n}{gcd(n, i)}\\
n = k\cdot gcd(n, i)
\end{align}
from which clearly follows that k divides n.
