Equal Powers of Finite Order Element

Theorem
Let $G$ be a group whose identity is $e$.

Let $g \in G$ be of finite order, and let $\left|{g}\right| = k$.

Then:
 * $g^r = g^s \iff k \mathop \backslash \left({r - s}\right)$.

Proof
First, suppose that $k \mathop \backslash \left({r - s}\right)$.

From the definition of divisor:
 * $k \mathop \backslash \left({r - s}\right) \implies \exists t \in \Z: r - s = k t$

So:
 * $g^{r - s} = g^{k t}$

Thus:

Now let $g^r = g^s$.

Then:
 * $g^{r-s} = g^r g^{-s} = g^s g^{-s} = e$

Now by the Division Theorem:
 * $r - s = q k + t$

for some $q \in \Z, 0 \le t < k$.

Thus:
 * $e = g^{r - s} = g^{k q + t} = \left({g^k}\right)^q g^t = e^q g^t = g^t$

So by the definition of $k$:
 * $\left({t < k}\right) \land \left({e = g^t}\right) \implies t = 0$

So:
 * $r - s = q k + 0 = q k \implies k \backslash \left({r - s}\right)$