Equal Powers of Finite Order Element

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

Let $g \in G$ be of finite order.

Let $\order g = k$.

Then:
 * $g^r = g^s \iff k \divides \paren {r - s}$

Necessary Condition
Suppose that $k \divides \paren {r - s}$.

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

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

Thus:

Sufficient Condition
Let $g^r = g^s$.

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

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

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

Thus:
 * $e = g^{r - s} = g^{k q + t} = \paren {g^k}^q g^t = e^q g^t = g^t$

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

So:
 * $r - s = q k + 0 = q k \implies k \divides \paren {r - s}$