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

Proof

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

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

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, we can say $$r - s = q k + t$$ 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 $$\left({t < k}\right) \and \left({e = g^t}\right) \implies t = 0$$ by the definition of $$k$$.

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