Element to Power of Multiple of Order is Identity

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

Let $$a \in G$$ have finite order such that $$\left|{a}\right| = k$$.

Then:
 * $$\forall n \in \Z: k \backslash n \iff a^n = e$$

Proof
Let $$k \in \N$$ be the smallest such that $$a^k = e$$ as per the hypothesis.


 * Let $$a^n = e$$.

Let $$n = q k + r, 0 \le r < k$$.

By Element to the Power of Remainder, $$a^r = a^n = e$$.

But $$0 \le r < k$$.

Since $$k$$ is the smallest such that $$a^k = e$$, we have that:
 * $$1 \le s < k \implies a^s \ne e$$

Thus $$r = 0$$, i.e. $$k \backslash n$$.


 * Now suppose $$k \backslash n$$.

Then $$\exists s \in \Z: n = s k$$.

So:

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