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: