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

where $k \mathrel \backslash n$ denotes that $k$ is a divisor of $n$.

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

Necessary Condition
Let $a^n = e$.

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

By Element to Power of Remainder:
 * $a^r = a^n = e$

But $0 \le r < k$.

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

Thus $r = 0$.

That is:
 * $k \mathrel \backslash n$

Sufficient Condition
Suppose $k \mathrel \backslash n$.

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

So: