# 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 $\order a = k$.

Then:

$\forall n \in \Z: k \divides n \iff a^n = e$

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

$a^r = a^n = e$

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:

$k \divides n$

### Sufficient Condition

Suppose $k \divides n$.

Then:

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

So:

 $\displaystyle a^n$ $=$ $\displaystyle a^{s k}$ $\displaystyle$ $=$ $\displaystyle \paren {a^k}^s$ $\displaystyle$ $=$ $\displaystyle e^s$ $\displaystyle$ $=$ $\displaystyle e$

$\blacksquare$