# Element to Power of Remainder

## Theorem

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

Let $a \in G$ have finite order such that $\order a = k$.

Then:

$\forall n \in \Z: n = q k + r: 0 \le r < k \iff a^n = a^r$

## Proof

Let $n \in \Z$.

We have:

 $\ds n$ $=$ $\ds q k + r$ $\ds \leadstoandfrom \ \$ $\ds n - r$ $=$ $\ds q k$ $\ds \leadstoandfrom \ \$ $\ds k$ $\divides$ $\ds \paren {n - r}$

The result follows from Equal Powers of Finite Order Element.

$\blacksquare$