Element to Power of Remainder

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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:

\(\displaystyle n\) \(=\) \(\displaystyle q k + r\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle n - r\) \(=\) \(\displaystyle q k\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle k\) \(\divides\) \(\displaystyle \paren {n - r}\)

The result follows from Equal Powers of Finite Order Element.

$\blacksquare$


Sources