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.

$\blacksquare$


Sources

  • 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Subgroups and Cosets: $\S 41$
  • 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 38.4$ Period of an element: $\text{(i)}$