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: $\text{(i)}$