Element to the Power of Multiple of Order

Theorem

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

Let $a \in G$ have finite order such that $\left|{a}\right| = k$.

Then:

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

Proof

Let $k \in \N$ be the smallest such that $a^k = e$ as per the hypothesis.

• Let $a^n = e$.

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

By Element to the Power of Remainder, $a^r = a^n = e$.

But $0 \le r < k$.

Since $k$ is the smallest such that $a^k = e$, we have that:

$1 \le s < k \implies a^s \ne e$

Thus $r = 0$, i.e. $k \backslash n$.

• Now suppose $k \backslash n$.

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

So:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle a^n\)	\(=\)	\(\displaystyle a^{s k}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({a^k}\right)^s\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle e^s\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle e\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

$\blacksquare$

Sources

• J.A. Green: Sets and Groups (1965)... (previous)... (next): $\S 5.4$: Example $101$
• Richard A. Dean: Elements of Abstract Algebra (1966): $\S 1.4$: Theorem $3 \ (2)$
• Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): $\S 38.4 \ \text{(ii)}$