Element to Power of Multiple of Order is Identity

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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:

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

where $k \divides n$ denotes that $k$ is a divisor of $n$.


Proof

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


Necessary Condition

Let $a^n = e$.

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

By Element to Power of Remainder:

$a^r = a^n = e$

But $0 \le r < k$.

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

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

Thus $r = 0$.

That is:

$k \divides n$


Sufficient Condition

Suppose $k \divides n$.

Then:

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

So:

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

$\blacksquare$


Sources