Element to Power of Group Order is Identity

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Theorem

Let $G$ be a group whose identity is $e$ and whose order is $n$.


Then:

$\forall g \in G: g^n = e$


Proof

Let $G$ be a group such that $\order G = n$.

Let $g \in G$ and let $\order G = k$.

From Order of Element Divides Order of Finite Group:

$k \divides n$

So:

$\exists m \in \Z_{>0}: k m = n$

Thus:

\(\displaystyle g^n\) \(=\) \(\displaystyle \paren {g^k}^m\) Powers of Group Elements: Product of Indices
\(\displaystyle \) \(=\) \(\displaystyle e^m\) Definition of Order of Group Element: $g^k = e$
\(\displaystyle \) \(=\) \(\displaystyle e\) Power of Identity is Identity

$\blacksquare$


Sources