Element to Power of Group Order is Identity

Theorem

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

Then:

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

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

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

$k \divides n$

So:

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

Thus:

$\ds g^n$

$=$

$\ds \paren {g^k}^m$

$\ds $

$=$

$\ds e^m$

Definition of Order of Group Element: $g^k = e$

$\ds $

$=$

$\ds e$

$\blacksquare$

1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem: Theorem $25.7$
1966: Richard A. Dean: Elements of Abstract Algebra ... (previous) ... (next): $\S 1.9$: Theorem $12$ Corollary
1978: John S. Rose: A Course on Group Theory ... (previous) ... (next): $0$: Some Conventions and some Basic Facts
1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $7$: Cosets and Lagrange's Theorem: Exercise $16 \ \text{(i)}$