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:
\(\ds g^n\) | \(=\) | \(\ds \paren {g^k}^m\) | Powers of Group Elements: Product of Indices | |||||||||||
\(\ds \) | \(=\) | \(\ds e^m\) | Definition of Order of Group Element: $g^k = e$ | |||||||||||
\(\ds \) | \(=\) | \(\ds e\) | Power of Identity is Identity |
$\blacksquare$
Sources
- 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 $16 \ \text{(i)}$