Boolean Group is Abelian
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Theorem
Let $G$ be a Boolean group.
Then $G$ is abelian.
Proof 1
By definition of Boolean group, all elements of $G$, other than the identity, have order $2$.
By Group Element is Self-Inverse iff Order 2 and Identity is Self-Inverse, all elements of $G$ are self-inverse.
The result follows directly from All Elements Self-Inverse then Abelian.
$\blacksquare$
Proof 2
Let $ a, b \in G$.
By definition of Boolean group:
- $\forall x \in G: x^2 = e$
where $e$ is the identity of $G$.
Then:
\(\ds a b\) | \(=\) | \(\ds a e b\) | Group Axiom $\text G 2$: Existence of Identity Element | |||||||||||
\(\ds \) | \(=\) | \(\ds a \paren {a b}^2 b\) | as $\forall x \in G: x^2 = e$ | |||||||||||
\(\ds \) | \(=\) | \(\ds a \paren {a b} \paren {a b} b\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {a a} \paren {b a} \paren {b b}\) | Group Axiom $\text G 1$: Associativity | |||||||||||
\(\ds \) | \(=\) | \(\ds a^2 \paren {b a} b^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds e \paren {b a} e\) | as $\forall x \in G: x^2 = e$ | |||||||||||
\(\ds \) | \(=\) | \(\ds b a\) |
Thus $a b = b a$ and therefore $G$ is abelian.
$\blacksquare$
Sources
- 1974: Thomas W. Hungerford: Algebra ... (previous) ... (next): $\S 1.1$: Exercise $13$
- 1978: John S. Rose: A Course on Group Theory ... (previous) ... (next): $0$: Some Conventions and some Basic Facts: Exercise $3$
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $6$: An Introduction to Groups: Exercise $9$
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 44.4$ Some consequences of Lagrange's Theorem
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $3$: Elementary consequences of the definitions: Exercise $1$
- 2008: Paul Halmos and Steven Givant: Introduction to Boolean Algebras ... (previous) ... (next): $\S 1$: Exercise $8$