Boolean Group is Abelian

Theorem

Let $G$ be a Boolean group.

Then $G$ is abelian.

By definition of Boolean group, all elements of $G$, other than the identity, have order $2$.

$\blacksquare$

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.

$\blacksquare$

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 $8$