Boolean Group is Abelian

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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.


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 $G$.


\(\ds a b\) \(=\) \(\ds a e b\) 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.