Boolean Group is Abelian/Proof 2

Theorem

Let $G$ be a Boolean group.

Then $G$ is abelian.

Proof

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.

$\blacksquare$