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 $G2$: Properties of Identity
\(\ds \) \(=\) \(\ds a \left({a b}\right)^2 b\) as $\forall x \in G: x^2 = e$
\(\ds \) \(=\) \(\ds a \left({a b}\right) \left({a b}\right) b\)
\(\ds \) \(=\) \(\ds \left({a a}\right) \left({b a}\right) \left({b b}\right)\) Group Axiom $G1$: Associativity
\(\ds \) \(=\) \(\ds a^2 \left({b a}\right) b^2\)
\(\ds \) \(=\) \(\ds e \left({b a}\right) e\) as $\forall x \in G: x^2 = e$
\(\ds \) \(=\) \(\ds b a\)

Thus $a b = b a$ and therefore $G$ is abelian.

$\blacksquare$


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