Abelian Group Induces Commutative B-Algebra

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Theorem

Let $\struct {G, \circ}$ be an abelian group whose identity element is $e$.

Let $*$ be the binary operation on $G$ defined as:

$\forall a, b \in G: a * b = a \circ b^{-1}$

where $b^{-1}$ is the inverse element of $b$ under the operation $\circ$.


Then the algebraic structure $\struct {G, *}$ is a commutative $B$-algebra.

That is:

$\forall a, b \in G: a * \paren {0 * b} = b * \paren {0 * a}$


Proof

From Group Induces $B$-Algebra, $\struct {G, *}$ is a $B$-Algebra.

As in the proof Group Induces $B$-Algebra, we let:

$0 := e$


Now we demonstrate $0$-commutativity.

Let $x, y \in G$:

\(\ds x * \paren {0 * y}\) \(=\) \(\ds x \circ \paren {e \circ y^{-1} }^{-1}\) Definition of $*$ and $0$
\(\ds \) \(=\) \(\ds x \circ \paren {y^{-1} }^{-1}\) Definition of Identity Element
\(\ds \) \(=\) \(\ds x \circ y\) Inverse of Group Inverse
\(\ds \) \(=\) \(\ds y \circ x\) Definition of Abelian Group
\(\ds \) \(=\) \(\ds y \circ \paren {x^{-1} }^{-1}\) Inverse of Group Inverse
\(\ds \) \(=\) \(\ds y \circ \paren {e \circ x^{-1} }^{-1}\) Definition of Identity Element
\(\ds \) \(=\) \(\ds y * \paren {0 * x}\) Definition of $*$ and $0$

Hence the result.

$\blacksquare$