# Abelian Group Induces Commutative B-Algebra

## Theorem

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

Let $*$ be the product inverse operation on $G$:

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

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.

$\blacksquare$