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 $x, y \in G$:

Hence the result.