Abelian Group Induces Commutative B-Algebra

Theorem
Let $\left({G, \circ}\right)$ 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 $\left({G, *}\right)$ is a commutative $B$-algebra.

That is:
 * $\forall a, b \in G: a * \left({0 * b}\right) = b * \left({0 * a}\right)$

Proof
From Group Induces $B$-Algebra, $\left({G, *}\right)$ 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.