Group Induces B-Algebra

Theorem
Let $\struct {G, \circ}$ be a 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 $B$-algebra.

Proof
We have that:
 * $\forall x \in G: x * x = x \circ x^{-1} = e$

by definition of inverse element.

Let $0 := e$.

Then it can be seen that:
 * $(\text A 0): \quad \exists 0 \in G$
 * $(\text A 1): \quad \forall x \in G: x * x = 0$

Next note that:
 * $0^{-1} = e^{-1} = e = 0$

and so:
 * $\forall x \in G: x * 0 = x \circ 0^{-1} = x \circ e = x$

demonstrating that:
 * $(\text A 2): \quad \forall x \in G: x * 0 = x$

Finally, let $x, y, z \in G$:

This demonstrates:
 * $(\text A 3): \quad \forall x, y, z \in G: \paren {x * y} * z = x * \paren {z * \paren {0 * y} }$

All axioms of the $B$-algebra have been demonstrated to hold.

Hence the result.