# 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

First note that by Group Axiom $\text G 0$: Closure:

$\forall a, b \in G: a * b \in G$

and so $B$-Algebra Axiom $(\text {AC})$ holds.

We have that:

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

by definition of inverse element.

Let $0 := e$.

Then:

$B$-Algebra Axiom $(\text A 0)$: $\quad \exists 0 \in G$
$B$-Algebra Axiom $(\text A 1)$: $\quad \forall x \in G: x * x = 0$

Note that:

 $\ds \exists 0 \in G: \,$ $\ds 0^{-1}$ $=$ $\ds e^{-1}$ Definition of $0$ $\ds$ $=$ $\ds e$ Inverse of Identity Element is Itself $\ds$ $=$ $\ds 0$ Definition of $0$

and so:

 $\ds \forall x \in G: \,$ $\ds x * 0$ $=$ $\ds x \circ 0^{-1}$ Definition of $*$ $\ds$ $=$ $\ds x \circ e$ Definition of $0$ $\ds$ $=$ $\ds x$ Definition of Identity Element

demonstrating:

$B$-Algebra Axiom $(\text A 2)$: $\quad \forall x \in G: x * 0 = x$

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

 $\ds \paren {x * y} * z$ $=$ $\ds \paren {x \circ y^{-1} } \circ z^{-1}$ Definition of $*$ $\ds$ $=$ $\ds x \circ \paren {y^{-1} \circ z^{-1} }$ Group Axiom $\text G 1$: Associativity $\ds$ $=$ $\ds x \circ \paren {z \circ y}^{-1}$ Inverse of Group Product $\ds$ $=$ $\ds x \circ \paren {z \circ \paren {y^{-1} }^{-1} }^{-1}$ Inverse of Group Inverse $\ds$ $=$ $\ds x * \paren {z * y^{-1} }$ Definition of $*$ $\ds$ $=$ $\ds x * \paren {z * \paren {0 \circ y^{-1} } }$ as $0 = e$ $\ds$ $=$ $\ds x * \paren {z * \paren {0 * y} }$ Definition of $*$

This demonstrates:

$B$-Algebra Axiom $(\text A 3)$: $\quad \forall x, y, z \in G: \paren {x * y} * z = x * \paren {z * \paren {0 * y} }$

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

Hence the result.

$\blacksquare$