Group Induces B-Algebra

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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 $x, y, z \in G$:

\(\ds \paren {x * y} * z\) \(=\) \(\ds \paren {x \circ y^{-1} } \circ z^{-1}\) Definition of $*$
\(\ds \) \(=\) \(\ds x \circ \paren {y^{-1} \circ z^{-1} }\) Group Axiom $\text G 1$: Associativity, as $G$ is a group
\(\ds \) \(=\) \(\ds x \circ \paren {z \circ y}^{-1}\) 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:

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


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

Hence the result.

$\blacksquare$