Group Induces B-Algebra

Theorem

Let $\left({G, \circ}\right)$ be a 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 $B$-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:

$(A0): \quad \exists 0 \in G$

$(A1): \quad \forall x \in G: x * x = 0$

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:

$(A2): \quad \forall x \in G: x * 0 = x$

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

\(\displaystyle \left({x * y}\right) * z\)

\(=\)

\(\displaystyle \left({x \circ y^{-1} }\right) \circ z^{-1}\)

by definition of $*$

\(\displaystyle \)

\(=\)

\(\displaystyle x \circ \left({y^{-1} \circ z^{-1} }\right)\)

$G$ is a group, so $\circ$ is associative

\(\displaystyle \)

\(=\)

\(\displaystyle x \circ \left({z \circ y}\right)^{-1}\)

Inverse of Group Product

\(\displaystyle \)

\(=\)

\(\displaystyle x \circ \left({z \circ \left({y^{-1} }\right)^{-1} }\right)^{-1}\)

Inverse of Group Inverse

\(\displaystyle \)

\(=\)

\(\displaystyle x * \left({z * y^{-1} }\right)\)

by definition of $*$

\(\displaystyle \)

\(=\)

\(\displaystyle x * \left({z * \left({0 \circ y^{-1} }\right) }\right)\)

as $0 = e$

\(\displaystyle \)

\(=\)

\(\displaystyle x * \left({z * \left({0 * y}\right) }\right)\)

by definition of $*$

This demonstrates:

$(A3): \quad \forall x, y, z \in G: \left({x * y}\right) * z = x * \left({z * \left({0 * y}\right)}\right)$

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

Hence the result.

$\blacksquare$