B-Algebra Induces Group

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

Let $*$ be the binary operation on $X$ defined as:


 * $\forall a, b \in X: a * b := a \circ \paren {0 \circ b}$

Then the algebraic structure $\struct {X, *}$ is a group such that:


 * $\forall x \in X: 0 \circ x$ is the inverse element of $x$ under $*$.

That is:
 * $\forall a, b \in X: a * b^{-1} := a \circ b$

Proof
Let $x, y, z \in X$:

We will show that $\struct {X, *}$ satisfies each of the group axioms in turn:

By definition of $*$, we have:


 * $x * y = x \circ \paren {0 \circ y}$

By Axiom $(AC)$ for $B$-algebras:


 * $x \circ \paren {0 \circ y} \in X$

Whence $x * y \in X$, and so $\paren {X, *}$ is a closed structure.

Thus it is seen that $*$ is associative.

Let $e := 0$; we will show that it is an identity element of $\struct {X, *}$.

Hence $0$ is an identity for $*$.

Let us prove that for all $x \in X$, $0 \circ x$ is an inverse element to $x$.

That is, each $x \in X$ has a unique inverse element $x^{-1}$ under $*$.

This inverse element is $0 \circ x$.

It follows that:

All the axioms have been shown to hold and the result follows.

Also see

 * Group Induces $B$-Algebra