B-Algebra Induces Group

Theorem
Let $\left({X, \circ }\right)$ be a $B$-algebra.

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


 * $\forall a, b \in X: a * b := a \circ \left({0 \circ b}\right)$

Then the algebraic structure $\left ({X, *}\right)$ 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 $\left ({X, *}\right)$ satisfies each of the group axioms in turn:

G0: Closure
By definition of $*$, we have:


 * $x * y = x \circ \left ({0 \circ y}\right)$

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


 * $x \circ \left({0 \circ y}\right) \in X$

Whence $x * y \in X$, and so $*$ is closed.

G1: Associativity
Thus it is seen that $*$ is associative.

G2: Identity
Let $e := 0$; we will show that it is an identity element of $\left ({X, *}\right)$.

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

G3: Invertibility
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