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:

Group Axiom $\text G 0$: Closure

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.

$\Box$

Group Axiom $\text G 1$: Associativity

 $\ds \paren {x * y} * z$ $=$ $\ds \paren {x \circ \paren {0 \circ y} } \circ \paren {0 \circ z}$ Definition of $*$ $\ds$ $=$ $\ds x \circ \paren {\paren {0 \circ z} \circ \paren {0 \circ \paren {0 \circ y} } }$ Axiom $(A3)$ for $B$-algebras $\ds$ $=$ $\ds x \circ \paren {\paren {0 \circ z} \circ y}$ Identity: $0 \circ \paren {0 \circ x} = x$ $\ds$ $=$ $\ds x \circ \paren {0 \circ \paren {y \circ \paren {0 \circ z} } }$ Axiom $(A3)$ for $B$-algebras $\ds$ $=$ $\ds x * \paren {y * z}$ Definition of $*$

Thus it is seen that $*$ is associative.

$\Box$

Group Axiom $\text G 2$: Existence of Identity Element

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

 $\ds x * e$ $=$ $\ds x \circ \paren {0 \circ 0}$ Definition of $*$ and $e$ $\ds$ $=$ $\ds x \circ 0$ Axiom $(A1)$ for $B$-algebras $\ds$ $=$ $\ds x$ Axiom $(A2)$ for $B$-algebras

 $\ds e * x$ $=$ $\ds 0 \circ \paren {0 \circ x}$ Definition of $*$ and $e$ $\ds$ $=$ $\ds x$ Identity: $0 \circ \paren {0 \circ x} = x$

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

$\Box$

Group Axiom $\text G 3$: Existence of Inverse Element

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

 $\ds x * \paren {0 \circ x}$ $=$ $\ds x \circ \paren {0 \circ \paren {0 \circ x} }$ Definition of $*$ $\ds$ $=$ $\ds x \circ x$ Identity: $0 \circ \paren {0 \circ x} = x$ $\ds$ $=$ $\ds 0$ Axiom $(A1)$ for $B$-algebras

 $\ds \paren {0 \circ x} * x$ $=$ $\ds \paren {0 \circ x} \circ \paren {0 \circ x}$ Definition of $*$ $\ds$ $=$ $\ds 0$ Axiom $(A1)$ for $B$-algebras

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

This inverse element is $0 \circ x$.

$\Box$

It follows that:

 $\ds a * b^{-1}$ $=$ $\ds a \circ \paren {0 \circ b^{-1} }$ Definition of $*$ $\ds$ $=$ $\ds a \circ \paren {0 \circ \paren {0 \circ b} }$ Definition of $b^{-1}$ $\ds$ $=$ $\ds a \circ b$ Identity: $0 \circ \paren {0 \circ x} = x$

$\Box$

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

$\blacksquare$