# 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:

### G0: 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.

$\Box$

### G1: Associativity

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

Thus it is seen that $*$ is associative.

$\Box$

### G2: Identity

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

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

 $\displaystyle e * x$ $=$ $\displaystyle 0 \circ \paren {0 \circ x}$ by definition of $*$ and $e$ $\displaystyle$ $=$ $\displaystyle x$ Identity: $0 \circ \paren {0 \circ x} = x$

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

$\Box$

### G3: Invertibility

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

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

 $\displaystyle \paren {0 \circ x} \circ x$ $=$ $\displaystyle \paren {0 \circ x} \circ \paren {0 \circ x}$ by definition of $*$ $\displaystyle$ $=$ $\displaystyle 0$ 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:

 $\displaystyle a * b^{-1}$ $=$ $\displaystyle a \circ \paren {0 \circ b^{-1} }$ by definition of $*$ $\displaystyle$ $=$ $\displaystyle a \circ \paren {0 \circ \paren {0 \circ b} }$ by definition of $b^{-1}$ $\displaystyle$ $=$ $\displaystyle a \circ b$ Identity: $0 \circ \paren {0 \circ x} = x$

$\Box$

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

$\blacksquare$