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 $\left ({X, *}\right)$ is a closed structure.

$\Box$

G1: Associativity

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

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

 $\displaystyle e * x$ $=$ $\displaystyle 0 \circ \left ({0 \circ x}\right)$ by definition of $*$ and $e$ $\displaystyle$ $=$ $\displaystyle x$ Identity: $0 \circ \left({0 \circ x} \right) = x$

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 $x$.

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

 $\displaystyle \left({0 \circ x}\right) \circ x$ $=$ $\displaystyle \left({0 \circ x}\right) \circ \left({0 \circ x}\right)$ 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:

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

$\Box$

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

$\blacksquare$