B-Algebra Induces Group

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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.

$\blacksquare$


Also see