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.
$\blacksquare$