B-Algebra Induces Group

From ProofWiki
Jump to navigation Jump to search

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.

$\blacksquare$


Also see