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.

Proof
Let $x, y, z \in X$:

We will show that $\left ({X, *}\right)$ satisfies each of the group axioms in turn:

G0: Closure

 * $x * y = x \circ \left ({0 \circ y}\right)$

As $\circ$ is a binary operation of $\left({X, \circ }\right)$.


 * $x \circ \left({0 \circ y}\right) \in X \implies x * y \in X$

G2: Identity
Let $e := 0$; we will show that it is an identity element of $\left ({X, *}\right)$.

G3: Invertibility
Let us prove that for all $x \in X$, $0 \circ x$ is an inverse to $x$.

Hence each $x \in X$ has a unique inverse under $*$.

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

Also see

 * Group Induces $B$-Algebra