# Commutative B-Algebra Induces Abelian Group

## Theorem

Let $\left({X, \circ }\right)$ be a commutative $B$-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 an abelian 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

From B-Algebra Induces Group, 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 $*$.

It remains to show that $*$ is a commutative operation.

Let $x, y \in X$:

\(\displaystyle x * y\) | \(=\) | \(\displaystyle x \circ \left({0 \circ y}\right)\) | by definition of $*$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle y \circ \left({0 \circ x}\right)\) | by definition of commutative $B$-algebra | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle y * x\) | by definition of $*$ |

Hence the result.

$\blacksquare$