Commutative B-Algebra Induces Abelian Group

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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.

$\blacksquare$


Also see