Commutative B-Algebra Induces Abelian Group
Jump to navigation
Jump to search
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 $x, y \in X$:
\(\ds x * y\) | \(=\) | \(\ds x \circ \left({0 \circ y}\right)\) | by definition of $*$ | |||||||||||
\(\ds \) | \(=\) | \(\ds y \circ \left({0 \circ x}\right)\) | by definition of commutative $B$-algebra | |||||||||||
\(\ds \) | \(=\) | \(\ds y * x\) | by definition of $*$ |
Hence the result.
$\blacksquare$