# Abelian Group Induces Commutative B-Algebra

## Theorem

Let $\left({G, \circ}\right)$ be an abelian group whose identity element is $e$.

Let $*$ be the binary operation on $G$ defined as:

$\forall a, b \in G: a * b = a \circ b^{-1}$

where $b^{-1}$ is the inverse element of $b$ under the operation $\circ$.

Then the algebraic structure $\left({G, *}\right)$ is a commutative $B$-algebra.

That is:

$\forall a, b \in G: a * \left({0 * b}\right) = b * \left({0 * a}\right)$

## Proof

From Group Induces $B$-Algebra, $\left({G, *}\right)$ is a $B$-algebra.

As in the proof Group Induces $B$-Algebra, we let:

$0 := e$

Now we demonstrate $0$-commutativity.

Let $x, y \in G$:

 $\displaystyle x * \left({0 * y }\right)$ $=$ $\displaystyle x \circ \left({e \circ y^{-1} }\right)^{-1}$ by definition of $*$ and $0$ $\displaystyle$ $=$ $\displaystyle x \circ (y^{-1})^{-1}$ by definition of identity element $\displaystyle$ $=$ $\displaystyle x \circ y$ Inverse of Group Inverse $\displaystyle$ $=$ $\displaystyle y \circ x$ by definition of abelian group $\displaystyle$ $=$ $\displaystyle y \circ \left({x^{-1} }\right)^{-1}$ Inverse of Group Inverse $\displaystyle$ $=$ $\displaystyle y \circ \left({e \circ x^{-1} } \right)^{-1}$ by definition of identity element $\displaystyle$ $=$ $\displaystyle y * \left({0 * x}\right)$ by definition of $*$ and $0$

Hence the result.

$\blacksquare$