Abelian Group Induces Commutative B-Algebra

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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 $x, y \in G$:

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

Hence the result.

$\blacksquare$