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 $x, y \in G$:
| \(\displaystyle x * \left({0 * y }\right)\) | \(=\) | \(\displaystyle x \circ \left({e \circ y^{-1} }\right)^{-1}\) | $\quad$ 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.
$\blacksquare$
