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 $x, y \in G$:
\(\ds x * \left({0 * y }\right)\) | \(=\) | \(\ds x \circ \left({e \circ y^{-1} }\right)^{-1}\) | by definition of $*$ and $0$ | |||||||||||
\(\ds \) | \(=\) | \(\ds x \circ (y^{-1})^{-1}\) | by definition of identity element | |||||||||||
\(\ds \) | \(=\) | \(\ds x \circ y\) | Inverse of Group Inverse | |||||||||||
\(\ds \) | \(=\) | \(\ds y \circ x\) | by definition of abelian group | |||||||||||
\(\ds \) | \(=\) | \(\ds y \circ \left({x^{-1} }\right)^{-1}\) | Inverse of Group Inverse | |||||||||||
\(\ds \) | \(=\) | \(\ds y \circ \left({e \circ x^{-1} } \right)^{-1}\) | by definition of identity element | |||||||||||
\(\ds \) | \(=\) | \(\ds y * \left({0 * x}\right)\) | by definition of $*$ and $0$ |
Hence the result.
$\blacksquare$