Abelian Group Induces Commutative B-Algebra
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Theorem
Let $\struct {G, \circ}$ be an abelian group whose identity element is $e$.
Let $*$ be the product inverse operation on $G$:
- $\forall a, b \in G: a * b = a \circ b^{-1}$
where $b^{-1}$ is the inverse element of $b$ under the operation $\circ$.
Then the algebraic structure $\struct {G, *}$ is a commutative $B$-algebra.
That is:
- $\forall a, b \in G: a * \paren {0 * b} = b * \paren {0 * a}$
Proof
From Group Induces $B$-Algebra, $\struct {G, *}$ is a $B$-Algebra.
As in the proof Group Induces $B$-Algebra, we let:
- $0 := e$
Now we demonstrate $0$-commutativity.
Let $x, y \in G$:
\(\ds x * \paren {0 * y}\) | \(=\) | \(\ds x \circ \paren {e \circ y^{-1} }^{-1}\) | Definition of $*$ and $0$ | |||||||||||
\(\ds \) | \(=\) | \(\ds x \circ \paren {y^{-1} }^{-1}\) | Definition of Identity Element | |||||||||||
\(\ds \) | \(=\) | \(\ds x \circ y\) | Inverse of Group Inverse | |||||||||||
\(\ds \) | \(=\) | \(\ds y \circ x\) | Definition of Abelian Group | |||||||||||
\(\ds \) | \(=\) | \(\ds y \circ \paren {x^{-1} }^{-1}\) | Inverse of Group Inverse | |||||||||||
\(\ds \) | \(=\) | \(\ds y \circ \paren {e \circ x^{-1} }^{-1}\) | Definition of Identity Element | |||||||||||
\(\ds \) | \(=\) | \(\ds y * \paren {0 * x}\) | Definition of $*$ and $0$ |
Hence the result.
$\blacksquare$