Group Induces B-Algebra
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Theorem
Let $\struct {G, \circ}$ be a 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 $B$-algebra.
Proof
First note that by Group Axiom $\text G 0$: Closure:
- $\forall a, b \in G: a * b \in G$
and so $B$-Algebra Axiom $(\text {AC})$ holds.
We have that:
- $\forall x \in G: x * x = x \circ x^{-1} = e$
by definition of inverse element.
Let $0 := e$.
Then:
- $B$-Algebra Axiom $(\text A 0)$: $\quad \exists 0 \in G$
- $B$-Algebra Axiom $(\text A 1)$: $\quad \forall x \in G: x * x = 0$
Note that:
| \(\ds \exists 0 \in G: \, \) | \(\ds 0^{-1}\) | \(=\) | \(\ds e^{-1}\) | Definition of $0$ | ||||||||||
| \(\ds \) | \(=\) | \(\ds e\) | Inverse of Identity Element is Itself | |||||||||||
| \(\ds \) | \(=\) | \(\ds 0\) | Definition of $0$ |
and so:
| \(\ds \forall x \in G: \, \) | \(\ds x * 0\) | \(=\) | \(\ds x \circ 0^{-1}\) | Definition of $*$ | ||||||||||
| \(\ds \) | \(=\) | \(\ds x \circ e\) | Definition of $0$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds x\) | Definition of Identity Element |
demonstrating:
- $B$-Algebra Axiom $(\text A 2)$: $\quad \forall x \in G: x * 0 = x$
Finally, let $x, y, z \in G$:
| \(\ds \paren {x * y} * z\) | \(=\) | \(\ds \paren {x \circ y^{-1} } \circ z^{-1}\) | Definition of $*$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds x \circ \paren {y^{-1} \circ z^{-1} }\) | Group Axiom $\text G 1$: Associativity | |||||||||||
| \(\ds \) | \(=\) | \(\ds x \circ \paren {z \circ y}^{-1}\) | Inverse of Group Product | |||||||||||
| \(\ds \) | \(=\) | \(\ds x \circ \paren {z \circ \paren {y^{-1} }^{-1} }^{-1}\) | Inverse of Group Inverse | |||||||||||
| \(\ds \) | \(=\) | \(\ds x * \paren {z * y^{-1} }\) | Definition of $*$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds x * \paren {z * \paren {0 \circ y^{-1} } }\) | as $0 = e$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds x * \paren {z * \paren {0 * y} }\) | Definition of $*$ |
This demonstrates:
- $B$-Algebra Axiom $(\text A 3)$: $\quad \forall x, y, z \in G: \paren {x * y} * z = x * \paren {z * \paren {0 * y} }$
All axioms of the $B$-algebra have been demonstrated to hold.
Hence the result.
$\blacksquare$