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
We have that:
- $\forall x \in G: x * x = x \circ x^{-1} = e$
by definition of inverse element.
Let $0 := e$.
Then it can be seen that:
- $(\text A 0): \quad \exists 0 \in G$
- $(\text A 1): \quad \forall x \in G: x * x = 0$
Next note that:
- $0^{-1} = e^{-1} = e = 0$
and so:
- $\forall x \in G: x * 0 = x \circ 0^{-1} = x \circ e = x$
demonstrating that:
- $(\text A 2): \quad \forall x \in G: x * 0 = x$
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, as $G$ is a group | |||||||||||
\(\ds \) | \(=\) | \(\ds x \circ \paren {z \circ y}^{-1}\) | 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:
- $(\text A 3): \quad \forall x, y, z \in G: \paren {x * y} * z = x * \paren {z * \paren {0 * y} }$
All axioms of the $B$-algebra have been demonstrated to hold.
Hence the result.
$\blacksquare$