Group Induced by B-Algebra Induced by Group
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Theorem
Let $\struct {S, \circ}$ be a group.
Let $\struct {S, *}$ be the $B$-algebra described on Group Induces $B$-Algebra.
Let $\struct {S, \circ'}$ be the group described on $B$-Algebra Induces Group.
Then $\struct {S, \circ'} = \struct {S, \circ}$.
Proof
Let $a, b \in S$.
It is required to show that:
- $a \circ' b = a \circ b$
To achieve this, recall that $\circ'$ is defined on $B$-Algebra Induces Group to satisfy:
- $a \circ' b = a * \paren {e * b}$
which, by the definition of $*$ on Group Induces $B$-Algebra comes down to:
\(\ds a \circ' b\) | \(=\) | \(\ds a \circ \paren {e \circ b^{-1} }^{-1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a \circ \paren {b^{-1} }^{-1}\) | Group Axiom $\text G 2$: Existence of Identity Element | |||||||||||
\(\ds \) | \(=\) | \(\ds a \circ b\) | Inverse of Group Inverse |
Hence the result.
$\blacksquare$