B-Algebra Induced by Group Induced by B-Algebra
Jump to navigation
Jump to search
Theorem
Let $\struct {S, *}$ be a $B$-algebra.
Let $\struct {S, \circ}$ be the group described on $B$-Algebra Induces Group.
Let $\struct {S, *'}$ be the $B$-algebra described on Group Induces $B$-Algebra.
Then $\struct {S, *'} = \struct {S, *}$.
Proof
Let $a, b \in S$.
It is required to show that:
- $a *' b = a * b$
To achieve this, recall that $*'$ is defined on Group Induces $B$-Algebra to satisfy:
- $a *' b = a \circ b^{-1}$
which, by the definition of $\circ$ on $B$-Algebra Induces Group comes down to:
\(\ds a *' b\) | \(=\) | \(\ds a * \paren {0 * b^{-1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a * \paren {0 * \paren {0 * b} }\) | $B$-Algebra Induces Group | |||||||||||
\(\ds \) | \(=\) | \(\ds a * b\) | Identity: $0 * \paren {0 * x} = x$ |
Hence the result.
$\blacksquare$