B-Algebra Induced by Group Induced by B-Algebra

From ProofWiki
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.

$\blacksquare$


Also see