Group Induced by B-Algebra Induced by Group

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.

$\blacksquare$

Also see

• $B$-Algebra Induced by Group Induced by $B$-Algebra