B-Algebra Induced by Group Induced by B-Algebra

Theorem
Let $\left({S, *}\right)$ be a $B$-algebra.

Let $\left({S, \circ}\right)$ be the group described on $B$-Algebra Induces Group.

Let $\left({S, *'}\right)$ be the $B$-algebra described on Group Induces $B$-Algebra.

Then $\left({S, *'}\right) = \left({S, *}\right)$.

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:

Hence the result.

Also see

 * Group Induced by $B$-Algebra Induced by Group