# Group Induced by B-Algebra Induced by Group

## Theorem

Let $\left({S, \circ}\right)$ be a group.

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

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

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

## 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:

$a \circ' b = a * \left({e * b}\right)$

which, by the definition of $*$ on Group Induces $B$-Algebra comes down to:

 $\displaystyle a \circ' b$ $=$ $\displaystyle a \circ \left({e \circ b^{-1} }\right)^{-1}$ $\displaystyle$ $=$ $\displaystyle a \circ \left({b \circ e}\right)$ Inverse of Group Product $\displaystyle$ $=$ $\displaystyle a \circ b$

Hence the result.

$\blacksquare$