# Group Induced by B-Algebra Induced by Group

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## 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:

\(\ds a \circ' b\) | \(=\) | \(\ds a \circ \left({e \circ b^{-1} }\right)^{-1}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds a \circ \left({b \circ e}\right)\) | Inverse of Group Product | |||||||||||

\(\ds \) | \(=\) | \(\ds a \circ b\) |

Hence the result.

$\blacksquare$