# B-Algebra Induced by Group Induced by B-Algebra

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## 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:

\(\ds a *' b\) | \(=\) | \(\ds a * \left({0 * b^{-1} }\right)\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds a * \left({0 * \left({0 * b}\right)}\right)\) | $B$-Algebra Induces Group | |||||||||||

\(\ds \) | \(=\) | \(\ds a * b\) | Identity: $0 * \left({0 * x}\right) = x$ |

Hence the result.

$\blacksquare$