# Group Induced by B-Algebra Induced by Group

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## 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.

$\blacksquare$