B-Algebra is Quasigroup
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Theorem
Let $\struct {X, \circ}$ be a $B$-algebra.
Then $\struct {X, \circ}$ is a quasigroup.
Proof
By the definition of a quasigroup it must be shown that $\forall x \in X$ the left and right regular representations $\lambda_x$ and $\rho_x$ are permutations on $X$.
As $\struct {X, \circ}$ is a magma:
- $\forall x \in X$ the codomain of $\lambda_x$ and $\rho_x$ is $X$.
Hence it is sufficient to prove that $\lambda_x$ and $\rho_x$ are bijections.
We have that:
- $B$-Algebras are left cancellable
- $B$-Algebras are right cancellable
- Cancellable elements have injective regular representations.
Therefore:
- $\forall x \in X$: $\lambda_x$ and $\rho_x$ are injective mappings.
We also have that regular representations of $B$-algebras are surjective.
Therefore:
- $\forall x \in X$: $\lambda_x$ and $\rho_x$ are both injective and surjective mappings.
Hence by definition are $\lambda_x$ and $\rho_x$ bijections for all $x \in X$.
$\blacksquare$
Also see
- Quasigroup is not necessarily B-Algebra where it is demonstrated that the converse does not necessarily hold.