B-Algebra is Quasigroup

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.