Quasigroup is not necessarily B-Algebra

Theorem

Let $\struct {S, \circ}$ be a quasigroup.

Then $\struct {S, \circ}$ is not necessarily a $B$-algebra.

Proof

Proof by Counterexample

From Group is Quasigroup we take an arbitrary small group.

Consider the Cayley table of the group of order $3$:

$\begin{array}{c|cccccc}

 & 0 & 1 & 2 \\

\hline 0 & 0 & 1 & 2 \\ 1 & 1 & 2 & 0 \\ 2 & 2 & 0 & 1 \\ \end{array}$

By inspection we see that $B$-Algebra Axiom $(\text A 2)$ does not hold as $1 \circ 1 \ne 0$.

$\blacksquare$

Also see

• $B$-Algebra is Quasigroup