Self-Distributive Quasigroup with at least Two Elements has no Identity

Theorem

Let $\struct {S, \odot}$ be a self-distributive quasigroup.

Let $S$ have at least $2$ elements.

Then $\struct {S, \odot}$ has no identity element.

Proof

Aiming for a contradiction, suppose $S$ has an identity element $e$ and another element $a$ such that $a \ne e$.

Recall the definition of quasigroup:

$\forall a, b \in S: \exists ! x \in S: x \circ a = b$

That is:

We have:

\(\ds a \circ a\)

\(=\)

\(\ds a\)

Self-Distributive Quasigroup is Idempotent

\(\ds e \circ a\)

\(=\)

\(\ds a\)

Definition of Identity Element

That is, there are two $x \in S$ such that $x \circ a = a$.

This contradicts our assertion that $\struct {S, \odot}$ is a quasigroup.

Hence there can be no such element.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.21 \ \text{(d)}$