Self-Distributive Quasigroup with at least Two Elements has no Identity
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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 $16.21 \ \text{(d)}$