Condition for Algebraic Structure to be Self-Distributive Quasigroup
Theorem
Let $\struct {S, \odot}$ be an algebraic structure.
Then:
- $\struct {S, \odot}$ is a self-distributive quasigroup
- for every $a \in S$, the left and right regular representations $\lambda_a$ and $\rho_a$ on $S$ are automorphisms of $\struct {S, \odot}$.
Proof
Sufficient Condition
Let $\struct {S, \odot}$ be a self-distributive quasigroup.
By definition of quasigroup, we have that $\lambda_a$ and $\rho_a$ are both permutations on $S$.
It remains for the morphism property to be demonstrated.
Indeed:
\(\ds \forall x, y \in S: \, \) | \(\ds \map {\lambda_a} x \odot \map {\lambda_a} y\) | \(=\) | \(\ds \paren {a \odot x} \odot \paren {a \odot y}\) | Definition of Left Regular Representation | ||||||||||
\(\ds \) | \(=\) | \(\ds a \odot \paren {x \odot y}\) | Definition of Self-Distributive Structure | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {\lambda_a} {x \odot y}\) | Definition of Left Regular Representation |
and:
\(\ds \forall x, y \in S: \, \) | \(\ds \map {\rho_a} x \odot \map {\rho_a} y\) | \(=\) | \(\ds \paren {x \odot a} \odot \paren {y \odot a}\) | Definition of Right Regular Representation | ||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x \odot y} \odot a\) | Definition of Self-Distributive Structure | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {\rho_a} {x \odot y}\) | Definition of Right Regular Representation |
Thus $\lambda_a$ and $\rho_a$ have been shown to be automorphisms of $\struct {S, \odot}$.
$\Box$
Necessary Condition
Let $\odot$ be such that for every $a \in S$, the left and right regular representations $\lambda_a$ and $\rho_a$ on $S$ are automorphisms of $\struct {S, \odot}$.
We have a fortiori that:
- $\lambda_a$ and $\rho_a$ are both permutations on $S$
- $\lambda_a$ and $\rho_a$ are both homomorphisms of $S$.
Hence by definition $\struct {S, \odot}$ is a quasigroup.
It remains to demonstrate self-distributivity.
Indeed:
\(\ds \forall x, y, z \in S: \, \) | \(\ds x \odot \paren {y \odot z}\) | \(=\) | \(\ds \map {\lambda_x} {y \odot z}\) | Definition of Left Regular Representation | ||||||||||
\(\ds \) | \(=\) | \(\ds \map {\lambda_x} y \odot \map {\lambda_x} z\) | as $\lambda_x$ is a homomorphism on $S$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x \odot y} \odot \paren {x \odot z}\) | Definition of Left Regular Representation |
and:
\(\ds \forall x, y, z \in S: \, \) | \(\ds \paren {x \odot y} \odot z\) | \(=\) | \(\ds \map {\rho_z} {x \odot y}\) | Definition of Right Regular Representation | ||||||||||
\(\ds \) | \(=\) | \(\ds \map {\rho_z} x \odot \map {\rho_z} y\) | as $\rho_z$ is a homomorphism on $S$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x \odot z} \odot \paren {y \odot z}\) | Definition of Right Regular Representation |
Thus $\struct {S, \odot}$ is a self-distributive quasigroup.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.21 \ \text{(b)}$