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 $\struct {S, \odot}$.
 * for every $a \in S$, the left and right regular representations $\lambda_a$ and $\rho_a$ on $S$ are automorphisms of $\struct {S, \odot}$.

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:

and:

Thus $\lambda_a$ and $\rho_a$ have been shown to be automorphisms of $\struct {S, \odot}$.

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 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:

and:

Thus $\struct {S, \odot}$ is a self-distributive quasigroup.