Algebras determined by their globals

Question

If $A= (A, f_1, f_2, ...f_n)$ is an algebra, then its global (sometimes referred to as complex algebras) $\mathcal{U}(A)$ is defined on the power set $\wp(A)$ in the usual way.

It is known that $\mathcal{U}(A) \cong \mathcal{U}(B)$ implies $A\cong B$ for the class of finite mono-unary algebras (Drapal 1985). For which other classes of algebras (esp groupoids) is the result true?

A. Drápal, Globals of unary algebras, Czechoslovak Math. J. 35(110) (1985), 52–58.

Answer 1

For which other classes of algebras (esp groupoids) is the result true?

One class of groupoids with this property is the class of groups considered as groupoids. That is, if ${\mathcal U}(G,\ast)\cong{\mathcal U}(H,\ast)$, then $(G,\ast)\cong (H,\ast)$.

To see this, assume that $f\colon {\mathcal U}(G,\ast)\to{\mathcal U}(H,\ast)$ is an isomorphism. Then

1. ${\mathcal U}(G,\ast)$ is a semigroup with unit element $\{1_G\}$. Since the unit element of a semigroup is a definable element, any isomorphism $f\colon {\mathcal U}(G,\ast)\to{\mathcal U}(H,\ast)$ must satisfy $f(\{1_G\})=\{1_H\}$.
2. An element $x\in {\mathcal U}(G,\ast)$ is a singleton subset of $G$ iff $x$ an invertible element (=unit) of the semigroup ${\mathcal U}(G,\ast)$. The set of units of ${\mathcal U}(G,\ast)$ will be mapped bijectively onto the set of units of ${\mathcal U}(H,\ast)$ by the isomorphism $f$.
3. The restriction of $f$ to the set of units/singletons determines an isomorphism from $(G,\ast)$ to $(H,\ast)$. That is, the function $f'\colon G\to H$ defined so that $f(\{g\}) = \{f'(g)\}$ is an isomorphism from $(G,\ast)$ to $(H,\ast)$.