Algebras determined by their globals 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.
 A: 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

*

*

* ${\mathcal U}(G,\ast)$ 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 $f(\{1_G\})=\{1_H\}$.


* 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 $f$.


*
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 $(H,\ast)$.



