# Can we equip the power set $P$ of any set $S$ with a binary operation such that $P$ becomes a group (with some restrictions)?

Settings. Let $$S$$ be a (not necessarily finite) set, and $$P$$ the power set of $$S$$ (i.e., $$P$$ is the set of all subsets of $$S$$). A binary operator $$*:P\times P\to P$$ is said to be elementary if it can be given in terms of the standard set operations: the union operator $$\cup$$, the intersection operator $$\cap$$, the set difference operator $$\setminus$$, the symmetric difference operator $$\triangle$$, and the complement operator $$(\_)^\complement$$.

Some Examples. This operator $$\star$$ is considered an elementary binary operator: $$A \star B:= \big((M\setminus A)\cup (B\cap N)\big)^{\complement}\triangle \Big(A\cup B^\complement\Big)\text{ for all }A,B\subseteq S\,,$$ where $$M$$ and $$N$$ are fixed subsets of $$S$$. On the other hand, if $$|S|=2$$, then this operator $$\bullet$$ is not an elementary binary operator: $$A\bullet B:=\left\{ \begin{array}{ll} S&\text{if }A\subseteq B\,,\\ \emptyset&\text{otherwise}\,, \end{array}\right.$$ where $$A,B\subseteq S$$ (a proof can be done by imitating this answer).

Question. For which groups $$G$$ of order $$2^{|S|}$$ does there exist an elementary binary operator $$*:P\times P\to P$$ such that $$(P,*)$$ is a group isomorphic to $$G$$? If the case where $$S$$ is an infinite set is too troublesome, then an answer in the case where $$S$$ is a finite set is very welcome.

Let $$n:=|S|$$. Write $$Z_k$$ for the cyclic group of order $$k$$.

Trivial Answer. When $$G\cong Z_2^n$$, then the binary operator $$\triangle$$ does the work. My conjecture is that there are no other groups.

Known Result. When $$|S|=2$$ and $$G\cong Z_4$$, then there does not exist such an elementary binary operator.

Let us identify $$P$$ with $$\{0,1\}^S$$ in the obvious way. Then an elementary operation is just one which is given by applying some binary operation $$\{0,1\}\times\{0,1\}\to\{0,1\}$$ coordinatewise. Indeed, it is clear that every elementary operation must have this form (since all the basic Boolean operations have this form), and conversely it is a simple exercise in Boolean algebra to build every binary operation on $$\{0,1\}$$ out of the basic Boolean operations.
So, $$P$$ must just be a product of copies of some binary operation on $$\{0,1\}$$. As long as $$S$$ is nonempty, this means $$P$$ will be a group iff the corresponding operation on $$\{0,1\}$$ makes it a group (and the case where $$S$$ is empty is trivial). But there is only one group operation on $$\{0,1\}$$ up to isomorphism, so $$P$$ can only be isomorphic to $$\mathbb{Z}_2^S$$. (In fact, there are only two possible group operations at all: the usual symmetric difference operation and symmetric difference conjugated by swapping $$0$$ and $$1$$, which in terms of sets is just the complement of the symmetric difference.)