If $G$ is a group, then $M(G)=2^G$ is has a monoid structure when we define $AB$ to be $\{ab|a\in A,b\in B\}$ and $1_{M(G)}=\{1\}$. How much of the structure of $G$ can be recovered by studying the structure of $M(G)?$ Is there any known example of a sensibly large class $\mathscr{C}$ of groups, for which the following implication holds?
For any $G,H\in \mathscr C,$ if $M(G)\cong M(H),$ then $G\cong H.$
Could you please give me any references concerning the monoid of subsets of a group?
Edit: I'll write here what I came up with after I asked the quesiton. Please comment on it as well.
I've found a funny structure that I think is isomorphic to $M(G).$ Let $B=(\{0,1\},∨,∧).$ It's a semiring in which infinite sums (and products, actually) are well-defined. If I'm not mistaken, we can define the "group semiring" $B[G]_\infty$ just like we define group rings, but without the restriction of finite supports. It looks like the multiplicative structure of $B[G]_\infty$ is isomorphic to $M(G).$ The additive structure of $B[G]_\infty$ seems to be isomorphic to $(M(G),\cup).$
If we do keep the restriction of finite supports and define $B[G]$ as we would a group ring, I believe it's equivalent to changing $2^G$ to the family of all finite subsets of $G$ in the definition of $M(G).$
I haven't written the proofs of these things down, but I will if anyone thinks it's a good idea.
Edit: I don't know how close I am, but for what it's worth. OK, so $M(G)$ has a structure of an additively idempotent semiring. Indeed, let's take $(M(G),\cup,\cdot,\emptyset,\{1\}).$ The additive structure is obviously that of a commutative idempotent monoid. The multiplicative structure is that of a monoid. The distributive laws hold because:
$\begin{eqnarray}\left(\forall A,B,C\in M(G)\right) \;\;\; A(B\cup C) &=& \{ax\,|\,a\in A\wedge x\in B\cup C\}=\{ax\,|\, a\in A\wedge (x\in B\vee x\in C)\}\\ &=& \{ax \,|\,(a\in A\wedge x\in B)\vee (a\in A\wedge x\in C)\} \\ &=& \{ax\,|\,a\in A \wedge x\in B\}\cup\{ax\,|\,a\in A \wedge x\in C\}\\ &=& AB\cup AC,\end{eqnarray}$
and analogously
$\left(\forall A,B,C\in M(G)\right) \;\;\; (B\cup C)A=BA\cup CA.$
Now, in this semiring there is the subset $G'=\left\{\{g\}\,|\,g\in G\right\}.$ The multiplicative structure of this subset is isomorphic to $G.$ Suppose we have the semiring $(M,+,\cdot,0,1)$ and we know that $M\cong M(G)$ for some group $G.$ (The isomorphism is between semirings, not monoids.) Then finding out the structure of $G$ is equivalent to recognizing $G'$ inside $M.$ But this is possible when we have the additive structure. $G'$ is the set of all elements $g$ such that the equation $g+x=g$ has exactly one solution, that is $0.$
This, if I'm not mistaken means that the semiring structure of $M(G)$ gives the structure of $G$ unanmbiguously. So perhaps, having the monoid $M(G),$ we could try proving that there is a unique additive operation that makes this monoid into an (additively idempotent) semiring?
Edit: This is to announce that I have asked a follow-up question on MO. It is whether the class of inverse semigroups is globally determined (as defined in my answer below). Inverse semigroups can be seen as generalized groups and I'd like to know if they retain this particular property of groups.
