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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.

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Your proposed identity element is not an identity element. Consider what happens when you multiply by $\emptyset$. Also, $G$ is otherwise faithfully embedded into $2^G$ as the singletons. –  Zhen Lin Jan 21 '12 at 17:31
Isn't $\{1\}\cdot\emptyset=\emptyset?$ –  user23211 Jan 21 '12 at 17:35
Oops, sorry, you're right. I must have been thinking of something else... –  Zhen Lin Jan 21 '12 at 17:38
You're right that $G$ is faithfully embedded into $2^G$. I didn't think of that... But it doesn't mean that the implication from my question holds for any class $\mathscr C$ of groups, right? –  user23211 Jan 21 '12 at 17:44
There are definitely group theoretic things you can tell from $M(G)$. For example, abelianess of $M(G)$ is equivalent to abelianess for $G$. As to your question, the class $\mathscr{C}$ of finite groups of squarefree order for which no prime divisor is equivalent to 1 modulo another has the desired property (obviously) since they are determined by their order which is recoverable from $M$. Of course, similarly, the class $\mathscr{C}$ of cyclic groups has this property. Have you tried computing examples? How does $M(\mathbb{Z})$ compare to $M(\mathbb{Q})$? How'bout their Grothendieck groups? –  Alex Youcis Jan 21 '12 at 18:29
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3 Answers

up vote 10 down vote accepted

I believe it is always true that $M(G)\cong M(H)$ implies $G\cong H$. Because let $\phi:\ M(G)\rightarrow M(H)$ be a monoid isomorphism. Note that the only elements $x\in M(G)$ such that there exists $y\in M(G)$ with $xy=1$ are the one-element subsets of $G$. That is, the invertible elements in $M(G)$ are the one-element subsets of $G$; and the monoid product in $M(G)$, restricted to these one-element subsets of $G$, is just the group product of $G$. Thus $\phi$ restricted to these invertible elements induces an isomorphism between $G$ and $H$.

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It turns out that there has been a large amount of research on the semigroups of all (finite/non-empty) subsets of a semigroup. For some reason the authors seem to prefer to consider only the non-empty subsets and I will stick to this convention in this post.

Since the question has been favorited by five people who aren't myself, I understand that the results obtained may be of interests to the community. I will post some things that I have found, especially terminology and references so anyone can do more searching by themself.

The semigroup of all non-empty subsets of a semigroup $S$ is called the power semigroup of $S,$ or the global of $S,$ and denoted by $\mathcal{P}(S).$ Semigroups $S_1,S_2$ are said to be globally isomorphic iff $\mathcal P(S_1)\cong \mathcal P (S_2).$ A class of semigroups $\mathscr C$ is said to be globally determined iff for any semigroups $S_1,S_2$ from $\mathscr C,$ if $S_1,S_2$ are globally isomorphic, then they are isomorphic.

The fact that the class of groups is globally determined was first noted by J. Shafer in /5/ and the reasoning was of course like Steve D's. The question whether the class of all semigroups is globally determined was posed in the 1960s and answered in the negative by E.M. Mogiljanskaja in /4/. The question has a positive answer for the class of full transformation semigroups (even more, any semigroup globally isomorphic to a full transformation semigroup must be isomorphic to it). This was proved by Vazenin in /6/. M. Gould and J. A. Iskra proved in /1/ that the class of finite simple semigroups is globally determined. It was later extended by Tamura to the classes of completely simple and completely 0-simple semigroups (please see the last link to find the definitions of these terms).

Gould, Iskra and C. Tsinakis later proved in /2/ another result, which needs a definition.

Defintion. An element $x$ of a semigroup $S$ is called irreducible iff for any $y,z\in S$ such that $x=yz,$ we have $x\in\{y,z\}.$

Note that $x$ is irreducible iff $S\setminus\{x\}$ is a subsemigroup of $S.$ Tamura and Shafer earlier termed such elements "prime". The theorem says what follows.

Theorem. The class of all completely regular periodic monoids in which the identity element is irreducible is globally determined.

Finally, Y. Kobayashi proved in /3/ that the class of all semilattices (treated as algebraic structures) is globally determined.


/1/ M. Gould, J.A. Iskra, Globally determined classes of semigroups, Semigroup Forum Vol. 28 (1984), 1-11.

/2/ M. Gould, J.A. Iskra, C. Tsinakis, Globals of completely regular periodic groups, Semigroup Forum Vol 29 (1984), 365-374.

/3/ Y. Kobayashi, Semilattices are globally determined, Semigroup Forum Vol. 29 (1984), 217-222.

/4/ E. M. Mogilianskaja, The solution to a problem of Tamura, Sbornik Naučnyh Trudov Leningrad. Gos. Ped. Inst.,"Modern Analysis and Geometry", (1972), 148-151.

/5/ J. Shafer, Note on Power Semigroups, Math. Japan. 12 (1967), 32.

/6/ Ju. M. Vazenin, On the global oversemigroup of a symmetric semigroup, Mat. Zap. Ural. Gos. Univ 9 (1874), 3-10.

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A lot is known on the monoid $\mathcal{P}(G)$ of subsets of a finite group $G$ (also called a power group). For instance, two subgroups of $G$ are conjugate if and only if they are $\mathcal{J}$-equivalent in $\mathcal{P}(G)$. For more details, see

J.-É. Pin, PG = BG, a success story, in NATO Advanced Study Institute Semigroups, Formal Languages and Groups, J. Fountain (éd.), 33-47, Kluwer academic publishers, (1995). http://www.liafa.univ-paris-diderot.fr/~jep/PDF/BGPG.pdf

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