Grothendieck group of the monoid of subsets in a group The subsets of a group form a monoid, with the product of two subsets $A$ and $B$ the subset of all products ab with $a \in A$, $b\in B$, and identity the trivial subgroup. I'm wondering what its Grothendieck group should be. Is there an easy way to see it? 
Addition: I'm not sure if I can make this a nontrivial question, but what if we suppose the group is abelian and disregard the empty set?
 A: Grothendieck groups are usually only defined for commutative monoids, and your construction is not a commutative monoid unless your underlying group is abelian.
Nonetheless, you can define the universal enveloping group of an arbitrary monoid (which agrees with the Grothendieck group in the commutative monoid case). One way to obtain it is to take a presentation for the monoid as a monoid, and to consider the group that is presented by the same set of generators and of relations (that is, the same presentation, but now read as if it is a group presentation).
The universal enveloping group of $M$, $M_{\rm gp}$, is a group, together with a monoid homomorphism $q\colon M\to M_{\rm gp}$, such that for any group $H$ and any monoid homomorphism $a\colon M\to H$, there exists a unique group homomorphism $f\colon M_{\rm gp}\to H$ such that $a=fq$. (In other words, $M_{\rm gp}$ is the image of $M$ under the left adjoint of the forgetful functor $\mathcal{G}roup\to\mathcal{M}onoid$). 
But in your situation, your monoid has a zero element: if you look at the collection of all subsets, then the empty set is a zero element. If you look at the collection of all nonempty subsets, then the whole group is a zero element. If you look at the collection of proper nonempty subsets, then it's not closed under multiplication (since you can have two proper subsets whose product equals the entire group).
And whenever you have a zero element, the universal enveloping group is trivial: suppose $M$ is a monoid with zero element $z$. If $H$ is a group, and $a\colon M\to H$ is a monoid homomorphism, then $a(z) = a(zz) = a(z)a(z)$, hence $a(z)=e$ must be the identity of $H$. But then for all $m\in M$ we have $a(m) = a(m)e = a(m)a(z) = a(mz) = a(z) = e$, so $a$ is the trivial map. 
Thus, the trivial group $\{1\}$ with the trivial map $q\colon M\to\{1\}$ has the desired universal property for $M_{\rm gp}$, so $M_{\rm gp}$ is trivial. 
A: Since $\emptyset\cdot A=\emptyset$ for all subsets $A\subseteq G$ of your group, the Grothendieck group of your monoid is trivial. 
