Grothendieck group of integer sets with Minkowski addition Consider a set $Z := \{X\in 2^\mathbb{Z}: \Vert X\Vert < \infty\}\setminus\{\emptyset\} = \{X\in 2^\mathbb{Z}: \Vert X\Vert\in \mathbb{N}_+\}$ and give it an operation of Minkowski addition, that is:
$$ A + B = \{a + b: a\in A \ \wedge \ b\in B\}. $$
This structure is associative, commutative and has an identity element $\{0\}$. If I'm not mistaken, then it's an abelian monoid. What I'm interested in, is its Grothendieck group. Let us have a(n) (equivalence) relation $\sim$ on $Z\times Z$ defined as:
$$ (A, B)\sim (C, D)\iff A+D+K = C + B + K $$
for some $K\in Z$. 
My question is: What is the structure of the group $\ Z\times Z/\sim\ $?
Is it described in more detail somewhere?
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Possible hints for me:  
There seems to be no set $K$ that would make, for example, $ (\{1,2\}, \{2,3\})$ and $ (\{-1,1\}, \{3,7\}) $
equivalent. If I hadn't excluded infinite sets and the empty one, I could now use $\mathbb{Z}$ (or $\emptyset$, if adding it was defined the way Minkowski did) for $K$ and thus obtain a trivial group.
Now it seems that there's no set $Y\in Z$ such that for any $X\in Z$, $\ Y+X = Y$.
So, first hint I'd appreciate, is to show me, if there really is no such $K$ that makes following statement true:
$$ \{4, 5, 8, 9\} + K = \{1, 2, 3, 4\} + K. $$
Or perhaps there is one...?
Probably in other words: does the cancellation law hold for this structure?
Secondly, the set of singletons $\{X\in 2^\mathbb{Z}: \Vert X\Vert = 1\} < Z$ is isomorphic with $\mathbb{Z}$. I'd say the same holds for $\{[(\{x\}, \{0\})]: x\in\mathbb{Z} \} < \ Z\times Z/\sim\ $, meaning the group really can't be trivial.  
On the other hand, $(\{1,2\}, \{2,3\}) \sim (\{3,4\}, \{4,5\})$ for $K = \{0\}$, so it's not isomorphic to $Z\times Z$ either (i.e. $\sim$ isn't an identity relation). I imagine a hint would be to show me how to think about, for example, $[\{0\}, \{1,2\}]$ ("negative $\{1,2\}$ set").
 A: The Grothendieck group of $Z$ is the free abelian group on 2 generators $\mathbb{Z}^2$.
To see this, first consider the equality:
$$A + \mathrm{Conv}(A) = \mathrm{Conv}(A)+\mathrm{Conv}(A)$$
with $\mathrm{Conv}(A)$ the interval defined by
$$\mathrm{Conv}(A) = \{ a_-,~ a_- +1,\dots,~ a_+ \}$$
with $a_-$ and $a_+$ the minimal resp. maximal element of $A$.
The equality above shows that $Z$ is not cancellative. The elements $A$  and $\mathrm{Conv}(A)$ determine the same elements in the Grothendieck group. So the Grothendieck group of $Z$ can be computed as the Grothendieck group of the submonoid $Z' \subseteq Z$ consisting of the intervals $[a,b] = \{a,~a+1,\dots,b\}$ for $a,b \in \mathbb{Z}$, $a \leq b$. It is easy to check that the Minkowski sum of two intervals $[a,b]$ and $[c,d]$ is given by $[a+c,b+d]$. This means that the map
$$Z' \longrightarrow \mathbb{Z}\times\mathbb{N},\qquad [a,b] \mapsto (a,b-a)$$
is an isomorphism of monoids.
So the Grothendieck group of $Z$ agrees with the Grothendieck group of $\mathbb{Z}\times\mathbb{N}$, which is $\mathbb{Z}^2$.
