Let be $A(\mathbb{Z})$ the set of subsets of 1 or 2 elements, like: $\{ 1 \}, \{ 1, 2 \}$.
I would like to know if we could prove there is no map $+ : A(\mathbb{Z}) \times A(\mathbb{Z}) \to A(\mathbb{Z})$ such that:
- $+$ is associative
- $+$ is commutative
Bonus:
- $+$ has a neutral element $e_{+}$
- $\forall x \in A(\mathbb{Z}), \exists y \in A(\mathbb{Z}), x + y = e_{+}$.
I'm trying to see if I can find a binary operation so that $(A(\mathbb{Z}), +)$ would be a group (but a monoid or weaker is acceptable), I'm open to add $\emptyset$ inside of $A$ if that makes it easier to form the structure.
The motivation for this question comes from the fact that: $(\mathbb{Z}^2, +)$ is not isomorph to $(\mathbb{Q}, +)$ due to the divisibility property of $\mathbb{Q}$ which $\mathbb{Z}^2$ does not have.
What I tried so far:
(1) Mix of $\max, \min$, average operations to create a sum-like operation.
(2) Study what kind of orders would happen to exist on $A(\mathbb{Z})$ and see if I could derive a contradiction using the fact that $\lvert A(\mathbb{Z}) \rvert = \lvert \mathbb{N} \rvert$.
You can give $A(\mathbb{Z})$ a group structure by transporting it across a bijection. For example, pick a bijection $f\colon A(\mathbb{Z})\to \mathbb{Z}$, and define $x+_fy = f^{-1}(f(x)+f(y))$. This makes $(A(\mathbb{Z}),+_f)$ isomorphic to $(\mathbb{Z},+)$.
You note that $\mathbb{Z}^2$ is not isomorphic to $\mathbb{Q}$ because it is not divisible. This is true with the usual group structure on $\mathbb{Z}^2$, but note that you can use the same trick to equip $\mathbb{Z}^2$ with a group structure under which it's isomorphic to $\mathbb{Q}$. Just pick a bijection $f\colon \mathbb{Z}^2\to \mathbb{Q}$ and transport the group structure as above.
In fact, this method can be used to equip any non-empty set with a group structure (even a divisible abelian group structure, if you like) see here. Interestingly, this statement requires the axiom of choice to prove, and in fact it's equivalent to the axiom of choice!