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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 $\lvert A(\mathbb{Z}) \rvert = \lvert \mathbb{N} \rvert$.

Raito
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  • What do you mean by "parts"? (Do you mean subsets?) – Shaun Mar 20 '19 at 13:30
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    @Shaun subsets, yes – Raito Mar 20 '19 at 13:34
  • @AlexKruckman Well, yes… It solves the problem haha, thank you, I didn't thought to transport the sum operation through a well-known set! Would you like to answer the question so I can close it and upvote your answer? – Raito Mar 20 '19 at 13:39

1 Answers1

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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!

Alex Kruckman
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