# All group structures on a set with cardinality $\aleph_0$

Assume we consider the additive group $$(\mathbb{Z}, 0, +)$$. I am wondering what other group structures are there with neutral element 0 fixed? Is there a way to classify them or find them all?

• It depends on what counts as a classification. But if this assumed classification is not very rough, the answer would include the classification of all finite groups, because a direct product of a finite group and $\mathbb{Z}$ is also countable. Sep 7, 2022 at 10:43
• Up to isomorphism there are continuously many (i.e. $2^{\aleph_0}$) countably infinite groups. Sep 7, 2022 at 10:51
• @PeterKropholler Could you tell me a little more about it, i.e., how we can get them all? Sep 7, 2022 at 13:07
• @tobias Every infinite countable group is in bijection with $\mathbb{Z}$, and this bijection can be arranged so that the identity of the group is mapped to $0$ (there is no group theory here). This gives a group structure on $\mathbb{Z}$ with identity $0$, one structure for each countable group. As Peter says, there are continuously many countable groups up to isomorphism, and hence this construction gives continuously many group structures on $\mathbb{Z}$ with your required property.
Sep 7, 2022 at 13:16
• Is this question really that unreasonable? Groups are coded as a subset of Baire space $\omega^{\omega^2}$ in the post, and isomorphism of groups is an equivalence relation. Classification of equivalence relations is a standard topic in descriptive set theory. (The first question is usually whether we can find a Borel invariant, i.e. a map from groups to $\mathbb{R}$ such that two groups are isomorphic iff they map to the same thing.) Sep 9, 2022 at 4:36

Let me setup the framework first. Fixing the underlying set of the group as $$\mathbb{N}$$, you can see the set of all countable groups with this fixed underlying set as $$\mathcal{G}=\{\bullet \in \mathcal{P}(\mathbb{N}^3): \bullet \text{ is a binary operation on } \mathbb{N} \text{ for which }(\mathbb{N},\bullet) \text{ is a group}\}$$ By putting the topology on $$\mathcal{P}(\mathbb{N}^3)$$ induced from the product topology of $$2^{\mathbb{N}^3}$$, you can form the Polish space $$\mathcal{G}$$ of countable groups. Then you can consider the isomorphism relation $$\cong$$ on $$\mathcal{G}$$ as a subset of the product space $$\mathcal{G} \times \mathcal{G}$$. There is a way to measure the relative complexity of measurable equivalence relations on Polish spaces using what is known as Borel reducibility.
Given Polish spaces $$X$$ and $$Y$$, we say that $$E \subseteq X \times X$$ is Borel reducible to $$F \subseteq Y \times Y$$ if there exists a Borel map $$f: X \rightarrow Y$$ such that $$x E y \Leftrightarrow f(x) F f(y)$$ for all $$x,y \in X$$. Intuitively speaking, $$E$$ being Borel reducible to $$F$$ means that classifying the elements of $$X$$ up to $$E$$-equivalence is no more difficult than classifying the elements of $$Y$$ up to $$F$$-equivalence. This is because, assuming that you know how to classify the elements of $$Y$$, you can classify the elements of $$X$$ by simply applying a "measurable computation" i.e. applying the function $$f$$.