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?

  • $\begingroup$ 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. $\endgroup$ Sep 7, 2022 at 10:43
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    $\begingroup$ Up to isomorphism there are continuously many (i.e. $2^{\aleph_0}$) countably infinite groups. $\endgroup$ Sep 7, 2022 at 10:51
  • $\begingroup$ @PeterKropholler Could you tell me a little more about it, i.e., how we can get them all? $\endgroup$
    – tobias
    Sep 7, 2022 at 13:07
  • $\begingroup$ @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. $\endgroup$
    – ADL
    Sep 7, 2022 at 13:16
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    $\begingroup$ 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.) $\endgroup$
    – Ville Salo
    Sep 9, 2022 at 4:36

1 Answer 1


There is a way to make a precise sense of the question of classifying all countable groups from the point of view of descriptive set theory.

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 $f$.

I will not go into further details since they require a longer answer but let me point out this: The isomorphism relation on the space of countable groups is what is known as a Borel complete relation. This means that if you could somehow classify all countable groups up to isomorphism, then I would be able to use your classification to classify all countable graphs, linear orders, Boolean algebras etc. Here you can consider any class of countable structures over a countable relational language satisfying a fixed set of sentences.

Here are some other classification problems that you would have solved if you could classify all countable groups up to isomorphism: Homeomorphisms of the Cantor set up to conjugacy, Polish ultrametric spaces up to isometry, Riemann surfaces up to conformal equivalence. These results that I mention are no easy results. For more details about the analysis of classification problems from the point of view of descriptive set theory, you can check out Su Gao's book Invariant Descriptive Set Theory.

All these results suggests that there is no reasonable way to classify all countable groups.


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