# What is the cardinality of the set of all groups (up to isomorphism) of countable order?

Apologies if this is a known result, I have looked around and could not find it.

There is a pretty vast literature on the number of groups up to isomorphism of order $$n$$, for any natural number $$n$$. This paper for instance classifies for which $$n$$ are there are one, two, or three groups of order $$n$$. However, I haven't been able to find any results, not even partial, on the cardinality of the number of groups up to isomorphism of order $$|\mathbb N|$$. Is this known, and if not are there partial results on it? I suspected that the number of abelian groups up to isomorphism would be known (as it is a well-known result for finite groups) but I cannot find the result, if it is published.

• What is meant with a "group of countable order" ? Commented Oct 4, 2023 at 16:36
• There are lots of them. $\mathbb{Z}$, $\mathbb{Z} \times \mathbb{Z}$, etc. Commented Oct 4, 2023 at 16:39
• Not an answer to your question but this begs the question of whether this is a well-defined set. The 'set of all groups' is not well-defined, for example. I'm sure this is known, so I wonder if a set theorist can chime in on this? Commented Oct 4, 2023 at 16:53
• @jpmacmanus for any cardinality, there is a set $X$ of groups of this cardinality s.t. any group of this cardinality is isomorphic to one of groups in $X$, and this set has cardinality at most $|X|^{|X|^2}$ (which, I think, is $2^{|X|}$ assuming choice, if $X$ is infinite). Commented Oct 4, 2023 at 21:50
• (sorry, in the comment above it should be "$X$ has a cardinality of at most $|Y|^{|Y|^2}$", where $Y$ is target cardinality of group) Commented Oct 4, 2023 at 23:43

There is at most continuum such groups: every group can be defined by function $$\mathbb N \times \mathbb N \to \mathbb N$$, and there are only continuum such functions.

There is also at least continuum such groups: for every subset of primes $$X$$, group $$\mathbb Z \oplus \bigoplus\limits_{x \in X} \mathbb Z_x$$ is countable, and groups for different $$X$$ are non-isomorphic: if two sets $$X_1$$ and $$X_2$$ differ in prime $$p$$, then one of corresponding groups includes element of order $$p$$, and other doesn't.

• You don't need the $\mathbb{Z}$ summand: just take infinite subsets of primes; there's already a continuum of them. Commented Oct 4, 2023 at 16:45
• We have a choice here: add $\mathbb Z$ as a summand, or talk about infintie subsets. The former is shorter to write so I chose it. Commented Oct 4, 2023 at 21:52