How many non-isomorphic abelian groups of order $\kappa$ are there for $\kappa$ infinite?

Question

Let $\kappa$ be an infinite cardinal. How many non-isomorphic abelian groups of order (cardinality) $\kappa$ are there?

For finite $\kappa,$ we can use the classification theorem and obtain the number of non-isomorphic abelian groups of this order in finite time. I don't think there is a classification theorem for general infinite abelian groups, but I think this cannot be a difficult problem, since cardinality questions are generally easier for infinite sets than for finite sets. Nevertheless, I have no idea how to solve it.

EDIT If this is actually difficult, perhaps this question isn't:

For a cardinal number $\kappa,$ is there always a cardinal $\lambda$ such that there are more non-isomorphic abelian groups of order $\lambda$ than those of order $\kappa\,?$

Answer 1

It's known by Fisher, Eklof and Shelah (see theorem 2.1 in [1]) that there are abelian groups which are stable but not superstable. Another well known result of Shelah (see [2]) is that, roughly speaking (again, check the paper below for details), if $T$ is not superstable, then $T$ has $2^{\lambda}$ non isomorphic models of cardinality $\lambda$. These two results combined imply the existence of $2^{\lambda}$ non-isomorphic abelian groups of cardinality $\lambda$.

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Bibliography:

1. J. T. Baldwin and Jan Saxl (1976). Logical stability in group theory. Journal of the Australian Mathematical Society (Series A), 21 , pp 267-276.

2. S. Shelah (1974). Why There Are Many Nonisomorphic Models for Unsuperstable Theories. Proceedings of the International Congress of Mathematicians (Vol I), pp 259-264.

Answer 2

To your edit, this may depend on the value of the continuum, let us fix $\kappa$ to be an infinite cardinal, and $\mathbb P$ the set of primes in $\mathbb N$.

Denote for $p$ prime, denote the group $A_p=\bigoplus\limits_{i<\kappa}\mathbb Z/p\mathbb Z$.

Now for every $I\subseteq\mathbb P$ denote the group $Z_I=\bigoplus\limits_{p\in I}A_p$. Its cardinality is $\kappa$. If $I\neq J$ then without loss of generality there is some $n\in I\setminus J$, therefore in $Z_I$ there is an element of order $n$, while in $Z_J$ there are none of order $n$.

This shows that there are at least continuum many non-isomorphic subgroups. And I haven't began meddling with free abelian groups and other strange animals. Moreover the larger $\kappa$ gets the more "smaller" groups we can use. I'd conjecture that the number either ends up $2^{\kappa}$ many, or its representation is independent of ZFC (it has no representation as a function of $\kappa$ in terms of $\aleph$ cardinals and the continuum function).

Answer 3

Here are two observations:

1) The number of isomorphism classes of abelian groups of infinite cardinality $\kappa$ is at most the number of "group laws" on a set of cardinality $\kappa$, which is at most the number of functions $\kappa \times \kappa \rightarrow \kappa$, which is $\kappa^{\kappa^2} = 2^{\kappa}$.

2) There are $2^{\aleph_0}$ abelian groups of cardinality $\aleph_0$. Indeed, let $\mathcal{P}$ be the prime numbers, and for any subset $S \subset \mathcal{P}$, let $G_S = \mathbb{Z} \oplus \bigoplus_{p \in S} \mathbb{Z}/p\mathbb{Z}$.

The argument of part 2) doesn't seem to adapt so well to higher cardinalities, but off the top of my head I will guess that there are $2^{\kappa}$ isomorphism classes of abelian groups of cardinality $\kappa$ for all $\kappa \geq \aleph_0$.

(Added: Well, 2) adapts to higher cardinalities in the following anemic sense: for any infinite cardinal $\kappa$, we can take $G_S = \bigoplus_{\kappa} \mathbb{Z} \oplus \bigoplus_{p \in S} \mathbb{Z}/p\mathbb{Z}$ to see that there are at least $2^{\aleph_0}$ isomorphism classes of cardinality $\kappa$ abelian groups. This is awfully similar to what Asaf wrote above; for some reason it took me a little while to process his answer. I also suspect that the question can be answered by bringing out the Ulm invariants and their possible values, but the number of people who are conversant which such things may be rather small. E.g., not me...)

Answer 4

The answer to the second (easier) question is yes, I believe. If I have made any serious mistakes in the following argument, comments are more than welcome.

In $\mathrm {ZFC}$ any set can be well-ordered, so we can assume all cardinals are aleph numbers. So, suppose for some cardinal number $\kappa$ we have $\aleph_\alpha$ non-isomorphic abelian groups of order $\kappa$. Let $\omega_\alpha$ be the initial ordinal of $\aleph_\alpha$ and let $\lambda>\max\lbrace\aleph_{\omega_{\alpha+1}},\kappa\rbrace$ be any cardinal number. For every ordinal number $\beta<\omega_{\alpha+1}$ define: $$A_\beta=(\bigoplus\limits_{i<\omega_\beta}\mathbb Z/2\mathbb Z)\oplus(\bigoplus\limits_{i<\lambda}\mathbb Z/3\mathbb Z)$$ Now, because of the second (bigger) summand, $A_\beta$ has order $\lambda$. Because of the first summand, $A_\beta$ contains exactly $\aleph_\beta$ elements of order $2$. Hence, for $\beta_1\neq\beta_2$, $A_{\beta_1}$ and $A_{\beta_2}$ cannot be isomorphic. Since there are $\aleph_{\alpha+1}>\aleph_{\alpha}$ such ordinals $\beta$, we have found a cardinal $\lambda>\kappa$ such that there are more non-isomorphic abelian groups of order $\lambda$ than of order $\kappa$.