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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 $\kappa\,?$

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I'd think this might be dependent on your choice of axioms (e.g., if you assume Martin's Axiom+$2^{\aleph_0}>\aleph_1$, then you have a non-free Whitehead group of size $\aleph_1$, while if you assume $\lozenge$ you don't). –  Asaf Karagila Mar 13 '12 at 11:17
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@AsafKaragila Thank you for the comment. This is above me, but I have added a new, possibly less problematic, question. –  user23211 Mar 13 '12 at 11:25
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There is a trivial upper bound: for infinite cardinality $\kappa$, there can be at most $\kappa$ groups by counting the number of possible multiplication tables. –  Johannes Kloos Mar 13 '12 at 11:42
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@JohannesKloos: How is that compatible with my answer for $\kappa=\omega$? –  Asaf Karagila Mar 13 '12 at 11:44
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@Johannes: Since a multiplication table is a function from $G \times G \rightarrow G$, the obvious upper bound on the number of multiplication tables is $\kappa^{\kappa^2} = 2^{\kappa}$ for infinite $\kappa$. I suspect that this upper bound is obtained for all infinite cardinals, and I am almost sure it is for $\kappa = \aleph_0$. –  Pete L. Clark Mar 13 '12 at 12:53
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4 Answers

up vote 4 down vote accepted

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 $\lambda$.


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.

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I can't find the first paper. Can you give a real citation and not a link? –  Asaf Karagila Mar 14 '12 at 11:25
    
The paper's title is "Logical stability in group theory", the authors are John Baldwin and Jan Saxl. You should find it easily on google. –  Haim Mar 14 '12 at 12:34
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This definitely answers the question. Still, I have to think that somewhat less artillery should suffice: e.g. a proof which uses abelian group theory only and does not appeal to work of the leading model theorist of our time. –  Pete L. Clark Mar 18 '12 at 5:40
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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).

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Thanks. So for any cardinal number $\kappa,$ I will find at least continuum groups of order $\kappa.$ But will I find a $\lambda$ such that there are (for example) at least $2^{\mathfrak c}$ abelian groups of this order? –  user23211 Mar 13 '12 at 11:56
    
@ymar: I've added my conjecture that you can always find $2^\kappa$ many. In particular for $\kappa>\mathfrak c$ your answer would be positive. –  Asaf Karagila Mar 13 '12 at 11:59
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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...)

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Are there arbitrarily large rings "like" $\mathbb{Z}$? "Like" means "your proof of $2)$ goes through identically using $R$ and $R/pR$ as $p$ ranges over the primes (or irreducibles, or whatever) of $R$". Perhaps an Lowenheim-Skolem type argument works? (I am just thinking out loud). –  Jason DeVito Mar 13 '12 at 14:03
    
@Jason: I think I may have an argument which by transfinite induction shows that indeed there are $2^\kappa$ many non-isomorphic groups. Alas, the server is being a jerk so I'll wait until it's working properly again before posting it. –  Asaf Karagila Mar 13 '12 at 14:30
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@AsafKaragila Please do! I was never taught transfinite induction but I have a book that explains it at length, so with its assistance, I hope I might be able to understand your argument. –  user23211 Mar 13 '12 at 17:03
    
@ymar: I don't think my argument carries over well enough. Perhaps with additional assumptions on the behavior of infinite sets it is possible to carry the argument - but I am not sure what those would be, or how to prove that. –  Asaf Karagila Mar 13 '12 at 21:45
    
That's too bad... :( Perhaps it would still be possible to answer the question in the edit? –  user23211 Mar 13 '12 at 21:51
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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 $\kappa$.

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Yes, this works to answer the "easier question": nicely done. –  Pete L. Clark Mar 14 '12 at 2:38
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