# On the Absoluteness of Almost-Free Groups

This past year has been my first as faculty at Baylor University. While here I have had the chance to work and collaborate with a remarkable graduate student, Alexandra Pasi-Fitzmaurice. She and her advisor, Daniel Herden, have been investigating a certain family of infinite abelian groups called $\aleph_1$-free groups, specifically looking at how these groups sit within the logic structures of the axioms of ZFC (what most other mathematicians would call “regular” mathematics). The two of them recently posted an article on the arXiv (“On the Absoluteness of $\aleph_1$-Freeness,” arXiv link: 2104.10341). There is a beautiful mixture of elegance and surprise to this problem which I’ve just fallen in love with. Lexi and Daniel were very generous when they gave me the opportunity to collaborate with them on an upcoming work which I’ll be spending a series of posts in the future to fleshing out.

This paper deals with $\aleph_1$-free groups. Recall that an (abelian) group $A$ is called free if it can be written as $A = \bigoplus_{i \in S} \mathbb{Z} e_i$ for distinct elements $e_i$ in $A$. An $\aleph_1$-free group is an abelian group of rank $\aleph_1$, that is, the dimension of the vector space $\mathbb Q \otimes A$ is $\aleph_1$, the first infinite cardinal greater than $\aleph_0$ ($\aleph_0$ is the cardinality of the natural numbers), such that any countable subgroup of $A$ is free. For an example of such a group, the most accessible one is the ‘Baer-Specker’ group, $\prod_{i \in \omega} \mathbb Z$, which fails to be free but is $\aleph_1$-free. From an algebraic point of view, the $\aleph_1$-free groups are remarkably complex. Through a combination of theorems of Corner, Dugas, and Göbel in the mid-80’s, it was established that any ring which has a free additive group can be realized as the endomorphism ring of an $\aleph_1$-free group.

On the other hand, if one views $\aleph_1$-free groups from a set theoretic point of view, these groups are rather simple. Up until this paper, the following theorem seems to be ‘folklore’ among the mathematicians studying infinite abelian groups.

Theorem: The notion of an $\aleph_1$-free group is absolute in ZFC.

Without getting into too many technical details, the absolute notions of ZFC are those that are resilient in the face of model extension. For example an ordered pair of two elements in the set $S$ remains an ordered pair no matter the transitive model extension, similarly the property of being an ordinal is an absolute notion; the set $\omega$ is absolute, and even $(\mathbb Z, + \cdot)$ is absolute. Many properties that are usually taken for granted are not absolute. Of note among these is the notion of uncountable cardinality. Given any transitive model $M$ of ZFC such that $\kappa \in M$ is an uncountable cardinal, there is a model extension of $M$ which puts the set $\kappa$ in one-to-one correspondence with $\aleph_0$, the first infinite (but still countable) cardinal. Essentially, in the model extension, the cardinality of $\kappa$ is “collapsed” to be countable.

No matter the model extension, absolute properties will not change. By analyzing the statements of various well-known theorems in abelian groups, most notably the Pontryagin Criterion, Pasi and Herden establish that $\aleph_1$-freeness is an absolute property. This is a very strong statement which can be used to simplify many proofs involving $\aleph_1$-free groups, for example.

Lemma: If $H$ and $G/H$ are $\aleph_1$-free for abelian groups $H \subseteq G$, then $G$ is $\aleph_1$-free.

Proof: Say that there is a countable, transitive model $M$ of ZFC for which $H, G \in M$. Then take a generic extension $N$ of $M$ such that $\aleph_1$ is countable in $N$. Then $H$ and $G$ are countable $\aleph_1$-free groups in $N$, thus they are free groups. Since $H$ being free and $G/H$ being free together mean that $G$ is free. Then we can pull $G$ back down to the model $M$ where $G$ is at least $\aleph_1$-free by absoluteness. So $G$ is free.

This is a really cool project and I’m looking forward to showing y’all what we’ve done as a follow-up to this.