# Simple/natural questions in group theory whose answers depend on set theory

Take these two questions:

1. "Given objects $$X$$, is there always a group $$(X, e, *)$$ with those objects?” (Ans: yes iff Axiom of Choice.)
2. "Take a group $$G$$, its automorphism group $${\rm Aut}(G)$$, the automorphism group of that $${\rm Aut}({\rm Aut}(G))$$, $${\rm Aut}({\rm Aut}({\rm Aut}(G)))$$, etc.: does this automorphism tower terminate (count it as terminating when successive groups are iso)?" (Ans, Hamkins: Yes, but the very same group can lead to towers with wildly different heights in different set theoretic universes.)

Now, these two questions should be readily understood by a student who has just met a small amount of group theory in an introductory course, though their answers depend on set theoretic ideas going far beyond the little bit that appears in their introductory text (e.g. Alan Beardon's first year Cambridge text Algebra and Geometry). The question arising:

What other questions in group theory are there that would also strike a near-beginning student as simple and natural, and similarly involve more or less significant amounts of set theory in their answers?

• Whitehead's problem (if every ses of Abelian groups $0\to\Bbb Z\to B\to A\to 0$ splits, must $A$ be free? Equivalently is every compact path connected abelian topological group a product of circles?) is a famous example, but I suppose it's too advanced for your question – Alessandro Codenotti Mar 5 '20 at 9:27
• Here's another one, which might fit your question better. For a group $G$ define its dual $G^\ast=\mathrm{Hom}(G,\Bbb Z)$, just like for vector spaces there is a canonical homomorphism $G\to G^{\ast\ast}$ given by $g\mapsto(f\mapsto f(g))$ and a group is called reflexive if this map is an isomorphism. Question: is every free Abelian group reflexive? Answer: yes iff there are no measurable cardinals. – Alessandro Codenotti Mar 5 '20 at 9:30
• Not really group theory (more general algebra) but one result that blew my mind is: if $R_n=\mathbb{R}[X_1,\ldots, X_n]$ is the ring of real polynomials in $n$-variables and $Q_n$ is its field of fractions, then the projective dimension of $Q_n$ over $R_n$ is $\min(n, t+1)$ where $t$ is such that $2^{\aleph_0}=\aleph_t$. In other word, the result (which is a single natural number) heavily depends on the continuum hypothesis: algebra.msri.org/ShowFile?aid=15.0 – freakish Mar 5 '20 at 10:06
• Both Shelah and Magidor love the connections of set theory to group theory. So you can find a lot of their work connecting set theory to group theory. – Asaf Karagila Mar 5 '20 at 15:15
• Related to my answer and @Asaf's comment, Eklof and Mekler asked in their book whether there is a reflexive group with cardinality the least measurable, a question which was answered affirmatively by Shelah, who also showed that it is consistent to have reflexive groups of arbitrarily large cardinality. As far as I know it is still open whether "there are reflexive groups of arbitrarily large cardinality" is a theorem of ZFC – Alessandro Codenotti Mar 5 '20 at 15:26

One of my favourite results, told to me by my advisor over coffee once.

Let $$G$$ be an abelian group. We say that it has a norm if there is a function $$\nu\colon G\to\Bbb R$$ whose behaviour is what you'd expect from "norm".

Say that a norm is discrete if its range in $$\Bbb R$$ is a discrete set.

Exercise. If $$G$$ is a free-abelian group, then it has a discrete norm.

Difficult theorem. If $$G$$ has a discrete norm, then it is free-abelian.

The only known proof uses Shelah's compactness theorem for singular cardinals. So quite significant heavy machinery from set theory and model theory combined.

Turning my comment into an answer. Given a group $$G$$ we can define its dual group $$G^\ast=\mathrm{Hom}(G,\Bbb Z)$$ and, just like for vector spaces, we get a canonical "evaluation" homomorphism $$G\to G^{\ast\ast}$$ given by $$g\mapsto(f\mapsto f(g))$$ and we call reflexive the groups for which this homomorphism is an isomorphism.

Theorem: Every free abelian group is reflexive iff there is no measurable cardinal.

I'll prove the $$\implies$$ direction, the other one is not as easy. Let $$\kappa$$ be measurable, and let $$\mathcal U$$ a $$\kappa$$-complete nonprincipal ultrafilter on $$\kappa$$.
Consider the free Abelian group $$\Bbb Z^{(\kappa)}=\bigoplus_{i<\kappa}\Bbb Z$$ with its standard basis $$\{e_\xi\}_{\xi<\kappa}$$ and note that $$\mathrm{Hom}(\Bbb Z^{(\kappa)},\Bbb Z)\simeq\Bbb Z^\kappa$$.
Consider the function $$\varphi\colon\Bbb Z^\kappa\to\Bbb Z$$ given by $$\varphi(x)=n$$ iff $$\{\xi\in\kappa\mid x(\xi)=n\}\in \mathcal U$$, we claim that this function is not in the image of the canonical homomorphism $$j\colon\Bbb Z^{(\kappa)}\to(\Bbb Z^{(\kappa)})^{\ast\ast}\simeq (\Bbb Z^\kappa)^\ast$$, indeed for every nonzero $$x\in\Bbb Z^{(\kappa)}$$ we have $$j(x)(e_\xi)=x(\xi)\neq 0$$ for some $$\xi$$, but $$\varphi(e_\xi)=0$$ for every $$\xi$$. (in the identification $$(\Bbb Z^{(\kappa)})^\ast\simeq \Bbb Z^k$$, $$e_\xi$$ corresponds to the projection $$\Bbb Z^\kappa\to\Bbb Z$$ on the $$\xi$$-th factor).

A reference for the proof of the other direction (actually of a much more general result which trivially implies the other direction) is Corollary III.3.8 in the beautiful book "Almost free modules: set theoretic methods", by Eklof and Mekler.

There's the Whitehead problem : let $$A$$ be an abelian group such that any extension $$0\to \mathbb Z\to K\to A\to 0$$ (with $$K$$ abelian) splits.

Is $$A$$ then necessarily a free abelian group ?

The question is famously independent of ZFC, so depends on set theory.

• Of course, indeed a famous case of independence. I'm not sure, though, that a first-year undergraduate who has only done a short algebra course is going to "see" the Whitehead problem as a natural question to raise (which is why I didn't mention it). – Peter Smith Mar 5 '20 at 12:02