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

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.