Worked solution: Shelah's proof that the Whitehead problem is independent of ZFC (1974)

A short exact sequence $0 \to \mathbb{Z} \to G \to W \to 0$ of abelian groups "splits" if there is a homomorphism $W \to G$ inverting the surjection $G \to W$, which forces $G \cong \mathbb{Z} \oplus W$. The group $\mathrm{Ext}^1(W, \mathbb{Z})$, from homological algebra, classifies all such extensions up to equivalence, with the zero element corresponding exactly to the splitting extension; so the sequence splits for every possible $G$ built this way if and only if $\mathrm{Ext}^1(W, \mathbb{Z}) = 0$.

Any free abelian group $W$ automatically satisfies $\mathrm{Ext}^1(W, \mathbb{Z}) = 0$: lifting each basis element of $W$ individually back into $G$ (any preimage works, since there are no relations to preserve) assembles into a full splitting map. J.H.C. Whitehead's question, posed in the 1950s, asks about the converse: is every abelian group $W$ with $\mathrm{Ext}^1(W, \mathbb{Z}) = 0$ (called a Whitehead group) automatically free?

This question sounds like pure algebra with no reference to set theory whatsoever — $\mathrm{Ext}^1$ is computed the same way whether or not one worries about uncountable cardinals. The next step shows the question is fully settled for small (countable) groups, which is exactly where the uncountable subtlety, and eventually set theory, will enter.