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

An abelian group $W$ of cardinality $\aleph_1$ is $\aleph_1$-free if it can be written as a continuous increasing chain $W = \bigcup_{\alpha<\omega_1} W_\alpha$ of countable subgroups, each $W_\alpha$ free, with $\bigcup_{\beta<\alpha} W_\beta = W_\alpha$ at limit stages $\alpha$. A standard theorem in the theory of Whitehead groups, using Stein's countable result together with a filtration argument, shows that every Whitehead group of cardinality $\aleph_1$ is $\aleph_1$-free; this should not be read as the claim that $\mathrm{Ext}^1=0$ is inherited by arbitrary subgroups without further argument.

$\aleph_1$-freeness alone does not imply global freeness — that is the crux of the whole problem. A well-known criterion (essentially due to Pontryagin, adapted to this uncountable setting) says an $\aleph_1$-free group $W$ is actually free exactly when the filtration $\langle W_\alpha \rangle$ can be chosen so that $W_{\alpha+1}/W_\alpha$ is free for a "club" (closed unbounded) set of stages $\alpha$ — that is, when the way each new floor sits on top of the previous ones is uniformly well-behaved.

Whether such a uniformly well-behaved filtration can always be found for a group with $\mathrm{Ext}^1(W, \mathbb{Z}) = 0$ turns out to be undecidable from $\mathrm{ZFC}$ alone: the next two steps show it can always be arranged under $V = L$, and can genuinely fail under $\mathrm{MA} + \neg\mathrm{CH}$.