Since you mentioned Whitehead's problem here is another interesting independence example.

For a group $G$ define its dual $G^\ast$ to be $\mathrm{Hom}(G,\Bbb Z)$. Like in the vector spaces case we get a canonical evaluation homomorphism $j\colon G\to G^{\ast\ast}$ given by $g\mapsto(f\mapsto f(g))$ and we call a group reflexive if $j$ is an isomorphism.

Now let $G$ be free abelian, must it be reflexive? The answer is provably positive for all "small" free abelian groups in $\mathsf{ZFC}$, in fact it is positive for all free abelian groups iff there is no measurable cardinal.