Why exactly is Whitehead's problem undecidable. I'm trying to  get a deeper understanding of Whitehead's problem.
It is possible to construct a group of cardinality $\aleph_1$ that satisfies Chase's condition, and is not free. This group is Whitehead when we add Martin's axiom to ZFC, it is not Whitehead when we add $Z=L$ to ZFC. Why is it not possible to check if the group is Whitehead within ZFC? I heard someone say that we can't know it in ZFC, because checking if every homomorphism with kernel isomorphic to $\mathbb{Z}$ onto the group splits, would take an uncountable amount of steps, and when a proof can be given in uncountable steps, it can also be given in countable steps, but this is not the case here. This is merely what I recall from what I heard, can someone either explain me or point me to literature on the subject? I am quite familiar with Eklof's essay, people who answer me can refer to it. 
I know that when we add Martin's axiom, we can prove that the non-free Chase group is Whitehead, we can do it because Martin's axiom allows us to connect homomorphisms with a countable domain to form a big splitting homomorphism. 
 A: Because uncountable cardinals are unwieldy. 
Note that the countable case is provable, and if you have seen the proof that $\lozenge$ implies the Whitehead conjecture is true, then you would have seen that to some extent the proof is somewhat similar.
The major difference is that when we deal with the countable case we can do it with a simple induction. In the uncountable case we need to go to transfinite induction, which passes through limit cases. Lots and lots of limit cases. And it turns out that these limit cases don't "go down easily".
I don't think it's something you can easily understand without understanding the proofs for both directions of the Whitehead conjecture's independence, as well other independence proofs. In particular, I think, that the independence of the axiom of countable choice is relevant here. In both cases we seem to want to "glue together" provable bits (in the Whitehead case, every countable Whitehead group is free; in the countable choice case, every finite subset has a choice function). But in both cases the induction doesn't quite catch the limit case.
I hope that this can be of some help.
A: In some sense (I think), you have, in your question, enough information to adequately answer it yourself.  When you wrote me (in response to my question as to what deeper knowledge you were seeking regarding Whitehead's problem), "why is it not possible to check if the group is Whitehead in ZFC", I am going to infer that "the group" you had in mind is the group you spoke about in your question, the "group of cardinality $\aleph_1$ that satisfies Chase's condition, and is not free" that is "Whitehead when we add Martin's axiom" (and $2^{\aleph_0}$ $\gt$ $\aleph_1$) and "is not Whitehead when we add V=L" to ZFC.
To show that you have, in fact, answered your own question, I would like to present a mathematical 'thought experiment' regarding the aforementioned group. The gist of the experiment is to use the following theorem, due to Solovay and Tennenbaum
Assume GCH [in V, the 'universe' formed by the 'cumulative hierarchy'--my comment] and let $\kappa$ be a regular cardinal greater than $\aleph_1$.  There exists a c.c.c. notion of forcing P such that the generic extension V[G] by P satisfies Martin's Axiom and $2^{\aleph_0}$=$\kappa$
to relate the domain of sets in which ZF+V=L holds to the domain of sets in which ZFC+Martin's Axiom+$2^{\aleph_0}$ $\gt$$\aleph_1$ holds.  That this theorem does is clear by the fact that ZF+V=L $\vdash$GCH so that the domain of sets in which ZFC+Martin's axiom +$2^{\aleph_0}$ $\gt$$\aleph_1$ is a forcing extension of the domain of sets in which ZF+V=L holds, both domains via the c.c.c. having the same cardinals and cofinalities.
Now according to your question, the group of cardinality $\aleph_1$ which is not free and satisfies Chase's condition, is not Whitehead in the domain of sets in which ZFC+V=L holds. Since this is a domain in which ZFC holds, by constructing this group in ZFC+V=L, you 'checked'  that his group is not Whitehead in this domain (it should be noted that in the Eklof paper you mention in your question the construction of the group A of cardinality $\aleph_1$ satisfying Chase's condition but was not free was carried out entirely in ZFC).  Now by the theorem of Solovay and Tennenbaum one can use the notion of forcing P (which satisfies the c.c.c.) to construct from the domain of sets which satisfies ZFC+V=L the domain of sets satisfying ZFC+Martin's Axiom +$2^{\aleph_0}$ $\gt$ $\aleph_1$ and since Goedel proved that L is a subdomain of sets in ZFC (actually ZF) in which not only the axioms of ZF hold but AC and GCH hold as well, one might rightfully infer that the domain of sets in which ZFC+Martin's Axiom +$2^{\aleph_0}$ $\gt$ $\aleph_1$ exists as well and reconstruct the group A in this domain so it is Whitehead--by construction you have checked that A is Whitehead within this domain in which ZFC holds.  This shows that it is impossible to check whether the group A you speak about in your question is or is not Whitehead in ZFC because ZFC in itself does not contain sufficient information (i.e. the extra axioms, by the 'facts' each of the domains prove about the group A, the axioms are incompatible) to decide the question.  
