Understanding why this "obvious" approach to Schreier-Nielsen's theorem doesn't work

Question

While talking about various proofs of Nielsen-Schreier Theorem, someone asked a question and I couldn't come up with a satisfying answer on the spot. I was discussing how there's a nice proof using algebraic topology and then there are algebraic proof which are not too difficult but certainly not as visual. But then someone said why doesnt "this" work, "this" being the following argument :

A free group of rank $\alpha$ ($\alpha$ being a cardinal number, not necessarily finite) is unique up to isomorphism so we can reason about them using the usual construction of a group of words on some set of size $\alpha$. Suppose $F$ is free, and $H\subset F$ is a subgroup. Assume $H$ is not free, then there's some relation satisfied by the generators, some reduced word that happens to equal the identity of $H$. But $H$ being a subgroup of $F$, this means that there is such a relation between elements of $F$, which is not possible since $F$ is free. My first objection was that a group could very well have a presentation with generators and a bunch of relations and still be free, my main example was the trivial group, as well as taking the group with presentation $\left\langle a,b,c\vert b=c\right\rangle$. But this was somewhat unsatisfactory for my interlocutor, because the exemples were a bit trivial and they were asking if they could simply just say "let $H$ be a non trivial subgroup of $F$, and assume it to not be free" and go on with their argument. I don't think the argument is correct still, for essentially the same reasons, the argument essentially relies on an equivalence "not free" $\iff$ "there is some 'non trivial' relation between elements of the group" for some meaning on non trivial.

Answer 1

Here's what does not make sense in this argument. You wrote:

Assume $H$ is not free, then there's some relation satisfied by the generators, some reduced word that happens to equal the identity of $H$.

The statement there's some relation satisfied by the generators presumes that a generating set of $H$ has been specified. But no such generating has been specified. Perhaps one might say "Pick a generating set of $H$. Assuming $H$ is not free, then there's some relation..." This is exactly what you showed with your presentation of the trivial group, but you didn't go far enough. Every group has a generating set which is not a free basis of that group --- even a free group has a generating set which is not a free basis. So if this argument were to work, it would essentially be proving that every group is not free, which is nonsense.

Quantifiers matter. The proper formulation of the "generator/relator" definition of a free group is an existential statement:

A group $H$ is free if there exists a subset $S \subset H$ which is a free basis for $H$, meaning that $S$ generates $H$ and every reduced word of $S$ represents a non-identity element of $H$.

So if someone claims to be proving that the subgroup $H<F$ is free using an argument by contradiction, the starting point of that argument is the statement that $H$ does not have a free basis, which translates into the following sentence:

Assume that $H$ is not free, then for every generating subset $S \subset H$ there's some reduced word in the generators of $S$ that happens to equal the identity of $H$.