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A subgroup of a free group is free. The proof I'm familiar with uses covering space theory. But today someone mentioned to me that if a free group $F$ had a non-free subgroup (i.e. a subgroup subject to some relation), then certainly that relation would still hold in $F$, contradicting the freeness of $F$.

This proof seems far too simple, so what has gone wrong? Please answer as if I have just learned about free groups, conceiving of them simply as groups which aren't subject to any relations.

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  • $\begingroup$ why would that relation hold in F? For that to happen wouldn't be required that the sub-group is the whole F? $\endgroup$
    – Francesco Peña-Garcia
    Commented Nov 23 at 2:43
  • $\begingroup$ If a word in letters from a set $S$ (where $F=\text{FreeGroup}(S)$) equals the identity in the subgroup $H \subseteq F$, then it also equals the identity in $F$, no? $\endgroup$
    – JMM
    Commented Nov 23 at 2:48
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    $\begingroup$ By the way - the proof you mention regarding covering space theory can easily be written down combinatorially. That is to define a $1$- cell complex, you need to define an interval, which requires defining the real numbers. However the proof that subgroups of free groups are free can be written down without defining a real number (they are just a useful visual aid). On the other hand there are results in combinatorial group theory, where topological arguments appear unavoidable: mathoverflow.net/questions/482713/… $\endgroup$
    – tkf
    Commented Nov 23 at 23:41

3 Answers 3

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A definition helps.

To say $F$ is free means that, for a specifically chosen generating set, there aren't any relations amongst the elements of that generating set (except for "trivial" relations which reduce to the identity by successive cancellation of a generator and its inverse). We could say that $F$ is freely generated by that generating set, or that the generating set is a free basis for $F$. But keep in mind: not every generating set of $F$ is a free basis. For instance, if $F$ has free basis $a,b$, then if you set $c=ab$ then $a,b,c$ is still a generating set, but it has the nontrivial relation $abc^{-1}$.

So, now take an arbitrary subgroup $A < F$. To prove that $A$ is free, you must prove that $A$ has a free basis. That.... is not obvious. Even if $A$ were free, you wouldn't know it from a badly chosen generating set. That's the hard work: finding a free basis of an arbitrary subgroup.

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    $\begingroup$ Any proof that ignores definitions is fake. $\endgroup$
    – Lee Mosher
    Commented Nov 23 at 3:56
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    $\begingroup$ You can't just make up some pithy phrase that is sorta-kinda like the definition, and then ignore the definition and depend on that pithy phrase instead. $\endgroup$
    – Lee Mosher
    Commented Nov 23 at 3:57
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    $\begingroup$ Keep in mind: there are relations in free groups. Free bases have cancellation relators. Generating sets that are not free bases have other relators. To characterize a non-free subgroup as being a "subgroup subject to a relator", however pithy, is ignoring the definitions. $\endgroup$
    – Lee Mosher
    Commented Nov 23 at 4:16
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    $\begingroup$ @FrancescoPeña-Garcia: Here’s what’s specifically wrong in the OP’s suggested “proof”. Suppose $S \leq F$. If some non-trivial relation holds in $S$ — say there are elements $a, b \in S$ with $ab = ba$ — then certainly that relation still holds in $F$. But this doesn’t contradict freeness unless $a$, $b$ are generators of $F$; for instance, if $F$ is freely generated by $x$, and $a = x^2$, $b = x^3$, then certainly $ab = x^5 = ba$. So the wrong proof overlooks the fact that characterisations of freeness as “no non-trivial relations” must also include “…between the generators”. $\endgroup$
    – Peter LeFanu Lumsdaine
    Commented Nov 23 at 10:51
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    $\begingroup$ Or, to simply repeat what I said in my post, if $F$ is freely generated by $a,b$, and $c=ab$, then certainly $abc^{-1}$ is a relation. $\endgroup$
    – Lee Mosher
    Commented Nov 23 at 15:32
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This fake proof would work for every algebraic theory. But for most algebraic theories, subalgebras of free algebras don't have to be free.

For example, submonoids of free commutative monoids are not necessarily free. An example is $(\mathbb{N} \setminus \{1\},+,0) \subseteq (\mathbb{N},+,0)$, which is the submonoid generated by $2,3$. This is the commutative monoid with the presentation $\langle x,y : 3x = 2y \rangle$.

The additive variant of this is the non-free subring $\mathbb{Z}[X^2,X^3] \subseteq \mathbb{Z}[X]$.

Also submodules of free modules don't have to be free. This is true over PID but for most rings it's wrong.

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  • $\begingroup$ Also a countable chain can be embedded as a sublattice of a free lattice but the chain is hardly a free lattice. $\endgroup$
    – user14111
    Commented Nov 24 at 10:42
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In fact, the Nielsen-Schreier theorem to which you refer requires the axiom of choice.

Historically it came after the analog for free abelian groups by Dedekind (which you can find in Lang), and is even more difficult.

At the risk of being a little redundant after @Lee's answer, let's work in our old friend $\Bbb Z. $ Of course it's a special case. All the (non-trivial) subgroups are infinite cyclic. But take any two elements which generate such a subgroup, and of course we have relations. The point is they don't generate a (cyclic) subgroup freely.

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