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There are two proofs of Nielsen-Schreier that I know of. The theorem states that every subgroup of a free group is free. The first proof uses topology and covering space theory and is rather elegant. The second uses combinatorial techniques on a free group of words with no relations.

Is there a more algebraic proof which somehow just uses the universal property of free groups and maybe other properties of groups that are proved more “algebraically”?

I’m interested because groups are defined purely algebraically by equations, and some proofs that a subgroup of a free abelian group is free abelian have a far more algebraic flavour. So perhaps there is some proof of Nielsen-Schreier that also has a more algebraic flavour?

- Does this group presentation define a nontrivial group?
- Abelianization of free group is the free abelian group
- Prove that $PSL(2,\mathbb{Z})$ is free product of $C_2$ and $C_3$
- Free Group generated by S is actually generated by S.
- I don't understand what a “free group” is!
- Epimorphisms from a free group onto a free group

Ideally I would like a proof that does not involve combinatorial properties of a group of words on generators; in other words preferably no facts from combinatorial group theory.

- Let G be a nonabelian group of order $p^3$, where $p$ is a prime number. Prove that the center of $G$ is of order $p$.
- Automorphism group of the Alternating Group - a proof
- What are the symmetries of a colored rubiks cube?
- If $F(a, b)=\langle a, B\rangle$ then $B=a^ib^{\epsilon}a^j$: a neat proof?
- Elementary manipulation with elements of group
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- Sort-of-simple non-Hopfian groups
- Why can quotient groups only be defined for subgroups?
- Isomorphism of $\mathbb{Z}/n \mathbb{Z}$ and $\mathbb{Z}_n$
- Is $(\mathbb{Q},+)$ the direct product of two non-trivial subgroups?

So the question can be marked off as answered…

I don’t think there can be a “purely functorial” proof that a subgroup of a free group is free (that is, a proof using just the universal property of the free group). If there were such a proof, one would naturally expect that it can be applied to work for the relatively free groups in any variety of groups. But the only varieties of groups in which subgroups of free groups are always free are the variety of all groups, the variety of all abelian groups, and the varieties of abelian groups of prime exponent. So this argues strongly against the existence of such a proof.

As long as I’m writing this as an answer, I’ll note that the technical name for such varieties is *Schreier varieties.* That is, a variety $\mathfrak{V}$ of algebras (in the sense of universal algebra) is said to be a *Schreier variety* if and only if every subalgebra of a free $\mathfrak{V}$-algebra is itself a free $\mathfrak{V}$-algebra. The proof that the only Schreier varieties of groups are the ones listed above is due to Peter Neumann and James Wiegold in *Schreier varieties of groups*, Math. Z. **85** (1964) 392-400. An alternate proof was given by Peter Neumann and Mike Newman, *On Schreier varieties of groups*, Math. Z. **98** (1967) 196-199.

A proof can also be found in Hanna Neumann’s book **Varieties of Groups**, in section 4.3.

Recently emerged a purely algebraic proof (author’s claim) employing diagram chasing of some of the wreath product’s **functorial** properties. By authors Ribes, L.; Steinberg, B. under the title: “A wreath product approach to classical subgroup theorems”. Enseign. Math. (2) 56 (2010), no. 1-2, 49–72., also available at the arXiv: Ribes, L.. They effectively use the universal property definition and are capable of proving the Kurosh’s Subgroup Theorem as well.

There is an algebraic version of the topological proof using **covering morphisms of groupoids**, due to Philip Higgins,

Higgins, P.~J. “Presentations of groupoids, with applications to groups”.

*Proc. Cambridge Philos. Soc.* 60 (1964) 7–20.

and the downloadable

Higgins, P.J. *Notes on categories and groupoids*, Mathematical Studies, Volume 32.

Van Nostrand Reinhold Co. London (1971); Reprints in Theory and Applications of

Categories, No. 7 (2005) pp 1–195.

He uses the solution of the word problem for free groupoids, though I think this can be avoided by using the (functorial) fact that if $p:G \to H$ is a covering morphism (actually fibration is sufficient) of groupoids, then the pullback functor $p^*: Gpds/H \to Gpds/G$ has a right adjoint and so preserves colimits. This result has other applications.

Addition 27 January, 2014: For the last paragraph, I can now refer you to my answer to this mathoverflow question. The point is that a subgroup $H$ of a group $G$ defines a covering morphism of groupoids $p:Tr(G,H) \to G$ where the first groupoid is the action groupoid of $G$ on the cosets of $H$. So the vertex groups of $Tr(G,H)$ are all isomorphic to $H$. You can then use the colimit preserving properties of $p_*$ as described in the answer to prove that if $G$ is a free group, then $Tr(G,H)$ is a free groupoid. It is also connected (= transitive). So choosing a maximal tree shows that any of its vertex groups are free groups. What is called a *Schreier transversal* is just a maximal tree in the groupoid $Tr(G,H)$.

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