Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

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.

share|cite|improve this question
I'm scratching my head... I don't think I know of any non-trivial property of free groups that can be proved using the definition via the universal property. – HJRW Mar 14 '11 at 18:24
I don't think there can be a proof that the subgroup of a free group is free using only the universal property. If there were such a proof, I would expect it to work for the relatively free groups in any variety, but the only varieties of groups in which subgroups of free groups are free are the variety of all groups, the variety of all abelian groups, and the varieties of abelian groups of prime exponent. – Arturo Magidin Mar 14 '11 at 18:28
Free groups are defined universally, but in what sense are arbitrary subgroups of free groups defined universally? @Henry: you can prove that they are infinite, non-isomorphic, and non-abelian for $n \ge 2$ using only the universal property. – Qiaochu Yuan Mar 14 '11 at 18:33
I guess you guys are right. Free groups realized as a group of words is a pretty natural way to think of free groups so it's not surprising that the only algebraic proof is the combinatorial one. Thank you all for your comments. – Jason Polak Mar 14 '11 at 19:43
Qiaochu - OK, if you think those properties are non-trivial. – HJRW Mar 15 '11 at 0:02
up vote 5 down vote accepted

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.

share|cite|improve this answer
i would like to stress that the new Ribes and Steinberg's treatment proceed to the Universal property's direct verification. – janmarqz Jan 9 '14 at 16:37

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.

share|cite|improve this answer
those maths continues to evolve – janmarqz Jan 8 '14 at 3:58
check my schematical solution…. Comment. – janmarqz Aug 23 '14 at 16:53

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 $Tr(G,H)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.