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The following questions popped out while I was preparing a course on profinite groups.

Closed subgroups of free profinite groups are not necessarily profinite free (e.g. the p-sylow subgroups, or the kernel of the map on the maximum p quotient, and many more other examples) thus the Nielsen-Schreier theorem fails in the profinite category.

Nevertheless, Nielsen-Schreier theorem carries over for open subgroups. The proofs I found (in Field Arithmetic by Fried-Jarden and in Profinite Groups by Ribes-Zalesskii) use the construction of free profinite groups as restricted completion of free abstract groups AND the Schreier basis of a finite indexed subgroup of a free abstract group. By restricted I mean that if X is a basis of a free abstract group, then the completion is w.r.t. the family of finite index normal subgroups that contain all but finitely many elements of X.

First question: can one avoid the use of the Schreier basis in proving that an open subgroup of free profinite is free profinite?

Note that in the finitely generated case the restricted completion is the same as the profinite completion, thus one does not need to use the Schreier basis in this case. Therefore it suffices to affirmatively answer the following.

Second question: is N-S for open subgroups of finitely generated free profinite groups implies N-S for open subgroups of non-finitely generated free profinite groups?

I apologize that the question became a bit lengthy...

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up vote 6 down vote accepted

Luis Ribes and I gave a proof of Nielsen-Schreier for open subgroups of free profinite groups that avoids using completions and the discrete Nielsen-Schreier theorem (well, actually our proof does both at once). We use wreath products instead. But we do use the Schreier basis. The ArXiv version is http://arxiv.org/pdf/0812.0027 and the final version is in l'enseignement mathématique.

Edit I think one could avoid the Schreier basis with our technique by using embedding problems like we do for quasi-free.

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I know about this paper. As you mention it uses the Schreier transversal. In the quasi free case you only need to solve split embedding problems. So in order for this to work, you need to prove the an open subgroup of a projective group is projective. This is also carried out via wreath products. –  Lior Bary-Soroker Apr 27 '12 at 20:00
    
Exactly, so you can basically do it all with wreath products and embedding theorems. –  Benjamin Steinberg Apr 27 '12 at 21:55
    
I should mention the wreath product proof of projectivity is due to Cossey, Kegel and Kovacs. –  Benjamin Steinberg Apr 27 '12 at 21:56
    
right, I couldn't remembered the names when I wrote it. I'm looking for a proof that will be easy to present in class, and CKK seems too far off. –  Lior Bary-Soroker Apr 28 '12 at 17:52
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Gee, I found their proof pretty easy. In fact John Rhodes and I rediscovered their proof unawares while working on profinite semigroups. In any event I think Ribes gives a simpler version of their proof in the Appendix to the second edition of Ribes and Zalesskii. If not look at my paper with Rhodes blms.oxfordjournals.org/content/40/3/375.abstract The proof there is at most one page. –  Benjamin Steinberg Apr 29 '12 at 0:48
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