Open subgroups of free profinite groups - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T05:23:05Z http://mathoverflow.net/feeds/question/95384 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/95384/open-subgroups-of-free-profinite-groups Open subgroups of free profinite groups Lior Bary-Soroker 2012-04-27T18:17:50Z 2012-04-27T19:27:50Z <p>The following questions popped out while I was preparing a course on profinite groups. </p> <p>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.</p> <p>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.</p> <blockquote> <p>First question: can one avoid the use of the Schreier basis in proving that an open subgroup of free profinite is free profinite?</p> </blockquote> <p>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. </p> <blockquote> <p>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?</p> </blockquote> <p>I apologize that the question became a bit lengthy...</p> http://mathoverflow.net/questions/95384/open-subgroups-of-free-profinite-groups/95391#95391 Answer by Benjamin Steinberg for Open subgroups of free profinite groups Benjamin Steinberg 2012-04-27T19:07:45Z 2012-04-27T19:27:50Z <p>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 <a href="http://arxiv.org/pdf/0812.0027" rel="nofollow">http://arxiv.org/pdf/0812.0027</a> and the final version is in l'enseignement mathématique.</p> <p><strong>Edit</strong> I think one could avoid the Schreier basis with our technique by using embedding problems like we do for quasi-free. </p>