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A finitely generated profinite group $G$ is said to satisfy Schreier's formula if for every open subgroup $L \leq_o G$ we have $d(L) = (d(G)-1)[G:L] + 1$. Here $d$ stands for the smallest cardinality of a (topological) generating set of a group.

A group $G$ is called supersolvable if if there exists a normal series: $$ \{1\} = H_0 \lhd H_1 \lhd \dots \lhd H_{n-1} \lhd H_{n} = G$$ such that each $H_{i+1}/H_i$ is cyclic and $H_i \lhd G$.

A group is called prosupersolvable if it is an inverse limit of finite supersolvable groups.

Let $p$ be a prime number. As $p$-groups are supersolvable, finitely generated free pro-$p$ groups satisfy Schreier's formula. Is this essentially the only example?


  • Is every finitely generated prosupersolvable group $G$ satisfying Schreier's formula is virtually a free pro-$p$ group for some prime $p$?

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  • $\begingroup$ I studied this 10 years ago with Auinger. I believed we proved it has an open normal free pro-p subgroup such that the quotient is a finite abelian group of exponent dividing p-1 but I will double check $\endgroup$ – Benjamin Steinberg Jan 8 '15 at 18:17
  • $\begingroup$ @BenjaminSteinberg I think that your arguments work only for varieties of groups. $\endgroup$ – Pablo Jan 8 '15 at 18:33
  • $\begingroup$ The paper only gives the proof for relatively free groups because we were interested in that. My memory seems to be we had a more general proof that we left out because we had a slicker proof for the relatively free case. The problem is I don't recall it yet from 10 years ago! $\endgroup$ – Benjamin Steinberg Jan 8 '15 at 18:46
  • $\begingroup$ At the moment I only remember the proof if there are only finitely many prime divisors of the order of the group. $\endgroup$ – Benjamin Steinberg Jan 8 '15 at 18:59
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My vague memory is that we proved with Auinger 10 years ago that a freely indexed pro-supersolvable group is virtually pro-p. We were only interested in the case of relatively free groups, which appears in http://link.springer.com/article/10.1007%2Fs00208-006-0767-2

This case admits a number of simplifications and the published version is very different from the first proof we had, which I believe worked for what you want. I can't remember the details so let me outline a special case.

Suppose first that $G$ is finitely generated (and not pro-cyclic), freely indexed and pro-supersolvable with order divisible by only finitely many primes. Then by an old result of Oltikar and Ribes http://projecteuclid.org/euclid.pjm/1102806646 the Frattini subgroup of $G$ is open. The Frattini subgroup is also pro-nilpotent. An open normal subgroup of a freely indexed group is again freely indexed. A theorem of Lubotzky says that a pro-nilpotent freely indexed (and not procyclic) group is free pro-p for some prime p. Thus a freely indexed pro-supersolvable group with order divisible by only finitely many primes has Frattini subgroup open and free pro-p for some prime p.

So one wants to show in the general case that a freely indexed pro-supersolvable groups has only finitely many prime divisors of its order.

We gave a geometric consequence for the Cayley graph of a freely indexed profinite group. Namely, any closed connected (in the sense of Ribes and Guldenhuys) subgraph of the Cayley graph contains each edge between two of its vertices. Using this, we essentially showed that such a profinite group is not a subdirect product of two profinite groups. This is how we showed a relatively free pro-supersolvable group which is freely indexed has only finitely many prime divisors. I am not sure if one can use this in general. I'd have to reread the paper more carefully or dig up some ancient versions of the paper on long lost computers to see if we really did do the general case at one point.

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  • $\begingroup$ Thank you very much for the detailed explaination, I was aware of the reduction to the case of infinitely many primes dividing the order. There do exist freely indexed prosolvable groups with infinitely many prime divisors, constructed by Lubotzky and v.d. Dries. But their example is not prosupersolvable... $\endgroup$ – Pablo Jan 8 '15 at 19:06
  • $\begingroup$ Supersolvable is crucial here. Every finite supersolvable group is a subdirect product of groups with a normal p-subgroups corresponding quotient an abelian group of order dividing p-1. This is what we used for the relatively free case. $\endgroup$ – Benjamin Steinberg Jan 8 '15 at 19:21
  • $\begingroup$ What is a relatively free profinite group? $\endgroup$ – Pablo Jan 8 '15 at 19:24
  • $\begingroup$ It means it is the pro-C completion of a free group with respect to a variety of finite groups in the sense of the book of Ribes and Zalesski. Or it is the free object in the class of profinite groups satisfying some profinite identity. $\endgroup$ – Benjamin Steinberg Jan 8 '15 at 20:03
  • $\begingroup$ So in this case I suspect that every occurrence of prosolvable in your answer should be prosupersolvable. $\endgroup$ – Pablo Jan 8 '15 at 20:18

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