Let $G$ be a finitely generated profinite group, and $H \leq_o G$. We say that $H$ satisfies Schreier's formula in $G$ if $d(H) - 1 = (d(G)-1)[G:H]$. We say that $G$ satisfies Schreier's formula if every open subgroup satisfies Schreier's formula in it. For $H \leq_c G$, we say that $H$ satisfies Schreier's formula in $G$ if every subgroup $H \leq L \leq_o G$ satisfies Schreier's formula in $G$. Note that $H$ satisfies Schreier's formula in $G$ if and only if there exists a descending chain of open subgroups, each satisfying Schreier's formula in $G$, intersecting in $H$.

The interest in these notions came to me when I was trying to prove that certain subgroups of free profinite groups are free profinite(the analogue for abstract free groups is settled by the famous Nielsen–Schreier theorem, while in the profinite case there are various counterexamples and sufficient conditions but no general theorem which would clarify the situation). The connection stems from the following proposition:

Let $F$ be a free finitely generated profinite group and $H \leq_c F$. Suppose that there exists $N \lhd_c F$ contained in $H$, such that $H/N$ does not satisfy Schreier's formula in $F/N$. Then, $H$ is a free profinite group.

This seems very promising to me because the prototype of a profinite group which does satisfy Schreier's formula is a rather "general" group like the free pro-$\mathcal{C}$ group for an appropriate class of finite groups(e.g. all groups, solvable groups, $p$-groups) so in "most" cases the proposition can be applied. Moreover, it can be shown that in the pro-$p$ case, satisfying Schreier's formula is equivalent to pro-$p$ freeness. On the other hand, I am aware of the fact that a profinite group can satisfy the formula even if it is not free for any class of finite groups. Nevertheless, I think that, in some sense, such a group must "contain" an "essential part" of a free profinite group. So, my first question is:

1. What can be said about a finitely generated profinite group which satisfies Schreier's formula?

Maybe such a group has a free pro-$\mathcal{C}$ group as a subgroup?

It turns out that an (abstract) finitely generated residually finite group satisfies Schreier's formula if and only if it is a free group as shown in: Schreier's index formula. A consequence of this is that if a profinite group $G$ is a profinite completion, then it (and probably every closed subgroup of it as suggested by: Schreier's formula and descending chains) doesn't satisfy Schreier's formula unless it is a free profinite group. This gives a necessery condition for a profinite group to be a completion of an abstract group, and may help answer the (open) question of describing the profinite completions inside the family of profinite groups.

On the other extreme, I find it quite hard to prove that a rather restricted profinite group won't satisfy Schreier's formula, even if it is very far from being free in any sense. Here are some questions that I am very interested in:

2.a. Let $G$ be a finitely generated profinite group, $N_1 , N_2 \lhd_c G$ with $N_1 \cap N_2 = {1}$, and $H \leq_c G$ not containing $N_i$, $i=1,2$. Can $H$ satisfy Schreier's formula in $G$?

I can prove that the answer is "no" under the additional assumption that $N_1N_2 = G$ (in this case $G$ is a direct product). The question is inspired mostly by Haran's diamond theorem proved in: http://www.math.tau.ac.il/~haran/Papers/twin.pdf.

I call a profinite group $G$ Noetherian, if evrey ascending chain of closed subgroups of it is eventually stable. A profinite group $G$ is called just-infinite if every nontrivial $M \lhd_c G$ is open.

2.b Let $G$ be a finitely generated Noetherian/just-infinite (or even both) profinite group. Can $G$ (or any closed subgroup of it) satisfy Schreier's formula?

Following: http://arxiv.org/pdf/1203.4217v2.pdf, I define the generalized derived subgroup $D(G)$ of a profinite group $G$ as the intersection of all open normal subgroups $N$ of $G$ with $G/N$ either abelian or simple. Profinite groups for which some finite iteration of $D()$ is $\{1\}$ are said to have finite abelian-simple length.

2.c Let $G$ be a finitely generated profinite group of finite abelian-simple length. Can $G$ (or, more importantly, any closed subgroup of it) satisfy Schreier's formula?

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    $\begingroup$ You might look at my paper with Karl Auinger on Varieties of supersolvable groups with the M. Hall property. We consider several strengthenings of the Schreier property still enjoyed by free profinite groups and can prove something thing about groups with them. If memory serves Zalesskii gave us an example of a profinite group with a tree-like Cayley graph in our sense with no free pro-C subgroup for any extension-closed C but I forget the details and it is not in the paper. Tree-like Cayley graph is stronger than Schreier. $\endgroup$ – Benjamin Steinberg May 23 '14 at 0:56
  • $\begingroup$ Here is a link link.springer.com/article/10.1007%2Fs00208-006-0767-2 $\endgroup$ – Benjamin Steinberg May 23 '14 at 0:58

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