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For a group $G$ we denote by $d(G)$ the cardinality of a smallest set of generators.

A finitely generated group $G$ is said to satisfy Schreier's formula if for every subgroup $H \subseteq G$ of index $k \in \mathbb{N}$ we have: $d(H) - 1 = k(d(G) - 1)$. If this equality does not hold, we say that $H$ violates Schreier's formula in $G$. For example, a finitely generated free group satisfies Schreier's formula.

Let $F$ be a free group on a finite number of generators, and $r \in \mathbb{N}$. Denote by $F^{(r)}$ the $r$-th term in the derived series of $F$. Set $Q_{r} = F/F^{(r)}$.

It has already been shown in: Schreier's index formula that $Q_r$ doesn't satisfy Schreier's formula, and here I ask for a generalization.

Is it true that every descending chain of subgroups of $Q_r$ eventually violates Schreier's formula?

More formally let, $H_1 \supseteq H_2 \supseteq \dots$ be an infinite countable properly descending chain of subgroups of $Q_r$. Must there be some $n \in \mathbb{N}$ such that $H_n$ violates Schreier's formula in $Q_r$?

Note that the case $r=1$ is trivial because every subgroup of finite index violates Schreier's formula, but the case $r=2$ is already nontrivial.

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  • $\begingroup$ Are your subgroups in the chain finite index with trivial intersection? If so the proof I gave in your old question works with some minor changes. $\endgroup$
    – Benjamin Steinberg
    Commented May 19, 2014 at 12:03
  • $\begingroup$ If the chain is finite index then by factoring out by the largest normal subgroup contained in the intersection you can again apply my old argument. $\endgroup$
    – Benjamin Steinberg
    Commented May 19, 2014 at 13:27
  • $\begingroup$ The chain is assumed to be built of finite index subgroups. The intersection must not be trivial but the largest normal subgroup contained in it may well be trivial. So your method solves it in the general case? After factoring the intersection does not have to become trivial. $\endgroup$
    – Pablo
    Commented May 19, 2014 at 14:10
  • $\begingroup$ I think so. These subgroups still separate points after killing the normal core. $\endgroup$
    – Benjamin Steinberg
    Commented May 19, 2014 at 14:24
  • $\begingroup$ Can you give here a full argument again please? I am not sure I see how $K'$ is to be chosen. Is there any reference to a book/notes where these notions are covered? I am not sure I fully understand all the definitions. Thanks a lot!!! $\endgroup$
    – Pablo
    Commented May 19, 2014 at 14:44

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