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A finitely generated group G is said to satisfy Schreier's index formula if for every subgroup H of index k in G we have: d(H) - 1 = k(d(G) - 1). For example, a finitely generated free group satisfies Schreier's index formula.

For a group G we denote by G' its commutator subgroup.

Let F be a free group on a finite number of generators. Does F/F'' satisfy Schreier's index formula? What about the next terms in the derived series of F?

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    $\begingroup$ Does $d$ denote the minimal number of generators of a group? $\endgroup$
    – HenrikRüping
    Commented Apr 14, 2014 at 12:05
  • $\begingroup$ Yes it does denote that. $\endgroup$
    – Pablo
    Commented Apr 14, 2014 at 13:29
  • $\begingroup$ I believe only a free group satisfies Schreier's index formula and that this was first observed by van den Dries and Lubotzsky, who proved a pro-p analogue. $\endgroup$
    – Benjamin Steinberg
    Commented Apr 14, 2014 at 16:02
  • $\begingroup$ Let me hedge and assume the group is residually finite. $\endgroup$
    – Benjamin Steinberg
    Commented Apr 14, 2014 at 16:10
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    $\begingroup$ @HJRW, this is why I hedged to the rf case. $\endgroup$
    – Benjamin Steinberg
    Commented Apr 15, 2014 at 1:31

2 Answers 2

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I claim that a residually finite, finitely generated group $G$ satisfies Schreier's formula iff it is free.

Here is the idea. Let $X$ be the generating set of $G$ of size $d(G)$. Suppose $G$ is not free on that generating set. Suppose $K$ is the presentation complex for $G$ wrt generators $X$ and some set of relations.

Then the Cayley graph of $G$ wrt to $X$ contains a simple loop $\ell$ at 1. Here simple means it visits no vertex twice except when the loop closes.

By residual finiteness we can find a finite cover $K'$ of $K$ in which the image $\ell'$ of $\ell$ is still a simple loop. I claim the finite index subgroup of $G$ represented by $K'$ based at the start of $\ell'$ fails Schreier's formula. Take a spanning tree for the 1-skeleton of $K'$ that contains all edges of the simple loop $\ell'$ except one. Then the usual method to compute the fundamental group of a graph would include $\ell'$ as part of a basis. But $\ell'$ is trivial in the fundamental group of $K'$ being the image of a loop in the Cayley graph of $G$. This shows the rank of the corresponding finite index subgroup of $G$ dropped.

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  • $\begingroup$ If you take the quotient of F by the intersection of all the terms in its derived series you get a finitely generated residually finite group, which is not free but satisfies Schreier's index formula. Does this contradict your claim? $\endgroup$
    – Pablo
    Commented Apr 14, 2014 at 19:59
  • $\begingroup$ The intersection of all the terms of the derived series is trivial because the free group is residually a finite p-group and hence residually solvable. So you do get a free group and there is no contradiction. $\endgroup$
    – Benjamin Steinberg
    Commented Apr 14, 2014 at 20:17
  • $\begingroup$ Ok, I think I've got you. What if instead F is the free profinite group on a finite number of generators? In this case the aforementioned intersection is not trivial. $\endgroup$
    – Pablo
    Commented Apr 15, 2014 at 7:27
  • $\begingroup$ Free pro-solvable groups and free pro-p groups satisfy Schreier's formula. $\endgroup$
    – Benjamin Steinberg
    Commented Apr 15, 2014 at 13:02
  • $\begingroup$ Suppose $H$ is a given subgroup containing the k-th derived subgroup of $F$. Must there be a finite index subgroup above $H$ violating Schreier's formula? I'm interested in this in the profinite case too. $\endgroup$
    – Pablo
    Commented Apr 17, 2014 at 13:05
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Benjamin Steinberg has now answered this in general. I used a computer calculation to get an explicit counterexample for $F/F''$. Let $F$ be free on $\{a,b\}$ and let $K$ be the kernel of the homomorphism $F \to S_3$ with $a \mapsto (1,2,3)$, $b \mapsto (1,2)$. So $|F:K|=6$ and $d(K)=7$.

Then $$[[a,b],[a^{-1},b^{-1}]], a^3, b^2, a^{-1} b a^{-1} b^{-1}, (a b)^2, b^{-1} a^{-1} b a b a^{-1} b^{-1}, b^{-1} a^{-1} b a b a b a b a^{-1} b^{-1}]$$ is a free basis of $K$, and since its first element lies in $F''$, it needs fewer than $7$ generators modulo $F''$.

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