A converse to Schreier's index-rank formula

TL;DR

This paper provides a proof that a finitely generated, residually finite group is free if its subgroups of finite index satisfy Schreier's index-rank relation, addressing a question raised in prior research.

Contribution

It offers a proof confirming that the index-rank relation characterizes free groups among finitely generated, residually finite groups, filling a gap in the literature.

Findings

Confirmed the characterization of free groups via Schreier's relation

Provided a proof addressing an open question from previous work

Clarified the conditions under which a group is free

Abstract

In "Subgroups of free profinite groups and large subfields of Q" (Israel J. Math. 39 (1981), no. 1-2, pages 25-45; MR 617288) A. Lubotzky and L. van den Dries raise the question whether a finitely generated, residually finite group is necessarily free if the rank function on its subgroups of finite index satisfies Schreier's well-known index rank relation (see Question 2 on p. 34). I answered this question in 1980 but, so far, I have not published my answer. This note fills the omission; it gives an amended and abridged version of my original proof.