False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28

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FALSE: the raw Schreier generators are always a free basis

Statement

The full list of Schreier generators attached to a transversal is always a free basis.

Facts & Assumptions

Given: The false claim above.

[L1]

Schreier generators are the elements $s (t, x) = t x {\overline{t x}}^{- 1}$ (Schreier generators in the right-coset convention).

[L2]

A Schreier system is a transversal closed under initial segments (Schreier transversals and Schreier systems).

[L3]

Nielsen-Schreier keeps only the nontrivial generators from a Schreier system (Under the stated choice boundary, every subgroup of a free group is free with its nontrivial Schreier generators as a basis).

Refutation

technique · direct

1.1

Let $H \leq F (a, b)$ be the index-two subgroup consisting of words with even exponent sum in $a$ , and use the Schreier system $T = {1, a}$ for its two right cosets.

L2givenconstruct

2.1

The raw Schreier generators are $s (1, a) = 1$ , $s (1, b) = b$ , $s (a, a) = a^{2}$ , and $s (a, b) = a b a^{- 1}$ . So the full list already contains the identity element.

L1step 1.1algebra

3.1

A free basis cannot contain the identity, whereas [L3] keeps only the nontrivial generators and thereby produces the actual basis ${b, a^{2}, a b a^{- 1}}$ . Hence the raw list is not always a free basis.

L3step 2.1∎

Depends on

Used by

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Sources

M. I. Kargapolov and Ju. I. Merzljakov, Fundamentals of the Theory of Groups (standard reference, not scraped)