CounterexampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28

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An arbitrary transversal need not give the reduced Schreier basis

Statement refuted

Any transversal of right cosets automatically yields the reduced Schreier basis.

Facts & Assumptions

Given: The false claim above.

[L1]

For this counterexample, if $T$ is any right transversal containing $1$ , define its raw transversal elements by $r_{T} (t, x) = t x {\overline{t x}}^{- 1}$ . When $T$ is a Schreier system, these are the Schreier generators of Schreier generators in the right-coset convention.

[L2]

A Schreier system is stronger than an arbitrary transversal: it must be closed under initial segments (Schreier transversals and Schreier systems).

[L3]

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

Counterexample

technique · direct

1.1

Let $H \leq F (a, b)$ be the subgroup of words with even exponent sum in $a$ . Its two right cosets are $H$ and $H a$ . The set $T = {1, a b}$ is a transversal, but it is not a Schreier system because the initial segment $a$ of $a b$ is not in $T$ .

L2givenconstruct

2.1

Using [L1], the nontrivial raw transversal elements are $a b^{- 1} a^{- 1}$ , $b$ , $a b a$ , and $a b a^{- 1}$ . Indeed, the last one is $r_{T} (a b, b) = a b^{2} (a b)^{- 1} = a b a^{- 1}$ . This list is redundant because $a b^{- 1} a^{- 1} = (a b a^{- 1})^{- 1}$ .

L1step 1.1algebra

3.1

By [L3], the reduced Schreier basis is guaranteed only for Schreier systems. Step 2.1 shows that the arbitrary transversal ${1, a b}$ instead gives a redundant list, so it does not yield the reduced Schreier basis. The Schreier initial-segment condition is load-bearing.

L3step 2.1∎

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