LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-28

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Schreier rewriting is invariant under free reduction

Statement

Let $F (X)$ be a free group, let $H \leq F (X)$ , let $T$ be a Schreier system, and let $τ$ be its Schreier rewriting map. If $u$ and $v$ are freely equivalent words on $X ⊔ X^{- 1}$ , then their Schreier rewrites are freely equivalent words in the Schreier generators:

$τ (u) \sim_{f r e e} τ (v) .$

In particular, the two rewrites represent the same element of the subgroup.

Facts & Assumptions

Given: A free group $F (X)$ , a subgroup $H \leq F (X)$ , a Schreier system $T$ , its rewriting map $τ$ , and freely equivalent words $u$ and $v$ on $X ⊔ X^{- 1}$ .

[L1]

Elementary cancellations delete adjacent inverse pairs $a a^{- 1}$ or $a^{- 1} a$ (Words in an alphabet with formal inverses, elementary cancellation, and reduced words).

[L2]

The rewrite $τ (w)$ is obtained by tracking the successive coset representatives of the prefixes of $w$ ; a letter $x$ contributes $s (t_{j - 1}, x)$ and a letter $x^{- 1}$ contributes $s (t_{j}, x)^{- 1}$ (The Schreier rewriting map).

Proof

technique · direct

1.1L1L2givenalgebra

It is enough to treat one elementary cancellation. By symmetry it suffices to consider $u = p x x^{- 1} q$ and $v = p q$ with $x \in X$ . Let $t = \overline{p}$ and $u_{1} = \overline{p x}$ . In the rewrite of $u$ , the letter $x$ contributes $s (t, x)$ and the following letter $x^{- 1}$ contributes $s (t, x)^{- 1}$ , so these two adjacent letters freely cancel.

2.1L1L2step 1.1∎

After those two letters are read, the current coset is again $H p$ , so the successive representatives used for the remaining suffix $q$ are exactly the same whether one starts from $u$ or from $v$ . Thus one elementary free cancellation turns $τ (u)$ into $τ (v)$ . Repeating this argument along a finite chain of elementary cancellations and reverse insertions proves that the two rewrites are freely equivalent, and hence represent the same subgroup element.

Depends on

Used by

Dependency tree · two levels

5 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Roger C. Lyndon and Paul E. Schupp, Combinatorial Group Theory (standard reference, not scraped)
M. I. Kargapolov and Ju. I. Merzljakov, Fundamentals of the Theory of Groups (standard reference, not scraped)