LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-01

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A relator set and its symmetrisation have the same normal closure

Statement

Let $R$ be a cyclically reduced relator set and let $R^{sym}$ be its symmetrisation. Then $R$ and $R^{sym}$ have the same normal closure in the free group on the generators.

Facts & Assumptions

Given: A cyclically reduced relator set $R$ in a free group $F (X)$ , and its symmetrisation $R^{sym}$ .

[F1]

The normal closure of a subset $S \subseteq F (X)$ is the smallest normal subgroup of $F (X)$ containing $S$ (The normal closure of a subset of a group).

[L1]

Every element of $R^{sym}$ is either a cyclic conjugate of a member of $R$ or of its inverse (The symmetrisation of a relator set closes under inverses and cyclic conjugates).

Proof

technique · direct

1.1

Let $N = ⟨ ⁣ ⟨ R ⟩ ⁣ ⟩$ . Because $N$ is normal by [F1], it contains $r^{- 1}$ whenever it contains $r$ , and it contains $u r u^{- 1}$ for every $u \in F (X)$ . Hence [L1] implies that every element of $R^{sym}$ already lies in $N$ . Therefore $⟨ ⁣ ⟨ R^{sym} ⟩ ⁣ ⟩ \subseteq N$ .

F1L1given

1.2

Every relator of $R$ belongs to $R^{sym}$ by definition, so the normal closure of $R^{sym}$ contains $R$ . By the minimality clause of [F1], $N \subseteq ⟨ ⁣ ⟨ R^{sym} ⟩ ⁣ ⟩$ .

F1L1given

2.1

The two containments from steps 1.1 and 1.2 are equalities, so the normal closures agree.

step 1.1step 1.2∎

Depends on

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Sources

GAP SmallCancellation manual, Chapter 1: Small Cancellation Theory — the classical conditions (standard reference, not scraped)
Jay Williams, Universal Countable Borel Quasi-Orders (standard reference, not scraped)
Nicholas Touikan, An Introduction to Combinatorial and Geometric Group Theory, Section 3.5 (standard reference, not scraped)
Clara Löh, Geometric Group Theory: An Introduction, Section 7.4.1 (standard reference, not scraped)