Alphabeta Math
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 Rsym⁡ be its symmetrisation. Then R and Rsym⁡ 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 Rsym⁡.

[F1]

The normal closure of a subset S⊆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 Rsym⁡ 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
1.1F1L1given

Let N=⟨ ⁣⟨R⟩ ⁣⟩. Because N is normal by [F1], it contains r−1 whenever it contains r, and it contains uru−1 for every u∈F(X). Hence [L1] implies that every element of Rsym⁡ already lies in N. Therefore ⟨ ⁣⟨Rsym⁡⟩ ⁣⟩⊆N.

1.2F1L1given

Every relator of R belongs to Rsym⁡ by definition, so the normal closure of Rsym⁡ contains R. By the minimality clause of [F1], N⊆⟨ ⁣⟨Rsym⁡⟩ ⁣⟩.

2.1step 1.1step 1.2∎

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

Depends on

Used by

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Sources