PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The normal closure of $R$ is the set of finite products of conjugates of elements of $R$ and their inverses

Statement

Let $G$ be a group and $R \subseteq G$ . Then

$⟨ ⁣ ⟨ R ⟩ ⁣ ⟩_{G} = {g_{1} r_{1}^{ε_{1}} g_{1}^{- 1} \dots g_{n} r_{n}^{ε_{n}} g_{n}^{- 1} : n \in N, g_{i} \in G, r_{i} \in R, ε_{i} \in {1, - 1}} .$

For $n = 0$ the displayed product is the identity. Replacing every conjugator $g_{i}$ by $g_{i}^{- 1}$ gives the equivalent convention $g_{i}^{- 1} r_{i}^{ε_{i}} g_{i}$ .

Facts & Assumptions

Given: A group $G$ , a subset $R \subseteq G$ , and the set $P$ of displayed finite products.

[F1]

Group multiplication is associative: $(x y) z = x (y z)$ for all $x, y, z \in G$ (Group and abelian group).

[F2]

A subset of a group is a subgroup when it contains the identity and is closed under products and inverses (Subgroup).

[F3]

A subgroup $N \leq G$ is normal when $g N g^{- 1} = N$ for every $g \in G$ (Normal subgroup: invariance under conjugation).

[L1]

The normal closure of $R$ is the smallest normal subgroup of $G$ containing $R$ (The normal closure of a subset of a group).

[L2]

For group elements $g, h$ , $(g h)^{- 1} = h^{- 1} g^{- 1}$ (In a group $e^{- 1} = e$ , $(g^{- 1})^{- 1} = g$ and $(g h)^{- 1} = h^{- 1} g^{- 1}$ , the order of the last product being essential).

Proof

technique · direct

1.1

The empty product puts the identity in $P$ ; concatenating two finite products keeps them in $P$ ; and [L2] shows that the inverse of a product is the reverse product of factors $(g r^{ε} g^{- 1})^{- 1} = g r^{- ε} g^{- 1}$ . Thus $P$ is a subgroup of $G$ by [F2].

F1F2L2

1.2

Each $r \in R$ is the one-factor product $e r e^{- 1}$ , so $R \subseteq P$ .

given

1.3

Conversely, the normal subgroup $⟨ ⁣ ⟨ R ⟩ ⁣ ⟩_{G}$ contains every $r_{i}^{\pm 1}$ and, by normality, every conjugate $g_{i} r_{i}^{\pm 1} g_{i}^{- 1}$ ; subgroup closure then contains every finite product in $P$ , including the empty product, so $P \subseteq ⟨ ⁣ ⟨ R ⟩ ⁣ ⟩_{G}$ .

F2F3L1

2.1

For $h \in G$ , conjugating a displayed product by $h$ replaces each factor $g_{i} r_{i}^{ε_{i}} g_{i}^{- 1}$ by $(h g_{i}) r_{i}^{ε_{i}} (h g_{i})^{- 1}$ ; hence $h P h^{- 1} \subseteq P$ . Applying the same inclusion with $h^{- 1}$ and conjugating by $h$ gives the reverse inclusion, so $h P h^{- 1} = P$ and [F3] makes $P$ normal.

F1F3step 1.1

3.1

Since $P$ is a normal subgroup containing $R$ , minimality in [L1] gives $⟨ ⁣ ⟨ R ⟩ ⁣ ⟩_{G} \subseteq P$ .

L1step 1.2step 2.1

4.1

The inclusions of steps 3.1 and 1.3 give the displayed equality.

step 3.1step 1.3∎

Depends on

Used by

Dependency tree · two levels

12 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

Nicholas Touikan, An Introduction to Combinatorial and Geometric Group Theory, §1.4 (standard reference, not scraped)
M. Brittenham, Group presentations (standard reference, not scraped)

The normal closure of RR is the set of finite products of conjugates of elements of RR and their inverses