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PropositionStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11
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The normal closure of RR is the set of finite products of conjugates of elements of RR and their inverses

Statement

Let GG be a group and RGR\subseteq G. Then

 ⁣R ⁣G={g1r1ε1g11gnrnεngn1:nN, giG, riR, εi{1,1}}.\langle\!\langle R\rangle\!\rangle_G=\left\{g_1r_1^{\varepsilon_1}g_1^{-1}\cdots g_nr_n^{\varepsilon_n}g_n^{-1}:n\in\mathbb N,\ g_i\in G,\ r_i\in R,\ \varepsilon_i\in\{1,-1\}\right\}.

For n=0n=0 the displayed product is the identity. Replacing every conjugator gig_i by gi1g_i^{-1} gives the equivalent convention gi1riεigig_i^{-1}r_i^{\varepsilon_i}g_i.

Facts & Assumptions

Given: A group GG, a subset RGR\subseteq G, and the set PP of displayed finite products.

[F1]

Group multiplication is associative: (xy)z=x(yz)(xy)z=x(yz) for all x,y,zGx,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 NGN\leq G is normal when gNg1=NgNg^{-1}=N for every gGg\in G (Normal subgroup: invariance under conjugation).

[L1]

The normal closure of RR is the smallest normal subgroup of GG containing RR (The normal closure of a subset of a group).

Proof

technique · direct
1.1

The empty product puts the identity in PP; concatenating two finite products keeps them in PP; and [L2] shows that the inverse of a product is the reverse product of factors (grεg1)1=grεg1(gr^\varepsilon g^{-1})^{-1}=gr^{-\varepsilon}g^{-1}. Thus PP is a subgroup of GG by [F2].

F1F2L2
1.2

Each rRr\in R is the one-factor product ere1ere^{-1}, so RPR\subseteq P.

given
1.3

Conversely, the normal subgroup  ⁣R ⁣G\langle\!\langle R\rangle\!\rangle_G contains every ri±1r_i^{\pm1} and, by normality, every conjugate giri±1gi1g_ir_i^{\pm1}g_i^{-1}; subgroup closure then contains every finite product in PP, including the empty product, so P ⁣R ⁣GP\subseteq\langle\!\langle R\rangle\!\rangle_G.

F2F3L1
2.1

For hGh\in G, conjugating a displayed product by hh replaces each factor giriεigi1g_ir_i^{\varepsilon_i}g_i^{-1} by (hgi)riεi(hgi)1(hg_i)r_i^{\varepsilon_i}(hg_i)^{-1}; hence hPh1PhPh^{-1}\subseteq P. Applying the same inclusion with h1h^{-1} and conjugating by hh gives the reverse inclusion, so hPh1=PhPh^{-1}=P and [F3] makes PP normal.

F1F3step 1.1
3.1

Since PP is a normal subgroup containing RR, minimality in [L1] gives  ⁣R ⁣GP\langle\!\langle R\rangle\!\rangle_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

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Sources