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LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11
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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.

CG(x)C_G(x) and NG(H)N_G(H) are subgroups of GG

Statement

For every group GG, element xGx\in G, and subgroup HGH\le G, both the centralizer CG(x)C_G(x) and the normalizer NG(H)N_G(H) are subgroups of GG.

Facts & Assumptions

Given: A group GG, an element xGx\in G, and a subgroup HGH\le G.

[L1]

The centralizer is CG(x)={gG:gx=xg}C_G(x)=\{g\in G:gx=xg\} (The conjugacy class ClG(x)\operatorname{Cl}_G(x) and centralizer CG(x)C_G(x) of an element).

[L2]

The normalizer is NG(H)={gG:gHg1=H}N_G(H)=\{g\in G:gHg^{-1}=H\} (The normalizer NG(H)={gG:gHg1=H}N_G(H)=\{g\in G:gHg^{-1}=H\} of a subgroup).

Proof

technique · direct
1.1

The identity commutes with xx. If a,bCG(x)a,b\in C_G(x), then b1b^{-1} commutes with xx, and hence (ab1)x=a(b1x)=a(xb1)=x(ab1)(ab^{-1})x=a(b^{-1}x)=a(xb^{-1})=x(ab^{-1}); [L3] gives CG(x)GC_G(x)\le G.

L1L3
1.2

The identity normalizes HH. If a,bNG(H)a,b\in N_G(H), then b1Hb=Hb^{-1}Hb=H, and therefore (ab1)H(ab1)1=a(b1Hb)a1=aHa1=H(ab^{-1})H(ab^{-1})^{-1}=a(b^{-1}Hb)a^{-1}=aHa^{-1}=H.

L2
2.1

Applying [L3] to step 1.2 gives NG(H)GN_G(H)\le G, so both asserted sets are subgroups.

step 1.1step 1.2L3

Depends on

Used by

Cited to discharge well-definedness by The conjugacy class Cl_G(x) and centralizer C_G(x) of an element and The normalizer N_G(H)={g∈ G:gHg⁻¹=H} of a subgroup.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 15 results over 9 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources