Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck 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.

CG(x) and NG(H) are subgroups of G

Statement

For every group G, element x∈G, and subgroup H≤G, both the centralizer CG(x) and the normalizer NG(H) are subgroups of G.

Facts & Assumptions

Given: A group G, an element x∈G, and a subgroup H≤G.

[L1]

The centralizer is CG(x)={g∈G:gx=xg} (The conjugacy class Cl⁡G(x) and centralizer CG(x) of an element).

[L2]

The normalizer is NG(H)={g∈G:gHg−1=H} (The normalizer NG(H)={g∈G:gHg−1=H} of a subgroup).

[L3]

A nonempty subset S is a subgroup if ab−1∈S whenever a,b∈S (One-step subgroup test: a nonempty H⊆G is a subgroup iff gh−1∈H for all g,h∈H; the identity and the inverses of H are then those of G).

Proof

technique · direct
1.1L1L3

The identity commutes with x. If a,b∈CG(x), then b−1 commutes with x, and hence (ab−1)x=a(b−1x)=a(xb−1)=x(ab−1); [L3] gives CG(x)≤G.

1.2L2

The identity normalizes H. If a,b∈NG(H), then b−1Hb=H, and therefore (ab−1)H(ab−1)−1=a(b−1Hb)a−1=aHa−1=H.

2.1step 1.1step 1.2L3∎

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

Depends on

Dependency tree · 2 levels

Sources