Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck 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.

Orbit-stabiliser: G/GxGxG/G_x\to G\cdot x, gGxgxgG_x\mapsto g\cdot x, is a well-defined bijection

Statement

Let GG act on XX and let xXx\in X. The rule

Φ:G/GxGx,Φ(gGx)=gx,\Phi:G/G_x\longrightarrow G\cdot x,\qquad \Phi(gG_x)=g\cdot x,

is well-defined and bijective. Thus every orbit is naturally in bijection with the left cosets of its stabilizer.

Facts & Assumptions

Given: A left action of a group GG on a set XX and a point xXx\in X.

[L1]

The orbit and stabilizer are Gx={gx:gG}G\cdot x=\{g\cdot x:g\in G\} and Gx={gG:gx=x}G_x=\{g\in G:g\cdot x=x\} (The orbit GxG\cdot x and stabilizer GxG_x of a point in a group action).

[L2]

The stabilizer GxG_x is a subgroup of GG (The stabilizer GxG_x is a subgroup of GG).

[L3]

For a subgroup HGH\le G, the left coset represented by gg is gH={gh:hH}gH=\{gh:h\in H\} (Left and right cosets gHgH and HgHg of a subgroup).

[L4]

For HGH\le G, one has gH=hHgH=hH exactly when g1hHg^{-1}h\in H (xaHx\in aH iff a1xHa^{-1}x\in H, and aH=bHaH=bH iff a1bHa^{-1}b\in H).

[L5]

A function is bijective exactly when it is injective and surjective (Injection, surjection, bijection).

Proof

technique · constructive
1.1

Define Φ(gGx)=gx\Phi(gG_x)=g\cdot x. If gGx=hGxgG_x=hG_x, then g1hGxg^{-1}h\in G_x by [L4], so (g1h)x=x(g^{-1}h)\cdot x=x by [L1], and the action law gives hx=g((g1h)x)=gxh\cdot x=g\cdot((g^{-1}h)\cdot x)=g\cdot x; hence Φ\Phi is well-defined.

L1L2L3L4L5construct
2.1

Every yGxy\in G\cdot x has the form y=gx=Φ(gGx)y=g\cdot x=\Phi(gG_x) by [L1], so Φ\Phi is surjective.

step 1.1L1
3.1

If Φ(gGx)=Φ(hGx)\Phi(gG_x)=\Phi(hG_x), then gx=hxg\cdot x=h\cdot x, so (g1h)x=x(g^{-1}h)\cdot x=x and g1hGxg^{-1}h\in G_x; [L4] gives gGx=hGxgG_x=hG_x, so Φ\Phi is injective and therefore bijective.

step 1.1L1L4L5discharge-construct

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 25 results over 14 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