Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02
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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Coset multiplication (gH)(hH)=ghH is well defined if and only if H is normal

Statement

Let H≤G. The rule on left cosets

(aH)(bH):=abH

is independent of the representatives a and b if and only if H⊴G.

Facts & Assumptions

Given: A group G and a subgroup H≤G.

[F1]

The proposed coset product sends the pair (aH,bH) to abH (The quotient group G/N and coset product (gN)(hN)=ghN).

[L1]

A subgroup H is normal if and only if g−1Hg⊆H for every g∈G (Equivalent characterisations of a normal subgroup by conjugates and left and right cosets).

[L2]

For left cosets, aH=a′H if and only if a−1a′∈H (x∈aH iff a−1x∈H, and aH=bH iff a−1b∈H).

[F2]

A subgroup contains products of its elements (Subgroup).

Proof

technique · direct
1.1

Suppose H⊴G and aH=a′H, bH=b′H. By [L2], write a′=ah1 and b′=bh2 with h1,h2∈H. Then (ab)−1a′b′=b−1h1bh2∈H by [L1] and [F2], so [L2] gives abH=a′b′H. Hence [F1] is independent of both representatives.

givenF1L1L2F2algebra
1.2

Conversely, suppose [F1] is well defined. For h∈H and g∈G, the equal cosets H=eH=hH give the same product with gH, so gH=(eH)(gH)=(hH)(gH)=hgH.

givenF1
2.1

The equality gH=hgH gives g−1hg∈H by [L2]. Thus g−1Hg⊆H for every g, and [L1] gives H⊴G.

step 1.2L1L2
3.1

Step 1.1 proves sufficiency and steps 1.2 and 2.1 prove necessity, establishing the biconditional.

step 1.1step 1.2step 2.1∎

Depends on

Used by

Cited to discharge well-definedness by The quotient group G/N and coset product (gN)(hN)=ghN.

Dependency tree · two levels

13 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