Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02
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.

Every subgroup of index two is normal

Statement

If H≤G and [G:H]=2, then H⊴G.

Facts & Assumptions

Given: A group G and a subgroup H≤G with [G:H]=2.

[F1]

The index [G:H] is the cardinality of the left-coset set G/H when that set is finite (The coset set G/H and the index [G:H] of a subgroup).

[L1]

The distinct left cosets of H partition G (The left cosets of a subgroup partition the group).

[L2]

The rule gH↦Hg−1 is a bijection from the left cosets of H to its right cosets (Inversion induces a bijection gH↦Hg−1 from left cosets to right cosets).

[L3]

For g∈G, one has gH=H if and only if g∈H; the corresponding right-coset statement follows from Hg=H if and only if g∈H (x∈aH iff a−1x∈H, and aH=bH iff a−1b∈H).

[L4]

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

Proof

technique · direct
1.1

If g∉H, then gH≠H by [L3]. Since [F1] and the hypothesis give exactly two left cosets, [L1] shows that H and gH are disjoint and cover G, so gH=G∖H.

givenF1L1L3
1.2

By [L2] there are exactly two right cosets. If g∉H, then Hg≠H by [L3]; the same elementary coset argument shows that distinct right cosets are disjoint and cover G, so Hg=G∖H.

givenL2L3algebra
2.1

If g∈H, then gH=H=Hg by [L3]; if g∉H, steps 1.1 and 1.2 give gH=G∖H=Hg. Thus gH=Hg for every g∈G, and [L4] gives H⊴G.

step 1.1step 1.2L3L4∎

Depends on

Used by

Dependency tree · two levels

16 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