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

If H≤G and N⊴G, then HN is a subgroup and H∩N⊴H

Statement

If H≤G and N⊴G, then HN is a subgroup and H∩N⊴H.

Here HN:={hn:h∈H, n∈N}.

Facts & Assumptions

Given: A subgroup H≤G and a normal subgroup N⊴G.

[L1]

A nonempty subset closed under xy−1 is a subgroup (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).

[L2]

Normality means gNg−1=N for every g∈G (Normal subgroup: invariance under conjugation).

[L3]

A subgroup is normal if its conjugates by ambient elements lie in it (Equivalent characterisations of a normal subgroup by conjugates and left and right cosets).

[L4]

(ab)−1=b−1a−1 in a group (In a group e−1=e, (g−1)−1=g and (gh)−1=h−1g−1, the order of the last product being essential).

Proof

technique · direct
1.1L1L2L3L4givenalgebra

The identity lies in HN; for h1n1,h2n2∈HN, put m=n1n2−1∈N, so (h1n1)(h2n2)−1=h1mh2−1=h1h2−1(h2mh2−1)∈HN.

2.1step 1.1L1L2L3L4givenalgebra

Thus [L1] gives HN≤G; moreover for h∈H and x∈H∩N, both hxh−1∈H and hxh−1∈N, so it lies in H∩N.

3.1step 2.1∎

The conjugation closure in step 2.1 gives H∩N⊴H.

Depends on

Dependency tree · 2 levels

Sources