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.

[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).

Proof

technique · direct
1.1

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.

L1L2L3L4givenalgebra
2.1

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.

step 1.1L1L2L3L4givenalgebra
3.1

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

step 2.1∎

Depends on

Used by

Dependency tree · two levels

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Sources