LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-13

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.

Invariants for a group extension compose

Statement

For an extension $1 \to N \to G \overset{π}{\to} Q \to 1$ and a left $G$ -module $M$ , the subgroup $M^{N}$ has the well-defined action $π (g) m = g m$ , and $(M^{N})^{Q} = M^{G}$ naturally as abelian groups.

Facts & Assumptions

Given: The extension and module above.

[F1]

Invariants are elements fixed by every element of the acting group (The invariants functor).

[F2]

The kernel $N$ is normal and $G / N ≅ Q$ (In a group extension the kernel is normal and the quotient recovers the base).

Proof

1.1

If $m \in M^{N}$ , $g \in G$ and $n \in N$ , then $n (g m) = g (g^{- 1} n g) m = g m$ by normality. Thus $M^{N}$ is $G$ -stable. If $g^{'} = g n$ has the same image in $Q$ , then $g^{'} m = g n m = g m$ . The proposed action is independent of a lift, and its identity and product laws follow from those of the $G$ -action. No simultaneous selection of lifts is needed.

F1F2

2.1

Every $G$ -fixed element is $N$ -fixed and is fixed by each quotient element acting as in step 1.1. Conversely, if $m \in (M^{N})^{Q}$ , then $g m = π (g) m = m$ for every $g \in G$ . This proves equality in both directions. A $G$ -linear map sends fixed elements to fixed elements and respects the quotient action, so the equality is natural. It also applies to $M = 0$ , $N = 1$ and $Q = 1$ .

F1step 1.1∎

Depends on

Used by

LHS defines low-degree group cohomology False statement
N-invariants send injective G-modules to Q-acyclics Lemma
Lyndon-Hochschild-Serre spectral sequence Theorem

Dependency tree · two levels

6 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

Weibel, 6.8.2 (standard reference, not scraped)