LemmaStatement: AI-generatedProof: AI-generatedPipeline-generatedverified 2026-09-24 (gpt-6-sol)

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.

A normal quasisimple subgroup and a subnormal subgroup

Statement

Let $Q ⊴ G$ be quasisimple and let $S$ be a subnormal subgroup of $G$ . Then either $Q \leq S$ or $[Q, S] = 1$ .

Facts & Assumptions

Given: $Q, S, G$ as in the statement.

[F1]

A subnormal subgroup has a finite chain $S = S_{0} ⊴ S_{1} ⊴ \dots ⊴ S_{d} = G$ . (Subnormal and normal series, factors, refinements, and equivalence)

[F2]

If a normal subgroup of a quasisimple group is proper, it lies in the center. A perfect group is centralized by a subgroup whose conjugation commutators all lie in that center. (Quasisimple normal intersections and perfect central actions)

Proof

1.1

Choose a subnormal chain from F1. If $Q \leq S_{0} = S$ , the first alternative holds. Otherwise, because $Q \leq S_{d} = G$ , there is an index $j < d$ such that $Q \leq S_{j + 1}$ but $Q ≰ S_{j}$ . We only select an index from the given finite chain.

F1given

2.1

Both $Q$ and $S_{j}$ are normal in $S_{j + 1}$ , since $Q ⊴ G$ and $S_{j} ⊴ S_{j + 1}$ . Thus $Q \cap S_{j}$ is a proper normal subgroup of $Q$ , and F2 puts it in $Z (Q)$ . For $q \in Q$ and $s \in S_{j}$ , their commutator lies in both $Q$ and $S_{j}$ , hence in $Z (Q)$ . Since $S_{j}$ normalizes the perfect group $Q$ , the second clause of F2 gives $[Q, S_{j}] = 1$ . As $S \leq S_{j}$ , it follows that $[Q, S] = 1$ . No finiteness of $G$ beyond the finite subnormal chain is needed.

F2step 1.1algebra∎

Depends on

Used by

Distinct components commute Lemma

Dependency tree · two levels

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