LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-12

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.

Green exceptional family containment and fusion

Statement

Use the finite group, $P$ , $H$ , and families of Green exceptional intersection families. For $Q \leq P$ , the following are equivalent:

$Q \leq_{G} X, Q \leq_{H} X, Q \leq_{H} Y .$

Every member of $X$ is proper in $P$ , and $P \in Z$ . If $Q, R \in Z$ and $^{g} Q = R$ , then $g \in H$ . No member of $Y$ contains an $H$ -conjugate of a member of $Z$ .

Facts & Assumptions

Given: The above data, in particular $Q \leq P \leq H$ and $N_{G} (P) \leq H$ .

[F1]

The families and existential conjugate-containment convention are those of Green exceptional intersection families.

Proof

1.1

Suppose $^{g} Q \leq P \cap^{s} P$ for $s \notin H$ . If $g \in H$ , this already witnesses $Q \leq_{H} X$ . If $g \notin H$ , the first containment gives $Q \leq^{g^{- 1}} P$ ; combining with $Q \leq P$ gives $Q \leq P \cap^{g^{- 1}} P \in X$ , witnessed using the identity conjugator in $H$ . Thus $Q \leq_{G} X$ implies $Q \leq_{H} X$ .

F1givenalgebra

1.2

If $P \cap^{s} P = P$ , then $P \leq^{s} P$ and the two finite groups have equal orders. Hence $P =^{s} P$ , so $s \in N_{G} (P) \leq H$ . This contradicts $s \notin H$ . Each $X$ -member therefore has order strictly less than $∣ P ∣$ and cannot contain any conjugate of $P$ . Thus $P \in Z$ .

F1givenalgebra

2.1

Since $P \leq H$ , every $P \cap^{s} P$ is contained in $H \cap^{s} P$ . Consequently an $H$ -conjugate containment in $X$ is one in $Y$ . Conversely, if $^{h} Q \leq H \cap^{s} P$ with $h \in H$ and $s \notin H$ , then $Q \leq^{h^{- 1} s} P$ . As $h^{- 1} s \notin H$ and $Q \leq P$ , this gives $Q \leq P \cap^{h^{- 1} s} P \in X$ . In particular $Q \leq_{G} X$ . Together with 1.1 this proves all three equivalences.

F1step 1.1givenalgebra

3.1

If $^{g} Q = R$ for $Q, R \in Z$ and $g \notin H$ , then $R \leq P$ and $R \leq^{g} P$ give $R \leq P \cap^{g} P \in X$ . This contradicts $R \in Z$ , proving the asserted fusion statement. If an $H$ -conjugate of $Q \in Z$ lay in a $Y$ -member, step 2.1 would give $Q \leq_{G} X$ , another contradiction.

F1step 2.1givenalgebra

4.1

When $H = G$ both exceptional families are empty: all three containment statements are false, and all conjugators belong to $H$ . When $P = 1$ the normalizer hypothesis forces this same case. For $Q = 1$ and $H \neq G$ , each exceptional family has a member and all three containments hold, so $1 \notin Z$ . The proof uses only finite subgroup containments and their displayed witnesses, with no choice principle and no assertion that a $Y$ -member has smaller order than $P$ . This completes the claims. [F1, step 1.1, step 2.1, step 1.2, step 3.1] QED

Depends on

Green exceptional intersection families

Used by

Dependency tree · two levels

3 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

Saunders, Modular Representation Theory, Lemmas 4.18–4.19 and 4.35–4.38, Theorem 4.34 (standard reference, not scraped)
Lassueur–Farrell, Chapter 7, §29, Theorem 29.4 and proof (standard reference, not scraped)