TheoremStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-26

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.

Iwasawa's simplicity criterion for primitive actions

Statement

Let $G$ act faithfully and primitively on $Ω$ , and fix $α \in Ω$ . Write $G_{α} : = {g \in G : g \cdot α = α} .$ Assume $A ⊴ G_{α}$ is nontrivial and abelian, and that the conjugates ${g A g^{- 1} : g \in G}$ generate $G$ .

Then every nontrivial normal subgroup $N ⊴ G$ contains the commutator subgroup $[G, G]$ . In particular, if $G = [G, G]$ , then $G$ is simple.

Facts & Assumptions

Given: A faithful primitive action of $G$ on $Ω$ , a point $α \in Ω$ , a nontrivial abelian normal subgroup $A ⊴ G_{α}$ , and the conjugates of $A$ generate $G$ .

[L1]

In a faithful primitive action, every nontrivial normal subgroup is transitive (Normal subgroups of a primitive action are transitive or lie in the kernel).

[L2]

The commutator subgroup $[G, G]$ is the subgroup generated by all commutators $[g, h] = g h g^{- 1} h^{- 1}$ (Commutators $[g, h] = g h g^{- 1} h^{- 1}$ and the commutator subgroup $[G, G]$ ).

[L3]

A normal subgroup satisfies $g N g^{- 1} = N$ for every $g \in G$ (Normal subgroup: invariance under conjugation).

Proof

technique · direct

1.1

Let $N ⊴ G$ be nontrivial. By [L1], $N$ is transitive on $Ω$ . Hence for every $g \in G$ there is $n \in N$ with $n \cdot α = g \cdot α$ , and then $g^{- 1} n \in G_{α}$ . So $G = N G_{α}$ .

L1choose

2.1

Fix $g \in G$ , and write $g = n h$ with $n \in N$ and $h \in G_{α}$ as in step 1.1. Because $A ⊴ G_{α}$ , one has $h A h^{- 1} = A$ . For $a \in A$ , the element $n a n^{- 1} a^{- 1}$ lies in $N$ by [L3], so $n a n^{- 1} = (n a n^{- 1} a^{- 1}) a \in N A$ . Therefore $g A g^{- 1} = n (h A h^{- 1}) n^{- 1} = n A n^{- 1} \subseteq N A .$

step 1.1L3algebra

3.1

The conjugates of $A$ generate $G$ by hypothesis, and step 2.1 puts each of them inside $N A$ . Hence $G = N A$ . Modulo $N$ , this says $G / N$ is generated by the image of $A$ ; since $A$ is abelian, $G / N$ is abelian.

step 2.1

4.1

Because $G / N$ is abelian, every commutator of $G$ lies in $N$ . By [L2], the subgroup they generate is $[G, G]$ , so $[G, G] \leq N$ .

L2step 3.1

5.1

Step 4.1 holds for every nontrivial normal subgroup $N ⊴ G$ . Therefore if $G = [G, G]$ , every nontrivial normal subgroup contains all of $G$ and is equal to $G$ . So $G$ is simple.

step 4.1∎

Depends on

Used by

The finite Iwasawa criterion Corollary
Projective linear actions and Iwasawa's hypotheses Example
FALSE: transitivity alone forces nontrivial normal subgroups to be transitive False statement

Dependency tree · two levels

9 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

K. Conrad, Transitive Group Actions (standard reference, not scraped)
P. J. Cameron, Classical Groups, Sections 2.3-2.4 (standard reference, not scraped)