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

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FALSE: every nontrivial normal subgroup of a faithful primitive group is regular

Statement

Every nontrivial normal subgroup of a faithful primitive group is regular.

Facts & Assumptions

Given: The natural action of $S_{n}$ on ${1, \dots, n}$ for $n \geq 4$ .

[L1]

The natural action of $S_{n}$ is sharply $n$ -transitive, and hence $n$ -transitive; $k$ -transitivity implies every lower transitivity level (The natural actions of symmetric and alternating groups, k-transitivity implies k-homogeneity and lower transitivity).

[L2]

For every natural $n$ , the alternating group $A_{n}$ is a normal subgroup of $S_{n}$ ( $A_{n}$ is normal in $S_{n}$ ; for $n \geq 2$ , $2 ∣ A_{n} ∣ = n!$ , while $A_{n} = S_{n}$ for $n = 0, 1$ ).

[L3]

Every doubly transitive action is primitive (Every doubly transitive action is primitive).

Refutation

technique · direct

1.1

By [L2], $A_{n}$ is normal in $S_{n}$ , and it is nontrivial because it contains the $3$ -cycle $(1 2 3)$ . The natural $S_{n}$ -action is faithful, and [L1] makes it doubly transitive, hence primitive by [L3].

L1L2L3algebra

2.1

The element $(1 2 3) \in A_{n}$ fixes the point $4$ , so the action of $A_{n}$ is not free and therefore not regular. Thus $A_{n}$ is a nontrivial normal subgroup of a faithful primitive action that is not regular.

step 1.1algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

15 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

J. S. Milne, Group Theory, Chapter 4 (standard reference, not scraped)