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TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-26
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Normal subgroups of a primitive action are transitive or lie in the kernel

Statement

Let G act primitively on Ω, and let N⊴G be a normal subgroup. Then either N acts transitively on Ω, or every element of N fixes every point of Ω. In other words, every normal subgroup of a primitive action is either transitive or contained in the action kernel.

In particular, if the action is faithful and N≠1, then N is transitive.

Facts & Assumptions

Given: A primitive action of G on Ω and a normal subgroup N⊴G.

[L1]

A primitive action is a transitive action whose only block systems are the singleton partition and the one-block partition (Primitive and imprimitive transitive actions).

[L2]

Partitions into blocks are exactly the G-invariant equivalence relations (G-invariant block systems are exactly the invariant equivalence relations).

[L3]

A normal subgroup satisfies gNg−1=N for every g∈G (Normal subgroup: invariance under conjugation).

Proof

technique · direct
1.1givenalgebra

Define x∼Ny when y=n⋅x for some n∈N. This is an equivalence relation because 1∈N, inverses in N reverse the relation, and products in N compose it.

2.1L3step 1.1

The relation ∼N is G-invariant: if y=n⋅x with n∈N, then for every g∈G one has g⋅y=(gng−1)⋅(g⋅x), and [L3] puts gng−1 back in N.

3.1L1L2step 2.1

By [L2], the ∼N-classes form a block system. Since the action is primitive, [L1] makes that block system either the one-block partition or the singleton partition.

4.1step 3.1

In the one-block case, every point lies in the N-orbit of every other point, so N is transitive. In the singleton case, every N-orbit has one point, so each n∈N fixes every point of Ω.

5.1step 4.1∎

If the action is faithful and N≠1, the second case of step 4.1 is impossible. Hence a nontrivial normal subgroup of a faithful primitive action is transitive.

Depends on

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Sources