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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 NG 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 N1, then N is transitive.

Facts & Assumptions

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

[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 gNg1=N for every gG (Normal subgroup: invariance under conjugation).

Proof

technique · direct
1.1

Define xNy when y=nx for some nN. This is an equivalence relation because 1N, inverses in N reverse the relation, and products in N compose it.

givenalgebra
2.1

The relation N is G-invariant: if y=nx with nN, then for every gG one has gy=(gng1)(gx), and [L3] puts gng1 back in N.

L3step 1.1
3.1

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.

L1L2step 2.1
4.1

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 nN fixes every point of Ω.

step 3.1
5.1

If the action is faithful and N1, the second case of step 4.1 is impossible. Hence a nontrivial normal subgroup of a faithful primitive action is transitive.

step 4.1

Depends on

Used by

Dependency tree · two levels

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Sources