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Normal subgroups of a primitive action are transitive or lie in the kernel
Statement
Let act primitively on , and let be a normal subgroup. Then either acts transitively on , or every element of fixes every point of . 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 , then is transitive.
Facts & Assumptions
Given: A primitive action of on and a normal subgroup .
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).
Partitions into blocks are exactly the -invariant equivalence relations (G-invariant block systems are exactly the invariant equivalence relations).
A normal subgroup satisfies for every (Normal subgroup: invariance under conjugation).
Proof
Define when for some . This is an equivalence relation because , inverses in reverse the relation, and products in compose it.
The relation is -invariant: if with , then for every one has , and [L3] puts back in .
By [L2], the -classes form a block system. Since the action is primitive, [L1] makes that block system either the one-block partition or the singleton partition.
In the one-block case, every point lies in the -orbit of every other point, so is transitive. In the singleton case, every -orbit has one point, so each fixes every point of .
If the action is faithful and , the second case of step 4.1 is impossible. Hence a nontrivial normal subgroup of a faithful primitive action is transitive.
Depends on
Used by
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Sources
- J. S. Milne, Group Theory, Chapter 4 (standard reference, not scraped)
- K. Conrad, Transitive Group Actions (standard reference, not scraped)