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A subgroup containing the normalizer of a Sylow subgroup is self-normalizing
Statement
Let be a Sylow -subgroup of a finite group . If , then . In particular, . See Sylow II: in a finite group every -subgroup lies in a conjugate of any Sylow -subgroup, and the Sylow -subgroups form a single conjugacy class.
Facts & Assumptions
Given: The hypotheses and objects in the Statement.
Let be finite, let be a Sylow -subgroup, and let be a -subgroup. There is with . In particular, for every Sylow -subgroup there is with , so the Sylow -subgroups form one conjugacy class. (Sylow II: in a finite group every -subgroup lies in a conjugate of any Sylow -subgroup, and the Sylow -subgroups form a single conjugacy class).
Let be a subgroup (def-subgroup). The normalizer of in is Thus exactly when the conjugation automorphism preserves setwise (thm-conjugation-is-an-automorphism). The subgroup property is proved in lem-centralizers-and-normalizers-are-subgroups. (The normalizer of a subgroup).
Proof
For and , the groups and are Sylow in .
Conjugate them inside ; the resulting element puts in .
Specialize to to obtain . This proves the stated claim.
Depends on
Used by
Dependency tree · two levels
8 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
- Keith Conrad, Consequences of the Sylow Theorems, Sections 1-5 (standard reference, not scraped)