Characteristic subgroup of Sylow subgroup is weakly closed iff it is normal in every Sylow subgroup containing it
Suppose is a finite group, is a prime number, is a -Sylow subgroup (?), and is a Characteristic subgroup (?) of . Then, is a Weakly closed subgroup (?) in (relative to ) if and only if is a normal subgroup in every -Sylow subgroup containing it.
- Weakly closed implies conjugation-invariantly relatively normal in finite group
- Sylow implies order-conjugate: For a prime , any two -Sylow subgroups are conjugate.
- Sylow implies WNSCDIN
- WNSCDIN implies every normalizer-relatively normal conjugation-invariantly relatively normal subgroup is weakly closed
Given: A finite group , a prime , a -Sylow subgroup of . A characteristic subgroup of .
Weakly closed implies normal in every Sylow subgroup containing it
To prove: Given that is weakly closed in , and is a -Sylow subgroup of containing , is normal in .
Proof (quick version): This follows from facts (1) and (2). By fact (2), and are conjugate, and by fact (1), is normal in . (Note that this part does not use the assumption that is a characteristic subgroup of ).Proof (hands-on): PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]
Normal in every Sylow subgroup containing it implies weakly closed
To prove: Given that is normal in every -Sylow subgroup of containing it, is weakly closed in . In other words, if is such that , then .
Proof (quick version): The proof follows from facts (3) and (4), and the observation (again stemming from fact (2)) that the conjugates of a particular -Sylow subgroup are precisely all the -Sylow subgroups.Proof (hands-on): PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]