Conjugacy-closed abelian Sylow implies retract
This article gives the statement, and possibly proof, of a normal p-complement theorem: necessary and/or sufficient conditions for the existence of a Normal p-complement (?). In other words, it gives necessary and/or sufficient conditions for a given finite group to be a P-nilpotent group (?) for some prime number .
View other normal p-complement theorems
Suppose is a finite group, and is a prime dividing the order of . Further, suppose is a -Sylow subgroup, and is Abelian as a group, and is also conjugacy-closed: any two elements of that are conjugate in are conjugate in . (Note that this effectively means that no two distinct elements of are conjugate in ).
Then, there exists a normal p-complement: a normal subgroup such that , and is trivial. In other words, is a retract of .
- Burnside's normal p-complement theorem: Burnside's normal p-complement theorem states that if a Sylow subgroup is central in its normalizer, it is a retract. The proof of Burnside's normal p-complement theorem relies on this fact, along with a local conjugacy-determination property in the normalizer.
- Conjugacy-closed and Sylow implies retract: A stronger version of the theorem with the assumption of Abelianness dropped. The proof is somewhat harder in this case.
- Conjugacy-closed abelian Hall implies retract
- Focal subgroup theorem
- Any Abelian group is a direct product of its Sylow subgroups. In particular, a Sylow subgroup of an Abelian group is a direct factor, and has a normal complement.
Given: A finite group , a -Sylow subgroup that is conjugacy-closed in and abelian.
To prove: There exists a normal -complement .
- : By fact (1), is generated by elements of the form where are conjugate in . Since is conjugacy-closed in , this is the same as saying that are conjugate in , so .
- is trivial: This follows from the previous step, and the observation that is abelian.
- Consider the quotient by , and denote the image of as . Since , the abelianization of , is an Abelian group, and is a Sylow subgroup, there exists a normal complement . Let be the inverse image of under the quotient map. Clearly, , since it contains and its image is the whole of . Further, is trivial, because does not intersect , and the images and intersect trivially. Thus, is the required normal complement.