Conjugacy-closed abelian Sylow implies retract

Statement
Suppose $$G$$ is a finite group, and $$p$$ is a prime dividing the order of $$G$$. Further, suppose $$P$$ is a $$p$$-Sylow subgroup, and $$P$$ is Abelian as a group, and is also conjugacy-closed: any two elements of $$P$$ that are conjugate in $$G$$ are conjugate in $$P$$. (Note that this effectively means that no two distinct elements of $$P$$ are conjugate in $$G$$).

Then, there exists a normal p-complement: a normal subgroup $$N$$ such that $$NP = G$$, and $$N \cap P$$ is trivial. In other words, $$P$$ is a retract of $$G$$.

Applications

 * 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.

Stronger facts

 * 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

Opposite facts

 * Conjugacy-closed and Hall not implies retract

Facts used

 * 1) uses::Focal subgroup theorem
 * 2) 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.

Proof
Given: A finite group $$G$$, a $$p$$-Sylow subgroup $$P$$ that is conjugacy-closed in $$G$$ and abelian.

To prove: There exists a normal $$p$$-complement $$N$$.

Proof:


 * 1) $$P \cap [G,G] = [P,P]$$: By fact (1), $$P \cap [G,G]$$ is generated by elements of the form $$x^{-1}y$$ where $$x,y \in P$$ are conjugate in $$G$$. Since $$P$$ is conjugacy-closed in $$G$$, this is the same as saying that $$x,y$$ are conjugate in $$P$$, so $$P \cap [G,G] = [P,P]$$.
 * 2) $$P \cap [G,G]$$ is trivial: This follows from the previous step, and the observation that $$P$$ is abelian.
 * 3) Consider the quotient by $$[G,G]$$, and denote the image of $$P$$ as $$\overline{P}$$.  Since $$G/[G,G]$$, the abelianization of $$G$$, is an Abelian group, and $$\overline{P}$$ is a Sylow subgroup, there exists a normal complement $$\overline{N}$$. Let $$N$$ be the inverse image of $$\overline{N}$$ under the quotient map. Clearly, $$NP = G$$, since it contains $$[G,G]$$ and its image is the whole of $$G/[G,G]$$. Further, $$N \cap P$$ is trivial, because $$P$$ does not intersect $$[G,G]$$, and the images $$\overline{P}$$ and $$\overline{N}$$ intersect trivially. Thus, $$N$$ is the required normal complement.