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 $p$.
View other normal p-complement theorems

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 $G$.

Related facts

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. 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.