Conjugacy-closed abelian Hall implies retract

Statement

Suppose $G$ is a finite group and $H$ is an abelian conjugacy-closed Hall subgroup of $G$. Then, $H$ is a retract of $G$. In other words, $H$ possesses a normal complement $N$ in $G$: a normal subgroup $N$ of $G$ such that $N\cap H$ is trivial and $NH=G$.

Related facts

• Conjugacy-closed abelian Sylow implies retract
• Conjugacy-closed and Sylow implies retract
• Conjugacy-closed nilpotent Hall implies retract
• Conjugacy-closed and Hall not implies retract
• Burnside's normal p-complement theorem
• Thompson's normal p-complement theorem

Facts used

1. Conjugacy-closed and Hall implies commutator subgroup equals intersection with whole commutator subgroup (this is in turn an immediate corollary of the analogue of focal subgroup theorem for Hall subgroups.

Proof

Given: A finite group $G$ and an abelian conjugacy-closed Hall subgroup $H$ of $G$.

To prove: $H$ has a normal complement in $G$.

Proof:

1. $H\cap [G,G]$ is trivial: This follows from fact (1), and the given fact that $H$ is abelian.
2. $H$ has a normal complement: Consider the quotient by $[G,G]$, and denote the image of $H$ as ${\overline {H}}$. Since $G/[G,G]$, the abelianization of $G$, is an Abelian group, and ${\overline {H}}$ 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, $NH=G$, since it contains $[G,G]$ and its image is the whole of $G/[G,G]$. Further, $N\cap H$ is trivial, because $P$ does not intersect $[G,G]$, and the images ${\overline {H}}$ and ${\overline {N}}$ intersect trivially. Thus, $N$ is the required normal complement.