# 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 $NH = G$.

## 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.