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) uses::Conjugacy-closed and Hall implies commutator subgroup equals intersection with whole commutator subgroup (this is in turn an immediate corollary of the uses::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.