Schur-Zassenhaus theorem

Statement
Let $$G$$ be a finite group and $$H$$ be a normal Hall subgroup (viz a Hall subgroup which is also a normal subgroup). Then,


 * 1) $$H$$ is a permutably complemented subgroup. In other words, there exists a subgroup $$K$$ such that $$H \cap K$$ is trivial and $$HK = G$$.
 * 2) Any two complements to $$H$$ are conjugate subgroups

The second statement does not yet have a direct elementary proof; rather, there is a proof assuming that either $$H$$ or $$K$$ is solvable. However, a well-known corollary of the Feit-Thompson theorem tells us that this is always true.

Proof breakup

 * 1) Existence of a complement:
 * 2) All complements are conjugate: