Schur-Zassenhaus theorem

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This article states and (possibly) proves a fact that involves two finite groups of relatively prime order, requiring the additional datum that at least one of them is solvable. Due to the Feit-Thompson theorem, we know that for two finite groups of relatively prime orders, one of them is solvable. Hence, the additional datum of solvability can be dropped. However, the proof of the Feit-Thompson theorem is considered heavy machinery.
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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: For full proof, refer: Normal Hall implies permutably complemented
  2. All complements are conjugate: For full proof, refer: Hall retract implies order-conjugate