Hall subgroup

A Hall subgroup has order coprime to its index. For a set of primes $\pi$, a Hall $\pi$-subgroup has order using only primes of $\pi$ and index using only primes outside $\pi$. Thus its order contains the full prime-power contributions belonging to $\pi$. A Sylow subgroup is the single-prime instance. Hall subgroup existence in soluble groups generalizes Sylow existence to every prime set.

A finite normal subgroup whose order is coprime to its index has a group complement: if $N◃G$ is a Hall subgroup, then $G=NH$ and $N\cap H=1$ for some subgroup $H$. Complement conjugacy holds when either $N$ or $G/N$ is solvable; the existence assertion does not require that extra hypothesis.

For any prime set $\pi$, every finite soluble group has Hall $\pi$-subgroups, any two are conjugate, and every $\pi$-subgroup lies in one. Induction uses an elementary abelian minimal normal subgroup $N$ of characteristic $p$. If $p\in \pi$, the preimage of a quotient Hall subgroup is itself a Hall subgroup. If $p\notin \pi$, coprime splitting over an elementary abelian normal subgroup gives a conjugate class of complements in that preimage. The same complement conjugacy in the preimage of a smaller quotient subgroup gives the embedding assertion.

Every finite soluble group has a Hall subgroup for each prime set. Induct on the group order using an elementary abelian minimal normal subgroup. In the coprime hard case, lift a minimal normal subgroup of the quotient, choose its Sylow subgroup, and apply the Frattini argument. A proper normalizer reduces the order; a normal Sylow subgroup allows induction in its quotient. This proves existence without assuming an independent complement theorem.