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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.

A Hall subgroup for a prime set $\pi$ is a subgroup $H\le G$ whose order has only prime divisors in $\pi$ and whose index has no prime divisor in $\pi$:Here one is allowed in either class, and ${\pi }^{\prime }$ is the complementary set of primes. Equivalently $|H|$ contains the complete prime-power contribution to $|G|$ for each prime in $\pi$.