Existence of pi-subgroups for all prime sets pi is equivalent to existence of p-complements for all primes p

Statement
The following are equivalent for a finite group $$G$$:


 * 1) $$G$$ has $$\pi$$-Hall subgroups for all prime sets $$\pi$$. (Note that we can restrict attention to only those prime sets that are contained in the set of prime divisors of the order of $$G$$, because other primes are irrelevant to the definition of Hall subgroup and can be ignored).
 * 2) $$G$$ has p-complements for all prime numbers $$p$$. (Note that we can restrict attention to only those primes that divide the order of $$G$$, because for other primes, $$G$$ itself is a p-complement).

Related facts
The following two facts together show that the equivalent conditions that are the subject of this page occur precisely in finite solvable groups:


 * ECD condition for pi-subgroups in finite solvable groups shows that in a finite solvable group, these equivalent conditions are satisfied.
 * Hall's theorem shows that if these equivalent conditions are satisfied in a finite group, then it must be solvable.

(1) implies (2)
This follows directly: for each prime $$p$$, take $$\pi$$ as the set of all prime divisors of the order of $$G$$ other than $$p$$. Use (1) to demonstrate the existence of a $$\pi$$-Hall subgroup, which is therefore also a $$p$$-complement.

(2) implies (1)
Given: A finite group $$G$$, a set $$\pi$$ of prime divisors of the order of $$G$$. $$G$$ has a $$p$$-complement for every prime $$p$$ dividing the order of $$G$$.

To prove: $$G$$ has a $$\pi$$-Hall subgroup.

Proof: