Hall subgroups need not exist

Statement
Let $$G$$ be a finite group and $$\pi$$ be a set of prime numbers. Then, there need not exist a $$\pi$$-fact about::Hall subgroup of $$G$$. In other words, there need not exist a subgroup whose order and index are relatively prime, and where all prime factors of the order are in $$\pi$$.

Facts about Sylow subgroups

 * Sylow subgroups exist

More facts about existence of Hall subgroups

 * Hall subgroups exist in finite solvable
 * Hall's theorem on solvability: This states that Hall subgroups of all possible orders exist for a finite group if and only if the finite group is a finite solvable group.

Proof using Hall's theorem
The proof directly follows from Hall's theorem, and the fact that there exist finite groups that are not solvable; for instance, the alternating group of degree five.

A concrete example
Let $$G = A_5$$ be the alternating group of degree five. Consider the prime set $$\pi = \{ 2, 5 \}$$. For a $$\pi$$-Hall subgroup to exist, it must have order equal to $$20$$. However, $$A_5$$ has no subgroup of order $$20$$.

There are many ways of seeing this. For instance, let's assume we know that A5 is simple. Then, a subgroup of order $$20$$ gives an action of $$G$$ on the coset space of that subgroup (which has size three), yielding a nontrivial homomorphism from $$G$$ to the symmetric group on three elements. This contradicts the simplicity of $$G$$.