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 $\pi$-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 $\pi$.

Proof

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

Further information: Alternating group:A5

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 $G$.