Hall subgroups need not exist

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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.

Related facts

Facts about Sylow subgroups

More facts about existence of Hall subgroups

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.