Hall implies join of Sylow subgroups

Statement
Any Hall subgroup of a finite group can be expressed as a join of Sylow subgroups.

Facts used

 * 1) uses::Sylow subgroups exist
 * 2) uses::Sylow of Hall implies Sylow
 * 3) uses::Lagrange's theorem

Proof
Given: A finite group $$G$$, a Hall subgroup $$H$$.

To prove: $$H$$ is a join of Sylow subgroups.

Proof: Let $$\pi = \{ p_1, p_2, \dots, p_r \}$$ be the set of prime divisors of the order of $$H$$. For each $$p_i \in \pi$$, let $$P_i$$ be a $$p_i$$-Sylow subgroup of $$H$$. Such a $$P_i$$ exists by fact (1), and $$P_i$$ is also Sylow in $$G$$ by fact (2).

Now, the join of the $$P_i$$s is contained in $$H$$, because each $$P_i$$ is contained in $$H$$. On the other hand, the order of the join of the $$P_i$$s must be a multiple of the order of each $$P_i$$ by Lagrange's theorem, and hence it must be a multiple of their lcm. But the lcm of the orders of the $$P_i$$s is the order of $$H$$, forcing the join of the $$P_i$$s to equal $$H$$.