Hall implies join of Sylow subgroups

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., Hall subgroup) must also satisfy the second subgroup property (i.e., join of Sylow subgroups)
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Statement

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

Facts used

1. Sylow subgroups exist
2. Sylow of Hall implies Sylow
3. 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 $H$.