Hall implies join of Sylow subgroups

From Groupprops
Jump to: navigation, search
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)
View all subgroup property implications | View all subgroup property non-implications
Get more facts about Hall subgroup|Get more facts about join of Sylow subgroups


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


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_is is contained in H, because each P_i is contained in H. On the other hand, the order of the join of the P_is 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_is is the order of H, forcing the join of the P_is to equal H.