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Hall's theorem



Suppose G is a finite group such that for any subset \pi of the set of prime divisors of G, G has a \pi-Hall subgroup. Then, G is a Solvable group (?) (specifically, G is a Finite solvable group (?)).

Related facts

Facts used


Case that the order has one or two prime factors

If the order has two or fewer prime factors, fact (1) tells us that the group is solvable. (If the order has only one prime factor, the group is in fact nilpotent).

Case that the order has three or more prime factor

We prove this claim by induction on the number of prime factors of the order. Note that the cases of one or two prime factors have already been dealt with.

Suppose the order of G is p_1^{k_1}p_2^{k_2} \dots p_r^{k_r} where p_i are distinct primes.

  1. There exist p_i-complements for each prime p_i: In other words, for each p_i, there exists a subgroup A_i whose index is p_i^{k_i}. This follows from the assumption that there exist Hall subgroups of all possible orders.
  2. For any subset \pi of \{ p_1, p_2, \dots, p_r \}, the intersection of the subgroups A_i with p_i \in \pi, is a \pi'-Hall subgroup: PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]
  3. Each A_i satisfies the hypothesis of containing Hall subgroups of all possible orders: This follows from step (2).
  4. A_i is solvable for each i: This follows from the induction hypothesis.
  5. G is solvable: This follows from fact (2), and the observation that when r \ge 3, we have a collection of at least three solvable subgroups of pairwise coprime indices.