Pi-dominating pi-subgroup implies pi-Hall

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This article gives a proof/explanation of the equivalence of multiple definitions for the term order-dominating Hall subgroup
View a complete list of pages giving proofs of equivalence of definitions

Statement

Suppose G is a finite group, H is a subgroup of G, and \pi is a set of primes such that:

  • The set of prime divisors of the order of H is in \pi.
  • Given any \pi-subgroup K of G (i.e., any subgroup for which all prime divisors of its order are in \pi), there exists g \in G such that gKg^{-1} \le H.

Then, H is a \pi-Hall subgroup: in particular, its index is relatively prime to \pi, or equivalently, its order is the unique largest \pi-number dividing the order of G.

Note that this also shows that a subgroup is \pi-dominating for a set of primes \pi iff it is an order-dominating Hall subgroup.

Facts used

  1. Sylow subgroups exist
  2. Lagrange's theorem

Proof

Given: A finite group G, a \pi-dominating subgroup H.

To prove: H is \pi-Hall.

Proof: It suffices to show that for every p \in \pi, the largest power of p dividing the order of G also divides the order of H.

For this, let P be a p-Sylow subgroup (existence follows from fact (1)). The order of P is the largest power of p dividing the order of G. By the assumption, some conjugate gPg^{-1} is in H. The order of gPg^{-1} equals the order of P, and by fact (2), this divides the order of H. Thus, the order of H is a multiple of the largest power of p dividing the order of G, and this completes the proof.