Pi-dominating pi-subgroup implies pi-Hall

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.