Hall and central factor implies direct factor

This article gives a proof/explanation of the equivalence of multiple definitions for the term Hall direct factor
View a complete list of pages giving proofs of equivalence of definitions

Statement

Any Hall subgroup of a finite group that is a central factor of the group, is also a direct factor.

Facts used

Second isomorphism theorem
Normal Hall implies permutably complemented (this is the first half of the Schur-Zassenhaus theorem)

Proof

Given: A finite group $G$ , a Hall subgroup $H$ such that $H C_{G} (H) = G$ .

To prove: $H$ is a direct factor of $G$ .

Proof: Consider the subgroup $C_{G} (H)$ . The subgroup $Z (H) = H \cap C_{G} (H)$ is central in $C_{G} (H)$ . Also, by the second isomorphism theorem, $C_{G} (H) / (H \cap C_{G} (H)) ≅ H C_{G} (H) / H = G / H$ , so $Z (H)$ is a Hall subgroup of $C_{G} (H)$ .

Thus, $Z (H)$ is a central Hall subgroup of $C_{G} (H)$ .

In particular $Z (H)$ is normal Hall in $C_{G} (H)$ , so it possesses a complement, say $K$ , in $C_{G} (H)$ . Clearly, $H$ and $K$ permute element-wise, because $K \leq C_{G} (H)$ , and $H \cap K$ is trivial, because $K \cap (H \cap C_{G} (H))$ is trivial by construction. Finally, $Z (H) K = C_{G} (H)$ by construction, so $H K = H (Z (H) K) = H C_{G} (H) = G$ , and so, $H$ and $K$ are complements.

Thus: $H$ and $K$ commute element-wise, generate the whole group, and intersect trivially, making $G$ an internal direct product of $H$ and $K$ , and thus making $H$ a direct factor of $G$ .