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

1. Second isomorphism theorem
2. 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 $HC_{G}(H)=G$.

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, $C_{G}(H)/(H\cap C_{G}(H))\cong HC_{G}(H)/H=G/H$, 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 $HK=H(Z(H)K)=HC_{G}(H)=G$, 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 $G$.