# 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 $K \le C_G(H)$, 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 $G$.