Hall and central factor implies direct factor

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

Facts used

 * 1) uses::Second isomorphism theorem
 * 2) uses::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 $$G$$.