Special case of Schur-Zassenhaus theorem The theorem of Schur-Zassenhaus says that 

if $G$ is a finite group, and $H$ is a normal subgroup, such that $|H|$ and $|G/H|$ are relatively prime, then $G$ contains a subgroup $K$ of order equal to $|G/H|$.

(this subgroup $K$ will then be obviously a complement of $H$, i.e. $K\cap H=1$)
If we see the proof, then the non-trivial part comes in the case when $H$ is abelian.
Suppose we put one more extra condition: $H$ is central in $G$. Then is it easy to prove the theorem without (co)homological methods? Putting this a a question:

Let $G$ be a finite group and $H$ be a subgroup in center of $G$ such that $|H|$ and $|G/H|$ are co-prime. Then prove that $G$ contains a subgroup of order equal to $|G/H|$.


Burnside's normal complement theorem states that if a Sylow $p$-subgroup is central in its normalizer, then it has a normal complement. This result can be proven with transfer and a proof can be found in many textbooks on basic group theory.
Apply this to Sylow $p$-subgroups of $H$ (which are in this case Sylow $p$-subgroups of $G$ as well) to find a normal complement for $H$.
