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 $|G/H|$.

 A: 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 $H$.
A: This is not really an answer, but I am going to write out what I think is the standard averaging proof and show that it is already quite short. I leave you the exercise of checking that it gets even a little shorter if you assume $H$ (soon called $A$) is central. The real reason I am writing this is that I plan to teach it next term, and I want to practice.
I'm going to change to notation I prefer. Let $A$ be an abelian group, written additively, and let $G$ be a finite group such that multiplication by $|G|$ is invertible on $A$. Let 
$$1 \to A \overset{\exp}{\longrightarrow} \tilde{G} \to G \to 1$$
be a short exact sequence, where $\tilde{G}$ and $G$ are written multiplicatively. We must split the map $\tilde{G} \to G$. 
We write $\log$ for the inverse of $\exp$, and $\rho$ for the action of $G$ on $A$. So $\tilde{g} \exp(a)  =  \exp(\rho(g) a)\tilde{g}$ for any $\tilde{g} \in \tilde{G}$ with image $g \in G$.
Begin with a set-theoretic splitting $s: G \to \tilde{G}$. For $g$ and $h$ in $G$, let $\phi(g,h)$ be the unique element of $A$ such that
$$s(gh) = s(g) \exp(\phi(g,h)) s(h).$$
For any $g_1$, $g_2$, $h$ all in $G$, we have
$$s(g_1 g_2 h) = s(g_1 g_2) \exp(\phi(g_1 g_2,h)) s(h) = s(g_1) \exp(\phi(g_1, g_2 h)) s(g_2) \exp(\phi(g_2, h)) s(h)$$
so
$$s(g_1 g_2) \exp(\phi(g_1 g_2,h)) = s(g_1) \exp(\phi(g_1, g_2 h)) \cdot s(g_2) \exp(\phi(g_2, h)). \quad (\ast)$$
Consider the general relationship
$$s(g_1 g_2) \exp(h_3) = s(g_1) \exp(h_1) s(g_2) \exp(h_2)$$
which we can rewrite as
$$h_3 - \rho(g_2)^{-1}\cdot h_1 - h_2 = \log (s(g_1) s(g_2) s(g_1 g_2)^{-1}).$$ 
The latter form is clearly an affine linear equation in $h_1$, $h_2$, $h_3$, so the average of any collection of solutions is another solution.
We therefore average $(\ast)$ over $h \in G$ to get
$$s(g_1 g_2) \exp \left( \frac{1}{|G|} \sum_{h \in G} \phi(g_1 g_2,h) \right) = s(g_1) \exp \left( \frac{1}{|G|} \sum_{h \in G} \phi(g_1,h) \right) \cdot s(g_2) \exp \left( \frac{1}{|G|} \sum_{h \in G} \phi(g_2,h) \right).$$
In other words, $g \mapsto s(g) \exp \left( \tfrac{1}{|G|} \sum_{h \in G} \phi(g,h) \right)$ is a splitting of $\tilde{G} \to G$. $\square$
