If an extension is both central and split it is a direct product. Why? Is there any quick way to see that if an extension is both central and split it's basically a direct product? Or rather, a semi-direct product is a direct product iff the corresponding extension is split. I saw this stated in one of my lecture notes without any proof. (Note that I do not know any cohomology.)
 A: Assume we have an short exact sequence
$$ 1\to A\stackrel i\to B\stackrel p\to C\to 1$$
that is split, i.e., we also have $C\stackrel s\to B$ with $p\circ s=\operatorname{id}_C$, and is central, then we can define $A\oplus C\stackrel f\to B$ by sending $(a,c)\mapsto i(a)s(c)$. This is a homomorphism because centrality gives us $$i(a)s(c)\cdot i(a')s(c')=i(a)i(a')s(c)s(c')=i(aa')s(cc').$$
We also have a $B\stackrel g\to A\oplus C$ that sends $b\mapsto (i^{-1}(bs(p(b))^{-1}),p(b))$, which you can check similarly to actually be a homomorphism. Finally, $f$ and $g$ can be verified to be inverses of each other
A: Consider the extension (short exact sequence) $1 \to A \to G \to B \to 1$. 
We already know that split extension can be written as semi-direct products, i.e., $G = A \rtimes_\varphi B$ where $\varphi$ is the homomorphism $\varphi: B \to \mathrm{Aut}(A)$. 
If the extension is now also central, the elements of $A$ must be injectively embedded into the center of the group $G$. Hence the elements of the form $(a, e) \in A \rtimes_{\varphi} B$ (s.t. $a \in A$) necessarily commute with all elements of $A \rtimes_{\varphi} B$. However, we also have (for all $b\in B$):
$$(a, e).(e, b) = (a\varphi(e)(e), b) = (a, b)$$
and 
$$(e, b).(a, e) = (e\varphi(b)(a), b) = (\varphi(b)(a), b)$$
For these two to be equivalent, $\varphi(b)$ is necessarily the identity and since this has to hold true for all $b\in B$, the homomorphism $\varphi$ is trivial. This implies that the semi-direct product is, in fact, a direct product in this context.  
