Eilenberg-Maclane space $K(G\rtimes H, 1)$ for a semi-direct product. We know that $K(G\times H, 1)=K(G,1)\times K(H,1)$. Do we know something like this for a semi-direct product, where $K(G,1)$ denotes the Eilenberg-Maclane space.
 A: $K(G \rtimes H, 1)$ fits into a fibration sequence
$$K(G, 1) \to K(G \rtimes H, 1) \to K(H, 1).$$
So for example one can access the homology and cohomology using the Serre spectral sequence. See this answer for some context. 
A: Here are some details on how to get the fibration in Qiaochu Yuan's answer:$\newcommand{\Z}{\mathbb{Z}}$
Let $H \curvearrowright E_H$ act freely on a weakly contractible space $E_H$, and let the exact sequence of groups $1 \to G \to G \rtimes H  \xrightarrow{\phi} H \to 1$ be given.  Associated to the  map $\phi : G \rtimes H \to H$, there is an action $G \curvearrowright E_H$.  The induced map $\Phi: E_H/G \to E_H/H$ given by $Ge \to He$ is well defined because $\phi(G)\subset H$. 
Now make the additional assumption that the groups are discrete so that $Z G=BG$, $Z  G \rtimes H=B G \rtimes H $, $Z H=BH$ :  After all  $\pi_1 B(G \rtimes H)=G \rtimes H$ and same for $G, H$.   $\pi_k(B(\text{any of the groups }))=0$ for $k>2$(since they are homotopy groups of contractible spaces).
Turn the map $\mathbb{Z}G \rtimes H   \to \mathbb{Z}H $ into a fibration with fiber $F$ via a space homotopy equivalent to a $\mathbb{Z} G \rtimes H$.  This will be another model for $\mathbb{Z}G \rtimes H$. The homotopy long exact sequence for the fibration $F \hookrightarrow \mathbb{Z} G \rtimes H  \to\mathbb{Z} H$ implies that $\pi_k(F)=0$ for $k>1$.  Since the induced map $\pi_1 (B(G \rtimes H)) \to \pi_1(BH)$ is exactly $\phi$, and is therefore surjective, we get the top row and the horizontal isomorphisms of the following commutative diagram   $\require{AMScd} \begin{CD}
1 @>>>\pi_1(F) @>>> \pi_1(\mathbb{Z}G \rtimes H) @>\phi>> \pi_1(\mathbb{Z}H) @>>> 1\\
@|@. @| @| @|\\
1 @>>>\pi_1(\mathbb{Z} G ) @>\psi >> \pi_1(\mathbb{Z} G \rtimes H) @>>> \pi_1(\mathbb{Z} H) @>>> 1
\end{CD}$
We can define a map from $\pi_1( \Z G) \to \pi_1(F)$ by chasing the diagram because $\phi \circ (\text{ equals sign }) \circ \psi=0$.  Since we defined the map $\pi_1( \Z G) \to \pi_1(F)$ by chasing the diagram, the following diagram commutes:  
$ \begin{CD}
1 @>>>\pi_1(F) @>>> \pi_1(\mathbb{Z}G \rtimes H) @>\phi>> \pi_1(\mathbb{Z}H) @>>> 1\\
@|@AAA @| @| @|\\
1 @>>>\pi_1(\mathbb{Z} G ) @>\psi >> \pi_1(\mathbb{Z} G \rtimes H) @>>> \pi_1(\mathbb{Z} H) @>>> 1
\end{CD}$
By the five lemma, the fiber $F$ is $ZG$ up to weak homotopy equivalence.
Summarizing, we get the fibration in Qiaochu Yuan's answer $\Z G \hookrightarrow \Z G\rtimes H \to \Z H$.
I have not shown that the fiber will be a CW complex.  Notice that I didn't assume that the map on $\pi_1$ was induced topologically.  But because I showed that it has the same homotopy groups as an eilenberg mclane space, the postnikov invariants vanish and I have shown that there is indeed a weak homotopy equivalence between the fiber and $\Z G$. But it is possible to show that the fiber is a CW complex in which case I can say that it has the homotopy type of a $\Z G$.
I'll put something on the cohomology calculation in a little bit.  There are way too many  ...by a spectral sequence argument....  and to the best of my knowledge no calculations of this type done on all of mathstackexchange.
