Eilenberg-Maclane space $K(G\rtimes H, 1)$ for a semi-direct product.

Question

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.

Answer 1

$K(G \rtimes H, 1)$ 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.

Answer 2

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.