TITLE: Classifying space of semidirect product of groups
QUESTION [7 upvotes]: Assume that $G$ and $H$ are two groups and $G\rtimes _\phi H$ is their semidirect product. My question is, how does the classifying space $B(G\rtimes_\phi H)$ of $G\rtimes _\phi H$ relate to $BG$ and $BH$?
In particular, we consider a split exact sequence
$$
1 \longrightarrow T^m \longrightarrow G_1  \longrightarrow F\longrightarrow1
$$
where $F$ is a finite group. It is well known that $BT^m=K(\mathbb Z^m,2)$ and $BF=K(F,1)$. How can we express $BG_1$？

REPLY [8 votes]: I am adding my comment as an answer.
Every extension of groups $1 \to H \to G \to K \to 1$ corresponds to a fibration $$BH \to BG \to BK,$$ or a little more precisely at the space level $$EG/H \to (EG \times EK)/G \to EK/K.$$
The content of your question is what one can say when the extension is a semidirect product. Then $BG \to BK$ admits a section, as well as $BH \to BG$. 
This means that the row $E_2^{*,0}$ and column $E_2^{0,*}$ survive to the $E_\infty$ page of the spectral sequence, as the map $H^*(BK) \to H^*(BG)$ is injective while the map $H^*(BG) \to H^*(BH)$ is surjective; these maps factor through the aforementioned row and column, respectively. (This can be used to good effect in simple cases when there is nothing else of interest in the SS.) 
It's not clear to me what else one could reasonably say at this level of generality.