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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 $BG_1$

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    $\begingroup$ Do you mean $K(\Bbb Z^m, 1)$? $\endgroup$ – Ali Caglayan Feb 9 '19 at 23:53
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    $\begingroup$ 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. Then $E^{*,0}_2$ survives to the $E_\infty$ page of the corresponding LHSSS. I don't know what else you could (or would want) to say. $\endgroup$ – Mike Miller Feb 10 '19 at 0:34
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    $\begingroup$ @AliCaglayan If $T^m$ means the $m$--torus $(S^1)^m$, then indeed $BT^m = (BS^1)^m = (\mathbb{C} P^\infty)^m = K(\mathbb{Z}^m, 2)$. Note that the question doesn't specify whether we're discussing discrete groups or more general topological groups. $\endgroup$ – Dan Ramras Feb 10 '19 at 4:36
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    $\begingroup$ @DanRamras For what it's worth, the projection $G_1 \to F$ in the original version of the question (at the time of Ali's comment) had kernel $\Bbb Z^m$ instead of $T^m$. $\endgroup$ – Mike Miller Feb 10 '19 at 6:23
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    $\begingroup$ @Totoro this deserves to be a new question, and sounds like the thing Ronnie Brown might know the answer to; the answer in your case probably has to do with crossed module homomorphisms. Note already the difficulty in determining homomorphisms into a semidirect product just from the homomorphisms into a factor. Further, even for the very simple case $G_1 = O(2)$, the answer to your question is somewhat complicated: bundles are classified by a class $w \in H^1(X;\Bbb Z/2)$ and the "twisted Euler class" $e \in H^2(X;\Bbb Z_w)$ living in cohomology with $w$-twisted local coefficients. $\endgroup$ – Mike Miller Feb 11 '19 at 0:11
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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.

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