# Classifying space of semidirect product of groups

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 $$BG_1$$

• Do you mean $K(\Bbb Z^m, 1)$? – Ali Caglayan Feb 9 '19 at 23:53
• 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. – Mike Miller Feb 10 '19 at 0:34
• @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. – Dan Ramras Feb 10 '19 at 4:36
• @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 $T^m$. – Mike Miller Feb 10 '19 at 6:23
• @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. – Mike Miller Feb 11 '19 at 0:11

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.)