I want to show that $B(NT) \simeq (BT)_{hW}$, where $NT$ is the normalizer of the maximal torus, $W$ is the Weyl group (that is $W = NT/T$) and $(BT)_{hW}$ is the homotopy orbit/Borel construction, $(BT)_{hW} = EW \times_W BT$.
What I did: B(NT) can be modeled by $E(NT) \times_{NT} E(NT/T)$, since $E(NT/T)$ is contractible. Note that $E(NT)/T \simeq BT$, as $T$ is a normal subgroup. Hence, $E(NT) \times_{NT} E(NT/T) \simeq E(NT) \times_{NT/T} E(NT/T) \simeq BT \times_W E(W)$. This last step, namely $E(NT) \times_{NT} E(NT/T) \simeq E(NT) \times_{NT/T} E(NT/T)$, should be correct but I am having difficults to make it precise in order to really convince myself. How could one explain it?
Is there another way of showing this? Maybe using the fact that from the exact sequence $1 \to T \to NT \to W \to 1$ there is a fibration sequence $BT \to BNT \to BW$?
Many thanks!
This is true for any normal (closed) subgroup $N$ of a (topological) group $G$. In general, if $G$ acts freely on $X$, then $X/G\simeq X_{hG}$ (see equivariant cohomology in case of free actions (basic question)). We can take then take $(EG)/N$ as a model for $BN$ . The residual action of $G/N$ on $(EG)/N$ is free, therefore we obtain $BN_{hG/N}\simeq (EG/N)/(G/N)\cong EG/G = BG$