I would like to know what are the group cohomology classes $H^d[U(1)\rtimes Z_2, Z]$ and $H^d[U(1)\rtimes Z_2, Z_T]$, and/or how to calculate them.
It can be shown that $H^d[U(1), Z]$ is $Z$ for even $d$ and 0 for odd $d$.
Here $\rtimes$ is the semidirect product: $T U T = U^{-1}$ for $U \in U(1)$, where
$T$ is the generator of $Z_2$.
In $H^d[U(1)\rtimes Z_2, Z]$, the module $Z$ is the trivial module.
In $H^d[U(1)\rtimes Z_2, Z_T]$, the module $Z_T$ is still the Abelian group $Z$,
but $U(1)\rtimes Z_2$ has a non-trivial action on $Z_T$:
$(U,1) a = a$ and $(U,T) a = -a$, $a \in Z_T$.
Thanks!
Best Answer
The group $U(1) \rtimes \mathbb{Z}/2$ you describe is nothing but the group $O(2)$ (as $U(1) = SO(2)$).
As such I think one can see the spectral sequence for the extension does collapse, and one obtains $$H^*(BO(2);\mathbb{Z}) = \mathbb{Z}[x_2, x_3, x_4]/(2x_2, 2x_3, x_3^2-x_2x_4).$$ Here we can take $x_2 = \beta(w_1)$ and $x_3 = \beta(w_2)$, the Bocksteins on the Stiefel--Whitney classes, and $x_4 = p_1$ the Pontrjagin class.
If you allow me to write $\mathbb{Z}^-$ for your $\mathbb{Z}_T$, there is a short exact sequence of $\mathbb{Z}[\mathbb{Z}/2]$-modules $$0 \to \mathbb{Z} \overset{1 \mapsto 1+T}\to \mathbb{Z}[\mathbb{Z}/2] \to \mathbb{Z}^- \to 0$$ and $H^*(BO(2); \mathbb{Z}[\mathbb{Z}/2]) = H^*(BSO(2);\mathbb{Z}) = \mathbb{Z}[e_2]$ is a polynomial algebra on the Euler class. The long exact sequence on cohomology gives the following short exact sequences, once you know that $x_4 \mapsto e_2^2$ under $H^*(BO(2); \mathbb{Z}) \to H^*(BO(2); \mathbb{Z}[\mathbb{Z}/2])$ (since $p_1 = e^2 \in H^4(BSO(2);\mathbb{Z})$): \begin{eqnarray*} 0 \to H^{4i}(BO(2); \mathbb{Z}^-) \to H^{4i+1}(BO(2); \mathbb{Z}) \to 0 \\ 0 \to H^{4i+1}(BO(2); \mathbb{Z}^-) \to H^{4i+2}(BO(2); \mathbb{Z}) \to 0 \\ 0 \to \mathbb{Z}\langle e_2^{2i+1} \rangle \to H^{4i+2}(BO(2); \mathbb{Z}^-) \to H^{4i+3}(BO(2); \mathbb{Z}) \to 0\\ 0 \to H^{4i+3}(BO(2); \mathbb{Z}^-) \to H^{4i+4}(BO(2); \mathbb{Z}) \to \mathbb{Z}\langle e_2^{2i+2}\rangle \to 0\\ \end{eqnarray*} One can read off the groups $H^*(BO(2);\mathbb{Z}^-)$ from these short exact sequences.
(It is probably more efficient to describe $H^*(BO(2);\mathbb{Z}^-)$ as a module over $R := H^*(BO(2);\mathbb{Z})$.)