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!



Calculate the  group cohomology classes $H^d[U(1)\rtimes Z_2, Z]$ and   $H^d[U(1)\rtimes Z_2, Z_T]$
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 $R := H^\*(BO(2);\mathbb{Z})$.)


