A version of the Universal Coefficient Theorem that relates the integer cohomology of a group $G$ to its cohomology with coefficients in an abelian group $M$ is as follows:
$H^n(G,M) = H^n(G,\mathbb Z) \otimes M \quad \times \quad\text{Tor}_1^{\mathbb Z}(H^{n+1}(G,\mathbb Z), M)$
It is assumed in this expression that $G$ acts trivially on coefficients. Now suppose $G$ is an extension of the group $\mathbb Z_2$ by some group $G_0$ and $M = \mathbb Z_2$. I am interested in finding the cohomology of $G$ with coefficients in the module $\mathbb Z^{sgn}$, whose coefficients are twisted by the $\mathbb Z_2$ factor, while $G_0$ has trivial action. I wanted to know if the following statement, which looks like the statement of the UCT, is actually valid:
$H^n(G,\mathbb Z_2) = H^n(G,\mathbb Z^{sgn}) \otimes \mathbb Z_2 \quad \times \quad\text{Tor}_1^{\mathbb Z}(H^{n+1}(G,\mathbb Z^{sgn}), \mathbb Z_2)$
(So just to be clear, the l.h.s. has trivial action on $\mathbb Z_2$, while the r.h.s. has nontrivial action on $\mathbb Z^{sgn}$. I have checked this by hand for some simple examples and it works out, but I don't have any proof. )