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. )