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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. )

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    $\begingroup$ This is true; you can use the short exact sequence $0 \to \Bbb Z^{sgn} \to \Bbb Z^{sgn} \to \Bbb Z/2 \to 0$ to show it. $\endgroup$ Nov 29, 2018 at 15:23
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    $\begingroup$ @TylerLawson How is this SES used? If I apply $H^\ast(G,\cdot)$ to it, I'm not sure how to identify $Ker\big(H^{n+1}(G,\mathbb{Z}^{sgn})\to H^{n+1}(G,\mathbb{Z}^{sgn})\big)$ in the LES with the desired Tor term (similarly for $Coker$ applied to the map on $H^n$). I think the desired Tor term can be identified with the kernel of the map $\mathbb{Z}^{sgn}\otimes H^{n+1}(G,\mathbb{Z}^{sgn})\to\mathbb{Z}^{sgn}\otimes H^{n+1}(G,\mathbb{Z}^{sgn})$ obtained by tensoring the map $\mathbb{Z}^{sgn}\to\mathbb{Z}^{sgn}$ with $H^{n+1}(G,\mathbb{Z}^{sgn})$, but I'm not sure how these all relate. $\endgroup$ Dec 3, 2018 at 15:50
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    $\begingroup$ Ah, perhaps the tensored map $H^{n+1}(G,\mathbb{Z}^{sgn})\otimes\mathbb{Z}^{sgn}\to H^{n+1}(G,\mathbb{Z}^{sgn})\otimes\mathbb{Z}^{sgn}$ is the same as the induced map $H^{n+1}(G,\mathbb{Z}^{sgn})\to H^{n+1}(G,\mathbb{Z}^{sgn})$ by "pushing" the $\mathbb{Z}_2$-action of the $\mathbb{Z}^{sgn}$ factor onto the $H^{n+1}(G,\mathbb{Z}^{sgn})$ factor, then $H^{n+1}(G,\mathbb{Z}^{sgn})\otimes\mathbb{Z}\cong H^{n+1}(G,\mathbb{Z}^{sgn})$. $\endgroup$ Dec 3, 2018 at 16:12
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    $\begingroup$ @ChrisGerig Because the map $\Bbb Z^{sgn} \to \Bbb Z^{sgn}$ is multiplication-by-2, the induced map $H^*(G;\Bbb Z^{sgn}) \to H^*(G;\Bbb Z^{sgn})$ is multiplication-by-2. The cokernel is the tensor product, and the kernel is the Tor-group. This recovers the natural short exact sequence from the UCT. $\endgroup$ Dec 3, 2018 at 20:57

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