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I am working in a problem group cohomology and nailed it down to compute a cohomology using an spectral sequence argument. The situation is as follows:

Let $G = C_4 = \langle \sigma \rangle$ be the cyclic group of order $4$, $k = \mathbb{F}_2$ and $M^*=M^0 \oplus M^1 \oplus M^2$ be a graded module where $M^1 = M^2 = k$ is the trivial $G$-module and $M^1 = k \oplus k$ be the $G$-module where $\sigma(a,b) = (b,a)$.

To compute $H^*(G, M^*)$, we can use a spectral sequence argument with $E_2^{p,q} =H^p(G,M^q)$. Since $M^0$ and $M^2$ are trivial, $E_2^{p,q} = H^p(G,k) \otimes M^q$ for $q=0,2$.

When $q = 1$, we can use Shapiro's lemma to show that $E_2^{p,1} = H^p(C_2;k) \cong k[t]$ where $C_2 = \langle \sigma^2 \rangle$.

However, I have no leads on how to compute the differential $d_2:E_2^{p,1} = H^p(C_2,k) \rightarrow E_2^{p+2,0} = H^{p+2}(C_4,k)$.

I appreciate any input in this computation, or if there is any alternative to describe $H^*(G,M^*)$.

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  • $\begingroup$ Is $M^*$ supposed to be a complex or just a graded module? If it is just a graded module there is nothing left to show. $\endgroup$ Mar 31, 2021 at 18:23
  • $\begingroup$ $M^*$ shows up as the cohomology of a space $X$ with an action of $G$. Probably I should rephrase the problem, as this computation is the group cohomology of $G$ with coefficients in the cochains of $X$ (or the $G$-equivariant cohomology of $X$). $\endgroup$
    – C. Zhihao
    Mar 31, 2021 at 19:10
  • $\begingroup$ Are you trying to compute the cohomology of the homotopy quotient $X/G$? I think the answer then should depend on what X actually is (and not only on the cohomology of X as a G-module) $\endgroup$
    – user42024
    Mar 31, 2021 at 23:17
  • $\begingroup$ The cohomology of a cyclic group G with coefficients any G-module N is well-known. Why can't you just use the formulars for H*(G,N)? $\endgroup$
    – user120513
    Apr 1, 2021 at 4:29
  • $\begingroup$ @user42024 Yes that is what I am trying to compute. I'll see if can bring the topology of $ X$. $\endgroup$
    – C. Zhihao
    Apr 1, 2021 at 13:36

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