Let $m \geq 2$ be a positive integer and $C_m$ be a finite cyclic group of order $m$.
My question: How to compute the cohomology groups $H^i(C_m \times C_n, A)$ for $i > 0$ in the following two cases:
- $A = \mathbb{Z}$ is a trivial $C_m \times C_n$-module,
- $m=n$ is even and $A = \mathbb{Z}$ with an action of $C_m$ given by $k \cdot z = (-1)^k z$.
EDIT Sept. 14: I have almost figure out the result via Approach 4 , after having been collaborating with my roommate, by taking a particular resolution. I shall sort them out in a few weeks and post it as an answer. Thank you all for your attention!
My attempts: I have obtained a bunch of partial results, but failed to the general result.
Attempt 1: (Kunneth formula) In the first case when $A=\mathbb{Z}$ is trivial, one may apply the Kunneth formula to get the cohomology group, but the discussion is too tedious, full of mess with parity on $m$ and $n$. More unsatisfactory to me is that the Kunneth's formula in my mind is the Exercise 6.1.8 in Weibel's book:
Kunneth's formula: There is a split short exact sequence $$ 0 \rightarrow \bigoplus_{p+q=n} H^p(G,\mathbb{Z}) \otimes H^q(H,\mathbb{Z}) \rightarrow H^n(G \times H, \mathbb{Z}) \rightarrow \bigoplus_{p+q=n+1} \mathrm{Tor}_{1}^{\mathbb{Z}}(H^p(G,\mathbb{Z}),H^q(H,\mathbb{Z})). $$ where $\mathbb{Z}$ above are trivial modules.
, which only treats the trivial coefficient case.
Attempt 2: (Hochschild-Serre spectral sequence) Taking $G=C_m \times C_n$ and $H=C_n$, we have the convergent spectral sequence $$ \mathcal{E}: E_2^{p,q} = H^p(C_m, H^q(C_n, \mathbb{Z})) \Rightarrow H^{p+q}(C_m \times C_m, \mathbb{Z}). $$ Since $G$ is an abelian group and $\mathbb{Z}$ is a trivial module, the conjugation actions of $G/H \cong C_m$ on the cohomology groups $H^q(C_n, \mathbb{Z})$ are trivial. So we explicitly compute the $E_2$-page as $$ \begin{matrix} \mathbb{Z}/n & 0 & \mathbb{Z}/d & 0 & \mathbb{Z}/d & \cdots \\ 0 & 0 & 0 & 0 & 0 & \cdots \\ \mathbb{Z}/n & 0 & \mathbb{Z}/d & 0 & \mathbb{Z}/d & \cdots \\ 0 & 0 & 0 & 0 & 0 & \cdots \\ E_2^{0,0}=\mathbb{Z} & 0 & \mathbb{Z}/m & 0 & \mathbb{Z}/m & \cdots \\ \end{matrix} $$ Then $E_{\infty}^{p,q} = E_{2}^{p,q}$ by carefully checking the direction of arrows. Then we know directly:
- $H^0(C_m \times C_n, \mathbb{Z}) = \mathbb{Z}$,
- $H^{\text{odd}}(C_m \times C_n, \mathbb{Z}) = 0$.
But for cohomology group of even degree, I can only obtain a filtration of the cohomology: $$ H^{2i}(C_m \times C_n, \mathbb{Z}) \supseteq \ast \supseteq \ast \cdots \supseteq \ast = 0 $$ where the subquotients are $\mathbb{Z}/m$, $\mathbb{Z}/d$ for $(i-1)$ times and one $\mathbb{Z}/n$. Then if everything splits, then $$ H^{2i}(C_m \times C_n, \mathbb{Z}) = \mathbb{Z}/m \oplus (\mathbb{Z}/d)^{i-1} \oplus \mathbb{Z}/n. $$ But how can I see the splitting of this filtration?
Attempt 3: (Inflation-restriction exact sequence) Actually the inf-res sequence is derived from the above spectral sequence, yet I still tried to gain something from it. We can only use the $n=1$ version of the inf-res sequence, which gives $$ 0 \rightarrow H^1(C_n, \mathbb{Z}) \xrightarrow{\mathrm{inf}} H^1(C_m \times C_n, \mathbb{Z}) \xrightarrow{\mathrm{res}} H^1(C_m, \mathbb{Z})^{C_n} \xrightarrow{\mathrm{transgression}} H^2(C_n, \mathbb{Z}) \rightarrow \cdots. $$ This turns into $$ 0 \rightarrow 0 \xrightarrow{\mathrm{inf}} H^1(C_m \times C_n, \mathbb{Z}) \xrightarrow{\mathrm{res}} 0 \xrightarrow{\mathrm{transgression}} 0. $$ hence $$ H^1(C_m \times C_n, \mathbb{Z}) \cong 0. $$ This can be seen from the spectral sequence above. So unfortunately, I got nothing new.
(EDIT: I have corrected the calculation here, as I mixed up the homology and cohomology of cyclic groups)
These are all my attempts by now, yet unfortunately they all fail to give a complete result, or even a promising method to treat the second case where $\mathbb{Z}$ is not a trivial module.
So is there any way to settle down these cohomology groups? Or could someone provide some reference on the result of the cohomology group?
Thank you so much for answering and commenting! :)
EDIT: The comment by Dietrich Burde is quite useful, but I would like to see a more direct approach since this question only appears in a final exam of an elementary course on homological algebra, so it wouldn't require so much machinery that hard to establish during a two-hour exam.
So Approach 4: can we create an explicit resolution to do the job? Yet the standard resolution is quite complicated. :(
Here I found a resolution provided by Sasha: https://mathoverflow.net/questions/36730/describe-the-second-cohomology-group-h2z-n-times-z-n-k . But I cannot found the pattern for the general term of the resolution.