# group cohomology equivalent to topological singular cohomology

Let $$G=<\sigma>$$ be a cyclic group of order $$n$$. For any $$\mathbb{Z}[G]$$ module $$M$$ it is known that the group cohomology

$$H^i(G, M) = \begin{cases} M^G &\text{ if } i = 0 \\ M^G/NM &\text{ if } i \text{ odd} \\ ker(N)/(\sigma -1)M &\text{ if i even} \end{cases}$$

where $$N$$ is the norm map appearing in the free resolution of $$\mathbb{Z}$$. I am wondering if there is a similar description for the group $$G \times G$$ and any $$\mathbb{Z}[G\times G]$$ module $$M$$.

In particular, if $$G$$ is a cyclic group of order $$2$$, $$H^i(G,M) = M^G/NM$$ for all $$i>0$$. I am wondering if in the case of $$H^*(G\times G, M)$$ the cohomology is also a quotient of $$H^*(G; k) \otimes_k M^{G\times G}$$ where $$M$$ is now a $$k[G\times G]$$ module and $$k$$ is a field with two elements.

• Is $M$ a $G$-module or a $G\times G$-module ? In the first case what does $H^i(G\times G, M)$ mean; and in the second one, what does $M^G$ mean ? – Max Feb 18 at 9:09
• I will clarify it. Thanks. – C. Zhihao Feb 18 at 14:59
• This is not true. Try the trivial module and apply the Kunneth theorem. – user98602 Feb 18 at 15:26
• does not the Kunneth theorem apply for cohomologies of the form $H^*(G \times H; M \otimes N)$? – C. Zhihao Feb 18 at 16:07
• @MikeMiller I now get what you meant, Thanks for pointing that out. I will formulate an output that will make more sense. – C. Zhihao Feb 19 at 20:14