# Calculation of Group Cohomology of $\mathbb{Z}/2\mathbb{Z}$ over $\mathbb{Z}$

I am trying to learn some group cohomology and I'm starting to get my head around the theory, but I find it hard to find some explicit examples of the calculation of group cohomology of some small finite groups. For example, I think it would help my understanding to see a calculation of the Tate cohomology groups of $\mathbb{Z}/2\mathbb{Z}$ over $\mathbb{Z}$.

On wikipedia it says that $$\hat{H}^p(\mathbb{Z}/2\mathbb{Z};\mathbb{Z})=\begin{cases} 0, &p\text{ odd;}\\ \mathbb{Z}/2\mathbb{Z}, &p\text{ even}.\end{cases}$$

Could someone show me how to compute this by explicit calculation?

Thanks!

• $\mathbb{Z}/2\mathbb{Z}$ is a $\mathbb{Z}$-module, but the converse is not true. Hence $H^0(\mathbb{Z}/2\mathbb{Z};\mathbb{Z})=(\mathbb{Z}/2\mathbb{Z})^\mathbb{Z}=0$. Right? – Hebe Dec 17 '15 at 15:41
• Any group can be given the trivial action. – Zhen Lin Dec 17 '15 at 15:53
• @Hebe If $\mathbb{Z}$ weren't a $\mathbb{Z}/2\mathbb{Z}$ module, $H^*(\mathbb{Z}/2\mathbb{Z}; \mathbb{Z})$ wouldn't even be defined. (And $\mathbb{Z}$ even has two $\mathbb{Z}/2\mathbb{Z}$ actions, the trivial action, and the sign action). – Najib Idrissi Dec 17 '15 at 15:58
• @NajibIdrissi Oh, yes! Hence the cohomology depends on the action, right? – Hebe Dec 17 '15 at 16:05
• @Hebe Yes, but if nothing special is written it's generally assumed to be the trivial action (and the cohomology groups listed in the question are indeed those with coefficients in $\mathbb{Z}$ with the trivial action). – Najib Idrissi Dec 17 '15 at 16:06