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?


  • $\begingroup$ $\mathbb{Z}/2\mathbb{Z}$ 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? $\endgroup$ – Hebe Dec 17 '15 at 15:41
  • $\begingroup$ Any group can be given the trivial action. $\endgroup$ – Zhen Lin Dec 17 '15 at 15:53
  • 1
    $\begingroup$ @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). $\endgroup$ – Najib Idrissi Dec 17 '15 at 15:58
  • $\begingroup$ @NajibIdrissi Oh, yes! Hence the cohomology depends on the action, right? $\endgroup$ – Hebe Dec 17 '15 at 16:05
  • $\begingroup$ @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). $\endgroup$ – Najib Idrissi Dec 17 '15 at 16:06

The action of any group G (here G = Z/2Z) on Z is always implicitly supposed to be the trivial action. If G is cyclic, there is a theorem saying that Tate cohomology is periodic, with period 2, more precisely H^p(G,M) is isomorphic to H^p+2(G,M), so it suffices to compute H^0 and H^1, which are given explicitly by H^0(G,M)=M^G/N(M) and H^(G,M)=KerN/(s-1)M. Here s denotes a generator of G and N, the norm (or trace) map, is defined by N(m)=sum of all s^i(m). This gives your answer.


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