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Let $V$ be the Klein four group $\{1,\sigma,\tau,\sigma \tau\}$. Consider the $V$-module $A$ which is the cokernel of the map $\mathbb{Z} \to \mathbb{Z}[V]$ which sends $1$ to $1 + \sigma + \tau + \sigma \tau$, where we are considering the action of $V$ on $\mathbb{Z}[V]$ by translation. This induces an action of $V$ on $A$.

How to calculate the cohomology groups $H^i(V,A)$ for $i = 1,2,3$ for this action?

I thought about using the associated long exact sequence in cohomology, but I'm not sure which cohomology groups in this long exact sequence are zero and which aren't...

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up vote 3 down vote accepted

TGhe cohomology groups $H^p(V,\mathbb Z[V])$ are zero when $p>0$, because the module $\mathbb Z[V]$ is coinduced (it is obvious that it is induced, and since $V$ is finite induced modules and coinduced modules coincide) This implies that the long exact sequence corresponding to your short exact sequence $$0\to\mathbb Z\to\mathbb Z[V]\to A\to 0$$ splits into isomorphisms $$H^p(V,A)\cong H^{p+1}(V,\mathbb Z), \qquad\forall p\geq1,$$ and an exact sequence $$0\to H^0(V,\mathbb Z)\to H^0(V,\mathbb Z[V])\to H^0(V,A)\to H^1(V,\mathbb Z)\to 0$$

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So if I get this correctly, this gives $H^i(V,A) \cong H^{i + 1}(V,\mathbb{Z})$ for all $i \geq 1$? This reduced the problem to calculating cohomology for the trivial action on $\mathbb{Z}$? – Evariste Nov 30 '11 at 0:41
ok, and this gives $H^1(V,A) = (\mathbb{Z}/2)^2$, $H^2(V,A) = \mathbb{Z}/2$ and $H^3(V,A) = (\mathbb{Z}/2)^3$, if I'm right. – Evariste Nov 30 '11 at 0:53

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