Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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...

share|improve this question

1 Answer 1

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 $$0\to H^0(V,\mathbb Z)\to H^0(V,\mathbb Z[V])\to H^0(V,A)\to H^1(V,\mathbb Z)\to 0$$

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.