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Let $G=H\times J$, where $H\cong J\cong C_2$ (cyclic group of order 2). Let $M \cong \mathbb{Z}$ be a $G$-module via "trivial $H$-action and negation $J$-action". My question is "What are the group cohomologies $H^*(G,M)$?"

I tried to compute them via the Hochschild-Serre spectral sequence $E_2^{p,q}=H^p(J,H^q(H,M))$. The computation of each term is easy, but I don't have any information about arrows and cannot determine $H^*(G,M)$.

More generally, let $G=C_k\rtimes C_2$ be a dihedral group and $M \cong \mathbb{Z}$ a $G$-module via "trivial $C_k$-action and negation $C_2$-action". Can one compute the group cohomologies $H^*(G,M)$?

I would appreciate it if you could provide an explicit computation, or a reference where I can find one. Thank you very much in advance.

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Perhaps by looking at the long exact sequence of cohomology groups induced by the short exact sequence of coefficient modules $0\to M\to M\to \mathbb{Z}/2\to 0$ (where the first map is multiplication by $2$) you will be able to decide which, if any, of the differentials are non-zero. –  Mark Grant Sep 7 '13 at 18:34
    
Thank you for the comment Mark, but I do not see why the long exact sequence gives information about the differential of the spectral sequence. –  Callum P Dunne Sep 7 '13 at 20:21

2 Answers 2

up vote 5 down vote accepted

See Kuenneth-formula for group cohomology with nontrivial action on the coefficient

Let $C=C_2$ be the cyclic group of order two, $\def\ZZ{\mathbb Z}\ZZ$ the trivial module over $C$ and $S$ the $C$-module which is $\ZZ$ as an abelian group with with the non-trivial action of $C$. You want to compute $H^\bullet(C\times C,M)$ with $M=\ZZ\otimes S$, and the Künneth formula tells us that there is a short exact sequence $$0\to (H^\bullet(C,\ZZ)\otimes H^\bullet(C,S))^p\to H^p(C\times C,M)\to (Tor^\ZZ_1(H^\bullet(C,\ZZ),H^\bullet(C,S))^{p+1}\to0$$

You can easily compute $H^\bullet(C,-)$ with coefficients in every $C$-module and the Tor is also easy to compute. If I am not making too many mistakes, then \begin{gather} (H^\bullet(C,\ZZ)\otimes H^\bullet(C,S))^{p}=(Tor_1(H^\bullet(C,\ZZ),H^\bullet(C,S))^{p}=0 \end{gather} for all even $p\geq0$, and \begin{gather} (H^\bullet(C,\ZZ)\otimes H^\bullet(C,S))^{p}=(\ZZ/2\ZZ)^r,\\ (Tor_1(H^\bullet(C,\ZZ),H^\bullet(C,S))^{p}=(\ZZ/2\ZZ)^{r-1} \end{gather} for all odd $p=2r-1\geq0$. If follows that in the short exact sequenc above exactly one of the first and third terms are not zero for a given $p$, so we get an isomorphism in all cases.


For the crossed product $G=C_k\rtimes C_2$ with coefficients $S=\mathbb Z$ with $C_k$ acting trivially and $C_2$ changing sizes. Notice that if $k$ is odd, then the H-L-S spectral sequence has second page $H^p(C_2,H^q(C_k,S))$, and this is zero except when $q=0$ and $p$ is odd, so this gives us the desired result in this case.

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I was not aware that Kuenneth-formula holds when one of the modules is trivial. Thank you very much, Marinano! –  Callum P Dunne Sep 8 '13 at 9:24

This is not a complete answer, but it's too long for a comment. Here is what I would try (for $G=C_p\rtimes C_2$ for any $p$): your $M$ sits in the exact sequence $$ 0\rightarrow M\rightarrow \mathbb{Z}[G/C_p] \rightarrow \boldsymbol{1} \rightarrow 0, $$ where $\boldsymbol{1}$ denotes the free $\mathbb{Z}$-rank 1 module with trivial $G$-action. This gives rise to the long exact sequence $$ 0\rightarrow M^G=0\rightarrow \boldsymbol{1}\stackrel{m}{\rightarrow}\boldsymbol{1}\rightarrow H^1(G,M)\rightarrow H^1(G,\mathbb{Z}[G/C_p])\rightarrow \\ \rightarrow H^1(G,\boldsymbol{1})\rightarrow H^2(G,M)\rightarrow H^2(G,\mathbb{Z}[G/C_p])\rightarrow \boldsymbol{1}\rightarrow \ldots $$ When you write out the map $m$, you see that it is multiplication by 2 (that's because $\mathbb{Z}[G/C_p]$ is actually indecomposable, i.e. involutions reflect this rank 2 lattice in a diagonal).

Also Shapiro's lemma gives you that $H^1(G,\mathbb{Z}[G/C_p]) \cong H^1(C_p,\boldsymbol{1})={\rm Hom}(C_p,\mathbb{Z}) = 0$, and similarly $H^1(G,\boldsymbol{1}) = 0$. Now here are some miscellaneous observations.

  • $H^1(G,M)\cong{\rm coker}\; m \cong \mathbb{Z}/2\mathbb{Z}$.
  • Explicit calculations will give you $H^2(G,\mathbb{Z}[G/C_p])\cong H^2(C_p,\boldsymbol{1})\cong \mathbb{Z}/p\mathbb{Z}$, while $H^2(G,\boldsymbol{1})\cong \mathbb{Z}/2\mathbb{Z}$, so $$ H^2(G,M)\cong {\rm ker}(\mathbb{Z}/p\mathbb{Z}\rightarrow \mathbb{Z}/2\mathbb{Z}). $$ This is clearly equal to $\mathbb{Z}/p\mathbb{Z}$ when $p$ is odd; according to MAGMA it's also true when $p=2$, but I guess to see that you would actually have to write out the map $H^2(C_p,\boldsymbol{1})\rightarrow H^2(G,\boldsymbol{1})$.

  • You should be able to get quite a lot of information about higher $H^i$ if you carefully follow through the quotient map and Shapiro and write down the maps $H^i(C_p,\boldsymbol{1})\rightarrow H^i(G,\boldsymbol{1})$.

  • MAGMA can compute some small cohomology groups (up to $H^3$). For example here is the code for computing cohomology of this particular module:

    G:=DihedralGroup(5);
    M:=GModule(G, [Matrix([[1]]),Matrix([[-1]])]);
    MC:=CohomologyModule(G,M);
    CohomologyGroup(MC,3); //Computes $H^3$

    MAGMA also allows you to play around with concrete co-cycles and thereby write down connecting homomorphisms between these guys.

Of course, you can do the same thing starting with the sequence $0\rightarrow \boldsymbol{1}\rightarrow \mathbb{Z}[G/C_p]\rightarrow M\rightarrow 0$.

Sorry, that's all I have to offer for now.

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Thank you for sharing the idea, Alex. This will help me a lot. –  Callum P Dunne Sep 8 '13 at 9:22

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