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I'm wondering if the group cohomology of a finite group $G$ can be made nontrivial with a nice choice of a finite $G$-module M. In other words, given a finite group $G$ and a number $n$, does there exist a finite $G$-module $M$ such that $H^n(G,M)$ is non-zero?

I would also be interested in the special case that $G$ is a finite $p$-group and n = 2. Can I always get $H^2(G,M) \ne 0$ for some finite $M$?

Thanks for your help.

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If $G$ is a finite $p$-group, then $H^2(G,\mathbb{F}_p)$ is always non-trivial. –  user641 Oct 6 '13 at 1:49
    
Thanks Steve! I just found that Kenneth Brown's Cohomology of Groups also has this fact -- if $G$ is a finite $p$-group then $H^i(G,\mathbb{Z}_p)$ is non-trivial for each $i$. –  user60593 Oct 6 '13 at 2:46

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Yes, for each $n\ge 0$ there is a $G$-module $M$ (depending on $n$) such that $H^n(G,M)\neq 0$ (provided $G\neq 1$ finite). .

Such an $M$ can be constructed by induction:

  1. First note that $H^i(G,F)=0$ for each free $\mathbb{Z}G$-module $F$ and all $i>0$ by Shapiro's lemma and Brown, VIII, 5.2.

  2. Next, show $H^1(G,I_G)=\mathbb{Z}/|G|$ where $I_G \trianglelefteq \mathbb{Z}G$ is the augmentation ideal (and $G$-action is given by multiplication in $\mathbb{Z}G$).

  3. Let $n\ge 2$ and suppose $N$ is a $G$-module such that $H^{n-1}(G,N) \neq 0$. Choose a short exact sequence $0 \to M \to F \to N\to 0$ of $G$-modules with $F$ free. Then, by the long exact sequence in cohomology and 1. we obtain the exact sequence $$0=H^{n-1}(G,F) \to H^{n-1}(G,N) \to H^n(G,M) \to H^n(G,F)=0,$$ i.e. $H^n(G,M)\cong H^{n-1}(G,N)\neq 0$.

Note: By starting with $I_G$ you can even arrange $H^n(G,M)=\mathbb{Z}/|G|$.

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Note that the construction above yields a finitely generated $M$ if one starts with $I_G$: If $N$ is supposed to be f.g. then $F$ can be choosen of finite rank and hence $M$ is f.g. because $\mathbb{Z}G$ is Noetherian. –  Ralph Oct 6 '13 at 18:08
    
The question asks for a finite $G$-module $M$. –  Yassine Guerboussa Oct 6 '13 at 18:48
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Well, in ring theory a finite module usually means finitely generated. Anyway. Let $M \le F$ be f.g. as above with $H^n(G,M)=\mathbb{Z}/|G|$ and let $p$ be a prime divisor of $|G|$. Since $M$ is (as abelian group) a finitely generated abelian group, $M/pM$ is a finite abelian group and a $G$-module. Now the long exact cohomology sequence corresponding to the short exact sequence of $G$-modules $0 \to M \xrightarrow[]{p} M \to M/pM\to 0$ shows $H^n(G,M/pM)\neq 0$. –  Ralph Oct 6 '13 at 19:04
    
I'm not very familiar with this, but it is not clear to me how one can deduce from the long exact cohomology sequence, and the non triviality of $H^n(G,M)$ (or $H^{n+1}(G,M)$), that $H^n(G,M/pM)$ is not trivial. I will be pleased if you can explain this more. –  Yassine Guerboussa Oct 6 '13 at 20:03
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I'm happy to do so: A part of the long exact cohomology sequence is the exact sequence $H^n(G,M) \xrightarrow[]{p} H^n(G,M) \to H^n(G,M/pM) \to \cdots$. Now suppose $H^n(G,M/pM)=0$. Using $H^n(G,M)=\mathbb{Z}/|G|$ we have the exact sequence $\mathbb{Z}/|G| \xrightarrow[]{p} \mathbb{Z}/|G| \to 0$, i.e. multiplication by $p$ is surjective. However, this isn't possible since $p$ divides $|G|$. –  Ralph Oct 6 '13 at 20:14

It is well known that, If $G$ and $M$ are finite $p$-groups, $H^2(G,M)=0$ is equivalent to $H^i(G,M)=0$ for all $i \geq 0$.

If one take any $M$ on which $G$ acts trivially, then $H^1(G,M) \cong Hom(G,M)$, which is not trivial unless $M$ or $G$ are trivial.

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Thanks for the answer Yassine- do you know of any corresponding result for finite groups? Even knowing if $H^2(G,M)$ is non-trivial for clever choice of $M$ and finite $G$ would be very interesting to me. –  user60593 Oct 6 '13 at 13:21
    
If I remember well, the property of having a trivial cohomology group in one dimension implies that all of theme are trivial, is special to finite p-groups. For the second, I don't really know. –  Yassine Guerboussa Oct 6 '13 at 13:28
    
@user60593 : The following link may help you journals.cambridge.org/action/…;, unfortunately I can not access the paper. –  Yassine Guerboussa Oct 6 '13 at 13:45
    
Actually, this solves the problem in all dimensions, one just need to solve it in dimension 1 (of course for finite p-groups) –  Yassine Guerboussa Oct 6 '13 at 19:07
    
@Ralph : I don't understand why you deleted the right comment, and leaved the wrong one. –  Yassine Guerboussa Oct 6 '13 at 19:48

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