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I hope this question is not unreasonably broad. It is about calculating or at least bounding the cardinality of cohomology groups in case they are finite.

Let us assume we are given a group $G$ and a $\mathbb{Z}G$-module $A$. Then we can define the cohomology groups of $G$ as $$H^n(G,A) = \mathop{Ext}\nolimits^n_{\mathbb{Z}G}(\mathbb{Z}, A).$$

I am interested in techniques for calculating or bounding the cardinality of the cohomology groups in case they are finite.

One simple example can be given for cyclic groups. If $G$ is cyclic of order $k$ and $A$ is a trivial $\mathbb{Z}G$-module (i.e., $gm = m$ for all $g \in G$ and $m \in \mathbb{Z}$), then $H^0(G,A) = A$, $H^{2n-1}(G,A) = A[k]$, and $H^{2n}(G,A) = A/kA$, where $n \geq 1$. So this means, in particular, that $|H^{2n}(G,\mathbb{Z})| = k$. But the above theorem gives much more then what I ask for, namely the structure of $H^n(G,A)$. This is of course very well as long as I can get the cardinality from it.

So my question is: What techniques are known for calculating or bounding the cardinality of $H^n(G,A)$ in case it is a finite group?

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There are silly bounds, assuming some information on the coefficients. For example, taking $A=\mathbb Z$ one has that the complex which computes cohomology in terms of the bar resolution is made up of free abelian groups of a rank one can make precise, so each cohomology group is generated by at most that number of elements; since we know multiplication by $|G|$ is zero on cohomology, this gives a bound. – Mariano Suárez-Alvarez Feb 17 '13 at 2:55
(And that is enough to show that the Poincare-Hilbert series of cohomology coverges, say) – Mariano Suárez-Alvarez Feb 17 '13 at 3:05
@Martin Brandenburg: I only claim that $|H^{2n}(G,\mathbb{Z})| = |\mathbb{Z}/k\mathbb{Z}| = k$ and did not intend to suggest that $|A/kA| = k$ or $\leq k$ holds in general. – Gregor Samsa Feb 17 '13 at 8:34

There are some interesting results for simple faithful modules for finite groups for $n=1$ and 2.


M. Aschbacher and R. Guralnick, Some applications of the first cohomology group. J. Algebra 90 (1984), 446–460,

it is proved that, for a simple faithful module $M$ for a finite group $G$, we have $|H^1(G,M)| < |M|$.


Guralnick, R. M.; Kantor, W. M.; Kassabov, M.; Lubotzky, A. Presentations of finite simple groups: a quantitative approach. J. Amer. Math. Soc. 21 (2008), no. 3, 711–774,

it is proved that, for simple faithful modules $M$ for a finite group $G$ defined over a field, the dimension of $H^2(G,M)$ is at most 18.5 times the dimension of $M$.

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A particular paper that comes to mind is:
On orders and vanishing of integral cohomology groups, Angelina Chin
Here she obtains a recurrence relation involving the orders of the cohomology groups of a finite group $G$ whose quotient by a normal subgroup is cyclic.

1) Using Hopf's formula or $H^2(G,\mathbb{Z})\cong Hom(G,\mathbb{Q}/\mathbb{Z})$ should get you the orders of groups for 2nd cohomology.
2) You can check to see if your group has periodic cohomology (period $d$), and then $|H^n(G,\mathbb{Z})|=|H^{n+d}(G,\mathbb{Z})|=|G|$ for some $n$.
3) (Trivially) You can check to see if your group is cohomologically trivial, and then its cohomology vanishes.
4) For abelian groups you can probably get some information from the explicit calculation of $H_*(G,\mathbb{Z}_p)$, given in Theorem V.6.6 of Ken Brown's Cohomology of Groups textbook.

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In 2): not all cohomology groups of a periodic group have the same order. For example, the oddly-indexed cohomology groups of a finite cyclic groups with values in $\mathbb Z$ are trivial. – Mariano Suárez-Alvarez Feb 17 '13 at 2:52
Right, forgot to include "there exists $n$" – Chris Gerig Feb 17 '13 at 4:36
Ah. In fact, you can always take $n=d$. – Mariano Suárez-Alvarez Feb 17 '13 at 4:39
Thanks for your answer, Chris. The paper by Chin seems to be quite helpful. – Gregor Samsa Feb 17 '13 at 21:22

A rather less naive bound on the sizes of cohomology than the one I sketched in a comment to the question follows from the famous result of Venkov that if $G$ is a finite subgroup of $U(n)$, then $H^*(G,\mathbb F_p)$ has rational Hilbert series of the form $\frac{r(t)}{\prod_{i=1}^n(1-t^{2i})}$ with $r\in\mathbb Z[t]$, and such that the pole at $1$ has order equal to the $p$-rank of $G$. This gives very good bounds, although for each prime at a time.

To deal with the case of more general coefficients, there is the whole theory of complexity of modules.

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