# Group Cohomology Vs Profinite group Cohomology

What is the difference of the group cohomology and the profinite group cohomology? I think one reason is that in the profinite group situation, the G-module must be continuous. Does any other difference?

I also want to find an profinite group whose group cohomology is not the profinite group cohomology. If there is not such a group, we need not to define the profinite group cohomology.

• My impression is that in a suitable context, they are the same. That context is topological groups. To a topological group, one mat construct $EG\to BG$ in a variety of ways (I like the simplicial bar construction my self). Then regular group (co)homology would be the (co)homology of $BG$ when $G$ is given the discrete topology and profinite group (co)homology is the (co)homology of $BG$ when $G$ is profinite. – Baby Dragon May 1 '13 at 13:59
• What do you mean by "what is the difference"? – Bruno Joyal Nov 22 '13 at 4:40
• Thanks for answering this question, I have a simple example at 4.2.4 in the book: Gille P, Szamuely T. Central simple algebras and Galois cohomology[J]. 2006. – Strongart Sep 15 '18 at 14:00

## 2 Answers

You could of course start from any topological group G acting continuously on a topological module M and define the cohomology H*(G, M) by means of continuous cochains. But what is sought actually is a "cohomological functor", i.e. a machinery which transforms a short exact sequence of modules into an infinite exact sequence of coh. groups. If your group G is profinite and your module M is discrete (this is equivalent to saying that M is the union of all the submodules fixed by the open subgroups of G), then the above construction works perfectly, what you obtain is "Galois cohomology", which is essential e.g. in infinite Galois theory. But in number theory, some deep problems require a profinite G (e.g. a Galois group) acting continuously on a compact M (e.g. the p-adic integers), and here you no longer have a coh. functor.

If you take a prime $p$ and a non-commutative finitely generated free pro-$p$ group $F$, then the second continuous cohomology group with coefficients in $\mathbb Z/p$ is trivial $$H^2_{\sf cont}(F,\mathbb Z/p)=0$$ but the discrete cohomology group is uncountable (in particular, it is nontrivial) $$H^2_{\sf disc}(F,\mathbb Z/p)\ne 0.$$

see this or this