Quantcast
Channel: Relations between the cohomology of discrete groups and of profinite groups - MathOverflow
Viewing all articles
Browse latest Browse all 2

Relations between the cohomology of discrete groups and of profinite groups

$
0
0

Let $G$ be a discrete group and $K$ be the profinite completion of $G$. Let $C_K$ denote the category of contionuous $K$-modules and ${C_K}'$ denotes category of finite continuous $K$-modules. Now for $A \in {C_K}'$, I am trying to prove the restriction map\begin{equation} H^q(K,A) \rightarrow H^q(G,A) \end{equation} is surjective for all $q \leq n$ if for all $x \in H^q(G,A)$, $1\leq q\leq n$, there exist an $A' \in C_K$ (not necessarily finite) such that $x$ maps to $0$ in $H^q(G,A')$. This is an exercise in Serre's Galois Cohomology. I am able to prove the statement is valid for $n=1$ without assuming the existence of $A'$. For arbitrary $n$, my idea is using the long exact sequence of cohomology groups obtained from \begin{equation} 0 \rightarrow A \rightarrow A' \rightarrow A'/A \rightarrow 0 \end{equation} and doing induction on $n$. However since $A'$ is not necessarily finite this argument does not work. Is there any way of replacing $A'\in C_K$ with some abelian group in ${C_K}'$ ? or am I missing something ? Thanks,
Ozlem


Viewing all articles
Browse latest Browse all 2

Trending Articles