Let $G$ be a profinite group, $A$ be a discrete $G$-module, and $n>0$ be an integer.
Why is the cohomology group $H^n(G;A)$ a torsion abelian group?
Here $H^n$ denotes the continuous cohomology groups. This thread is related, but I didn't find the answer to my question.
— I know that any (continuous) cocycle $f : G^n \to A$ has finite image, for $G$ is compact and $A$ is discrete. If the subgroup generated by the image of $f$ inside $A$ is also finite (say of cardinality $k$), then $k \cdot f = 0 : G^n \to A$ so that the class of $f$ in $H^n(G;A)$ has order at most $k$. If $f$ is also a group morphism, which holds if $n=1$ and $A$ is a trivial $G$-module, then the image of $f$ inside $A$ is already a subgroup, so the aforementioned condition is satisfied.
– But in general, we only want to find a multiple of $f$ which is a coboundary (without this multiple to be the zero map itself, as it was the case above). I'm not sure how to proceed. Is there may a smarter way to do it?
Thank you!