I had (perhaps very elementary) doubt in the understanding of the computation of first homology group of a finite group over a divisible group.
Let $\pi$ be a finite group of order $n$ and $D$ be a divisible torsion-free group. Here action of $\pi$ on $D$ is not given, so I then assumed that it is any action.
Let $f\colon \pi\rightarrow D$ be a $1$-cocycle. It is then asserted that $nf$ is a $1$-coboundry, which I do not understand.
This may be trivial too, but (I am learning homology now) I confused at this point. Any hint for this?
Le $f$ be a $1$-cocycle from $\pi$ to $D$. Let $n$ be the cardinal of $\pi$. Remark that for $g\in \pi$ :
$$g\cdot \sum_{h\in H}f(h)=\sum_{h\in H}(f(gh)-f(g))=\sum_{h\in H}f(gh)-n\cdot f(g)=\sum_{h\in H}f(h)-n\cdot f(g) $$
That is if $x:= \sum_{h\in H}f(h)\in D$ then $g\cdot x-x=n\cdot f(g)$. When $n\cdot f$ is always a 1-coboundary ($D$ being divisible or not...).
Consequence : $H^1(\pi,D)$ is of $n$-torsion. If $D$ is divisible then this implies that $H^1=0$.
By the way, you can find the same trick to show that whenever $f\in Z^m(\pi,D)$ then $n\cdot f$ is a $k$-coboundary (however this is more difficult to prove... with this trick, you might want to show directly that $H^k(\pi,D)$ is of $n$-torsion using restriction-corestriction kind of proof).