OK, I'm just learning about cohomology of groups, and I want to make sure I'm not missing anything in this question:
Let $n$ be an abelian group, let $G$ be a group, let $\tau$ be an action of $G$ on $N$ by automorphisms, and let $n>0$ be an integer. Prove that if every element of $N$ has finite order dividing an integer $m$, then every member of $H^n(G,N)$ has finite order dividing $m$.
To my mind, every element of $H^n(G,N)$ is an $N$-valued function (mod $B^n$, and belonging to $Z^n$). So if $m \cdot \nu = 0$ for all $\nu \in N$, then certainly $(g_1, \dotsc, g_n) \mapsto m \cdot f(g_1, \dotsc, g_n)$ is the zero function...right? Am I missing something?
You are not missing anything, seeing $H^n(G,N)$ through the inhomogenous cochain (i.e. functions from $G^n$ to $N$ verifying some equations mod $B^n$) allows you to make this proof.
A far more intersting result is the following : if $G$ is a group of finite order and $n>0$ then any $f\in Z^n(G,N)$ verifies that $|G|.f\in B^n(G,N)$ so that every element of $H^n(G,N)$ has finite order dividing $|G|$.
Hint : given $f\in Z^n(G,N)$ we set :
$$\nu(g_1,...,g_{n-1}):=\sum_{g\in G}f(g_1,...,g_{n-1},g) $$
And then compute :
$$\partial^{n-1}(\nu)(g_1,...,g_n) ...$$
I leave the (heavy) calculation to you and juste remark that if $|G|$ is finite and mutliplication by $|G|$ is a bijection in $N$ then you will have that $H^n(G,N)=0$ if $n>0$. In particular, this is true if $N$ is $\mathbb{Q}$-vectorial space. So it gives easily a cancellation of group cohomology (which appears to be a good one, it applies sometimes).