Let $G$ be a finite group. If $M$ is a free $\mathbf{Z}[G]$-module, then $H^1(G',M) = H^2(G',M) = 0$ for all subgroups $G' \subset G$. Are there any other modules, free of finite rank over $\mathbf{Z}$ but not over $\mathbf{Z}[G]$, with this property?



What if the low-degree cohomology of a $G$-module and all its restrictions vanish?
I recommend reading Ken Brown's chapter on "Cohomologically Trivial Modules" from his book *Cohomology of Groups*. In particular, you should note that there is no need for only low-degrees: If $M$ is a free $\mathbb{Z}G$-module then $H^i=0$ for all $i>0$. 


Now for general $M$, your conditions on the (low-degree) cohomologies implies the much stronger condition that $M$ is cohomologically trivial, i.e. $H^i(G',M)=H\_i(G',M)=0$ for all $G'\subseteq G$ and all $i>0$. But if $M$ is cohomologically trivial and $\mathbb{Z}$-free, then it is in fact $\mathbb{Z}G$-projective. And if you're $\mathbb{Z}G$-projective, then you're cohomologically trivial. So the desired set of $G$-modules consists of the $\mathbb{Z}G$-projective ones.


