Suppose $G$ is a finite Galois group, and $M$ is an infinite $G$-module. When can I say that $H^1(G, M)$ is finite?
I know this not true in general. Is it true under certain assumptions on $M$?


To be more precise, can I say that $H^1(G, V/T)$ is finite? where $V$ is finite-dimensional $\mathbb{Q}\_p$-representation of $G$, $T$ denotes a lattice in $V$ which is invariant under the action of $G$, and $G$ is an inertia group of a prime in a finite extension of $\mathbb{Q}$.



Finiteness of cohomology group
To summarize the discussion from the comments as an answer:


For a finite group and a finitely generated $R[G]$-module $M$, the groups $H^\*(G;M)$ are computed by a chain complex of finitely generated $R$-modules, so under reasonable assumptions on $R$ the cohomology groups are finitely generated as $R$-modules. Also, they are torsion in positive degrees (independent of any assumption on the coefficients, this is a transfer argument). It follows that the positive-degree $H^\*(G; \mathbb{Z}\_p^n)$ are finite abelian $p$-groups, and that the positive-degree $H^\*(G; \mathbb{Q}\_p^n)$ vanish. The desired finiteness of $H^\*(G; \mathbb{Q}\_p^n / \mathbb{Z}\_p^n)$ now follows from the long exact sequence.


