Background:
I want to consider relative group cohomology: the construction is as follows. I have a subgroup $H\subseteq G$ (and note that I don't want to assume that $H$ is normal in $G$), and a $\mathbb Z[G]$-module $M$. Then we have the standard chain complxes $C^\ast(G;M)$ and $C^\ast(H,M)$, and there is a natural morphism $C^\ast(G,M)\to C^\ast(H,M)$, which induces the "restriction homomorphism" on group cohomology $\operatorname{res}:H^\ast(G,M)\to H^\ast(H,M)$. Let us define the "relative group cohomology" as the cohomology of the chain complex which fits into the exact sequence:
$$
0\to C^\ast(G,H;M)\to C^\ast(G;M)\to C^\ast(H;M)\to 0
$$
(i.e. $C^\ast(G,H;M)$ is defined to be the kernel of the second map). I haven't ever heard of these "relative group cohomology" groups, but it seems like a very natural idea to me, and what I'm trying to do is define algebraically the cellular cohomology groups $H^\ast(K(G,1),K(H,1);M)$ (in case we have a $K(H,1)$ which is naturally a subcomplex of a $K(G,1)$). If anyone has a good reference for these I'd like to know! Note that by definition, the relative group cohomology groups $H^\ast(G,H;M)$ fit into a natural long exact sequence:
$$
\cdots\to H^\ast(G,H;M)\to H^\ast(G;M)\to H^\ast(H;M)\to\cdots
$$
and this is what one would expect the cellular cohomology groups $H^\ast(K(G,1),K(H,1);M)$ to satisfy. If I've messed this construction up, please tell me.
Question:
How do I understand the relative group cohomology in terms of derived functors? We know that $H^\ast(G;M)=\operatorname{Ext}^\ast_{\mathbb Z[G]}(\mathbb Z,M)$ and $H^\ast(H;M)=\operatorname{Ext}^\ast_{\mathbb Z[H]}(\mathbb Z,M)$. But since these are $\operatorname{Ext}$'s in different categories, it doesn't seem clear how to fit a third into the exact sequence. What I'd like is some $\operatorname{Ext}$ definition of the relative group cohomology groups I've defined above.
More Info:
I've tried the following, but it seems to give the "wrong" answer. We can get everything into the same category by observing that $M^H=\operatorname{Hom}_{\mathbb Z[G]}(\mathbb Z[G/H],M)$, and thus the cohomology is given by $H^\ast(H;M)=\operatorname{Ext}^\ast(\mathbb Z[G/H],M)$ (from now on, all $\operatorname{Ext}$'s are in the category of $\mathbb Z[G]$-modules). Furthermore (and correct me if I am wrong), the restriction homomorphism $H^\ast(G,M)\to H^\ast(H,M)$ is induced by the "sum coefficients" morphism $\mathbb Z[G/H]\to\mathbb Z$ (giving the map $\operatorname{Ext}^\ast(\mathbb Z,M)\to\operatorname{Ext}^\ast(\mathbb Z[G/H],M)$). So, now it looks like we get what we want, but now comes a surprise. The "first argument"s of the $\operatorname{Ext}$'s fit into a short exact sequence:
$$
0\to\ker\to\mathbb Z[G/H]\to\mathbb Z\to 0
$$
and thus we have a long exact sequence of $\operatorname{Ext}$:
$$
\cdots\to\operatorname{Ext}^\ast(\mathbb Z,M)\to\operatorname{Ext}^\ast(\mathbb Z[G/H],M)\to\operatorname{Ext}^\ast(\ker,M)\to\cdots
$$
But now it looks like $\operatorname{Ext}^\ast(\ker,M)$ is not giving the relative group cohomology groups we want: the long exact sequence isn't the same as the one above, it's gotten flipped around. I guess this doesn't entirely disqualify the construction, since perhaps we have $H^\ast(G,H;M)=\operatorname{Ext}^{\ast-1}(\ker,M)$, but in this case I'd still like an explanation for why this dimension shifting happens.
Best Answer
Your argument is correct, and you do get that the relative group cohomology in your terms is a shift of these Ext-groups.
One reason why the dimension-shifting behavior occurs is that, with original gradings, you can't possibly have that $H^i(G,H;M)$ are the derived functors of $H^0(G,H;M)$. In fact, $C^0(G,H;M)$ is always the zero group, and so $H^0(G,H;M)$ is always zero; its derived functors are zero.
In this case, what your argument shows is that $H^{i+1}(G,H;M)$ is the i'th derived functor of $H^1(G,H;M)$.
If I might be permitted to wax philosophical, once one has moved to chain complexes the general yoga of triangulated categories says that one shouldn't really worry about the distinction between a kernel and the shift of a cokernel. Up to weak equivalence of chain complexes, one can always move between these.