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.