Free and surface groups cohomology What is a good reference for results on cohomology of finite rank free groups and surface groups with group ring coefficients? 
I am interested in the case when the group acts on its group ring via conjugation. In particular, as for free groups I am interested in how the first cohomology of the group and its infinite index subgroups are related (how big the kernel of the restriction map is). Are there explicit ways to find these kernels (maybe in some special cases)? And also what is known about the cohomology groups of groups of compact surfaces (again with group ring coefficients and action by conjugation)? What relationships between cohomology of a free group and cohomology of a surface group can be derived from considering this surface group as a quotient group of the corresponding free group?
I don't know much about the subject but it would be very helpful to get the whole picture of these things before reading abstract proofs. 
Added on 02.03.2012:
Well,I don't know why I decided to restrict myself to normal subgroups of a free group only
(see comment below). Actually, it would be interesting to get any information on subgroups which are not normal too. Initially, I was interested in a question which could be formulated in terms of group theory but the question itself (or rather its infinitesimal version) led me to cohomology. 
Dealing with cohomology, my first desire as a newbie was to see some good correspondence between subgroups of a free group and kernels of restrictions. Pretty soon I found that this is far from being Galois correspondence. Many subgroups have the same restriction kernel and such subgroups may even be not commensurable.
 A: Concerning your question on the relation between the cohomology of the free group and the surface group: 
Let $F$ be a free group, $N \trianglelefteq F$ and $M$ an $F/N$-module. Consider $M$ as $F$-module via $F \to F/N$. Then the seven-term-exact sequence yields the exact sequence 
$$0 \to H^1(F/N;M) \xrightarrow[]{inf} H^1(F\;;M) \xrightarrow[]{res}H^1(N;M).$$ 
Thus $H^1(F/N;M)$ is a subgroup of $H^1(F\;;M)$, namely the kernel of the restriction resp. the image of the inflation. 
If we interpret $H^1(F\;;M)$ in terms of derivations then $H^1(F/N;M)$ consists of those classes that are represented by derivations $F \to M$ such that $f|N = 0$. Moreover, if $N$ is the normal closure of $R \subseteq F$ then $f|N = 0$ is equivalent to $f|R=0$.
Example: $F=F_{2n}$ and $R=\lbrace [x_1,x_2] \cdots [x_{2n-1},x_{2n}]\rbrace$, i.e. $F/N$ is the fundamental group of a closed oriented surface of genus $n$. Take $\mathbb{Z}$-coefficients with trivial action. Now derivations are linear und hence vanish on commutators. Consequently $H^1(F/N; \mathbb{Z}) = H^1(F\;;\mathbb{Z}) = \mathbb{Z}^{2n}$. (One can also derive a formula for general coefficients, but it isn't very handy). 
Concerning the restriction homomorphism of a free group: By the Nielsen-Schreier theorem, a subgroup of a free group is again free. This reduces the problem to compute the restriction between two free groups. This can easily been done by using the free resolutions from Brown, I (4.4) and computing a chain map between the resolutions. Since I guess that you are primarily interested in surface groups that can be treated with help of Lyndon's paper or by the method above, I leave out details. 
