After our detour to learn some representation theory, we now return to homological algebra and apply our earlier definition of the Ext groups.  Given a finite group $G$, we start by attaching another group ring to it which superficially looks a lot like $\Bbb CG$ but in fact has very different properties.  This is the {\sl integral group ring} $\Bbb ZG$, defined in the same way as $\Bbb CG$, except that the coefficients $k_g$ of the group elements $g$ now lie in $\Bbb Z$ rather than $\Bbb C$.  Modules over this ring are then the same as linear actions of $G$ on abelian groups $A$ (obeying the same axioms as modules of $G$ over complex vector spaces, except that the scalars are now integers; we do not require that $A$ be free as a $\Bbb Z$-module).  The much looser structure of $\Bbb Z$-modules compared to vector spaces already has the important consequence that the notion of irreducibility is essentially useless in this context:  since for example $kA$ will always be a $G$-submodule for any
$G$-module $A$ and $k\in\Bbb Z$, regardless of how $G$ acts on $A$, hardly any $G$-modules will qualify as irreducible.  Accordingly there is no useful way to decompose $G$-modules as a direct sum of simpler pieces in general.  We focus attention on very special $G$-modules, namely the ones one which $G$ acts trivially, and study they fit into nontrivial modules.  More precisely, given a $G$-module $A$ we define its $n$th {\sl cohomology group $H^n(G,A)$ with coefficients in $A$} as
Ext_{{\Bbb Zg}^i(\Bbb Z,A)$, where the action of $G$ on $\Bbb Z$ is trivial.  The functor attaching
$H^n(G,A)$ to $A$ may be regarded as a higher right derived functor of the functor attaching to $A$ its submodule $A^G = \{a\in A:  ga = a, g\in G\}$ of the $G$-invariant elements of $A$; this functor is left but not right exact.  We can compute (as we would for any Ext group) by projectively resolving
$\Bbb Z$ and taking $G$-homomorphisms into $A$.  

As a first example, suppose $G=\Bbb Z_N$ is cyclic of order $N$, generated by say $x$.  Here the group ring $R=\Bbb ZG$ is the quotient $\Bbb Z[x]/(x^N - 1)$ of the polynomial ring $\Bbb Z[x]$ by the ideal generated by $x^N - 1$; note that this quotient is not a sum of matrix rings over $\Bbb Z$.  Here a surjective map from the projective $R$-module $R$ is given by the {\sl augmentation map}, sending $\sum_g k_g g$ to $\sum k_g$; its kernel is the ideal $(x-1)$ generated by $x-1$ in $R$.  We extend the surjective map $R\rightarrow\Bbb Z$ to a projective resolution
$R\rightarrow\cdots\rightarrow R\rightarrow\Bbb Z$ by multiplying by
$(x^N -1)/(x-1) = x^{N-1} + \ldots x + 1$ and $x-1$ alternately in $R$; accordingly (as when computing Ext$_{\Bbb Z_n}^i (\Bbb Z_d,M)$ earlier) we find that $H^0(G,A) = A^G$ (this much holds for any finite group $G$ and $G$-module $A$), while $H^i(G,A) = {}_{(x^n - 1)/(x-1) A/(1-x)A$ for $i$ odd, $H^i(G,A) = {}_{x-1} A/(x^N-1)/(x-1)$ for $i$ positive and even, where for any $r\in R$ we denote by
${}_r A,rA$ the respective $R$-submodules $\{a\in A: ra = 0\}$ and $\{ra:a\in A\}$.  

In general it will be very useful to know that there is a standard way to resolve $\Bbb ZG$ that works for any $G$; actually there are two ways to do this and we will need both of them for different purposes..  Begin by setting $R_n = \Bbb ZG^{\iotimes n}$, the $(n+1)$st tensor power of
$R=\Bbb ZG$, where the tensor product takes pace over $\Bbb Z$ (not $R$).  There are two ways to make $R_n$ into an $R$-module:  either have $R$ act just on the leftmost factor of $R$ (as we have done earlier for bimodules over general noncommutative rings), or have $R$ act on all the factors, via $g\cdot(g_0\otimes\cdots\otimes g_n = gg_0\otimes\cdots\otimes gg_n$; this second way is a very special construction that works only for group rings.  In both cases $R_n$ is free over $R$, with basis all the tensors $1\oitmes g_1\otimes\cdots\otimes g_n$, where the $g_i$ range over $G$.  For both actions we define $d_0':R\rightarrow\Bbb Z$ by the augmentation map used above.  For the second action we define $d'_n:R_n\rightarrow R_{n-1}$ via
$d_n'(g_0\otimes g_1\otimes\g_n) = \sum{i=0}^n (-1)^i (g_0\otimes\cdots\hat{g_i}\cdots\otimes g_n$, where the hat over $g_i$ means that this term is omitted.  It is very easy to see that the $d_n'$ are $G$-homomorphisms and $d_{n-1}' d_n' = 0$; if we denote by $s_n: R_n\rightarrow R_{n+1}$ the map sending $g_0\otimes\cdots g_n$ to $1\otimes g_0\otimes\cdots\otimes g_n$ then it is also not difficult to check that $s_{n-1} d_n' + d_{n+1} s_n = 1$, so that every element in the kernel of $d_{n-1}'$ lies in the range of $d_n'$.  Hence the $R_n$ and $d_n'$ define a projective resolution of $\Bbb ZG$ (with respect to the {\sl second} action of $R$ on $R_n$).  Turning now to the first action of $R$ on $R_n$, we get a more complicated formula for $d_n:R_n\rightarrow R_{n-1}$, defined by
$d_n(1\otimes g_1\otimes\cdots\otimes g_n) =
 g_1\otimes\cdots\otimes g_n +
\sum_{i=1}^{n-1} (-1)^i (1\otimes g_1\otimes\cdots\otimes g_{i-1\otimes g_ig_{i+1}\otimes g_{i+2}\cdots\otimes g_n)
+ (-1)^n(1\otimes g_1\otimes\cdots\otimes g_n$ for $n>0$, while $d_0 = d_0'$.  Now it is much more tedious to verify that $d_{n-1} d_n = 0$ and that the complex defined by the $d_n$ is exact, but fortunately the map sending $g_1\otimes\cdots g_n$ to 
$g_1\otimes g_1g_2\otimes\cdots\otimes g_1 g_2\ldots g_n$ commutes with the $G$-actions on the first and second complexes *and* the boundary maps, so the second complex, like the first one, is exact.  Taking homomorphisms into $A$ and recalling that $R_n$ is free with basis the tensors
 $1\otimes g_1\otimes\cdots\otimes g_n$, we see that a homomorphism $f$ from $R_{n+1}$ to $A$
 is completely determined by $f(1\otimes\cdots\otimes g_n$, which (adjusting our notation) we denote by $f(g_1,\ldots,g_n)$, so that we now regard $f$ as a function from $G\times\cdots G$ to $A$.  In this notation its coboundary $d^n f$ is defined by
$d^n(g_1,\ldots,g_{n+1} = g_1\cdot f(g_2,\ldots,g_{n+1})
+\sum_{i=1}^n (-1)^i f(g_1,\ldots,g_i g_{G,A)$ is the quotient of the kernel of $d^n$ by the image of
$d^{n-1}$; note that (like any Ext group) it is only a $\Bbb Z$-module and carries no action of $G$.  (One could write down the analogous formula for $H^n(G,A)$ using the other resolution of $\Bbb Z$, but the $G$-action in this case would give this formula a more awkward appearance.)