# Fundamentals of (Abelian) Group Cohomology

In this post we will talk about the basic theory of group cohomology, including the cohomology of profinite groups.

We will assume that the reader is familiar with the basic theory of derived functors as in, say, Weibel’s Homological Algebra.

# Group Cohomology

## The Category of G-Modules

Let $G$ be a group. We say that an abelian group $A$ is a $G$-module if it comes equipped with a $G$-action $G\times A\to A$ which is additive: $g(a+b)=ga+gb$ for all $a,b\in A$. Of course, as is the case in most parts of mathematics, this is not the only way to define a $G$-module. For example, the above action is clearly equivalent to equipping $G$ with a homomorphism $G\to\text{Aut}(A)$, where the automorphisms are group automorphisms.Perhaps less tautological, is the equivalence of $\mathbb{Z}[G]$-modules with $G$-modules.

Recall that $\mathbb{Z}[G]$, called the integral group ring over $G$, is the free abelian group on the set $G$, with multiplication given by linearly extending the group operations on $G$. So, for example, if $G=\mu_p(\mathbb{C})\cong \mathbb{Z}/p\mathbb{Z}$, the group of $p^{\text{th}}$-roots of unity in $\mathbb{C}$, then $\mathbb{Z}[G]$ is the free abelian group on $\{1,\ldots,\zeta^{p-1}\}$ (where $\zeta$ is a primitive $p^{\text{th}}$-roots of unity). The multiplication between the formal symbols $\zeta^i$ is given by

$\zeta^a\cdot\zeta^b=\zeta^{a+b\mod p}$

So, in this instance, we see that $\mathbb{Z}[\mu_p]$ is given merely by $\mathbb{Z}[T]/(T^p-1)$.

Now, the relation between $G$-modules and $\mathbb{Z}[G]$-modules is simple. Any $G$-module $A$ can be made into a $\mathbb{Z}[G]$-module by merely extending the action of $G$ on $A$ linearly. Conversely, given a $\mathbb{Z}[G]$-module $A$, the restriction of the action of the scalars $\mathbb{Z}[G]$ to the subset $G\subseteq\mathbb{Z}[G]$ defines a $G$-module structure.

There is a fancier way to understand this equivalence though. Namely, consider the group ring functor $\mathbf{Grp}\to\mathbf{Ring}$ given on objects by $G\mapsto \mathbb{Z}[G]$ and given on morphisms $f:G\to H$ by extending linearly:

$\displaystyle \sum_g z_g g\mapsto \sum_g z_g f(g)$

It turns out that the group ring functor is the left adjoint to the units functor $U:\mathbf{Ring}\to\mathbf{Grp}$ taking $R\to R^\times$. Thus, if $G$ is a group, and $R$ is a ring, we have that

$\text{Hom}_\mathbf{Ring}(\mathbb{Z}[G],R)=\text{Hom}_\mathbf{Grp}(G,R^\times)$

Thus, if $A$ is an abelian group, and $\text{End}(A)$ is its ring of group endomorphisms, then the above adjunction tells us that

$\text{Hom}_\mathbf{Ring}(\mathbb{Z}[G],\text{End}(A))=\text{Hom}_\mathbf{Grp}(G,\text{Aut}(A))$

But, a ring map $\mathbb{Z}[G]\to\text{End}(A)$ is the same thing as giving $A$ a $\mathbb{Z}[G]$-module structure, and as noted above, a group map $G\to\text{Aut}(A)$ is just defining a $G$-module structure on $A$. Thus, we see that these two notions are the same.

The important thing to note about this equivalence is that it tells us that the category $G\text{-}\mathbf{Mod}$ of $G$-modules is abelian, being equivalent to the category $\mathbb{Z}[G]\text{-}\mathbf{Mod}$ of (left) $\mathbb{Z}[G]$-modules. In the future, we shall freely pass between the two equivalent notions of $G$-modules, depending on which is more convenient. We shall denote the morphism sets in $G\text{-}\mathbf{Mod}$ either as $\text{Hom}_G(A,B)$ or $\text{Hom}_{\mathbb{Z}[G]}(A,B)$ depending on which is more illuminating.

Let us now consider some examples of $G$-modules which might be of interest to us:

Every abelian group $A$ is a $G$-module with the trivial structure given by $ga=a$ for all $a\in G$ (i.e. the homomorphism $G\to\text{Aut}(A)$ is trivial). We shall call a group with the trivial structure a trivial $G$-module. For all intents and purposes, a trivial $G$-module is just the same as an abelian group.

Let $L/K$ be a Galois extension of fields. Then, $L$ (thought of as an additive group) is a $\text{Gal}(L/K)$-module with the usual action $\sigma a:=\sigma(a)$ for $a\in L$ and $\sigma\in \text{Gal}(L/K)$. Similarly, $L^\times$ becomes a $\text{Gal}(L/K)$-module with the structure $\sigma a=\sigma(a)$ for $a\in L^\times$ and $\sigma\in\text{Gal}(L/K)$. Succinctly, this is because $\text{Gal}(L/K)$ is both a subgroup of $\text{Aut}(L)$ and $\text{Aut}(L^\times)$.

In a similar vein, let $L/K$ be a Galois extension of either local or global fields. Then, $\mathcal{O}_L$ and $\mathcal{O}_L^\times$ become $\text{Gal}(L/K)$-submodules of $L$ and $L^\times$ respectively.

Let $A$ be an abelian variety over $K$. Then, the group $A(\overline{K})$ becomes a module over the absolute Galois group $G_K:=\text{Gal}(K^\text{sep}/K)$ with the usual action on the coordinates.

Let $R$ be the integral polynomial ring $\mathbb{Z}[T_1,\ldots,T_n]$ in $n$-variables. Consider the action of the symmetric group $S_n$ on $R$ given by permuting the indeterminates. Clearly $S_n$ acts not only as group automorphism, but in fact ring automorphisms. Thus, we see that with this action $R$ becomes an $S_n$-module.

Let $A$ be some object over a field $k$ (e.g. a $k$-vector space or a $k$-algebra). Then, $A_K:=A\otimes_k K$, where $K/k$ is a Galois extension, is a $\text{Gal}(K/k)$-module with $\sigma(a\otimes x)=a\otimes\sigma(x)$.

Every representation one might have seen in a graduate algebra course, in other words a homomorphism $G\to \text{GL}(V)$, where $V$ is some $k$-space, usually $k=\mathbb{C})$, makes $V$ into a $G$-module. There are two equivalent ways of thinking about this. The easier of the two is just to note that $\text{GL}(V)\subseteq \text{Aut}(V)$ (where on the right we are just considering $V$ with its underlying group structure). Thus, a representation is, in particular, a homomorphism $G\to\text{Aut}(V)$. The other is to recall that giving a representation of $G$ on $V$ is the same as endowing $V$ with a $\mathbb{C}[G]$-module structure (this may have even been your definition of $\mathbb{C}$-representation. But, this then clearly also endows $V$ with a $\mathbb{Z}[G]$ ($\subseteq\mathbb{C}[G]$) structure. One can think about representation theory (at least of finite groups) as the study of linearized $G$-modules.

As you can see from the above examples, some of the most important examples of $G$-modules occur in the context of number theory.

## Group Cohomology: Motivation

Now, while $G$-modules are often useful objects themselves, very often we are more interested in a certain group associated to a $G$-module. In particular, if $A$ is a $G$-module, let us denote by $A^G$ the group of invariants of $A$, which is defined as follows:

$A^G=\{a\in A:ga=a\text{ for all }g\in G\}$

In other words, $A^G$ is just the set of fixed points of the $G$-action on $A$. Note that while $A^G$ does have an inherited $G$-module structure from $A$, this structure is trivial (it’s the largest such $G$-submodule of $A$) and so really carries no more structure than the underlying abelian group of $A$.

Usually, the group of invariants of a $G$-module has some sort of geometric or arithmetic meaning. The whole $G$-module is often something large we can get a hand on, and the set of $G$-modules is the object that we’re really after.

Consider the example of an abelian variety $A$ over $K$ as above. We are often interested in the set $A(K)$ of $K$-points, but can usually only get a hand on the full set $A(\overline{K})$ of $\overline{K}$-points. But, we can recover $A(K)$ as the set of $G_K$-fixed points of $A(\overline{K})$.

To be more concrete, $E$ may be an elliptic curve over $\mathbb{Q}$. The $\overline{\mathbb{Q}}$-points of $E$ are something we have a somewhat good understanding of. But, what we’re really interested in are the $E(\mathbb{Q})$-points, which are nothing but the $G_\mathbb{Q}$-fixed points of $E(\overline{\mathbb{Q}})$.

As another example, consider the case discussed above where $K/k$ is a Galois extension of fields, and $A$ is a $k$-object. Then, $A_K$ is a $K$-object which might be of a simpler nature. In simple cases, it is easy to show that if $G=\text{Gal}(K/k)$ acts on $A_K$ as above, then $A_K^G\cong A$ as a $k$-object.

Now, since the full $G$-modules are usually the objects for which we can more easily deal with, we can often times find relationships between them–ways of putting two simpler $G$-modules together to describe a possibly more complicated $G$-module. More rigorously, I mean that we are often times able to find exact sequences of $G$-modules:

$0\to A\to B\to C\to 0$

But, since it’s really the group of invariants of theses $G$-modules that we may care about, we would really like it to be true that $B^G$ can also be put together nicely from the modules $A^G$ and $C^G$. Or, said differently, we’d like the induced sequence

$0\to A^G\to B^G\to C^G\to 0$

to be exact. But, we don’t always get what we want. It turns out that, in general, this sequence need not be exact. In particular, it is always true that the sequence

$0\to A^G\to B^G\to C^G$

is exact, but it is not true that taking the invariant module preserves surjections.

As a quick example of this, consider the exact sequence of groups

$1\to \mu_n(\overline{\mathbb{Q}})\to \overline{\mathbb{Q}}^\times\xrightarrow{n}\overline{\mathbb{Q}}^\times\to 1$

where $n$ denotes the $n^{\text{th}}$-power map. Note that this is actually an exact sequence of $G_\mathbb{Q}$-modules. Passing to the $G_\mathbb{Q}$-invariants gives us

$1\to \mu_n(\mathbb{Q})\to \mathbb{Q}^\times\xrightarrow{n}\mathbb{Q}^\times$

where, of course, we have neglected to put a $1$ on the right hand side, because the $n^{\text{th}}$-power map, while surjective on $\overline{\mathbb{Q}}$, is not surjective on $\mathbb{Q}$.

This is where the general ideology of (right) derived functors comes into play. Namely, the functor $-^G$ of $G$-invariants $G\text{-}\mathbf{mod}\to\mathbf{Ab}$ is left exact, but not right exact. We can thus produce it’s associated right derived functors $R^i(-^G)$, which we will denote by $H^i(G,-)$. The general theory then tells us that for every exact sequence

$0\to A\to B\to C\to 0$

of $G$-modules, we get the long exact sequence

$0\to A^G\to B^G\to C^G\to H^1(G,A)\to H^1(G,B)\to H^1(G,C)\to\cdots$

thus, effectively, fixing the non-surjectivity by adding in the groups $H^i(G,A), H^i(G,B)$, and $H^i(G,C)$. These right derived functors, $H^i(G,-)$ will be what we define as the group cohomology functors.

## Group Cohomology: Definitions and Basic Theorems

So, now that we have (hopefully) been sufficiently motivated for the definition of group cohomology, we can dive right into it. As stated above, the fixed points functor $-^G$ is a left exact (necessarily additive!) functor from the abelian category $G\text{-}\mathbf{Mod}$ to $\mathbf{Ab}$. Since $G\text{-}\mathbf{Mod}$ is equivalent/isomorphic to the category $\mathbb{Z}[G]\text{-}\mathbf{Mod}$, and categories of modules always have enough injectives, we know that $G\text{-}\mathbf{Mod}$ also has enough injectives.

Thus, we are able to form the right derived functors $R^i(-^G)$. If $A$ is a $G$-module, we then define the $i^{\text{th}}$ (group) cohomology of $A$ with $G$-coefficients, denoted $H^i(G,A)$, to be $R^i(-^G)(A)$.

Just to recall, here is the general methodology to produce $H^i(G,A)$. First we find a resolution of $A$ by injective $G$-modules $0\to A\to I^{\bullet}$ (which we know exists by the above discussion), we then obtain $H^i(G,A)$ as the $i^{\text{th}}$ homology group of the deleted complex $0\to (I^\bullet)^G$.

Now, while this presentation of $H^i(G,-)$ is the most obvious in definition, there is an equivalent, but notationally different, way of defining $H^i(G,-)$. This different presentation comes from the realization that the functor $-^G$ is naturally isomorphic to another functor for which we already have well-understood theory of its derived functor. Namely, let $\mathbb{Z}$ have the trivial $G$-module structure, and define $h^\mathbb{Z}:= \text{Hom}_{\mathbb{Z}[G]}(\mathbb{Z},-)$ then:

Theorem 1: The two functors $-^G$ and $h^\mathbb{Z}$ are isomorphic.

Proof: Define the morphism $\alpha:h^\mathbb{Z}\implies -^G$ as follows. For any $G$-module $A$, define

$\alpha(A):h^\mathbb{Z}(A)\to A^G:\varphi\mapsto \varphi(1)$

To see that this is indeed well-defined (i.e. that $\alpha(A)(\varphi)\in A^G$), we merely note that for any $\varphi\in h^\mathbb{Z}(A)$ we have that

$g\varphi(1)=\varphi(g\cdot 1)=\varphi(1)$

which does, in fact, show that $\varphi(1)\in A^G$. Now, it’s clear that $\alpha(A)$ is a morphism of abelian groups. We claim that it is an isomorphism. The injectivity follows immediately from the fact that $1$ generates $\mathbb{Z}$ as a $\mathbb{Z}[G]$-module. For surjectivity, let $a\in A^G$ be arbitrary. Define $\varphi:\mathbb{Z}\to A$ by $\varphi(n)=na$. A quick check shows that $\varphi$ is a $G$-module map, and that $\alpha(A)(\varphi)=a$, giving the desired surjectivity of $\alpha(A)$.

It thus remains to check the coherence property of $\alpha$. Given $G$-modules $A$ and $B$, and a $G$-module map $f:A\to B$ we need to check that

$\begin{matrix}h^\mathbb{Z}(A) & \xrightarrow{h^\mathbb{Z}(f)} & h^\mathbb{Z}(B)\\ \downarrow & & \downarrow \\ A^G & \xrightarrow{f} & B^G\end{matrix}$

But, if $\varphi\in h^\mathbb{Z}(A)$, then the top and the right arrow give me $\varphi\mapsto f\circ\varphi\mapsto (f\circ \varphi)(1)$ and the left and then bottom arrow give me $\varphi\mapsto \varphi(1)\mapsto f(\varphi(1))$ and so commutativity is proven. $\blacksquare$

Now, while on the face this just seems neat “our functor is just a Hom functor!” it actually has a much more consequential corollary:

Corollary 2: Let $G$ be a group. Then, there are natural isomorphisms $H^i(G,-)\cong \text{Ext}^i_{\mathbb{Z}[G]}(\mathbb{Z},-)$.

Proof: This just follows since the functors $-^G$ and $h^\mathbb{Z}$ are isomorphic, and so for every $i\geqslant 0$

$H^i(G,-)=R^i(-^G)\cong R^i(h^\mathbb{Z})\cong \text{Ext}^i_{\mathbb{Z}[G]}(\mathbb{Z},-)$

as desired. $\blacksquare$

This is great since the theory of the $\text{Ext}$ functor is very well understood, all of which we can now bring to bear on our study of group cohomology!

## Group Cohomology: The Bar Resolution

One of the biggest advantages that Corollary 2 gives us, is the ability to create a very concrete description of the cohomology groups $H^i(G,A)$–in particular $H^1(G,A)$. The reason for this is simple. Now that we know that $H^\bullet(G,A)\cong \text{Ext}^\bullet_{\mathbb{Z}[G]}(\mathbb{Z},A)$ we have been granted a huge computational boon.

As of now, our technique for computing $H^i(G,A)$ has been to find an injective resolution for $A$, take the deleted resolution, apply $-^G$, and then take homology. With this approach, finding a uniform description of $H^i(G,A)$ sounds very difficult since there is no uniform injective resolution that works for all $A$ (duh!).

But, now that we know $H^\bullet(G,A)\cong\text{Ext}_{\mathbb{Z}[G]}(\mathbb{Z},A)$ we see that we now have a fighting chance of getting our desired uniform description. Indeed, while it seems fruitless to try and describe some universally nice injective resolution we know that to find $\text{Ext}^i_{\mathbb{Z}[G]}(\mathbb{Z},A)$ we can either find an injective resolution of $A$, or a projective resolution of $\mathbb{Z}$. In particular, we see that instead of having to find some general resolution which is amenable for all possible $A$, we can find one resolution for $\mathbb{Z}$, and see how the $A$ mash up against it.

This is where the bar resolution comes in. It’s a fairly canonical projective resolution $P_\bullet\to \mathbb{Z}\to 0$ which will be conducive to giving a straightforward presentation of the $H^i(G,A)$‘s.

Let us define $P_i$ to be the free $\mathbb{Z}$-module on $G^{i+1}$, where we give $P_i$ the structure of a $G$-module via extending linearly the “diagonal action” $g(g_0,\ldots,g_i)=(gg_0,\ldots,gg_i)$. We claim that $P_i$ is actually a free $\mathbb{Z}[G]$-module. More specifically:

Theorem 3: Let $P_i$ be as above. Then, $\{(1,g_1,\ldots,g_{i})\}$ is a basis for $P_i$ as a $\mathbb{Z}[G]$-module.

Proof: To see that they are $\mathbb{Z}[G]$-independent, suppose that

$\displaystyle \sum_{\mathbf{g}=(g_1,\ldots,g_i)\in G^i}(\sum_{h\in G} n_{\mathbf{g},h}h)(1,g_1,\ldots,g_i)=0$

then we’d have that

$\displaystyle \sum_{\mathbf{g}=(g_1,\ldots,g_i)\in G^i}\sum_{h\in G} n_{\mathbf{g},h} (h,g_1 h,\ldots,h g_i)=\sum_{(h,g_1,\ldots,g_i)\in G^i)}n_{\mathbf{g},h}(h,hg_1,\ldots,hg_n)=0$

But, note that $G^{i+1}\ni (h,g_1,\ldots,g_i)\to (h,hg_1,\ldots,hg_i)\in G^{i+1}$ is an injection, and thus we see that this is a linear coefficients of $\mathbb{Z}$-basis elements which gives $0$. Thus, each $n_{\mathbf{g},h}$ must be zero, and so each coefficient

$\displaystyle \sum_{h\in G}n_{\overline{g},h}h=0$

as desired.

To see that $\{(1,g_1,\ldots,g_i):(g_1,\ldots,g_i)\in G^i\}$ is a spanning set of $P_i$, it suffices to show each basis element $(g_1,\ldots,g_n)$ is a $\mathbb{Z}[G]$-linear combination of elements of this set. But, note that

$(g_1,\ldots,g_{i+1})=g_1(1,g_1^{-1}g_2,\ldots,g_1^{-1}g_{i+1})$

from where the conclusion follows. $\blacksquare$

With this theorem proven, we can unabashedly define $\mathcal{B}_i:=\{(1,g_1,\ldots,g_i):(g_1,\ldots,g_i)\in G^i\}$ to be the fundamental basis for $P_i$.

As the notation suggests, the $P_i$ will be the terms of our bar resolution $P_\bullet\to A\to 0$. So, we need to construct differentials $d_i:P_i\to P_{i-1}$. We do so by extending $\mathbb{Z}$-linearly the following

$\displaystyle d_i((g_0,\ldots,g_i))=\sum_{j=0}^{i}(-1)^{j} (g_0,\ldots,\widehat{g_j},\ldots,g_i)$

where, as usual, the hat denotes leaving off that element of $\{g_j\}$. One can easily check that this is, in fact, a map of $G$-modules.

Now, verifying that $d^2=0$ is very similar to the general such computation associated to simplicial complexes (e.g. the one for simplicial homology in algebraic topology)–one just makes a rearrangment of the terms so full cancellation occurs.

So, now we can finally define the bar resolution of $\mathbb{Z}$ with respect to $G$ to be the following resolution:

$\cdots\to P_2\xrightarrow{d_2}P_1\xrightarrow{d_1}P_0\xrightarrow{\varepsilon}\mathbb{Z}\to0$

where $\varepsilon$ is the so-called augmentation map given by

$\displaystyle \varepsilon:\sum_g n_g g\mapsto \sum_g n_g$

We have the following claim which, if there is any sense in the phrase “bar resolution”, is true:

Theorem 4: The bar resolution is a resolution.

The proof is elementary, and is available in any text containing a chapter on group cohomology. The main idea being to construct a null-homotopy of the resolution.

So, now that we know that $P_\bullet\to A\to 0$ really is a projective resolution of $\mathbb{Z}$ as a $G$-module, we know that

$H^i(G,A)\cong H_i(\text{Hom}_{\mathbb{Z}[G]}(P_\bullet,A))$

So, this is not very useful, unless we can nicely describe the complex

$0\to \text{Hom}_{\mathbb{Z}[G]}(P_0,A)\xrightarrow{d_0^\ast}\text{Hom}_{\mathbb{Z}[G]}(P_1,A)\xrightarrow{d_1^\ast}\text{Hom}_{\mathbb{Z}[G]}(P_2,A)\to\cdots$

But, of course, since we wouldn’t be talking about the bar resolution unless we could have slick description, we can infer that we must. There are two common explications of this above sequence. We give the first mostly for cultural reasons since it is occasionally mentioned, but it is the second that will be the real workhorse for us.

For the first construction, let us define the $i^{\text{th}}$ homogenous cochains associated to the pair $(G,A)$ as the group

$\widetilde{C}^i(G,A):=\left\{f\in\text{Map}(G^{i+1}, A):f((gg_0,\ldots,gg_i))=gf(g_0,\ldots,g_i)\right\}$

In other words, an $i^{\text{th}}$ homogenous cochain is just a (set!) map $G^{i+1}\to A$ which commutes with the diagonal action of $G$ on $G^{i+1}$. Of course, $\widetilde{C}^i(G,A)$ is a group with

$(f_1+f_2)((g_0,\ldots,g_i)=f_1(g_0,\ldots,g_i)+f_2(g_0,\ldots,g_i)$

which, as one can check, really does define another element of $\widetilde{C}^i(G,A)$.

To see why we care about these homogenous cochains, let’s consider what an element $f$ of $\text{Hom}_{\mathbb{Z}[G]}(P_i,A)$ looks like. Well, since $P_i$ is generated as a $G$-module by $\{(g_0,\ldots,g_i)\}$ we know that we can identify $f$ with its values on $G^{i+1}$. In this way we obtain an injection

$\text{Hom}_{\mathbb{Z}[G]}(P_i,A)\to \text{Map}(G^{i+1},A)$

which is easily seen to be a homomorphism. The question remains as to what the image of this map is. Since $f$ is a $G$-module map, we know that for any $g\in G$ and $(g_0,\ldots,g_i)$ we have that

$f((gg_0,\ldots,gg_i))=f(g(g_0,\ldots,g_{i}))=g f((g_0,\ldots,g_{i}))$

Thus, we see that the image of $\text{Hom}_{\mathbb{Z}[G]}(P_i,A)$ lies in $\widetilde{C}^i(G,A)$. Conversely, if $f\in\widetilde{C}^i(G,A)$ then we see that

$f((g_0,\ldots,g_{i}))=g_0f((1,g_0^{-1}g_1,\ldots,g_0^{-1}g_{i}))$

and so $f$ is the $\mathbb{Z}[G]$-linear extension of a map $\mathcal{B}_i\to A$, which we always know does give a well-defined $\mathbb{Z}[G]$-map $P_i\to A$ since $\mathcal{B}_i$ is a basis.

In this way, we obtain group isomorphisms

$\text{Hom}_{\mathbb{Z}[G]}(P_i,A)\to \widetilde{C}^i(G,A)$

from which we define the maps $\widetilde{\partial}^i:\widetilde{C}^i(G,A)\to\widetilde{C}^{i+1}(G,A)$ to be such that the following ladder diagram commutes:

$\begin{matrix}0 & \to & \text{Hom}_{\mathbb{Z}[G]}(P_0,A) & \xrightarrow{d_0^\ast} & \text{Hom}_{\mathbb{Z}[G]}(P_1,A) & \xrightarrow{d_1^\ast} & \text{Hom}_{\mathbb{Z}[G]}(P_2,A) & \to & \cdots\\ & & \downarrow ^{\approx} & & \downarrow ^{\approx} & & \downarrow ^{\approx} & &\\ 0 & \to & \widetilde{C}^0(G,A) & \xrightarrow{\widetilde{\partial}^0} & \widetilde{C}^1(G,A) & \xrightarrow{\widetilde{\partial}^1} & \widetilde{C}^2(G,A) & \to & \cdots\end{matrix}$

A moment’s thought shows that $\widetilde{\partial}_i(f)$, for $f\in \widetilde{C}^i(G,A)$ has the following explicit description

$\displaystyle \widetilde{\partial}_i(f)((g_0,\ldots,g_{i+1}))=\sum_{j=0}^{i+1}(-1)^j f(g_0,\ldots,\widehat{g_j},\ldots,g_{i+1})$

as one would expect.

Now, let us describe the second, more practical identification of $\text{Hom}_{\mathbb{Z}[G]}(P_\bullet,A)$. The basic idea is that while above we identified an element of $f\in\text{Hom}_{\mathbb{Z}[G]}(P_i,A)$ with its restriction to the generating set $G^{i+1}\subseteq P_i$, it is much more natural (as was even witnessed above by some comments) to identify $f$ with its restriction to the basis $\mathcal{B}_i$.

To this end, let us define the $i^{\text{th}}$ inhomogenous cochains of the pair $(G,A)$ to be the group

$C^i(G,A):=\text{Map}(G^i,A)$

where, as above, $C^i(G,A)$ has the group structure by adding maps pointwise. Since $\mathcal{B}_i$ is a basis for $P_i$ as a $\mathbb{Z}[G]$-module we obtain isomorphisms

$\text{Hom}_{\mathbb{Z}[G]}(P_i,A)\cong\text{Map}(\mathcal{B}_i,A)\cong \text{Map}(G^i,A)$

where the last isomorphism comes from the tautological bijection $\mathcal{B}_i\to G^i$.

We define differentials $\partial^i:C^i(G,A)\to C^{i+1}(G,A)$, as before, to be such that

$\begin{matrix}0 & \to & \text{Hom}_{\mathbb{Z}[G]}(P_0,A) & \xrightarrow{d_0^\ast} & \text{Hom}_{\mathbb{Z}[G]}(P_1,A) & \xrightarrow{d_1^\ast} & \text{Hom}_{\mathbb{Z}[G]}(P_2,A) & \to & \cdots\\ & & \downarrow ^{\approx} & & \downarrow ^{\approx} & & \downarrow ^{\approx} & &\\ 0 & \to & C^0(G,A) & \xrightarrow{\partial^0} & C^1(G,A) & \xrightarrow{\partial^1} & C^2(G,A) & \to & \cdots\end{matrix}$

commutes. Now, while it is less obvious than the case $\widetilde{\partial}^i$ what $\partial^i$ does explicitly, a quick computation shows that

$\displaystyle \partial^i(f)(g_1,\ldots,g_{i+1})=g_1 f(g_2,\ldots,g_{i+1})+\sum_{j=1}^{i}f(g_1,\ldots,g_jg_{j+1},\ldots,g_{i+1})+(-1)^{i+1}f(g_1,\ldots,g_{i+1})$

This shows the advantages of the homogenous cochains over the inhomogenous. While the groups of inhomogenous cochains are much simpler than their homogenous counterpart, this is made in exchange for a considerably more complicated differential. It turns out though that, for explicit computations, it is the simpler groups (i.e. the inhomogenous cochains) that we shall want.

## An Explicit Description of the First Cohomology Group

Now, from the construction of homogenous and inhomogenous cochains, and their differentials, we have the following:

Theorem 5: Let $G$ be a group, and $A$ a $G$-module. Then, for all $i\geqslant 0$

$H^i(G,A)\cong H_i(\widetilde{C}^\bullet(G,A))\cong H_i(C^\bullet(G,A))$

The most important consequence of  Theorem 5 is a very concrete way to deal with $H^1(G,A)$.

To make this description, all we need to do is think about what an element of $\ker\partial_1$ and an element of $\text{im}(\partial_0)$ look like. Well, given an element $\varphi\in C^1(G,A)$, which is just a map $\varphi:G\to A$, we know that

$\partial^1(\varphi)(g_1,g_2)=g_1\varphi(g_2)-\varphi(g_1g_2)+\varphi(g_1)$

So, $\varphi\in\ker\partial^1$ if and only if for all $g,h\in G$ we have that

$\varphi(gh)=g\varphi(h)+\varphi(g)$

where, it is important to note, that the addition is taking place in $A$ (i.e. it’s $A$‘s addition), but the multiplication is the $G$-action on $A$.

We call an element of $\ker\partial^1$crossed homomorphism $G\to A$. This is because if $A$ happens to be a trivial $G$-module, then a crossed homomorphism is just a group homomorphism, and otherwise it’s like a homomorphism that has been “twisted” (or crossed) by the $G$-action. It is worth noting that any crossed homomorphism must take the identity element $1\in G$ to the identity element $0\in A$ since

$\varphi(1)=\varphi(1^2)=1\varphi(1)+\varphi(1)=2\varphi(1)$

which, by subtraction, gives the desired result.

Now, we have left to describe $\text{im}(\partial^0)$. Well, an element $f\in C^0(G,A)$ is nothing but a map $f:G^0\to A$ or equivalently, since $G^0=\{\ast\}$, an element $a\in A$. So, we can identify $C^0(G,A)$ with $A$ itself. Then, applying $\partial_0$ to this element $f=a$ we get

$\partial^0(f)(g)=ga-a$

Thus, the elements of $\text{im}(\partial^0)$ are just the maps $G\to A$ in $C^1(G,A)$ defined by $g\mapsto ga-a$ for some fixed $a\in A$. We call an element of $\text{im}(\partial^0)$principal crossed homomorphism. Principal since it is obtained from just one element $a\in A$.

Thus, Theorem 5 allows us to conclude the following:

Theorem 6: Let $G$ be a group, and $A$ a $G$-module. Then,

$\displaystyle H^1(G,A)\cong\frac{\left\{\text{crossed homomorphisms }G\to A\right\}}{\{\text{principal crossed homomorphisms } G\to A}$

Now, as was mentioned above, if $A$ is a trivial $G$-module, then a crossed homomorphism is just a homomorphism. And, a principal crossed homomorphism is just the trivial map since in this case

$ga-a=a-a=0$

Thus, we have the following corrollary:

Corollary 7: If $G$ is a group, and $A$ is a trivial $G$-module, then

$H^1(G,A)\cong\text{Hom}_\mathbf{Grp}(G,A)$

This explicit characterization also allows us to nicely calculate $H^1(\mathbb{Z}/n\mathbb{Z},A)$ for any $\mathbb{Z}/n\mathbb{Z}$-module $A$.

To this end, let $G$ be a finite group, $A$ a $G$-module, and let us define the norm map $\text{Nm}_G:A\to A$ by

$\displaystyle \text{Nm}_G:a\mapsto \sum_{g\in G}ga$

Note that $\text{Nm}_G$ actually sends $A$ into $A^G$ by Cayley’s theorem. In some sense, $\text{Nm}_G$ is the “freest” possible way to turn a generic element of $A$ into a $G$-invariant one. This “averaging” technique should be very familiar to anyone who has studied representation theory.

With this definition of the norm map, we can now give a different, more useful description of $H^1(\mathbb{Z}/n\mathbb{Z},A)$. Because there are possible confusions between elements of $\mathbb{Z}/n\mathbb{Z}$ and generic elements of a group (e.g. $1$ vs. $1$!) we shall write $\mathbb{Z}/n\mathbb{Z}\cong \langle\sigma\rangle$. This notation is suggestive of future theorems as well (see Hilbert’s Theorem 90):

Theorem 8: Let $A$ be a $G:=\mathbb{Z}/n\mathbb{Z}$-module. Then,

$\displaystyle H^1(G,A)\cong \frac{\ker\text{Nm}_G}{(1-\sigma)A}$

Proof: From Theorem 6, we know that $H^1(G,A)$ is crossed homomorphisms modulo principal crossed homomorphisms. The basic idea is that any principal crossed homomorphism should entirely be determined by its image on $\varphi$. Indeed, while $\varphi$ may not be a homomorphism, the crossed homomorphism property implies that

$\varphi(\sigma^2)=\sigma\varphi(\sigma)+\varphi(\sigma)$

and

\begin{aligned}\varphi(\sigma^3) &=\sigma\varphi(\sigma^2)+\varphi(\sigma)\\ &=\sigma(\sigma\varphi(\sigma)+\varphi(\sigma)+\varphi(\sigma)\\ &=\sigma^2\varphi(\sigma)+\sigma\varphi(\sigma)+\varphi(\sigma)\end{aligned}

and more generally

$\varphi(\sigma^k)=\sigma^k\varphi(\sigma)+\cdots+\varphi(\sigma)=(\sigma^{k-1}+\cdots+\sigma^0)\varphi(\sigma)$

Thus, we see that the map

$F:\{\text{crossed homomorphisms }G\to A\}\to A:\varphi\mapsto\varphi(1)$

is injective. We also see that

$0=\varphi(1)=\varphi(\sigma^n)=(\sigma^{n-1}+\cdots+\sigma^0)\varphi(1)=\text{Nm}_G(\varphi(1))$

and so the image of $F$ lands inside of $\ker\text{Nm}_G$. Conversely, if $a\in\text{Nm}_G$, then

$\varphi:G\to A:\sigma^k\mapsto (\sigma^{k-1}+\cdots+\sigma^0)a$

does define a crossed homomorphism $G\to A$ which maps, under $F$, to $a$. Thus, we see that $F$ is an isomorphism. Thus, we may conclude that

$\displaystyle H^1(G,A)\cong \frac{\ker \text{Nm}_G}{F(\{\text{principal crossed homomorphisms }G\to A\})}$

But, under $F$, a principal homomorphisms $g\mapsto ga-a$ maps to $\sigma a-\sigma$. Thus,

$F(\{\text{principal crossed homomorphisms }G\to A\})=(1-\sigma)A$

and so the theorem follows. $\blacksquare$

## Coinduced Modules and Shapiro’s Lemma

We now move on to one of the most important computational, as well as theoretical, tools for group cohomology. The notion of the coinduced module. It will be a pivotal piece of machinery for moving between group cohomology over different groups.

The basic idea of coinduced modules is simple. Suppose that $G$ is a group, and $H\subseteq G$ a subgroup. We want a way to associate to an $H$-module $A$, some $G$-module $B$ which is “freest” in some sense. While there are many possible definitions of “freest”, perhaps the most natural given the context would be that the cohomology does not change. More precisely, that $H^i(H,A)\cong H^i(G,B)$ in some natural way. So, the obvious question now is how to construct this $B$ from our given $A$.

One way one might achieve constructing such a $B$, is asking something even stronger than just $H^i(H,A)\cong H^i(G,B)$. Note first that $\mathbb{Z}[G]$ is a free $\mathbb{Z}[H]$-module with basis $S$ any transversal of $G/H$. Thus, any projective resolution of $\mathbb{Z}$ as a $\mathbb{Z}[G]$-module, is also a projective resolution of $\mathbb{Z}$ as a $\mathbb{Z}[H]$ module. We can then further impose that $B$ be such that any time $P_\bullet\to \mathbb{Z}\to 0$ is a projective resolution of $\mathbb{Z}$ as a $\mathbb{Z}[G]$-module that

$\text{Hom}_{\mathbb{Z}[G]}(P_\bullet,B)\cong\text{Hom}_{\mathbb{Z}[H]}(P_\bullet,A)$

where this is an isomorphism of complexes. This then would imply that $H^i(G,B)\cong H^i(H,A)$ since

\begin{aligned}H^i(G,B) &=\text{Ext}^i_{\mathbb{Z}[G]}(\mathbb{Z},B)\\ &\cong H_i(\text{Hom}_{\mathbb{Z}[G]}(P_\bullet,B))\\ &\cong H_i(\text{Hom}_{\mathbb{Z}[H]}(P_\bullet,A))\\ &\cong \text{Ext}^i_{\mathbb{Z}[H]}(\mathbb{Z},A)\\ &=H^i(H,A)\end{aligned}

as desired.

Now, if we want this stronger condition to hold, it seems feasible that we’d want

$\text{Hom}_{\mathbb{Z}[G]}(M,B)\cong\text{Hom}_{\mathbb{Z}[H]}(M,A)$

for any $\mathbb{Z}[G]$-module $M$. But, by Yoneda’s lemma, this uniquely characterizes $B$. So, any module $B$ we can find which satisfies this property will be our desired module. One way of getting such a module is to recall the Tensor-Hom Adjunction, which says that if $R\to S$ is a ring map, then for any $S$-module $X$ and $R$-module $Y$ we have

$\text{Hom}_S(X,\text{Hom}_R(S,Y))=\text{Hom}_R(X\otimes_S S,Y)=\text{Hom}_R(X,Y)$

So, considering the ring map $\mathbb{Z}[H]\to\mathbb{Z}[G]$, the $\mathbb{Z}[G]$-module $M$, and the $\mathbb{Z}[H]$-module $A$, we have

$\text{Hom}_{\mathbb{Z}[G]}(M,\text{Hom}_{\mathbb{Z}[H]}(\mathbb{Z}[G],A))=\text{Hom}_{\mathbb{Z}[H]}(M\otimes_{\mathbb{Z}[G]}\mathbb{Z}[G],A)=\text{Hom}_{\mathbb{Z}[H]}(M,A)$

and so our desired module is $B=\text{Hom}_{\mathbb{Z}[H]}(\mathbb{Z}[G],A)$!

So, let us, now not so mysteriously, define for a $\mathbb{Z}[H]$-module $A$, the coinduced module, denoted $\text{Coind}^G_H(A)$, to be the $\mathbb{Z}[G]$-module $\text{Hom}_{\mathbb{Z}[H]}(\mathbb{Z}[G],A)$. We give $\text{Coind}^G_H(A)$ the $\mathbb{Z}[G]$-module structure, as usual, by defining $(gf)(x)=f(xg)$ for all $g,x\in G$.

Remark: It’s important to point out that there is another, dual, notion to the coinduced module called the induced module. It’s uses come when one is trying to study group homology. It’s definition, also dual to ours, is just $\text{Ind}^G_H(A):=A\otimes_{\mathbb{Z}[H]}\mathbb{Z}[G]$. It turns out that if $H$ is of finite index in $G$, then $\text{Ind}^G_H(A)\cong\text{Coind}^G_H(A)$ as $\mathbb{Z}[G]$-modules. For this reason, and perhaps because “coinduced” is less pleasant to say than “induced”, some people elect to call what we call the coinduced module, the induced module. This can be confusing if one is not cognizant of the difference.

So, now that we have defined the coinduced module, let’s show that it does, indeed, satisfy the desired property (although, a proof is contained in the above motivaion):

Theorem(Shapiro’s Lemma): Let $G$ be a group, $H\subseteq G$ a subgroup, and $A$ a $H$-module. Then, for all $i\geqslant 0$

$H^i(G,\text{Coind}^G_H(A))\cong H^i(H,A)$

Proof: Let’s first verify the case $i=0$. Which, amounts to showing that

$\text{Coind}^G_H(A)^G\cong A^H$

But, this is simple if we use our alternate description of the fixed submodule and the universal property which motivated us to define the coinduced module

$\text{Coind}^G_H(A)^G=\text{Hom}_{\mathbb{Z}[G]}(\mathbb{Z},\text{Coind}^G_H(A))\cong \text{Hom}_{\mathbb{Z}[H]}(\mathbb{Z},A)=A^H$

as desired.

Now, for the general case. Let $0\to A\to I^\bullet$ be an injective resolution of $A$ as a $\mathbb{Z}[H]$-module. We claim then that $0\to \text{Coind}^G_H(A)\to\text{Coind}^G_H(I^\bullet)$ is an injective resolution of $\text{Coind}^G_H(A)$. The fact that this sequence is exact follows from the fact that $\text{Codind}^G_H(-)=\text{Hom}_{\mathbb{Z}[H]}(\mathbb{Z}[G],-)$, which is an exact functor since $\mathbb{Z}[G]$ is a free $\mathbb{Z}[H]$-module. To see that the terms of $\text{Coind}^G_H(I^\bullet)$ are injective, we merely note that for each $j\geqslant 0$

$\text{Hom}_{\mathbb{Z}[G]}(-,\text{Coind}_H^G(I^j))\cong \text{Hom}_{\mathbb{Z}[H]}(-,I^j)$

which, by the injectivity of $I^j$, implies that $\text{Hom}_{\mathbb{Z}[G]}(-,\text{Coind}^G_H(I^j))$ is an exact functor, which implies that $\text{Coind}^G_H(I^j)$ is injective.

We thus finish by noting that

$H^i(H,A)\cong H_i((I^j)^H)\cong H_i(\text{Coind}^G_H(I^j)^G)\cong H^i(G,\text{Coind}^G_H(A))$

which is what we wanted to prove. $\blacksquare$
One important consequence of Shapiro’s lemma is the acyclicity of so called coinduced modules. For a group $G$, we call a $G$-module $A$ coinduced, without reference to any subgroup, if it is of the form $A\cong \text{Coind}^G_{\{1\}}(A_0)$ for some $\{1\}$-module (i.e. abelian group) $A_0$. For notational convenience, we shall denote $\text{Coind}^G_{\{1\}}(A_0)$ by just $\text{Coind}^G(A_0)$.

There is a more natural way to describe the modules $\text{Coind}^G(A_0)$ though. Namely, by definition $\text{Coind}^G(A_0)$ is just $\text{Hom}_{\mathbb{Z}}(\mathbb{Z}[G],A_0)$. Now, since $\mathbb{Z}[G]$ is a free $\mathbb{Z}$-module, this shows that, as a group, $\text{Coind}^G(A_0)$ is nothing more than $A_0^{\oplus G}$ (i.e. a direct sum of $A_0$‘s, indexed by the elements of $G$). The $G$-action on $A_0^{\oplus G}$ is nothing more than permuting the coordinates so that if $\displaystyle (a_g)$ is an element of $A_0^{\oplus G}$, then $g_0 (a_g)$ is the tuple of $A_0^{\oplus G}$ whose coordinate in the $g_0g$ position is just $a_g$. Moreover, it’s clear that every $G$-module of this form, $A_0^{\oplus G}$ for some abelian group $A_0$ is coinduced, being $\text{Coind}^G(A_0)$.

We now make precise what we mean by saying that coinduced modules are acyclic:

Theorem 9: Let $G$ be a group, and let $A_0$ be an abelian group. Then,

$H^i(G,\text{Coind}^G(A_0))=0$

for all $i>0$.

Proof: By Shapiro’s lemma we know that

$H^i(G,\text{Coind}^G(A_0))\cong H^i(\{1\},A_0)$

But, $H^i(\{1\},A_0)$ is trivial. To see this, note that $\mathbb{Z}[\{1\}]=\mathbb{Z}$, and so

$H^i(\{1\},A_0)\cong \text{Ext}^i_{\mathbb{Z}[\{1\}]}(\mathbb{Z},A_0)=\text{Ext}^i_{\mathbb{Z}}(\mathbb{Z},A)=0$

since $\mathbb{Z}$ is a free $\mathbb{Z}$-module, and $i>0$. $\blacksquare$

One nice thing that Theorem 9 allows us to discuss is the notion of weakly injective modules. In particular, if $G$ is a group, we say that a $G$-module is weakly injective if it is direct summand (as a $\mathbb{Z}[G]$-module!) of a coinduced module. But, since cohomology is additive, we may conclude by Theorem 9 that any weakly injective module is acyclic. Thus, we may use weakly injective modules as resolutions to compute cohomology.

Now, note that any coinduced $G$ module is also a coinduced $H$ module for any subgroup $H\subseteq G$ since

$\text{Hom}_{\mathbb{Z}}(\mathbb{Z}[G],A_0)=\text{Hom}_{\mathbb{Z}}(\mathbb{Z}[H]^{\oplus S},A_0)\cong \text{Hom}_{\mathbb{Z}}(\mathbb{Z}[H],A_0)^{S}\cong \text{Hom}_{\mathbb{Z}}(\mathbb{Z}[H],A_0^S)$

where $S$ is any set of coset representatives of $G/H$. From this, we may conclude that every weakly injective $G$-module is a weakly injective $H$-module. But, every injective $G$-module is weakly injective.

To see this, first let us note that every $G$-module can be embedded into a coinduced module. In fact, one can easily check that if $A$ is any $G$-module then the mapping

$\displaystyle A\to \text{Coind}^G(A):a\mapsto \left(\sum_{g\in G}a_g g\mapsto \sum_{g\in G}a_g ga\right)$

is an embedding of $G$-modules, where we consider $A$ just as an abelian group in $\text{Coind}^G(A)$. Thus, if we consider the short exact sequence of $G$-modules, where $I$ is injective,

$0\to I\to \text{Coind}^G(I)\to \text{Coind}^G(I)/I\to 0$

injectivity says that this sequence splits and so

$\text{Coind}^G(I)\cong I\oplus (\text{Coind}^G(I)/I)$

showing that injective $G$-modules are, in fact, weakly injective.

So, from our above discussion, we see that every injective $G$-module $I$ is weakly injective, and so acyclic as an $H$-module. Thus, we know that if $0\to A\to I^\bullet$ is a resolution of a $G$-module $A$ by injective $G$-modules, then it is also a resolution of $A$, thought of as an $H$-module, by acyclic $H$-modules. Thus, we may conclude the following:

Theorem 10: Let $G$ be a group and $H\subseteq G$ a subgroup. If $A$ is a $G$-module, and $0\to A\to I^\bullet$ is a resolution of $A$ by injective $G$-modules, then

$H^i(H,A)\cong H_i((I^\bullet)^H)$

While this doesn’t seem very important, it’s actually very computationally nice. If you want to find the cohomology of some $G$-module $A$, with coefficients in some various subgroups, you only have to find one injective resolution of $G$, which then will serve as a resolution with which you can compute cohomology with coefficients in any subgroup.

## Functorial Properties of Cohomology: Change of Groups

Now, built into the theory of derived functors is, well, the functorality. In particular, if $G$ is a group, and $A$ and $B$ are two $G$-modules, then any $G$-module map $A\to B$ induces group maps $H^i(G,A)\to H^i(G,B)$ for all $i\geqslant 0$. One may ask though if $H^i(G,A)$ is also functorial in the other entry. In particular, one might wonder if given a group map $\alpha:G\to G'$, whether this induces, in some way, group maps on cohomology groups.

To make any sense of this statement, we won’t just need a group map $\alpha:G'\to G$, but a map $\beta:A\to A'$, where $A$ is a $G$-module, and $A'$ is a $G$‘-module which, in some way, is coherent with $\alpha$. Rigorously, this means that

$\beta(\alpha(g')a)=g'\beta(a)$

for any $a\in A$ and $g'\in G'$. If $(\alpha:G'\to G,\beta:A\to A')$ satisfy this identity, we say that they are an intertwining pair.

Now, the point of an intertwining pair is that it allows us to induce a chain map  inhomogenous cochains of $(G,A)$ to the inhomogenous cochains of $(G',A')$. Let us denote the induced map $G'^i\to G^i$ given by $(g'_1,\ldots,g'_i)\mapsto (\alpha(g'_1),\ldots,\alpha(g'_i))$ by $\alpha^i$. Then, our desired chain map is as follows:

$\begin{matrix}0 & \to & \widetilde{C}^0(G,A) & \xrightarrow{\partial^0} & \widetilde{C}^1(G,A) & \xrightarrow{\partial^1} & \widetilde{C}^2(G,A) & \to & \cdots\\ & & \downarrow ^{f^0} & & \downarrow ^{f^1} & & \downarrow ^{f^2} & &\\ 0 & \to & \widetilde{C}^0(G',A') & \xrightarrow{\partial^0} & \widetilde{C}^1(G',A') & \xrightarrow{\partial^1} & \widetilde{C}^2(G',A') & \to & \cdots\end{matrix}$

where, of course, the $\partial^i$ on the top and bottom rows are in accordance to their respective pairs $(G,A)$ or $(G',A')$, and the $f^i$ are defined as follows:

$f^i:\varphi\mapsto \beta\circ\varphi\circ\alpha^i$

noting, of course, that $f^i(\varphi)$ really is a map $G'^i\to A'$. The verification that this is a chain map is elementary, and uses, in a pivotal way, the fact that $(\alpha,\beta)$ is an intertwining pair. Now, since $(f^i)$ is a chain map, we know that we get an induced map on cohomology $H^i(G,A)\to H^i(G',A')$ as we wanted.

The name of the game now is just to define, in several situations, natural intertwining pairs and discuss some if their properties:

First, let us consider a group $G$, along with a subgroup $H\subseteq G$, and a $G$-module $A$. We may then, of course, also consider $A$ as an $H$-module. Let’s then consider the intertwining pair $(\alpha,\beta)$ where $\alpha:H\to G$ is inclusion, and $\beta:A\to A$ is the identity map. By the above, we obtain morphisms

$\mathsf{Res}_i:H_i(G,A)\to H^i(G,A)$

called the restriction map. One can check that with respect to cohomology computed via the bar resolution, the restriction maps take a particularly simple, and name-worthy form. Namely, for $f:G^i\to A$ a cocycle in $C^i(G,A)$, one can check that $\mathsf{Res}_i$ maps $f$ to the cocycle $f\mid_{H^i}:H^i\to A$ in $C^i(H,A)$.

Next, we claim that we can describe the isomorphism $H^i(G,\text{Coind}^G_H(A))\to H^i(H,A)$ given to us by Shaprio’s Lemma by means of intertwining pairs. Let us first note that there is a natural map $\beta:\text{Coind}^G_H(A)\to A$ given by taking $\varphi$ to $\varphi(1)$. We claim that this map, along with $\alpha:H\to G$ being the inclusion, gives an intertwining pair. But, this ultimately just follows from the fact that each $\varphi$ is $\mathbb{Z}[H]$-linear, because then for any $h\in H$ and $\varphi\in\text{Coind}^G_H(A)$ we have that

$\beta(\alpha(h)\varphi)=\beta(h\varphi)=(h\varphi)(1)=\varphi(h)=h\varphi(1)=h\beta(\varphi)$

Thus, $(\alpha,\beta)$ really is an intertwining pair, and so we get an induced map

$H^i(G,\text{Coind}^G_H(A))\to H^i(H,A)$

We claim that this is the map coming from Shapiro’s Lemma. This is elementary though, and is left for you to check.

With this alternate way of describing Shapiro’s Lemma we can alternately describe the restriction maps now. Namely, consider the composition

$A\to \text{Coind}^G_H(A)\to A$

where $A$ in the second and third spots is thought about as an $H$-module, the first map is given by $a\mapsto (g\mapsto ga)$, and the second map is the map described in the previous example. The first map is clearly a map of $G$-modules, and so induces a map

$H^i(G,A)\to H^i(G,H)$

and the second map, just being the one coming from Shapiro’s Lemma, induces an isomorphism

$H^i(G,\text{Coind}^G_H(A))\xrightarrow{\approx}H^i(H,A)$

and so the composition of these maps gives rise to a morphism

$H^i(G,A)\to H^i(H,A)$

But, note that the composition $A\to\text{Coind}^G_H(A)\to A$ is defined by

$a\mapsto (\varphi:g\mapsto ga)\mapsto \varphi(1)=a$

and so the composition is also just the identity map on $A$, and so we recover that the composition

$H^i(G,A)\to H^i(G,\text{Coind}^G_H(A))\xrightarrow{\approx}H^i(H,A)$

is just the restriction map $\mathsf{Res}_i$.

We next describe the so-called “inflation maps”. Let’s suppose that $G$ is a group, $H$ a normal subgroup, and $A$ a $G$-module. Note that $A^H$ becomes a $G/H$-module by

$(g+H)a:= ga$

for any $g+H\in G/H$ and $a\in A^H$. It is precisely because all elements of $A^H$ are fixed by $H$ that this action is well-defined. Consider now the pair $(\alpha,\beta)$ where $\alpha:G\to G/H$ is the usual projection map, and $\beta:A^H\hookrightarrow A$ is the obvious inclusion. We claim that $(\alpha,\beta)$ is an intertwining pair. But, this amounts to the assertion that

$\beta(\alpha(g)a)=\beta((g+H)a)=\beta(ga)=ga=g\beta(a)$

for any $g\in G$ and $a\in A^H$. Thus, we obtain morphisms

$\mathsf{Inf}_i:H^i(G/H,A^H)\to H^i(G,A)$

for all $i\geqslant 0$, called the inflation maps.

The last important case of a “change of groups” morphism, comes as a sort of dual to the restriction maps. Suppose that $G$ is a group, and $H$ a subgroup of finite index. Let $A$ be a $G$-module. Let’s define a map

$\text{Coind}^G_H(A)=\text{Hom}_{\mathbb{Z}[H]}(\mathbb{Z}[G],A)\to\text{Hom}_{\mathbb{Z}[G]}(\mathbb{Z}[G],A)\cong A$

This map is defined by taking a map $\phi$ to the averaged map

$\displaystyle \widehat{\phi}(x):=\sum_{j=1}^{n}g_j \phi(g_j^{-1}x)$

where $\{g_1,\ldots,g_n\}$ are coset representatives–one can easily check by reindexing, and the fact that $\phi$ is $H$-linear, that $\widehat{\phi}$ is independent of the choice of coset reps. One also checks that $\widehat{\phi}$ is also $G$-linear as claimed. The fact that this is a morphism of $G$-modules is apparent, and thus we obtain morphisms

$\mathsf{Cor}_i:H^i(H,A)\cong H^i(G,\text{Coind}^G_H(A))\to H^i(G,A)$

these maps are called the corestriction maps.

One of the lowest hanging, but still juicy, fruit that comes from these definitions is the following seemingly innocuous theorem:

Theorem 11: Let $G$ be a group, and $H$ a subgroup of index $n$. Then, $\mathsf{Cor}_i\circ\mathsf{Res}_i$ is multiplication by $n$, for all $i\geqslant 0$.

Proof: The map $\mathsf{Res}_i$ is described as being the map induced by the composition

$A\to \text{Coind}^G_H(A)$

as described above, followed by Shapiro’s lemma. The map $\mathsf{Cor}_i$ is defined as being one induced by a map $\text{Coind}^G_H(A)\to A$, followed by Sharpiro’s lemma (but the one in the other direction–the inverse of the one in the previous sentence). Thus, the composition is just going to be the one induced by the composition

$A\to \text{Coind}_G^H(A)\to A$

But, thinking of $A$ As $\text{Hom}_G(G,A)$ this just takes $\phi$ to $\widehat{\phi}$ which, since $\phi$ is $G$-linear, is not just $n\phi$. Thus, the composition is just multiplication by $n$, and since cohomology is additive, the induced map is also just multiplication by $n$. $\blacksquare$

The utility of this theorem is the following corollary:

Corollary 12: If $G$ is a finite group of order $n$, then $H^i(G,A)$ is $n$-torsion for all $G$-modules $A$, and all $i\geqslant 0$. In particular, if $A$ is a finitely generated group, then $H^i(G,A)$ is finite.

Proof: For $i>0$ we know that $H^i(G,\text{Coind}^G(A))\cong H^i(\{1\},A)=\{0\}$ by Shapiro’s lemma. Thus, $\mathsf{Res}_i$ must be the zero map, having zero target. But, then

$0=\mathsf{Cor}_i(\mathsf{Res}_i(H^i(G,A)))=(\mathsf{Cor}_i\circ\mathsf{Res}_i)(A)=nA$

as desired.

For the finiteness of $H^i(G,A)$, merely note that by construction of the $H^i(G,A)$ (via the bar resolution, say) the fact that $A$ is a finitely generated group implies the same for $H^i(G,A)$. But, then $H^i(G,A)$ is a finitely generated torsion group, and so must be finite.$\blacksquare$

This corollary allows us to also make some immediate computations of cohomology groups. For example, if $A$ is $n$-divisible group (i.e. multiplication by $n$ is an automorphism $A\to A$), then $H^i(G,A)=0$ for any group $G$ of order $n$. Indeed, since $n:A\to A$ is an isomorphism, it induces the isomorphism $n:H^i(G,A)\to H^i(G,A)$. But, since $H^i(G,A)$ is $n$-torsion, this implies that $H^i(G,A)$ must be zero.