Group and Galois Cohomology — Chapter 1 Group cohomology

Chapter 1
Group cohomology

1.1. Group rings

Let $G$ be a group.

Definition 1.1.1. §

The group ring (or, more specifically, $ℤ$ -group ring) $ℤ [G]$ of a group $G$ consists of the set of finite formal sums of group elements with coefficients in $ℤ$

{\sum_{g \in G} a_{g} g ∣ a_{g} \in ℤ for all g \in G, almost all a_{g} = 0} .

with addition given by addition of coefficients and multiplication induced by the group law on $G$ and $ℤ$ -linearity. (Here, “almost all” means all but finitely many.)

In other words, the operations are

\sum_{g \in G} a_{g} g + \sum_{g \in G} b_{g} g = \sum_{g \in G} (a_{g} + b_{g}) g

and

(\sum_{g \in G} a_{g} g) (\sum_{g \in G} b_{g} g) = \sum_{g \in G} (\sum_{k \in G} a_{k} b_{k^{- 1} g}) g .

Remark 1.1.2. §

In the above, we may replace $ℤ$ by any ring $R$ , resulting in the $R$ -group ring $R [G]$ of $G$ . However, we shall need here only the case that $R = ℤ$ .

Definition 1.1.3. §

i.: The augmentation map is the homomorphism $𝜀 : ℤ [G] \to ℤ$ given by
$𝜀 (\sum_{g \in G} a_{g} g) = \sum_{g \in G} a_{g} .$
ii.: The augmentation ideal $I_{G}$ is the kernel of the augmentation map $𝜀$ .

Lemma 1.1.4. §

The augmentation ideal $I_{G}$ is equal to the ideal of $ℤ [G]$ generated by the set ${g - 1 ∣ g \in G}$ .

Proof.

Clearly $g - 1 \in \ker 𝜀$ for all $g \in G$ . On the other hand, if $\sum_{g \in G} a_{g} = 0$ , then

\sum_{g \in G} a_{g} g = \sum_{g \in G} a_{g} (g - 1) .

□

Definition 1.1.5. §

If $G$ is a finite group, we then define the norm element of $ℤ [G]$ by $N_{G} = \sum_{g \in G} g$ .

Remark 1.1.6. §

a.

We may speak, of course, of modules over the group ring $ℤ [G]$ . We will refer here to such $ℤ [G]$ -modules more simply as $G$ -modules. To give a $G$ -module is equivalent to giving an abelian group $A$ together with a $G$ -action on $A$ that is compatible with the structure of $A$ as an abelian group, i.e., a map

G \times A \to A, (g, a) \mapsto g \cdot a

satisfying the following properties:

(i): $1 \cdot a = a$ for all $a \in A$ ,
(ii): $g_{1} \cdot (g_{2} \cdot a) = (g_{1} g_{2}) \cdot a$ for all $a \in A$ and $g_{1}, g_{2} \in G$ , and
(iii): $g \cdot (a_{1} + a_{2}) = g \cdot a_{1} + g \cdot a_{2}$ for all $a_{1}, a_{2} \in A$ and $g \in G$ .

b.

A homomorphism $κ : A \to B$ of $G$ -modules is just a homomorphism of abelian groups that satisfies $κ (𝑔𝑎) = 𝑔𝜅 (a)$ for all $a \in A$ and $g \in G$ . The group of such homomorphisms is denoted by ${Hom}_{ℤ [G]} (A, B)$ .

Definition 1.1.7. §

We say that a $G$ -module $A$ is a trivial if $g \cdot a = a$ for all $g \in G$ and $a \in A$ .

Definition 1.1.8. §

Let $A$ be a $G$ -module.

i.

The group of $G$ -invariants $A^{G}$ of $A$ is given by

A^{G} = {a \in A ∣ g \cdot a = a for all g \in G, a \in A},

which is to say the largest submodule of $A$ fixed by $G$ .

ii.

The group of $G$ -coinvariants $A_{G}$ of $A$ is given by

A_{G} = A ∕ I_{G} A,

which is to say (noting Lemma 1.1.4) the largest quotient of $A$ fixed by $G$ .

Example 1.1.9. §

a.: If $A$ is a trivial $G$ -module, then $A^{G} = A$ and $A_{G} ≅𝐴$ .
b.: One has ${ℤ [G]}_{G} ≅ℤ$ . We have ${ℤ [G]}^{G} = (N_{G})$ if $G$ is finite and ${ℤ [G]}^{G} = (0)$ otherwise.

1.2. Group cohomology via cochains

The simplest way to define the $i$ th cohomology group $H^{i} (G, A)$ of a group $G$ with coefficients in a $G$ -module $A$ would be to let $H^{i} (G, A)$ be the $i$ th derived functor on $A$ of the functor of $G$ -invariants. However, not wishing to assume homological algebra at this point, we take a different tack.

Definition 1.2.1. §

Let $A$ be a $G$ -module, and let $i \geq 0$ .

i.: The group of $i$ -cochains of $G$ with coefficients in $A$ is the set of functions from $G^{i}$ to $A$ :
$C^{i} (G, A) = {f : G^{i} \to A} .$
ii.: The $i$ th differential $d^{i} = d_{A}^{i} : C^{i} (G, A) \to C^{i + 1} (G, A)$ is the map
$\begin{matrix} d^{i} (f) (g_{0}, g_{1}, \dots, g_{i}) = g_{0} \cdot f (g_{1}, \dots g_{i}) \\ + \sum_{j = 1}^{i} {(- 1)}^{j} f (g_{0}, \dots, g_{j - 2}, g_{j - 1} g_{j}, g_{j + 1}, \dots, g_{i}) + {(- 1)}^{i + 1} f (g_{0}, \dots, g_{i - 1}) . \end{matrix}$

We will continue to let $A$ denote a $G$ -module throughout the section. We remark that $C^{0} (G, A)$ is taken simply to be $A$ , as $G^{0}$ is a singleton set. The proof of the following is left to the reader.

Lemma 1.2.2. §

For any $i \geq 0$ , one has $d^{i + 1} \circ d^{i} = 0$ .

Remark 1.2.3. §

Lemma 1.2.2 shows that $C^{\cdot} (G, A) = (C^{i} (G, A), d^{i})$ is a cochain complex.

We consider the cohomology groups of $C^{\cdot} (G, A)$ .

Definition 1.2.4. §

Let $i \geq 0$ .

i.: We set $Z^{i} (G, A) = \ker d^{i}$ , the group of $i$ -cocycles of $G$ with coefficients in $A$ .
ii.: We set $B^{0} (G, A) = 0$ and $B^{i} (G, A) = im d^{i - 1}$ for $i \geq 1$ . We refer to $B^{i} (G, A)$ as the group of $i$ -coboundaries of $G$ with coefficients in $A$ .

We remark that, since $d^{i} \circ d^{i - 1} = 0$ for all $i \geq 1$ , we have $B^{i} (G, A) \subseteq Z^{i} (G, A)$ for all $i \geq 0$ . Hence, we may make the following definition.

Definition 1.2.5. §

We define the $i$ th cohomology group of $G$ with coefficients in $A$ to be

H^{i} (G, A) = Z^{i} (G, A) ∕ B^{i} (G, A) .

The cohomology groups measure how far the cochain complex $C^{\cdot} (G, A)$ is from being exact. We give some examples of cohomology groups in low degree.

Lemma 1.2.6. §

a.

The group $H^{0} (G, A)$ is equal to $A^{G}$ , the group of $G$ -invariants of $A$ .

b.

We have

Z^{1} (G, A) = {f : G \to A ∣ f (𝑔h) = 𝑔𝑓 (h) + f (g) for all g, h \in G}

and $B^{1} (G, A)$ is the subgroup of $f : G \to A$ for which there exists $a \in A$ such that $f (g) = 𝑔𝑎 - a$ for all $g \in G$ .

c.

If $A$ is a trivial $G$ -module, then $H^{1} (G, A) = Hom (G, A)$ .

Proof.

Let $a \in A$ . Then $d^{0} (a) (g) = 𝑔𝑎 - a$ for $g \in G$ , so $\ker d^{0} = A^{G}$ . That proves part a, and part b is simply a rewriting of the definitions. Part c follows immediately, as the definition of $Z^{1} (G, A)$ reduces to $Hom (G, A)$ , and $B^{1} (G, A)$ is clearly $(0)$ , in this case. □

We remark that, as $A$ is abelian, we have $Hom (G, A) = Hom (G^{ab}, A)$ , where $G^{ab}$ is the maximal abelian quotient of $G$ (i.e., its abelianization). We turn briefly to an even more interesting example.

Definition 1.2.7. §

A group extension of $G$ by a $G$ -module $A$ is a short exact sequence of groups

0 \to A \to_{}^{ι} E \to_{}^{π} G \to 1

such that, choosing any section $s : G \to E$ of $π$ , one has

s (g) 𝑎𝑠 {(g)}^{- 1} = g \cdot a

for all $g \in G$ , $a \in A$ . Two such extensions $E \to E^{'}$ are said to be equivalent if there is an isomorphism $𝜃 : E \to_{}^{\sim} E^{'}$ fitting into a commutative diagram

Equivalence of group extensions. A full diagram description follows.

Diagram description: Equivalence of group extensions

The two displayed rows are exact, and the squares commute.

Objects, listed by row and column:

Row 1, from left to right: column 1: 0; column 2: A; column 3: script E; column 4: G; column 5: 0.
Row 2, from left to right: column 1: 0; column 2: A; column 3: script E prime; column 4: G; column 5: 0.

Arrows and lines:

An arrow from 0 (row 1, column 1) to A (row 1, column 2), without a label.
An arrow from A (row 1, column 2) to script E, without a label.
Equality joins A (row 1, column 2) and A (row 2, column 2), without a label.
An arrow from script E to G (row 1, column 4), without a label.
An arrow from script E to script E prime, labelled theta.
An arrow from G (row 1, column 4) to 0 (row 1, column 5), without a label.
Equality joins G (row 1, column 4) and G (row 2, column 4), without a label.
An arrow from 0 (row 2, column 1) to A (row 2, column 2), without a label.
An arrow from A (row 2, column 2) to script E prime, without a label.
An arrow from script E prime to G (row 2, column 4), without a label.
An arrow from G (row 2, column 4) to 0 (row 2, column 5), without a label.

We denote the set of equivalence classes of such extensions by $E (G, A)$ .

We omit the proof of the following result, as it is not used in the remainder of these notes. We also leave it as an exercise to the reader to define the structure of an abelian group on $E (G, A)$ which makes the following identification an isomorphism of groups.

Theorem 1.2.8. §

The group $H^{2} (G, A)$ is in canonical bijection with $E (G, A)$ via the map induced by that taking a $2$ -cocycle $f : G^{2} \to A$ to the extension $E_{f} = A \times G$ with multiplication given by

(a, g) \cdot (b, h) = (a + 𝑔𝑏 + f (g, h), 𝑔h)

This identification takes the identity to the semi-direct product $A ⋉ G$ determined by the action of $G$ on $A$ .

One of the most important uses of cohomology is that it converts short exact sequences of $G$ -modules to long exact sequences of abelian groups. For this, in homological language, we need the fact that $C^{i} (G, A)$ provides an exact functor in the module $A$ .

Lemma 1.2.9. §

If $α : A \to B$ is a $G$ -module homomorphism, then for each $i \geq 0$ , there is an induced homomorphism of groups

α^{i} : C^{i} (G, A) \to C^{i} (G, B)

taking $f$ to $α \circ f$ and compatible with the differentials in the sense that

d_{B}^{i} \circ α^{i} = α^{i + 1} \circ d_{A}^{i} .

Proof.

We need only check the compatibility. For this, note that

\begin{matrix} d^{i} (α \circ f) (g_{0}, g_{1}, \dots, g_{i}) = g_{0} α \circ f (g_{1}, \dots g_{i}) \\ + \sum_{j = i}^{i} {(- 1)}^{j} α \circ f (g_{0}, \dots, g_{j - 2}, g_{j - 1} g_{j}, g_{j + 1}, \dots, g_{i}) + {(- 1)}^{i + 1} α \circ f (g_{0}, \dots, g_{i - 1}) \\ = α (d^{i} (f) (g_{0}, g_{1}, \dots, g_{i})), \end{matrix}

as $α$ is a $G$ -module homomorphism (the fact of which we use only to deal with the first term). □

In other words, $α$ induces a morphism of complexes $α^{\cdot} : C^{\cdot} (G, A) \to C^{\cdot} (G, B)$ . As a consequence, one sees easily the following

Notation 1.2.10. §

If not helpful for clarity, we will omit the superscripts from the notation in the morphisms of cochain complexes. Similarly, we will consistently omit them in the resulting maps on cohomology, described below.

Corollary 1.2.11. §

A $G$ -module homomorphism $α : A \to B$ induces maps

α^{*} : H^{i} (G, A) \to H^{i} (G, B)

on cohomology.

The key fact that we need about the morphism on cochain complexes is the following.

Lemma 1.2.12. §

Suppose that

0 \to A \to_{}^{ι} B \to_{}^{π} C \to 0

is a short exact sequence of $G$ -modules. Then the resulting sequence

0 \to C^{i} (G, A) \to_{}^{ι} C^{i} (G, B) \to_{}^{π} C^{i} (G, C) \to 0

is exact.

Proof.

Let $f \in C^{i} (G, A)$ , and suppose $ι \circ f = 0$ . As $ι$ is injective, this clearly implies that $f = 0$ , so the map $ι^{i}$ is injective. As $π \circ ι = 0$ , the same is true for the maps on cochains. Next, suppose that $f^{'} \in C^{i} (G, B)$ is such that $π \circ f^{'} = 0$ . Define $f \in C^{i} (G, A)$ by letting $f (g_{1}, \dots, g_{i}) \in A$ be the unique element such that

ι (f (g_{1}, \dots, g_{i})) = f^{'} (g_{1}, \dots, g_{i}),

which we can do since $im ι = \ker π$ . Thus, $im ι^{i} = \ker π^{i}$ . Finally, let $f^{″} \in C^{i} (G, C)$ . As $π$ is surjective, we may define $f^{'} \in C^{i} (G, B)$ by taking $f^{'} (g_{1}, \dots, g_{i})$ to be any element with

π (f^{'} (g_{1}, \dots, g_{i})) = f^{″} (g_{1}, \dots, g_{i}) .

We therefore have that $π^{i}$ is surjective. □

We now prove the main theorem of the section.

Theorem 1.2.13. §

Suppose that

0 \to A \to_{}^{ι} B \to_{}^{π} C \to 0

is a short exact sequence of $G$ -modules. Then there is a long exact sequence of abelian groups

0 \to H^{0} (G, A) \to_{}^{ι^{*}} H^{0} (G, B) \to_{}^{π^{*}} H^{0} (G, C) \to_{}^{δ^{0}} H^{1} (G, A) \to \dots .

Moreover, this construction is natural in the short exact sequence in the sense that any morphism

A morphism of short exact sequences of group modules. A full diagram description follows.

Diagram description: A morphism of short exact sequences of group modules

The two displayed rows are exact, and the squares commute.

Objects, listed by row and column:

Row 1, from left to right: column 1: 0; column 2: A; column 3: B; column 4: C; column 5: 0.
Row 2, from left to right: column 1: 0; column 2: A prime; column 3: B prime; column 4: C prime; column 5: 0.

Arrows and lines:

An arrow from 0 (row 1, column 1) to A, without a label.
An arrow from A to B, labelled iota.
An arrow from A to A prime, labelled alpha.
An arrow from B to C, labelled pi.
An arrow from B to B prime, labelled beta.
An arrow from C to 0 (row 1, column 5), without a label.
An arrow from C to C prime, labelled gamma.
An arrow from 0 (row 2, column 1) to A prime, without a label.
An arrow from A prime to B prime, labelled iota prime.
An arrow from B prime to C prime, labelled pi prime.
An arrow from C prime to 0 (row 2, column 5), without a label.

gives rise to a morphism of long exact sequences, and in particular, a commutative diagram

Naturality of the long exact cohomology sequence. A full diagram description follows.

Diagram description: Naturality of the long exact cohomology sequence

The two displayed rows are exact, and the squares commute.

Objects, listed by row and column:

Row 1, from left to right: column 1: ellipsis; column 2: H superscript (i)(G,A); column 3: H superscript (i)(G,B); column 4: H superscript (i)(G,C); column 5: H superscript (i plus 1)(G,A); column 6: ellipsis.
Row 2, from left to right: column 1: ellipsis; column 2: H superscript (i)(G,A prime ); column 3: H superscript (i)(G,B prime ); column 4: H superscript (i)(G,C prime ); column 5: H superscript (i plus 1)(G,A prime ); column 6: ellipsis.

Arrows and lines:

An arrow from ellipsis (row 1, column 1) to H superscript (i)(G,A), without a label.
An arrow from H superscript (i)(G,A) to H superscript (i)(G,B), labelled iota superscript (star).
An arrow from H superscript (i)(G,A) to H superscript (i)(G,A prime ), labelled alpha superscript (star).
An arrow from H superscript (i)(G,B) to H superscript (i)(G,C), labelled pi superscript (star).
An arrow from H superscript (i)(G,B) to H superscript (i)(G,B prime ), labelled beta superscript (star).
An arrow from H superscript (i)(G,C) to H superscript (i plus 1)(G,A), labelled delta superscript (i).
An arrow from H superscript (i)(G,C) to H superscript (i)(G,C prime ), labelled gamma superscript (star).
An arrow from H superscript (i plus 1)(G,A) to ellipsis (row 1, column 6), without a label.
An arrow from H superscript (i plus 1)(G,A) to H superscript (i plus 1)(G,A prime ), labelled alpha superscript (star).
An arrow from ellipsis (row 2, column 1) to H superscript (i)(G,A prime ), without a label.
An arrow from H superscript (i)(G,A prime ) to H superscript (i)(G,B prime ), labelled ( iota prime ) superscript (star).
An arrow from H superscript (i)(G,B prime ) to H superscript (i)(G,C prime ), labelled ( pi prime ) superscript (star).
An arrow from H superscript (i)(G,C prime ) to H superscript (i plus 1)(G,A prime ), labelled delta superscript (i).
An arrow from H superscript (i plus 1)(G,A prime ) to ellipsis (row 2, column 6), without a label.

Proof.

First consider the diagrams

Exact cochain sequences in adjacent degrees. A full diagram description follows.

Diagram description: Exact cochain sequences in adjacent degrees

The two displayed rows are exact, and the squares commute.

Objects, listed by row and column:

Row 1, from left to right: column 1: 0; column 2: C superscript (j)(G,A); column 3: C superscript (j)(G,B); column 4: C superscript (j)(G,C); column 5: 0.
Row 2, from left to right: column 1: 0; column 2: C superscript (j plus 1)(G,A); column 3: C superscript (j plus 1)(G,B); column 4: C superscript (j plus 1)(G,C); column 5: 0.

Arrows and lines:

An arrow from 0 (row 1, column 1) to C superscript (j)(G,A), without a label.
An arrow from C superscript (j)(G,A) to C superscript (j)(G,B), labelled iota.
An arrow from C superscript (j)(G,A) to C superscript (j plus 1)(G,A), labelled d superscript (j) subscript (A).
An arrow from C superscript (j)(G,B) to C superscript (j)(G,C), labelled pi.
An arrow from C superscript (j)(G,B) to C superscript (j plus 1)(G,B), labelled d superscript (j) subscript (B).
An arrow from C superscript (j)(G,C) to 0 (row 1, column 5), without a label.
An arrow from C superscript (j)(G,C) to C superscript (j plus 1)(G,C), labelled d superscript (j) subscript (C).
An arrow from 0 (row 2, column 1) to C superscript (j plus 1)(G,A), without a label.
An arrow from C superscript (j plus 1)(G,A) to C superscript (j plus 1)(G,B), labelled iota.
An arrow from C superscript (j plus 1)(G,B) to C superscript (j plus 1)(G,C), labelled pi.
An arrow from C superscript (j plus 1)(G,C) to 0 (row 2, column 5), without a label.

for $j \geq 0$ . Noting Lemma 1.2.12, the exact sequences of cokernels (for $j = i - 1$ ) and kernels (for $j = i + 1$ ) can be placed in a second diagram

Cochain quotients and cocycles for the snake lemma. A full diagram description follows.

Diagram description: Cochain quotients and cocycles for the snake lemma

The two displayed rows are exact, and the squares commute.

Objects, listed by row and column:

Row 1, from left to right: column 2: fraction (C superscript (i)(G,A)) over (B superscript (i)(G,A)); column 3: fraction (C superscript (i)(G,B)) over (B superscript (i)(G,B)); column 4: fraction (C superscript (i)(G,C)) over (B superscript (i)(G,C)); column 5: 0.
Row 2, from left to right: column 1: 0; column 2: Z superscript (i plus 1)(G,A); column 3: Z superscript (i plus 1)(G,B); column 4: Z superscript (i plus 1)(G,C).

Arrows and lines:

An arrow from fraction (C superscript (i)(G,A)) over (B superscript (i)(G,A)) to fraction (C superscript (i)(G,B)) over (B superscript (i)(G,B)), labelled iota.
An arrow from fraction (C superscript (i)(G,A)) over (B superscript (i)(G,A)) to Z superscript (i plus 1)(G,A), labelled d superscript (i) subscript (A).
An arrow from fraction (C superscript (i)(G,B)) over (B superscript (i)(G,B)) to fraction (C superscript (i)(G,C)) over (B superscript (i)(G,C)), labelled pi.
An arrow from fraction (C superscript (i)(G,B)) over (B superscript (i)(G,B)) to Z superscript (i plus 1)(G,B), labelled d superscript (i) subscript (B).
An arrow from fraction (C superscript (i)(G,C)) over (B superscript (i)(G,C)) to 0 (row 1, column 5), without a label.
An arrow from fraction (C superscript (i)(G,C)) over (B superscript (i)(G,C)) to Z superscript (i plus 1)(G,C), labelled d superscript (i) subscript (C).
An arrow from 0 (row 2, column 1) to Z superscript (i plus 1)(G,A), without a label.
An arrow from Z superscript (i plus 1)(G,A) to Z superscript (i plus 1)(G,B), labelled iota.
An arrow from Z superscript (i plus 1)(G,B) to Z superscript (i plus 1)(G,C), labelled pi.

(recalling that $B^{0} (G, A) = 0$ for the case $i = 0$ ), and the snake lemma now provides the exact sequence

H^{i} (G, A) \to_{}^{α^{*}} H^{i} (G, B) \to_{}^{β^{*}} H^{i} (G, C) \to_{}^{δ^{i}} H^{i + 1} (G, A) \to_{}^{α^{*}} H^{i + 1} (G, B) \to_{}^{β^{*}} H^{i + 1} (G, C) .

Splicing these together gives the long exact sequence in cohomology, exactness of

0 \to H^{0} (G, A) \to H^{0} (G, B)

being obvious. We leave naturality of the long exact sequence as an exercise. □

Remark 1.2.14. §

The maps $δ^{i} : H^{i} (G, C) \to H^{i + 1} (G, A)$ defined in the proof of theorem 1.2.13 are known as connecting homomorphisms. Again, we will often omit superscripts and simply refer to $δ$ .

Remark 1.2.15. §

A sequence of functors that take short exact sequences to long exact sequences (i.e., which also give rise to connecting homomorphisms) and is natural in the sense that every morphism of short exact sequences gives rise to a morphism of long exact sequences is known as a $δ$ -functor. Group cohomology forms a (cohomological) $δ$ -functor that is universal in a sense we omit a discussion of here.

1.3. Group cohomology via projective resolutions

In this section, we assume a bit of homological algebra, and redefine the $G$ -cohomology of $A$ in terms of projective resolutions.

For $i \geq 0$ , let $G^{i + 1}$ denote the direct product of $i + 1$ copies of $G$ . We view $ℤ [G^{i + 1}]$ as a $G$ -module via the left action

g \cdot (g_{0}, g_{1}, \dots, g_{i}) = (g g_{0}, g g_{1}, \dots, g g_{i}) .

We first introduce the standard resolution.

Definition 1.3.1. §

The (augmented) standard resolution of $ℤ$ by $G$ -modules is the sequence of $G$ -module homomorphisms

\dots \to ℤ [G^{i + 1}] \to_{}^{d_{i}} ℤ [G^{i}] \to \dots \to ℤ [G] \to_{}^{𝜀} ℤ,

where

d_{i} (g_{0}, \dots, g_{i}) = \sum_{j = 0}^{i} {(- 1)}^{j} (g_{0}, \dots, g_{j - 1}, g_{j + 1}, \dots, g_{i})

for each $i \geq 1$ , and $𝜀$ is the augmentation map.

At times, we may use $(g_{0}, \dots, \hat{g_{j}}, \dots, g_{i}) \in G^{i}$ to denote the $i$ -tuple excluding $g_{j}$ . To see that this definition is actually reasonable, we need the following lemma.

Proposition 1.3.2. §

The augmented standard resolution is exact.

Proof.

In this proof, take $d_{0} = 𝜀$ . For each $i \geq 0$ , compute

d_{i} \circ d_{i + 1} (g_{0}, \dots, g_{i + 1}) = \sum_{j = 0}^{i + 1} \sum_{\begin{array}{c} k = 0 \\ k \neq j \end{array}}^{i + 1} {(- 1)}^{j + k - s (j, k)} (g_{0}, \dots, \hat{g_{j}}, \dots, \hat{g_{k}}, \dots, g_{i + 1}),

where $s (j, k)$ is $0$ if $k < j$ and $1$ if $k > j$ . Each possible $(i - 1)$ -tuple appears twice in the sum, with opposite sign. Therefore, we have $d_{i} \circ d_{i + 1} = 0$ .

Next, define $𝜃_{i} : ℤ [G^{i}] \to ℤ [G^{i + 1}]$ by

𝜃_{i} (g_{1}, \dots, g_{i}) = (1, g_{1}, \dots, g_{i}) .

Then

\begin{array}{l} d_{i} \circ 𝜃_{i} (g_{0}, \dots, g_{i}) & = (g_{0}, \dots, g_{i}) - \sum_{j = 0}^{i} {(- 1)}^{j} (1, g_{0}, \dots, \hat{g_{j}}, \dots, g_{i}) \\ = (g_{0}, \dots, g_{i}) - 𝜃_{i - 1} \circ d_{i - 1} (g_{0}, \dots, g_{i}), \end{array}

which is to say that

d_{i} \circ 𝜃_{i} + 𝜃_{i - 1} \circ d_{i - 1} = {id}_{ℤ [G^{i}]} .

If $α \in \ker d_{i - 1}$ for $i \geq 1$ , it then follows that $d_{i} (𝜃_{i} (α)) = α$ , so $α \in im d_{i}$ . □

For a $G$ -module $A$ , we wish to consider the following complex

0 \to {Hom}_{ℤ [G]} (ℤ [G], A) \to \dots \to {Hom}_{ℤ [G]} (ℤ [G^{i + 1}], A) \to_{}^{D^{i}} {Hom}_{ℤ [G]} (ℤ [G^{i + 2}], A) \to \dots .

(1.3.1)

Here, we define $D^{i} = D_{A}^{i}$ by $D^{i} (φ) = φ \circ d_{i + 1}$ . We compare this with the complex of cochains for $G$ .

Theorem 1.3.3. §

The maps

ψ^{i} : {Hom}_{ℤ [G]} (ℤ [G^{i + 1}], A) \to C^{i} (G, A)

defined by

ψ^{i} (φ) (g_{1}, \dots, g_{i}) = φ (1, g_{1}, g_{1} g_{2}, \dots, g_{1} g_{2} \dots g_{i})

are isomorphisms for all $i \geq 0$ . This provides isomorphisms of complexes in the sense that $ψ^{i + 1} \circ D^{i} = d^{i} \circ ψ^{i}$ for all $i \geq 0$ . Moreover, these isomorphisms are natural in the $G$ -module $A$ .

Proof.

If $ψ^{i} (φ) = 0$ , then $φ (1, g_{1}, g_{1} g_{2}, \dots, g_{1} g_{2} \dots g_{i}) = 0$ for all $g_{1}, \dots, g_{i} \in G$ . Let $h_{0}, \dots, h_{i} \in G$ , and define $g_{j} = h_{j - 1}^{- 1} h_{j}$ for all $1 \leq j \leq i$ . We then have

φ (h_{0}, h_{1}, \dots, h_{i}) = h_{0} φ (1, h_{0}^{- 1} h_{1}, \dots, h_{0}^{- 1} h_{i}) = h_{0} φ (1, g_{1}, \dots, g_{1} \dots g_{i}) = 0 .

Therefore, $ψ^{i}$ is injective. On the other hand, if $f \in C^{i} (G, A)$ , then defining

φ (h_{0}, h_{1}, \dots, h_{i}) = h_{0} f (h_{0}^{- 1} h_{1}, \dots, h_{i - 1}^{- 1} h_{i}),

we have

φ (g h_{0}, g h_{1}, \dots, g h_{i}) = g h_{0} f ({(g h_{0})}^{- 1} g h_{1}, \dots, {(g h_{i - 1})}^{- 1} g h_{i}) = 𝑔𝜑 (h_{0}, h_{1}, \dots, h_{i})

and $ψ^{i} (φ) = f$ . Therefore, $ψ^{i}$ is an isomorphism of groups.

That $ψ^{\cdot}$ forms a map of complexes is shown in the following computation:

\begin{array}{l} ψ^{i + 1} (D^{i} (φ)) (g_{1}, \dots, g_{i + 1}) & = D^{i} (φ) (1, g_{1}, \dots, g_{1} \dots g_{i + 1}) \\ = φ \circ d_{i + 1} (1, g_{1}, \dots, g_{1} \dots g_{i + 1}) \\ = \sum_{j = 0}^{i + 1} {(- 1)}^{j} φ (1, g_{1}, \dots, \hat{g_{1} \dots g_{j}}, \dots, g_{1} \dots g_{i + 1}) . \end{array}

The latter term equals

\begin{matrix} g_{1} ψ^{i} (φ) (g_{2}, \dots, g_{i + 1}) + \sum_{j = 1}^{i} {(- 1)}^{j} ψ^{i} (φ) (g_{1}, \dots, g_{j - 1}, g_{j} g_{j + 1}, g_{j + 2}, \dots, g_{i + 1}) \\ + {(- 1)}^{i + 1} ψ^{i} (φ) (g_{1}, \dots, g_{i}), \end{matrix}

which is $d^{i} (ψ^{i} (φ))$ .

Finally, suppose that $α : A \to B$ is a $G$ -module homomorphism. We then have

α \circ ψ^{i} (φ) (g_{1}, \dots, g_{i}) = α \circ φ (1, g_{1}, \dots, g_{1} \dots g_{i}) = ψ^{i} (α \circ φ) (g_{1}, \dots, g_{i}),

hence the desired naturality. □

Corollary 1.3.4. §

The $i$ th cohomology group of the complex $({Hom}_{ℤ [G]} (ℤ [G^{i + 1}], A), D_{A}^{i})$ is naturally isomorphic to $H^{i} (G, A)$ .

In fact, the standard resolution is a projective resolution of $ℤ$ , as is a consequence of the following lemma and the fact that every free module is projective.

Lemma 1.3.5. §

The $G$ -module $ℤ [G^{i + 1}]$ is free.

Proof.

In fact, we have

ℤ [G^{i + 1}] ≅ ⨁_{(g_{1}, \dots, g_{i}) \in G^{i}} ℤ [G] (1, g_{1}, \dots, g_{i}),

and the submodule generated by $(1, g_{1}, \dots, g_{i})$ is clearly free. □

Remarks 1.3.6. §

a.

Lemma 1.3.5 implies that the standard resolution provides a projective resolution of $ℤ$ . It follows that

H^{i} (G, A) ≅ {Ext}_{ℤ [G]}^{i} (ℤ, A)

for any $G$ -module $A$ . Moreover, if $P_{\cdot} \to ℤ \to 0$ is any projective resolution of $ℤ$ by $G$ -modules, we have that $H^{i} (G, A)$ is the $i$ th cohomology group of the complex ${Hom}_{ℤ [G]} (P_{\cdot}, A)$ .

b.

By definition, ${Ext}_{ℤ [G]}^{i} (ℤ, A)$ is the $i$ th right derived functor of the functor that takes the value ${Hom}_{ℤ [G]} (ℤ, A)$ on $A$ . Note that ${Hom}_{ℤ [G]} (ℤ, A) ≅ A^{G}$ , the module of $G$ -invariants. Therefore, if $0 \to A \to I^{\cdot}$ is any injective $ℤ [G]$ -resolution of $A$ , then $H^{i} (G, A)$ is the $i$ th cohomology group in the sequence

0 \to {(I^{0})}^{G} \to {(I^{1})}^{G} \to {(I^{2})}^{G} \to \dots .

1.4. Homology of groups

In this section, we consider a close relative of group cohomology, known as group homology. Note first that $ℤ [G^{i + 1}]$ is also a right module over $ℤ [G]$ by the diagonal right multiplication by an element of $G$ . Up to isomorphism of $G$ -modules, this is the same as taking the diagonal left multiplication of $ℤ [G^{i + 1}]$ by the inverse of an element of $G$ .

Definition 1.4.1. §

The $i$ th homology group $H_{i} (G, A)$ of a group $G$ with coefficients in a $G$ -module $A$ is defined to be the $i$ th homology group $H_{i} (G, A) = \ker d_{i} ∕ im d_{i + 1}$ in the complex

\dots \to ℤ [G^{3}] \otimes_{ℤ [G]} A \to_{}^{d_{2}} ℤ [G^{2}] \otimes_{ℤ [G]} A \to_{}^{d_{1}} ℤ [G] \otimes_{ℤ [G]} A \to_{}^{d_{0}} 0

induced by the standard resolution.

We note that if $f : A \to B$ is a $G$ -module homomorphism, then there are induced maps $f_{*} : H_{i} (G, A) \to H_{i} (G, B)$ for each $i \geq 0$ .

Remark 1.4.2. §

It follows from Definition 1.4.1 that $H_{i} (G, A) ≅ {Tor}_{ℤ [G]}^{i} (ℤ, A)$ for every $i \geq 0$ , and that $H_{i} (G, A)$ may be calculated by taking the homology of $P_{\cdot} \otimes_{ℤ [G]} A$ , where $P_{\cdot}$ is any projective $ℤ [G]$ -resolution of $ℤ$ . Here, we view $P_{i}$ as a right $G$ -module via the action $x \cdot g = g^{- 1} x$ for $g \in G$ and $x \in X$ .

As a first example, we have

Lemma 1.4.3. §

We have natural isomorphisms $H_{0} (G, A) ≅ A_{G}$ for every $G$ -module $A$ .

Proof.

Note first that $ℤ [G] \otimes_{ℤ [G]} 𝐴≅𝐴$ , and the map $d_{1}$ under this identification is given by

d_{1} ((g_{0}, g_{1}) \otimes a) = (g_{0} - g_{1}) a .

Hence, the image of $d_{1}$ is $I_{G} A$ , and the result follows. □

As $A_{G} ≅ℤ \otimes_{ℤ [G]} A$ , we have in particular that $H_{i} (G, A)$ is the $i$ th left derived functor of $A_{G}$ . As with cohomology, we therefore have in particular that homology carries short exact sequences to long exact sequences, as we now spell out.

Theorem 1.4.4. §

Suppose that

0 \to A \to_{}^{ι} B \to_{}^{π} C \to 0

is a short exact sequence of $G$ -modules. Then there are connecting homomorphisms $δ_{*} : H_{i} (G, C) \to H_{i - 1} (G, A)$ and a long exact sequence of abelian groups

\dots \to H_{1} (G, C) \to_{}^{δ} H_{0} (G, A) \to_{}^{ι_{*}} H_{0} (G, B) \to_{}^{π_{*}} H_{0} (G, C) \to 0 .

Moreover, this construction is natural in the short exact sequence in the sense that any morphism of short exact sequences gives rise to a morphism of long exact sequences.

The following result computes the homology group $H_{1} (G, ℤ)$ , where $ℤ$ has the trivial $G$ -action.

Proposition 1.4.5. §

There are canonical isomorphisms $H_{1} (G, ℤ) ≅ I_{G} ∕ I_{G}^{2} ≅ G^{ab}$ , where $G^{ab}$ denotes the abelianization of $G$ , the latter taking the coset of $g - 1$ to the coset of $g \in G$ .

Proof.

Since $ℤ [G]$ is $ℤ [G]$ -projective, we have $H_{1} (G, ℤ [G]) = 0$ , and hence our long exact sequence in homology has the form

0 \to H_{1} (G, ℤ) \to H_{0} (G, I_{G}) \to H_{0} (G, ℤ [G]) \to H_{0} (G, ℤ) \to 0 .

Note that $H_{0} (G, I_{G}) ≅ I_{G} ∕ I_{G}^{2}$ and $H_{0} (G, ℤ [G]) ≅ℤ [G] ∕ I_{G} ≅ℤ$ via the augmentation map. Since $H_{0} (G, ℤ) ≅ℤ$ and any surjective map $ℤ \to ℤ$ (in this case the identity) is an isomorphism, we obtain the first isomorphism of the proposition.

For the second isomorphism, let us define maps $ϕ : G^{ab} \to I_{G} ∕ I_{G}^{2}$ and $ψ : I_{G} ∕ I_{G}^{2} \to G^{ab}$ . For $g \in G$ , we set

ϕ (g [G, G]) = (g - 1) + I_{G}^{2},

where $[G, G]$ denotes the commutator subgroup of $G$ . To see that this is a homomorphism on $G$ , hence on $G^{ab}$ , note that

(𝑔h - 1) + I_{G}^{2} = (g - 1) + (h - 1) + (g - 1) (h - 1) + I_{G}^{2} = (g - 1) + (h - 1) + I_{G}^{2}

for $g, h \in G$ .

Next, define $ψ$ on $α = \sum_{g \in G} a_{g} g \in I_{G}$ by

ψ (α + I_{G}^{2}) = \prod_{g \in G} g^{a_{g}} [G, G] .

The order of the product doesn’t matter as $G^{ab}$ is abelian, and $ψ$ is then a homomorphism if well-defined. It suffices for the latter to check that the recipe defining $ψ$ takes the generators $(g - 1) (h - 1)$ of $I_{G}^{2}$ for $g, h \in G$ to the trivial coset, but this follows as $(g - 1) (h - 1) = 𝑔h - g - h + 1$ , and for instance, we have

𝑔h \cdot g^{- 1} \cdot h^{- 1} \in [G, G] .

Finally, we check that the two homomorphisms are inverse to each other. We have

ϕ (ψ (α + I_{G}^{2})) = \sum_{g \in G} a_{g} (g - 1) + I_{G}^{2} = α + I_{G}^{2}

since $α \in I_{G}$ implies $\sum_{g \in G} a_{g} = 0$ , and

ψ (ϕ (g [G, G])) = ϕ ((g - 1) + I_{G}^{2}) = g [G, G] .

□

1.5. Induced modules

Definition 1.5.1. §

Let $H$ be a subgroup of $G$ , and suppose that $B$ is a $ℤ [H]$ -module. We set

\begin{array}{rcl} {Ind}_{H}^{G} (B) = ℤ [G] \otimes_{ℤ [H]} B & and & {CoInd}_{H}^{G} (B) = {Hom}_{ℤ [H]} (ℤ [G], B) . \end{array}

We give these $G$ -actions by

\begin{array}{rcl} g \cdot (α \otimes b) = (𝑔𝛼) \otimes b & and & (g \cdot φ) (α) = φ (α \cdot g) . \end{array}

We say that the resulting modules are induced and coinduced, respectively, from $H$ to $G$ .

Remark 1.5.2. §

What we refer to as a “coinduced” module is often actually referred to as an “induced” module.

We may use these modules to interpret $H$ -cohomology groups as $G$ -cohomology groups.

Theorem 1.5.3 (Shapiro’s Lemma). §

For each $i \geq 0$ , we have canonical isomorphisms

\begin{array}{rcl} H_{i} (G, {Ind}_{H}^{G} (B)) ≅ H_{i} (H, B) & and & H^{i} (G, {CoInd}_{H}^{G} (B)) ≅ H^{i} (H, B) \end{array}

that provide natural isomorphisms of $δ$ -functors.

Proof.

Let $P_{\cdot}$ be the standard resolution of $ℤ$ by $G$ -modules. Define

ψ_{i} : {Hom}_{ℤ [G]} (P_{i}, {CoInd}_{H}^{G} (B)) \to {Hom}_{ℤ [H]} (P_{i}, B)

by $ψ_{i} (𝜃) (x) = 𝜃 (x) (1)$ . If $𝜃 \in \ker ψ_{i}$ , then

𝜃 (x) (g) = (g \cdot 𝜃 (x)) (1) = 𝜃 (𝑔𝑥) (1) = 0 .

for all $x \in P_{i}$ and $g \in G$ , so $𝜃 = 0$ . Conversely, if $φ \in {Hom}_{ℤ [H]} (P_{i}, B)$ , then define $𝜃$ by $𝜃 (x) (g) = φ (𝑔𝑥)$ , and we have $ψ_{i} (𝜃) = φ$ .

As for the induced case, note that associativity of tensor products yields

P_{i} \otimes_{ℤ [G]} (ℤ [G] \otimes_{ℤ [H]} B) ≅ P_{i} \otimes_{ℤ [H]} B,

and $P_{i} = ℤ [G^{i + 1}]$ is free as a left $H$ -module, hence projective. (We leave it to the reader to check that usual $\otimes$ - $Hom$ adjunction can be similarly used to give a shorter proof of the result for cohomology.) □

In fact, if $H$ is of finite index in $G$ , the notions of induced and coinduced from $H$ to $G$ coincide.

Proposition 1.5.4. §

Suppose that $H$ is a subgroup of finite index in $G$ and $B$ is a $H$ -module. Then we have a canonical isomorphism of $G$ -modules

χ : {CoInd}_{H}^{G} (B) \to_{}^{\sim} {Ind}_{H}^{G} (B), χ (φ) = \sum_{\bar{g} \in H ∖ G} g^{- 1} \otimes φ (g),

where for each $\bar{g} \in H ∖ G$ , the element $g \in G$ is an arbitrary choice of representative of $\bar{g}$ .

Proof.

First, we note that $χ$ is a well-defined map, as

{(h𝑔)}^{- 1} \otimes φ (h𝑔) = g^{- 1} h^{- 1} \otimes h𝜑 (g) = g \otimes φ (g)

for $φ \in {CoInd}_{H}^{G} (B)$ , $h \in H$ , and $g \in G$ . Next, we see that $χ$ is a $G$ -module homomorphism, as

χ (g^{'} φ) = \sum_{\bar{g} \in H ∖ G} g^{- 1} \otimes φ (g g^{'}) = g^{'} \sum_{\bar{g} \in H ∖ G} {(g g^{'})}^{- 1} \otimes φ (g g^{'}) = g^{'} χ (φ)

for $g^{'} \in G$ . As the coset representatives form a basis for $ℤ [G]$ as a free $ℤ [H]$ -module, we may define an inverse to $χ$ that maps

\sum_{\bar{g} \in H ∖ G} g^{- 1} \otimes b_{g} \in {Ind}_{H}^{G} (B)

to the unique $ℤ [H]$ -linear map $φ$ that takes the value $b_{g}$ on $g$ for the chosen representative of $\bar{g} \in H ∖ G$ . □

In the special case of the trivial subgroup, we make the following definition.

Definition 1.5.5. §

We say that $G$ -modules of the form

\begin{array}{rcl} {Ind}^{G} (X) = ℤ [G] \otimes_{ℤ} X & and & {CoInd}^{G} (X) = {Hom}_{ℤ} (ℤ [G], X), \end{array}

where $X$ is an abelian group, are induced and coinduced $G$ -modules, respectively.

Remark 1.5.6. §

Note that Proposition 1.5.4 implies that the notions of induced and coinduced modules coincide for finite groups $G$ . On the other hand, for infinite groups, ${CoInd}^{G} (X)$ will never be finitely generated over $ℤ [G]$ for nontrivial $X$ , while ${Ind}^{G} (X)$ will be for any finitely generated abelian group $X$ .

Theorem 1.5.7. §

Suppose that $A$ is an induced (resp., coinduced) $G$ -module. Then we have $H_{i} (G, A) = 0$ (resp., $H^{i} (G, A) = 0$ ) for all $i \geq 1$ .

Proof.

Let $X$ be an abelian group. By Shapiro’s Lemma, we have

H_{i} (G, {Ind}^{G} (X)) = H_{i} ({1}, X)

for $i \geq 1$ . Since $ℤ$ has a projective $ℤ$ -resolution by itself, the latter groups are $0$ . The proof for cohomology is essentially identical. □

Definition 1.5.8. §

A $G$ -module $A$ such that $H^{i} (G, A) = 0$ for all $i \geq 1$ is called $G$ -acyclic.

We show that we may construct induced and coinduced $G$ -modules starting from abelian groups that are already equipped with a $G$ -action.

Remark 1.5.9. §

Suppose that $A$ and $B$ are $G$ -modules. We give ${Hom}_{ℤ} (A, B)$ and $A \otimes_{ℤ} B$ actions of $G$ by

\begin{array}{rcl} (g \cdot φ) (a) = 𝑔𝜑 (g^{- 1} a) & and & g \cdot (a \otimes b) = 𝑔𝑎 \otimes 𝑔𝑏, \end{array}

respectively.

Lemma 1.5.10. §

Let $A$ be a $G$ -module, and let $A^{\circ}$ be its underlying abelian group. Then

\begin{array}{rcl} {Hom}_{ℤ} (ℤ [G], A) ≅ {CoInd}^{G} (A^{\circ}) & and & ℤ [G] \otimes_{ℤ} 𝐴≅ {Ind}^{G} (A^{\circ}) . \end{array}

Proof.

We define

κ : {Hom}_{ℤ} (ℤ [G], A) \to {CoInd}^{G} (A^{\circ}), κ (φ) (g) = g \cdot φ (g^{- 1}) .

For $g, k \in G$ , we then have

(k \cdot κ (φ)) (g) = κ (φ) (𝑔𝑘) = 𝑔𝑘 \cdot φ (k^{- 1} g^{- 1}) = g \cdot (k \cdot φ) (g^{- 1}) = κ (k \cdot φ) (g),

so $κ$ is a $G$ -module homomorphism. Note that $κ$ is also self-inverse on the underlying set of both groups, so is an isomorphism. In the induced case, we define

ν : ℤ [G] \otimes_{ℤ} A \to {Ind}^{G} (A^{\circ}), ν (g \otimes a) = g \otimes g^{- 1} a .

For $g, k \in G$ , we now have

k \cdot ν (g \otimes a) = (𝑘𝑔) \otimes g^{- 1} a = ν (𝑘𝑔 \otimes 𝑘𝑎) = ν (k \cdot (g \otimes a)),

so $ν$ is a $G$ -module isomorphism with inverse $ν^{- 1} (g \otimes a) = g \otimes 𝑔𝑎$ . □

Remark 1.5.11. §

Noting the lemma, we will simply refer to ${Hom}_{ℤ} (ℤ [G], A)$ as ${CoInd}^{G} (A)$ and $ℤ [G] \otimes_{ℤ} A$ as ${Ind}^{G} (A)$ .

1.6. Tate cohomology

We suppose in this section that $G$ is a finite group. In this case, recall that we have the norm element $N_{G} \in ℤ [G]$ , which defines by left multiplication a map $N_{G} : A \to A$ on any $G$ -module $A$ . Its image $N_{G} A$ is the group of $G$ -norms of $A$ .

Lemma 1.6.1. §

The norm element induces a map ${\bar{N}}_{G} : A_{G} \to A^{G}$ .

Proof.

We have $N_{G} ((g - 1) a) = 0$ for any $g \in G$ and $a \in A$ , so the map factors through $A_{G}$ , and clearly $im N_{G} \subseteq A^{G}$ . □

Definition 1.6.2. §

We let ${\hat{H}}^{0} (G, A)$ (resp., ${\hat{H}}_{0} (G, A)$ ) denote the cokernel (resp., kernel) of the map in Lemma 1.6.1. In other words,

\begin{array}{rcl} {\hat{H}}^{0} (G, A) = A^{G} ∕ N_{G} A & and & {\hat{H}}_{0} (G, A) =_{N} A ∕ I_{G} A, \end{array}

where $_{N} A$ denotes the kernel of the left multiplication by $N_{G}$ on $A$ .

Example 1.6.3. §

Consider the case that $A = ℤ$ , where $ℤ$ is endowed with a trivial action of $G$ . Since $N_{G} : ℤ \to ℤ$ is just the multiplication by $| G |$ map, we have that ${\hat{H}}^{0} (G, ℤ) = ℤ ∕ | G | ℤ$ and ${\hat{H}}_{0} (G, ℤ) = 0$ .

Remark 1.6.4. §

In general, when we take cohomology with coefficients in a group, like $ℤ$ or $ℚ$ , with no specified action of the group $G$ , the action is taken to be trivial.

The Tate cohomology groups are an amalgamation of the homology groups and cohomology groups of $G$ , with the homology groups placed in negative degrees.

Definition 1.6.5. §

Let $G$ be a finite group and $A$ a $G$ -module. For any $i \in ℤ$ , we define the $i$ th Tate cohomology group by

{\hat{H}}^{i} (G, A) = {\begin{matrix} H_{- i - 1} (G, A) & if i \leq - 2 \\ {\hat{H}}_{0} (G, A) & if i = - 1 \\ {\hat{H}}^{0} (G, A) & if i = 0 \\ H^{i} (G, A) & if i \geq 1 . \end{matrix}

We have modified the zeroth homology and cohomology groups in defining Tate cohomology so that we obtain long exact sequences from short exact sequences as before, but extending infinitely in both directions, as we shall now see.

Theorem 1.6.6 (Tate). §

Suppose that

0 \to A \to_{}^{ι} B \to_{}^{π} C \to 0

is a short exact sequence of $G$ -modules. Then there is a long exact sequence of abelian groups

\dots \to {\hat{H}}^{i} (G, A) \to_{}^{ι} {\hat{H}}^{i} (G, B) \to_{}^{π} {\hat{H}}^{i} (G, C) \to_{}^{δ} {\hat{H}}^{i + 1} (G, A) \to \dots .

Moreover, this construction is natural in the short exact sequence in the sense that any morphism of short exact sequences gives rise to a morphism of long exact sequences.

Proof.

The first part follows immediately from applying the snake lemma to the following diagram, which in particular defines the map $δ$ on ${\hat{H}}^{- 1} (G, C)$ :

Joining homology and cohomology by the norm. A full diagram description follows.

Diagram description: Joining homology and cohomology by the norm

The two displayed rows are exact, and the squares commute.

Objects, listed by row and column:

Row 1, from left to right: column 1: ellipsis; column 2: H subscript (1)(G,C); column 3: H subscript (0)(G,A); column 4: H subscript (0)(G,B); column 5: H subscript (0)(G,C); column 6: 0.
Row 2, from left to right: column 2: 0; column 3: H superscript (0)(G,A); column 4: H superscript (0)(G,B); column 5: H superscript (0)(G,C); column 6: H superscript (1)(G,A); column 7: ellipsis.

Arrows and lines:

An arrow from ellipsis (row 1, column 1) to H subscript (1)(G,C), without a label.
An arrow from H subscript (1)(G,C) to H subscript (0)(G,A), labelled delta.
An arrow from H subscript (0)(G,A) to H subscript (0)(G,B), labelled iota subscript (star).
An arrow from H subscript (0)(G,A) to H superscript (0)(G,A), labelled bar of (N) subscript (G).
An arrow from H subscript (0)(G,B) to H subscript (0)(G,C), labelled pi subscript (star).
An arrow from H subscript (0)(G,B) to H superscript (0)(G,B), labelled bar of (N) subscript (G).
An arrow from H subscript (0)(G,C) to 0 (row 1, column 6), without a label.
An arrow from H subscript (0)(G,C) to H superscript (0)(G,C), labelled bar of (N) subscript (G).
An arrow from 0 (row 2, column 2) to H superscript (0)(G,A), without a label.
An arrow from H superscript (0)(G,A) to H superscript (0)(G,B), labelled iota subscript (star).
An arrow from H superscript (0)(G,B) to H superscript (0)(G,C), labelled pi subscript (star).
An arrow from H superscript (0)(G,C) to H superscript (1)(G,A), labelled delta.
An arrow from H superscript (1)(G,A) to ellipsis (row 2, column 7), without a label.

and the second part is easily checked. □

Tate cohomology groups have the interesting property that they vanish entirely on induced modules.

Proposition 1.6.7. §

Suppose that $A$ is an induced $G$ -module. Then ${\hat{H}}^{i} (G, A) = 0$ for all $i \in ℤ$ .

Proof.

By Theorem 1.5.7 and Proposition 1.5.4, it suffices to check this for $i = - 1$ and $i = 0$ . Let $X$ be an abelian group. Since ${ℤ [G]}^{G} = N_{G} ℤ [G]$ , we have

H^{0} (G, {Ind}^{G} (X)) = N_{G} ℤ [G] \otimes_{ℤ} X,

so ${\hat{H}}^{0} (G, {Ind}^{G} (X)) = 0$ by definition. We also have that

H_{0} (G, {Ind}^{G} (X)) = {(ℤ [G] \otimes_{ℤ} X)}_{G} ≅ℤ \otimes_{ℤ} 𝑋≅𝑋 .

(1.6.1)

Let $α = \sum_{g \in G} (g \otimes x_{g})$ be an element of ${Ind}^{G} (X)$ . Then

N_{G} α = N_{G} \otimes \sum_{g \in G} x_{g}

is trivial if and only if $\sum_{g \in G} x_{g} = 0$ , which by the identification in (1.6.1) is to say that $α$ has trivial image in $H_{0} (G, {Ind}^{G} (X))$ . Hence ${\hat{H}}_{0} (G, {Ind}^{G} (X)) = 0$ as well. □

The Tate cohomology groups can also be computed via a doubly infinite resolution of $G$ -modules. The proof of this is rather involved and requires some preparation.

Lemma 1.6.8. §

Let $X$ be a $G$ -module that is free of finite rank over $ℤ$ , and let $A$ be any $G$ -module. Then the map

ν : X \otimes_{ℤ} A \to {Hom}_{ℤ} ({Hom}_{ℤ} (X, ℤ), A), ν (x \otimes a) (φ) = φ (x) a

is an isomorphism of $G$ -modules.

Proof.

We note that

ν (g \cdot (x \otimes a)) (φ) = φ (𝑔𝑥) 𝑔𝑎,

while

(g \cdot ν (x \otimes a)) (φ) = 𝑔𝜈 (x \otimes a) (g^{- 1} φ) = (g^{- 1} φ) (x) 𝑔𝑎 = φ (𝑔𝑥) 𝑔𝑎,

so $ν$ is a homomorphism of $G$ -modules.

Let $x_{1}, \dots, x_{m}$ be any $ℤ$ -basis of $X$ , and let $x_{1}^{*}, \dots, x_{m}^{*}$ be the dual basis of ${Hom}_{ℤ} (X, ℤ)$ such that $x_{i}^{*} (x_{j}) = δ_{𝑖𝑗}$ for $1 \leq i, j \leq m$ . We define

ω : {Hom}_{ℤ} ({Hom}_{ℤ} (X, ℤ), A) \to X \otimes_{ℤ} A, ω (ψ) = \sum_{i = 1}^{m} x_{i} \otimes ψ (x_{i}^{*}) .

Then

(ν \circ ω) (ψ) (φ) = ν (\sum_{i = 1}^{m} x_{i} \otimes ψ (x_{i}^{*})) (φ) = \sum_{i = 1}^{m} φ (x_{i}) ψ (x_{i}^{*}) = ψ (\sum_{i = 1}^{m} φ (x_{i}) x_{i}^{*}) = ψ (φ) .

On the other hand,

(ω \circ ν) (x \otimes a) = \sum_{i = 1}^{m} x_{i} \otimes x_{i}^{*} (x) a = \sum_{i = 1}^{m} x_{i}^{*} (x) x_{i} \otimes a = x \otimes a .

Hence, $ν$ is an isomorphism. □

Lemma 1.6.9. §

Let $X$ and $A$ be $G$ -modules, and endow $X$ with a right $G$ -action by $x \cdot g = g^{- 1} x$ . Then we have a canonical isomorphism

X \otimes_{ℤ [G]} A \to_{}^{\sim} {(X \otimes_{ℤ} A)}_{G}

induced by the identity on $X \otimes_{ℤ} A$ .

Proof.

First, note that $X \otimes_{ℤ [G]} A$ is a quotient of the $G$ -module $X \otimes_{ℤ} A$ , which is endowed the diagonal left $G$ -action. By definition of the tensor product over $ℤ [G]$ , we have

g^{- 1} x \otimes a = x \cdot g \otimes a = x \otimes 𝑔𝑎,

so $G$ acts trivially on $X \otimes_{ℤ [G]} A$ , and hence the latter group is a quotient of ${(X \otimes_{ℤ} A)}_{G}$ . On the other hand, the $ℤ$ -bilinear map

X \times A \to {(X \otimes_{ℤ} A)}_{G}, (x, a) \mapsto x \otimes a

is $ℤ [G]$ -balanced, hence induces a map on the tensor product inverse to the above-described quotient map. □

We are now ready to prove the theorem.

Theorem 1.6.10. §

Let $P_{\cdot} \to_{}^{α} ℤ$ be a projective resolution by $G$ -modules of finite $ℤ$ -rank, and consider the $ℤ$ -dual $ℤ \to_{}^{\hat{α}} P_{*}^{\cdot}$ , where $P_{*}^{i} = {Hom}_{ℤ} (P_{i}, ℤ)$ for $i \geq 0$ and $G$ acts on $P_{*}^{i}$ by $(g \cdot φ) (x) = φ (g^{- 1} x)$ . Let $Q_{\cdot}$ be the exact chain complex

\dots \to P_{1} \to P_{0} \to_{}^{\hat{α} \circ α} P_{*}^{0} \to P_{*}^{1} \to \dots,

where $P_{0}$ occurs in degree $0$ . (That is, we set $Q_{i} = P_{i}$ for $i \geq 0$ and $Q_{i} = P_{*}^{- 1 - i}$ for $i < 0$ .) For any $G$ -module $A$ , the Tate cohomology group ${\hat{H}}^{i} (G, A)$ is the $i$ th cohomology group of the cochain complex $C^{\cdot} = {Hom}_{ℤ [G]} (Q_{\cdot}, A)$ .

Proof.

As $P_{i}$ is projective over $ℤ [G]$ , it is in particular $ℤ$ -free, so the $ℤ$ -dual sequence $ℤ \to P_{*}^{\cdot}$ is still exact. Let us denote the $i$ th differential on $P_{i}$ by $d_{i}$ and its $ℤ$ -dual by ${\hat{d}}^{i}$ . We check exactness at $Q_{0}$ and $Q_{- 1}$ . Let $β = \hat{α} \circ α$ . By definition, $im d_{1} = \ker α$ , and as $\hat{α}$ is injective, we have $im d_{1} = \ker β$ . Similarly, we have $\ker {\hat{d}}^{0} = im \hat{α}$ , and as $α$ is surjective, we have $\ker {\hat{d}}^{0} = im β$ . Therefore, $Q_{\cdot}$ is exact.

That ${Hom}_{ℤ [G]} (Q_{\cdot}, A)$ computes the Tate cohomology groups ${\hat{H}}^{i} (G, A)$ follows immediately from the definition for $i \geq 1$ . By Lemma 1.6.9, we have an isomorphism

P_{i} \otimes_{ℤ [G]} A \to_{}^{\sim} {(P_{i} \otimes_{ℤ} A)}_{G} .

By Proposition 1.6.7 and Lemma 1.5.10, we have that

{(P_{i} \otimes_{ℤ} A)}_{G} \to_{}^{{\bar{N}}_{G}} {(P_{i} \otimes_{ℤ} A)}^{G}

is an isomorphism as well, and following this by the restriction

{(P_{i} \otimes_{ℤ} A)}^{G} \to {Hom}_{ℤ} {(P_{*}^{i}, A)}^{G} = {Hom}_{ℤ [G]} (P_{*}^{i}, A)

of the map $ν$ of Lemma 1.6.8, we obtain in summary an isomorphism

χ_{i} : P_{i} \otimes_{ℤ [G]} A \to {Hom}_{ℤ [G]} (P_{*}^{i}, A) .

Next, we check that the maps $ν$ of Lemma 1.6.8 commute with the differentials on the complexes $P_{\cdot} \otimes_{ℤ} A$ and ${Hom}_{ℤ} (P_{*}^{\cdot}, A)$ , the former of which are just the tensor products of the differentials $d_{i}$ on the $P_{i}$ with the identity (then also denoted $d_{i}$ ), and the latter of which are double duals $\tilde{d_{i}}$ of the $d_{i}$ , i.e., which satisfy

\tilde{d_{i}} (ψ) (φ) = ψ (φ \circ d_{i}),

for $ψ \in {Hom}_{ℤ} (P_{*}^{i}, A)$ and $φ \in {Hom}_{ℤ} (P_{i - 1}, ℤ)$ . We have

(ν \circ d_{i}) (x \otimes a) (φ) = ν (d_{i} (x) \otimes a) (φ) = φ (d_{i} (x)) a .

On the other hand, we have

(\tilde{d_{i}} \circ ν) (x \otimes a) (φ) = \tilde{d_{i}} (ν (x \otimes a)) (φ) = ν (x \otimes a) (φ \circ d_{i}) = φ (d_{i} (x)) a,

as desired. Moreover, the $d_{i}$ commute with ${\bar{N}}_{G}$ on ${(P_{i} \otimes_{ℤ} A)}_{G}$ , being that they are $G$ -module maps. Hence, the maps $χ_{i}$ for all $i$ together provide an isomorphism of complexes. In particular, the $i$ th cohomology group of ${Hom}_{ℤ [G]} (Q_{\cdot}, A)$ is ${\hat{H}}^{i} (G, A)$ for all $i \leq - 2$ , and we already knew this for all $i \geq 1$ .

It remains to consider the cases $i = 0, - 1$ . We need to compute the cohomology of

P_{1} \otimes_{ℤ [G]} A \to P_{0} \otimes_{ℤ [G]} A \to_{}^{τ} {Hom}_{ℤ [G]} (P_{0}, A) \to {Hom}_{ℤ [G]} (P_{1}, A)

(1.6.2)

in the middle two degrees, and

τ (x \otimes a) (y) = (χ_{0} (x \otimes a) \circ \hat{α} \circ α) (y) = \sum_{g \in G} (\hat{α} \circ α) (y) (𝑔𝑥) 𝑔𝑎 = \sum_{g \in G} α (y) α (x) 𝑔𝑎,

noting that $α (𝑔𝑥) = α (x)$ for every $g \in G$ and $\hat{α} (n) (x) = 𝑛𝛼 (x)$ for every $n \in ℤ$ . On the other hand, viewing $α$ and $\hat{α}$ as inducing maps

\begin{array}{rcl} λ : P_{0} \otimes_{ℤ [G]} A \to A_{G} & and & \hat{λ} : A^{G} \to {Hom}_{ℤ [G]} (P_{0}, A), \end{array}

respectively, we have

\begin{matrix} (\hat{λ} \circ {\bar{N}}_{G} \circ λ) (x \otimes a) (y) = \hat{λ} ({\bar{N}}_{G} (α (x) a)) (y) = \hat{λ} (\sum_{g \in G} α (x) 𝑔𝑎) (y) \\ = \sum_{g \in G} \hat{α} (α (x)) (y) 𝑔𝑎 = \sum_{g \in G} α (x) α (y) 𝑔𝑎 . \end{matrix}

In other words, we have $τ = \hat{λ} \circ {\bar{N}}_{G} \circ λ$ . As the cokernel of the first map in (1.6.2) is $H_{0} (G, A) = A_{G}$ and the kernel of the last is $H^{0} (G, A) = A^{G}$ , with these identifications given by the maps $λ$ and $\hat{λ}$ respectively, we have that the complex given by $A_{G} \to_{}^{{\bar{N}}_{G}} A^{G}$ in degrees $- 1$ and $0$ computes the cohomology groups in question, as desired. □

As what is in essence a corollary, we have the following version of Shapiro’s lemma.

Theorem 1.6.11. §

Let $G$ be a finite group, let $H$ be a subgroup, and let $B$ be an $H$ -module. Then for every $i \in ℤ$ , we have canonical isomorphisms

\begin{array}{rcl} {\hat{H}}^{i} (G, {CoInd}_{H}^{G} (B)) ≅ {\hat{H}}^{i} (H, B) \end{array}

that together provide natural isomorphisms of $δ$ -functors.

Proof.

The proof is nearly identical to that of Shapiro’s lemma for cohomology groups. That is, we may simply use the isomorphisms induced by

ψ_{i} : {Hom}_{ℤ [G]} (Q_{i}, {CoInd}_{H}^{G} (B)) \to {Hom}_{ℤ [H]} (Q_{i}, B)

by $ψ_{i} (𝜃) (x) = 𝜃 (x) (1)$ , for $Q_{\cdot}$ the doubly infinite resolution of $ℤ$ for $G$ of Theorem 1.6.10. □

1.7. Dimension shifting

One useful technique in group cohomology is that of dimension shifting. The key idea here is to use the acyclicity of coinduced modules to obtain isomorphisms among cohomology groups.

To describe this technique, note that we have a short exact sequence

0 \to A \to_{}^{ι} {CoInd}^{G} (A) \to A^{*} \to 0,

(1.7.1)

where $ι$ is defined by $ι (a) (g) = a$ for $a \in A$ and $g \in G$ , and $A^{*}$ is defined to be the cokernel of $ι$ . We also have a short exact sequence

0 \to A_{*} \to {Ind}^{G} (A) \to_{}^{π} A \to 0,

(1.7.2)

where $π$ is defined by $π (g \otimes a) = a$ for $a \in A$ and $g \in G$ , and $A_{*}$ is defined to be the kernel of $π$ .

Remark 1.7.1. §

If we view $A$ as $ℤ \otimes_{ℤ} A$ , we see by the freeness of $ℤ$ as a $ℤ$ -module and the definition of ${Ind}^{G} (A)$ that $A_{*} ≅ I_{G} \otimes_{ℤ} A$ with a diagonal action of $G$ . Moreover, viewing $A$ as ${Hom}_{ℤ} (ℤ, A)$ , we see that $A^{*} ≅ {Hom}_{ℤ} (I_{G}, A)$ .

Proposition 1.7.2. §

With the notation as above, we have

\begin{array}{rcl} H^{i + 1} (G, A) ≅ H^{i} (G, A^{*}) & and & H_{i + 1} (G, A) ≅ H_{i} (G, A_{*}) \end{array}

for all $i \geq 1$ .

Proof.

By Lemma 1.5.10, we know that ${CoInd}^{G} (A)$ (resp., ${Ind}^{G} (A)$ ) is coinduced (resp., induced). The result then follows easily by Theorem 1.5.7 and the long exact sequences of Theorems 1.2.13 and 1.4.4. □

For Tate cohomology groups, we have an even cleaner result.

Theorem 1.7.3 (Dimension shifting). §

Suppose that $G$ is finite. With the above notation, we have

\begin{array}{rcl} {\hat{H}}^{i + 1} (G, A) ≅ {\hat{H}}^{i} (G, A^{*}) & and & {\hat{H}}^{i - 1} (G, A) ≅ {\hat{H}}^{i} (G, A_{*}) \end{array}

for all $i \in ℤ$ .

Proof.

Again noting Lemma 1.5.10, it follows from Theorem 1.5.4 and Proposition 1.6.7 that the long exact sequences associated by Theorem 1.6.6 to the short exact sequences in (1.7.1) and (1.7.2) reduce to the isomorphisms in question. □

This result allows us to transfer questions about cohomology groups in a certain degree to analogous questions regarding cohomology groups in other degrees. Let us give a first application.

Proposition 1.7.4. §

Suppose that $G$ is a finite group and $A$ is a $G$ -module. Then the groups ${\hat{H}}^{i} (G, A)$ have exponent dividing $| G |$ for every $i \in ℤ$ .

Proof.

By Theorem 1.7.3, the problem immediately reduces to proving the claim for $i = 0$ and every module $A$ . But for any $a \in A^{G}$ , we have $| G | a = N_{G} a$ , so ${\hat{H}}^{0} (G, A)$ has exponent dividing $| G |$ . □

This has the following important corollary.

Corollary 1.7.5. §

Suppose that $G$ is a finite group and $A$ is a $G$ -module that is finitely generated as an abelian group. Then ${\hat{H}}^{i} (G, A)$ is finite for every $i \in ℤ$ .

Proof.

We know that ${\hat{H}}^{i} (G, A)$ is a subquotient of the finitely generated abelian group $Q_{i} \otimes_{ℤ [G]} A$ of Theorem 1.6.10, hence is itself finitely generated. As it has finite exponent, it is therefore finite. □

We also have the following.

Corollary 1.7.6. §

Suppose that $G$ is finite. Suppose that $A$ is a $G$ -module on which multiplication by $| G |$ is an isomorphism. Then ${\hat{H}}^{i} (G, A) = 0$ for $i \in ℤ$ .

Proof.

Multiplication by $| G |$ on $A$ induces multiplication by $| G |$ on Tate cohomology, which is then an isomorphism. Since by Proposition 1.7.4, multiplication by $| G |$ is also the zero map on Tate cohomology, the Tate cohomology groups must be $0$ . □

1.8. Comparing cohomology groups

Definition 1.8.1. §

Let $G$ and $G^{'}$ be groups, $A$ a $G$ -module and $A^{'}$ a $G^{'}$ -module. We say that a pair $(ρ, λ)$ with $ρ : G^{'} \to G$ and $λ : A \to A^{'}$ group homomorphisms is compatible if

λ (ρ (g^{'}) a) = g^{'} λ (a)

for all $g^{'} \in G^{'}$ and $a \in A$ .

Compatible pairs are used to provide maps among cohomology groups.

Proposition 1.8.2. §

Suppose that $ρ : G^{'} \to G$ and $λ : A \to A^{'}$ form a compatible pair. Then the maps

C^{i} (G, A) \to C^{i} (G^{'}, A^{'}), f \mapsto λ \circ f \circ (ρ \times \dots \times ρ)

induce maps on cohomology $H^{i} (G, A) \to H^{i} (G^{'}, A^{'})$ for all $i \geq 0$ .

Proof.

One need only check that this is compatible with differentials, but this is easily done using compatibility of the pair. That is, if $f^{'}$ is the image of $f$ , then to show that

d^{i} f^{'} (g_{0}^{'}, \dots, g_{i}^{'}) = λ (d^{i} f (ρ (g_{0}^{'}), \dots, ρ (g_{i}^{'}))),

immediately reduces to showing that the first terms on both sides arising from the expression for the definition of the differential are equal. Since the pair is compatible, we have

g_{0}^{'} f^{'} (g_{1}^{'}, \dots g_{i}^{'}) = g_{0}^{'} λ (f (ρ (g_{1}^{'}), \dots, ρ (g_{i}^{'}))) = λ (ρ (g_{0}^{'}) f (ρ (g_{1}^{'}), \dots, ρ (g_{i}^{'}))),

as desired. □

Remark 1.8.3. §

Using the standard resolution, we have a homomorphism

{Hom}_{ℤ [G]} (ℤ [G^{i + 1}], A) \to {Hom}_{ℤ [G^{'}]} (ℤ [{(G^{'})}^{i + 1}], A^{'}), ψ \mapsto λ \circ ψ \circ (ρ \times \dots \times ρ)

attached to a compatible pair $(ρ, λ)$ that is compatible with the map on cochains.

Remark 1.8.4. §

Given a third group $G^{″}$ , a $G^{″}$ -module $A^{″}$ , and compatible pair $(ρ^{'}, λ^{'})$ with $ρ^{'} : G^{″} \to G^{'}$ and $λ^{'} : A^{'} \to A^{″}$ , we may speak of the composition $(ρ \circ ρ^{'}, λ^{'} \circ λ)$ , which will be a compatible pair that induces the morphism on complexes that is the composition of the morphisms arising from the pairs $(ρ, λ)$ and $(ρ^{'}, λ^{'})$ .

Example 1.8.5. §

In Shapiro’s Lemma, the inclusion map $H ↪ G$ and the evaluation at $1$ map ${CoInd}_{H}^{G} (B) \to B$ form a compatible pair inducing the isomorphisms in its statement.

We consider two of the most important examples of compatible pairs, and the maps on cohomology arising from them.

Definition 1.8.6. §

Let $H$ be a subgroup of $G$ . Let $A$ be a $G$ -module.

a.

Let $e : H ↪ G$ be the natural inclusion map. Then the maps

Res : H^{i} (G, A) \to H^{i} (H, A)

induced by the compatible pair $(e, {id}_{A})$ on cohomology are known as restriction maps.

b.

Suppose that $H$ is normal in $G$ . Let $q : G \to G ∕ H$ be the quotient map, and let $ι : A^{H} \to A$ be the inclusion map. Then the maps

Inf : H^{i} (G ∕ H, A^{H}) \to H^{i} (G, A)

induced by the compatible pair $(q, ι)$ are known as inflation maps.

Remark 1.8.7. §

Restriction of an $i$ -cocycle is just simply that, it is the restriction of the map $f : G^{i} \to A$ to a map $Res (f) : H^{i} \to A$ given by $Res (f) (h) = f (h)$ for $h \in H^{i}$ . Inflation of an $i$ -cocycle is just as simple: $Inf (f) (g) = f (\bar{g})$ , for $g \in G^{i}$ and $\bar{g}$ its image in ${(G ∕ H)}^{i}$ .

Example 1.8.8. §

In degree $0$ , the restriction map $Res : A^{G} \to A^{H}$ is simply inclusion, and the inflation map $Inf : {(A^{H})}^{G ∕ H} \to A^{G}$ is the identity.

Remarks 1.8.9. §

a.: Restriction provides a morphism of $δ$ -functors. That is, it provides a sequence of natural transformations between the functors $H^{i} (G, \cdot)$ and $H^{i} (H, \cdot)$ on $G$ -modules (which is to say that restriction commutes with $G$ -module homomorphisms) such that for any short exact sequence of $G$ -modules, the maps induced by the natural transformations commute with the connecting homomorphisms in the two resulting long exact sequences.
b.: We could merely have defined restriction for $i = 0$ and used dimension shifting to define it for all $i \geq 1$ , as follows from the previous remark.

Theorem 1.8.10 (Inflation-Restriction Sequence). §

Let $G$ be a group and $N$ a normal subgroup. Let $A$ be a $G$ -module. Then the sequence

0 \to H^{1} (G ∕ N, A^{N}) \to_{}^{Inf} H^{1} (G, A) \to_{}^{Res} H^{1} (N, A)

is exact.

Proof.

The injectivity of inflation on cocycles obvious from Remark 1.8.7. Let $f$ be a cocycle in $Z^{1} (G ∕ N, A^{N})$ . If $f (\bar{g}) = (g - 1) a$ for some $a \in A$ and all $g \in G$ , then $a \in A^{N}$ as $f (\bar{1}) = 0$ , so $Inf$ is injective. Also, note that $Res \circ Inf (f) (n) = f (\bar{n}) = 0$ for all $n \in N$ .

Let $f^{'} \in Z^{1} (G, A)$ and suppose $Res (f^{'}) = 0$ . Then there exists $a \in A$ such that $f^{'} (n) = (n - 1) a$ for all $n \in N$ . Define $k \in Z^{1} (G, A)$ by $k (g) = f^{'} (g) - (g - 1) a$ . Then $k (n) = 0$ for all $n \in N$ . We then have

k (𝑔𝑛) = 𝑔𝑘 (n) + k (g) = k (g)

for all $g \in G$ and $n \in N$ , so $k$ factors through $G ∕ N$ . Also,

k (g) = k (g \cdot g^{- 1} 𝑛𝑔) = k (𝑛𝑔) = 𝑛𝑘 (g) + k (n) = 𝑛𝑘 (g),

so $k$ has image in $A^{N}$ . Therefore, $k$ is the inflation of a cocycle in $Z^{1} (G ∕ N, A^{N})$ , proving exactness. □

In fact, under certain conditions, we have an inflation-restriction sequence on the higher cohomology groups.

Proposition 1.8.11. §

Let $G$ be a group and $N$ a normal subgroup. Let $A$ be a $G$ -module. Let $i \geq 1$ , and suppose that $H^{j} (N, A) = 0$ for all $1 \leq j \leq i - 1$ . Then the sequence

0 \to H^{i} (G ∕ N, A^{N}) \to_{}^{Inf} H^{i} (G, A) \to_{}^{Res} H^{i} (N, A)

is exact.

Proof.

Let $A^{*}$ be as in (1.7.1). By Theorem 1.8.10, we may assume that $i \geq 2$ . Since $H^{1} (N, A) = 0$ , we have an exact sequence

0 \to A^{N} \to {CoInd}^{G} {(A)}^{N} \to {(A^{*})}^{N} \to 0

(1.8.1)

in $N$ -cohomology. Moreover, noting Lemma 1.5.10, we have that

{CoInd}^{G} {(A)}^{N} ≅ {Hom}_{ℤ [N]} (ℤ [G], A^{\circ}) ≅ {Hom}_{ℤ} (ℤ [G ∕ N], A^{\circ}) ≅ {CoInd}^{G ∕ N} (A),

where $A^{\circ}$ is the abelian group $A$ with a trivial $G$ -action. Thus, the connecting homomorphism

δ^{i - 1} : H^{i - 1} (G ∕ N, {(A^{*})}^{N}) \to_{}^{\sim} H^{i} (G ∕ N, A^{N})

in the $G ∕ N$ -cohomology of (1.8.1) is an isomorphism for $i \geq 2$ .

Consider the commutative diagram

Dimension shifting for inflation and restriction. A full diagram description follows.

Diagram description: Dimension shifting for inflation and restriction

The two displayed rows are exact, and the squares commute.

Objects, listed by row and column:

Row 1, from left to right: column 1: 0; column 2: H superscript (i minus 1)(G / N,(A superscript (star)) superscript (N)); column 3: H superscript (i minus 1)(G,A superscript (star)); column 4: H superscript (i minus 1)(N,A superscript (star)).
Row 2, from left to right: column 1: 0; column 2: H superscript (i)(G / N,A superscript (N)); column 3: H superscript (i)(G,A); column 4: H superscript (i)(N,A).

Arrows and lines:

An arrow from 0 (row 1, column 1) to H superscript (i minus 1)(G / N,(A superscript (star)) superscript (N)), without a label.
An arrow from H superscript (i minus 1)(G / N,(A superscript (star)) superscript (N)) to H superscript (i minus 1)(G,A superscript (star)), labelled Inf.
An arrow from H superscript (i minus 1)(G / N,(A superscript (star)) superscript (N)) to H superscript (i)(G / N,A superscript (N)), labelled delta superscript (i minus 1).
An arrow from H superscript (i minus 1)(G,A superscript (star)) to H superscript (i minus 1)(N,A superscript (star)), labelled Res.
An arrow from H superscript (i minus 1)(G,A superscript (star)) to H superscript (i)(G,A), labelled delta superscript (i minus 1).
An arrow from H superscript (i minus 1)(N,A superscript (star)) to H superscript (i)(N,A), labelled delta superscript (i minus 1).
An arrow from 0 (row 2, column 1) to H superscript (i)(G / N,A superscript (N)), without a label.
An arrow from H superscript (i)(G / N,A superscript (N)) to H superscript (i)(G,A), labelled Inf.
An arrow from H superscript (i)(G,A) to H superscript (i)(N,A), labelled Res.

(1.8.2)

We have already seen that the leftmost vertical map in (1.8.2) is an isomorphism, and since ${CoInd}^{G} (A)$ is an coinduced $G$ -module, the central vertical map in (1.8.2) is an isomorphism. Moreover, as a coinduced $G$ -module, ${CoInd}^{G} (A)$ is also coinduced as an $N$ -module, and therefore the rightmost vertical map in (1.8.2) is also an isomorphism. Therefore, the lower row of (1.8.2) will be exact if the top row is. But the top row is exact by Theorem 1.8.10 if $i = 2$ , and by induction if $i > 2$ , noting that

H^{j - 1} (N, A^{*}) ≅ H^{j} (N, A) = 0

for all $j < i$ . □

We consider one other sort of compatible pair, which is conjugation.

Proposition 1.8.12. §

Let $A$ be a $G$ -module.

a.

Let $H$ be a subgroup of $G$ . Let $g \in G$ , and define $ρ_{g} : 𝑔𝐻 g^{- 1} \to H$ by $ρ_{g} (k) = g^{- 1} 𝑘𝑔$ for $k \in 𝑔𝐻 g^{- 1}$ . Define $λ_{g} : A \to A$ by $λ_{g} (a) = 𝑔𝑎$ . Then $(ρ_{g}, λ_{g})$ forms a compatible pair, and we denote by $g^{*}$ the resulting map

g^{*} : H^{i} (H, A) \to H^{i} (𝑔𝐻 g^{- 1}, A) .

We have $g_{1}^{*} \circ g_{2}^{*} = {(g_{1} \circ g_{2})}^{*}$ for all $g_{1}, g_{2} \in G$ .

b.

Suppose that $N$ is normal in $G$ . Then $H^{i} (N, A)$ is a $G$ -module, where $g \in G$ acts as $g^{*}$ . We refer to the above action as the conjugation action of $G$ . The conjugation action factors through the quotient $G ∕ N$ and turns $N$ -cohomology into a $δ$ -functor from the category of $G$ -modules to the category of $(G ∕ N)$ -modules.

c.

The action of conjugation commutes with restriction maps among subgroups of $G$ , which is to say that if $K \leq H \leq G$ and $g \in G$ , then the diagram

Conjugation commutes with restriction. A full diagram description follows.

Diagram description: Conjugation commutes with restriction

The structural squares and triangles displayed here commute.

Objects, listed by row and column:

Row 1, from left to right: column 1: H superscript (i)(H,A); column 2: H superscript (i)(K,A).
Row 2, from left to right: column 1: H superscript (i)(gHg superscript (minus 1),A); column 2: H superscript (i)(gKg superscript (minus 1),A).

Arrows and lines:

An arrow from H superscript (i)(H,A) to H superscript (i)(K,A), labelled Res.
An arrow from H superscript (i)(H,A) to H superscript (i)(gHg superscript (minus 1),A), labelled g superscript (star).
An arrow from H superscript (i)(K,A) to H superscript (i)(gKg superscript (minus 1),A), labelled g superscript (star).
An arrow from H superscript (i)(gHg superscript (minus 1),A) to H superscript (i)(gKg superscript (minus 1),A), labelled Res.

commutes.

Proof.

a.

First, we need check compatibility:

λ_{g} (ρ_{g} (h) a) = g \cdot g^{- 1} h𝑔𝑎 = h𝑔𝑎 = h λ_{g} (a) .

Next, we have

\begin{array}{rcl} λ_{g_{1} g_{2}} = λ_{g_{1}} \circ λ_{g_{2}} & and & ρ_{g_{1} g_{2}} = ρ_{g_{2}} \circ ρ_{g_{1}}, \end{array}

so by Remark 1.8.4, composition is as stated.

b.

Suppose that $κ : A \to B$ is a $G$ -module homomorphism. If $α \in H^{i} (N, A)$ is the class of $f \in Z^{i} (N, A)$ , then $κ^{*} \circ g^{*} (α)$ is the class of

(n_{1}, \dots, n_{i}) \mapsto κ (𝑔𝑓 (g^{- 1} n_{1} g, \dots, g^{- 1} n_{i} g)) = 𝑔𝜅 (f (g^{- 1} n_{1} g, \dots, g^{- 1} n_{i} g)),

and so has class $g^{*} \circ κ^{*} (α)$ .

Moreover, if

0 \to A \to_{}^{ι} B \to_{}^{π} C \to 0

is an exact sequence of $G$ -modules, then let

δ : H^{i} (N, C) \to H^{i + 1} (N, A)

Let $γ \in H^{i} (N, C)$ . We must show that $δ \circ g^{*} (γ) = g^{*} \circ δ (γ)$ . Since $ι$ and $π$ are $G$ -module maps, by what we showed above, we need only show that the differential on $C^{i} (N, B)$ commutes with the map $g^{\cdot}$ induced by $g$ on cochains. Let $z \in C^{i} (N, B)$ . Then

\begin{matrix} g^{\cdot} \circ d^{i} (z) (n_{0}, \dots, n_{i}) = g d^{i} (z) (g^{- 1} n_{0} g, \dots, g^{- 1} n_{i} g) \\ = n_{0} 𝑔𝑧 (g^{- 1} n_{0} g, \dots, g^{- 1} n_{i} g) + \sum_{j = 1}^{i} {(- 1)}^{j} f (g^{- 1} n_{0} g, \dots, g^{- 1} n_{j - 1} n_{j} g, \dots, g^{- 1} n_{i} g) \\ + {(- 1)}^{i + 1} f (g^{- 1} n_{0} g, \dots, g^{- 1} n_{i - 1} g) = d^{i} (g^{*} (z)) (n_{0}, \dots, n_{i}) . \end{matrix}

It only remains to show that the restriction of the action of $G$ on Tate cohomology to $N$ is trivial. This is easily computed on $H^{0}$ : for $a \in A^{N}$ and $n \in N$ , we have $n^{*} (a) = 𝑛𝑎 = a$ . In general, let $A^{*}$ be as in (1.7.1) for the group $G$ . The diagram

Conjugation and a connecting homomorphism. A full diagram description follows.

Diagram description: Conjugation and a connecting homomorphism

The structural squares and triangles displayed here commute.

Objects, listed by row and column:

Row 1, from left to right: column 1: H superscript (i)(N,A superscript (star)); column 2: H superscript (i plus 1)(N,A).
Row 2, from left to right: column 1: H superscript (i)(N,A superscript (star)); column 2: H superscript (i plus 1)(N,A).

Arrows and lines:

An arrow from H superscript (i)(N,A superscript (star)) (row 1, column 1) to H superscript (i plus 1)(N,A) (row 1, column 2), labelled delta.
An arrow from H superscript (i)(N,A superscript (star)) (row 1, column 1) to H superscript (i)(N,A superscript (star)) (row 2, column 1), labelled n superscript (star).
An arrow from H superscript (i plus 1)(N,A) (row 1, column 2) to H superscript (i plus 1)(N,A) (row 2, column 2), labelled n superscript (star).
An arrow from H superscript (i)(N,A superscript (star)) (row 2, column 1) to H superscript (i plus 1)(N,A) (row 2, column 2), labelled delta.

which commutes what we have already shown. Assuming that $n^{*}$ is the identity on $H^{i} (N, B)$ for every $G$ -module $B$ by induction (and in particular for $B = A^{*}$ ), we then have that $n^{*}$ is the identity on $H^{i + 1} (N, A)$ as well.

c.

Noting Remark 1.8.4, it suffices to check that the compositions of the compatible pairs in question are equal, which is immediate from the definitions.

□

We note the following corollary.

Corollary 1.8.13. §

The conjugation action of $G$ on $H^{i} (G, A)$ is trivial for all $i$ : that is,

g^{*} : H^{i} (G, A) \to H^{i} (G, A)

is just the identity for all $g \in G$ and $i \geq 0$ .

On homology, the analogous notion of a compatible pair is a pair $(ρ, λ)$ where $ρ : G \to G^{'}$ and $λ : A \to A^{'}$ are group homomorphisms satisfying

λ (𝑔𝑎) = ρ (g) λ (a)

(1.8.3)

for all $g \in G$ and $a \in A$ . These then provide morphisms

\tilde{ρ} \otimes λ : ℤ [G^{i + 1}] \otimes_{ℤ [G]} A \to ℤ [{(G^{'})}^{i + 1}] \otimes_{ℤ [G^{'}]} A^{'},

where $\tilde{ρ}$ is the induced map $ℤ [G^{i + 1}] \to ℤ [{(G^{'})}^{i + 1}]$ . By the homological compatibility of (1.8.3), these are seen to be compatible with the differentials, providing maps

H_{i} (G, A) \to H_{i} (G^{'}, A^{'})

for all $i \geq 0$ . As a consequence, we may make the following definition.

Definition 1.8.14. §

For $i \geq 0$ and a subgroup $H$ of $G$ , the corestriction maps

Cor : H_{i} (H, A) \to H_{i} (G, A)

are defined to be the maps induced by the compatible pair $(e, {id}_{A})$ , where $e : H \to G$ is the natural inclusion map.

Example 1.8.15. §

In degree $0$ , corestriction $Cor : A_{H} \to A_{G}$ is just the quotient map.

Definition 1.8.16. §

For $i \geq 0$ and a normal subgroup $H$ of $G$ , the coinflation maps

CoInf : H_{i} (G, A) \to H_{i} (G ∕ H, A_{H})

are defined to be the maps induces by the compatible pair $(q, π)$ , where $q : G \to G ∕ H$ and $π : A \to A_{H}$ are the quotient maps.

Remark 1.8.17. §

For a $G$ -module $A$ and any normal subgroup $H$ of $G$ , the sequence

H_{1} (H, A) \to_{}^{Cor} H_{1} (G, A) \to_{}^{CoInf} H_{1} (G ∕ H, A_{H}) \to 0 .

is exact.

Remark 1.8.18. §

For a subgroup $H$ of $G$ , the pair $(ρ_{g}^{- 1}, λ_{g})$ is a compatible pair for $H$ -homology, inducing conjugation maps

g_{*} : H_{i} (H, A) \to H_{i} (𝑔𝐻 g^{- 1}, A) .

If $H$ is a normal subgroup, then these again provide a $(G ∕ H)$ -action on $H_{i} (H, A)$ and turn $H$ -homology into a $δ$ -functor. Conjugation commutes with corestriction on subgroups of $G$ .

If $H$ is of finite index in $G$ , then we may define restriction maps on homology and corestriction maps on cohomology as well. If $G$ is finite, then we obtain restriction and corestriction maps on all Tate cohomology groups as well. Let us first explain this latter case, as it is a bit simpler. Take, for instance, restriction. We have

Res : H^{i} (G, A) \to H^{i} (H, A)

for all $i \geq 0$ , so maps on Tate cohomology groups for $i \geq 1$ . By Proposition 1.7.3, we have

{\hat{H}}^{i - 1} (G, A) ≅ {\hat{H}}^{i} (G, A_{*}),

and the same holds for $H$ -cohomology, as ${Ind}^{G} (A)$ is also an induced $H$ -module. We define

Res : {\hat{H}}^{i - 1} (G, A) \to {\hat{H}}^{i - 1} (H, A)

to make the diagram

Dimension shifting defines Tate restriction. A full diagram description follows.

Diagram description: Dimension shifting defines Tate restriction

The structural squares and triangles displayed here commute.

Objects, listed by row and column:

Row 1, from left to right: column 1: hat of (H) superscript (i minus 1)(G,A); column 2: hat of (H) superscript (i)(G,A subscript (star)).
Row 2, from left to right: column 1: hat of (H) superscript (i minus 1)(H,A); column 2: hat of (H) superscript (i)(H,A subscript (star)).

Arrows and lines:

An arrow from hat of (H) superscript (i minus 1)(G,A) to hat of (H) superscript (i)(G,A subscript (star)), labelled isomorphism symbol.
An arrow from hat of (H) superscript (i minus 1)(G,A) to hat of (H) superscript (i minus 1)(H,A), labelled Res.
An arrow from hat of (H) superscript (i)(G,A subscript (star)) to hat of (H) superscript (i)(H,A subscript (star)), labelled Res.
An arrow from hat of (H) superscript (i minus 1)(H,A) to hat of (H) superscript (i)(H,A subscript (star)), labelled isomorphism symbol.

commute.

If we wish to define restriction on homology groups when $G$ is not finite, we need to provide first a definition of restriction on $H_{0} (G, A)$ , so that we can use dimension shifting to define it for $H_{i} (G, A)$ with $i \geq 1$ . Similarly, we need a description of corestriction on $H^{0} (G, A)$ .

Definition 1.8.19. §

Suppose that $H$ is a finite index subgroup of a group $G$ and $A$ is a $G$ -module.

i.

Define

Res : H_{0} (G, A) \to H_{0} (H, A), x \mapsto \sum_{\bar{g} \in H ∖ G} g \cdot \tilde{x},

where $x \in A_{G}$ and $\tilde{x} \in A_{H}$ is any lift of it, and where $g$ denotes any coset representative of $\bar{g} \in H ∖ G$ .

ii.

Define

Cor : H^{0} (H, A) \to H^{0} (G, A), a \mapsto \sum_{\bar{g} \in G ∕ H} g \cdot a

where $a \in A^{H}$ and $g$ is as above.

Proposition 1.8.20. §

Let $G$ be a group and $H$ a subgroup of finite index. Then there are maps

\begin{array}{rcl} Res : H_{i} (G, A) \to H_{i} (H, A) & and & Cor : H^{i} (H, A) \to H^{i} (G, A) \end{array}

for all $i \geq 0$ that coincide with the maps of Definition 1.8.19 for $i = 0$ and that provide morphisms of $δ$ -functors.

Proof.

Again, we consider the case of restriction, that of corestriction being analogous. We have a commutative diagram with exact rows

Restriction on first homology via kernels. A full diagram description follows.

Diagram description: Restriction on first homology via kernels

The two displayed rows are exact, and the squares commute.

Objects, listed by row and column:

Row 1, from left to right: column 1: 0; column 2: H subscript (1)(G,A); column 3: H subscript (0)(G,A subscript (star)); column 4: H subscript (0)(G,Ind superscript (G)(A)).
Row 2, from left to right: column 1: 0; column 2: H subscript (1)(H,A); column 3: H subscript (0)(H,A subscript (star)); column 4: H subscript (0)(H,Ind superscript (G)(A)).

Arrows and lines:

An arrow from 0 (row 1, column 1) to H subscript (1)(G,A), without a label.
An arrow from H subscript (1)(G,A) to H subscript (0)(G,A subscript (star)), without a label.
A dashed arrow from H subscript (1)(G,A) to H subscript (1)(H,A), labelled Res.
An arrow from H subscript (0)(G,A subscript (star)) to H subscript (0)(H,A subscript (star)), labelled Res.
An arrow from H subscript (0)(G,A subscript (star)) to H subscript (0)(G,Ind superscript (G)(A)), without a label.
An arrow from H subscript (0)(G,Ind superscript (G)(A)) to H subscript (0)(H,Ind superscript (G)(A)), labelled Res.
An arrow from 0 (row 2, column 1) to H subscript (1)(H,A), without a label.
An arrow from H subscript (1)(H,A) to H subscript (0)(H,A subscript (star)), without a label.
An arrow from H subscript (0)(H,A subscript (star)) to H subscript (0)(H,Ind superscript (G)(A)), without a label.

which allows us to define restriction as the induced maps on kernels. For any $i \geq 2$ , we proceed as described above in the case of Tate cohomology to define restriction maps on the $i$ th homology groups.

That $Res$ gives of morphism of $δ$ -functors can be proven by induction using dimension shifting and a straightforward diagram chase and is left to the reader. □

Remark 1.8.21. §

Corestriction commutes with conjugation on the cohomology groups of subgroups of $G$ with coefficients in $G$ -modules. In the same vein, restriction commutes with conjugation on the homology of subgroups of $G$ with $G$ -module coefficients.

Corollary 1.8.22. §

Let $G$ be finite and $H$ a subgroup. The maps $Res$ and $Cor$ defined on both homology and cohomology above induce maps

\begin{array}{rcl} Res : {\hat{H}}^{i} (G, A) \to {\hat{H}}^{i} (H, A) & and & Cor : {\hat{H}}^{i} (H, A) \to {\hat{H}}^{i} (G, A) \end{array}

for all $i \in ℤ$ , and these provide morphisms of $δ$ -functors.

Proof.

The reader may check that $Res$ and $Cor$ defined in homological and cohomological degree $0$ , respectively, induce morphisms on the corresponding Tate cohomology groups. We then have left only to check the commutativity of one diagram in each case.

Suppose that

0 \to A \to_{}^{ι} B \to_{}^{π} C \to 0

is an exact sequence of $G$ -modules. For restriction, we want to check that

Restriction and the Tate connecting homomorphism. A full diagram description follows.

Diagram description: Restriction and the Tate connecting homomorphism

The structural squares and triangles displayed here commute.

Objects, listed by row and column:

Row 1, from left to right: column 1: hat of (H) superscript (minus 1)(G,C); column 2: hat of (H) superscript (0)(G,A).
Row 2, from left to right: column 1: hat of (H) superscript (minus 1)(H,C); column 2: hat of (H) superscript (0)(H,A).

Arrows and lines:

An arrow from hat of (H) superscript (minus 1)(G,C) to hat of (H) superscript (0)(G,A), labelled delta.
An arrow from hat of (H) superscript (minus 1)(G,C) to hat of (H) superscript (minus 1)(H,C), labelled Res.
An arrow from hat of (H) superscript (0)(G,A) to hat of (H) superscript (0)(H,A), labelled Res.
An arrow from hat of (H) superscript (minus 1)(H,C) to hat of (H) superscript (0)(H,A), labelled delta.

commutes. Let $c$ be in the kernel of $N_{G}$ on $C$ , and denote its image in ${\hat{H}}^{- 1} (G, C)$ by $\bar{c}$ . Choose $b \in B$ with $π (b) = c$ and considering $N_{G} b \in B^{G}$ , which is $ι (a)$ for some $a \in A^{G}$ . Then $δ (c)$ is the image of $a$ in ${\hat{H}}^{0} (G, A)$ . Then $Res (δ (\bar{c}))$ is just the image of $a$ in ${\hat{H}}^{0} (H, A)$ . On the other hand,

Res (\bar{c}) = \sum_{\bar{g} \in H ∖ G} g \tilde{c},

where $\tilde{c}$ is the image of $c$ in ${\hat{H}}^{- 1} (H, C)$ . We may lift the latter element to $\sum_{\bar{g}} 𝑔𝑐$ in the kernel of $N_{H}$ on $C$ and then to $\sum_{\bar{g}} 𝑔𝑏 \in B$ . Taking $N_{H}$ of this element gives us $N_{G} b$ , which is $ι (a)$ , and so $δ (Res (\bar{c}))$ is once again the image of $a$ in ${\hat{H}}^{0} (H, A)$ .

The case of corestriction is very similar, and hence omitted. □

The following describes an important relationship between restriction and corestriction.

Proposition 1.8.23. §

Let $G$ be a group and $H$ a subgroup of finite index. Then the maps $Cor \circ Res$ on homology, cohomology, and, when $G$ is finite, Tate cohomology, are just the multiplication by $[G : H]$ maps.

Proof.

It suffices to prove this on the zeroth homology and cohomology groups. The result then follows by dimension shifting. On cohomology we have the composite map

A^{G} \to_{}^{Res} A^{H} \to_{}^{Cor} A^{G},

where $Res$ is the natural inclusion and $Cor$ the map of Definition 1.8.19. For $a \in A^{G}$ , we have

\sum_{g \in G ∕ H} 𝑔𝑎 = [G : H] a,

as desired.

On homology, we have maps

A_{G} \to_{}^{Res} A_{H} \to_{}^{Cor} A_{G},

where $Res$ is as in Definition 1.8.19 and $Cor$ is the natural quotient map. For $x \in A_{G}$ and $\tilde{x} \in A_{H}$ lifting it, the element

\sum_{g \in H ∖ G} g \tilde{x}

has image $[G : H] x$ in $A_{G}$ , again as desired. □

Here is a useful corollary.

Corollary 1.8.24. §

Let $G_{p}$ be a Sylow $p$ -subgroup of a finite group $G$ , for a prime $p$ . Then the kernel of

Res : {\hat{H}}^{i} (G, A) \to {\hat{H}}^{i} (G_{p}, A)

has no elements of order $p$ .

Proof.

Let $α \in {\hat{H}}^{i} (G, A)$ with $p^{n} α = 0$ for some $n \geq 0$ . Then $Cor (Res (α)) = [G : G_{p}] α$ , but $[G : G_{p}]$ is prime to $p$ , hence $Cor (Res (α))$ is nonzero if $α \neq 0$ , and therefore $Res (α)$ cannot be $0$ unless $α = 0$ . □

We then obtain the following.

Corollary 1.8.25. §

Let $G$ be a finite group. For each prime $p$ , fix a Sylow $p$ -subgroup $G_{p}$ of $G$ . Fix $i \in ℤ$ , and suppose that

Res : {\hat{H}}^{i} (G, A) \to {\hat{H}}^{i} (G_{p}, A)

is trivial for all primes $p$ . Then ${\hat{H}}^{i} (G, A) = 0$ .

Proof.

The intersection of the kernels of the restriction maps over all $p$ contains no elements of $p$ -power order for any $p$ by Corollary 1.8.24. So, if all of the restriction maps are trivial, the group ${\hat{H}}^{i} (G, A)$ must be trivial. □

Finally, we remark that we have conjugation Tate cohomology, as in the cases of homology and cohomology.

Remark 1.8.26. §

Suppose that $G$ is finite and $H$ is a subgroup of $G$ . The conjugation maps on $H^{0} (H, A)$ and $H_{0} (H, A)$ induce maps on ${\hat{H}}^{0} (H, A)$ and ${\hat{H}}_{0} (H, A)$ , respectively, and so we use the conjugation maps on homology and cohomology to define maps

g^{*} : {\hat{H}}^{i} (H, A) \to {\hat{H}}^{i} (𝑔𝐻 g^{- 1}, A) .

for all $i$ . Again, these turn Tate cohomology for $H$ into a $δ$ -functor from $G$ -modules to $(G ∕ H)$ -modules when $H$ is normal in $G$ . Conjugation commutes with restriction and corestriction on subgroups of $G$ .

1.9. Cup products

We consider the following maps on the standard complex $P_{\cdot}$ :

κ_{i, j} : P_{i + j} \to P_{i} \otimes_{ℤ} P_{j}, κ_{i, j} (g_{0}, \dots, g_{i + j}) = (g_{0}, \dots, g_{i}) \otimes (g_{i}, \dots, g_{i + j}) .

That is, there is a natural map

{Hom}_{ℤ [G]} (P_{i}, A) \otimes_{ℤ} {Hom}_{ℤ [G]} (P_{j}, B) \to {Hom}_{ℤ [G]} (P_{i} \otimes_{ℤ} P_{j}, A \otimes_{ℤ} B)

defined by

φ \otimes φ^{'} \mapsto (α \otimes β \mapsto φ (α) \otimes φ^{'} (β)) .

Composing this with the map induced by precomposition with $κ_{i, j}$ gives rise to a map

{Hom}_{ℤ [G]} (P_{i}, A) \otimes_{ℤ} {Hom}_{ℤ [G]} (P_{j}, B) \to_{}^{\cup} {Hom}_{ℤ [G]} (P_{i + j}, A \otimes_{ℤ} B),

and we denote the image of $φ \otimes φ^{'}$ under this map by $φ \cup φ^{'}$ . Let us summarize this.

Definition 1.9.1. §

Let $φ \in {Hom}_{ℤ [G]} (P_{i}, A)$ and $φ^{'} \in {Hom}_{ℤ [G]} (P_{j}, B)$ . The cup product $φ \cup φ^{'} \in {Hom}_{ℤ [G]} (P_{i + j}, A \otimes_{ℤ} B)$ is defined by

(φ \cup φ^{'}) (g_{0}, \dots, g_{i + j}) = φ (g_{0}, \dots, g_{i}) \otimes φ^{'} (g_{i}, \dots, g_{i + j}) .

Lemma 1.9.2. §

Let $φ \in {Hom}_{ℤ [G]} (P_{i}, A)$ and $φ^{'} \in {Hom}_{ℤ [G]} (P_{j}, B)$ . Then

D_{A \otimes B}^{i + j} (φ \cup φ^{'}) = D_{A}^{i} (φ) \cup φ^{'} + {(- 1)}^{i} φ \cup D_{B}^{j} (φ^{'}),

where the differentials $D^{i}$ are as in (1.3.1).

Proof.

We compute the terms. We have

\begin{matrix} D_{A \otimes B}^{i + j} (φ \cup φ^{'}) (g_{0}, \dots, g_{i + j + 1}) = \sum_{k = 0}^{i} {(- 1)}^{k} φ (g_{0}, \dots, \hat{g_{k}}, \dots, g_{i + 1}) \otimes φ^{'} (g_{i + 1}, \dots, g_{i + j + 1}) \\ + \sum_{k = i + 1}^{i + j + 1} {(- 1)}^{k} φ (g_{0}, \dots, g_{i}) \otimes φ^{'} (g_{i}, \dots, \hat{g_{k}}, \dots, g_{i + j + 1}), \end{matrix}

while

(D_{A}^{i} (φ) \cup φ^{'}) (g_{0}, \dots, g_{i + j + 1}) = \sum_{k = 0}^{i + 1} {(- 1)}^{k} φ (g_{0}, \dots, \hat{g_{k}}, \dots, g_{i + 1}) \otimes φ^{'} (g_{i + 1}, \dots, g_{i + j + 1})

(1.9.1)

and

(φ \cup D_{B}^{j} (φ^{'})) (g_{0}, \dots, g_{i + j + 1}) = \sum_{k = i}^{j + i + 1} {(- 1)}^{k - i} φ (g_{0}, \dots, g_{i}) \otimes φ^{'} (g_{i}, \dots, \hat{g_{j}}, \dots, g_{i + j + 1}) .

(1.9.2)

As ${(- 1)}^{i + 1} + {(- 1)}^{i} = 0$ , the last term in (1.9.1) cancels with the ${(- 1)}^{i}$ times the first term in (1.9.2). The equality of the two sides follows. □

Remark 1.9.3. §

On cochains, we can define cup products

C^{i} (G, A) \otimes_{ℤ} C^{j} (G, B) \to_{}^{\cup} C^{i + j} (G, A \otimes_{ℤ} B)

of $f \in C^{i} (G, A)$ and $f^{'} \in C^{j} (G, B)$ by

(f \cup f^{'}) (g_{1}, g_{2}, \dots g_{i + j}) = f (g_{1}, \dots, g_{i}) \otimes g_{1} g_{2} \dots g_{i} f^{'} (g_{i + 1}, \dots, g_{i + j}) .

To see that these match up with the previous definition, note that if we define $φ$ and $φ^{'}$ by

\begin{array}{rcl} φ (1, g_{1}, \dots, g_{1} \dots g_{i}) = f (g_{1}, \dots, g_{i}) & and & φ^{'} (1, g_{1}, \dots, g_{1} \dots g_{j}) = f^{'} (g_{1}, \dots, g_{j}), \end{array}

then

\begin{array}{l} (f \cup f^{'}) (g_{1}, \dots, g_{i + j}) & = φ (1, g_{1}, \dots, g_{1} \dots g_{i}) \cup g_{1} g_{2} \dots g_{i} φ (1, g_{i + 1}, \dots, g_{i + 1} \dots g_{i + j + 1}) \\ = φ (1, g_{1}, \dots, g_{1} \dots g_{i}) \cup φ (g_{1} \dots g_{i}, g_{1} \dots g_{i + 1}, \dots, g_{1} \dots g_{i + j + 1}) \\ = (φ \cup φ^{'}) (1, g_{1}, \dots, g_{1} \dots g_{i + j + 1}), \end{array}

so the definitions agree under the identifications of Theorem 1.3.3. As a consequence of Lemma 1.9.2, the cup products on cochains satisfy

d_{A \otimes B}^{i + j} (f \cup f^{'}) = d_{A}^{i} (f) \cup f^{'} + {(- 1)}^{i} f \cup d_{B}^{j} (f^{'}) .

(1.9.3)

Lemma 1.9.4. §

The sequences

\begin{array}{rcl} 0 \to A \otimes_{ℤ} B \to {CoInd}^{G} (A) \otimes_{ℤ} B \to A^{*} \otimes_{ℤ} B \to 0 & (1.9.4) \\ 0 \to A_{*} \otimes_{ℤ} B \to {Ind}^{G} (A) \otimes_{ℤ} B \to A \otimes_{ℤ} B \to 0 & (1.9.5) \end{array}

are exact for any $G$ -modules $A$ and $B$ .

Proof.

Since the augmentation map $𝜀$ is split over $ℤ$ , it follows using Remark 1.7.1 that the sequences (1.7.1) and (1.7.2) are split as well. It follows that the sequences in the lemma are exact. □

Theorem 1.9.5. §

The cup products of Definition 1.9.1 induce maps, also called cup products,

H^{i} (G, A) \otimes_{ℤ} H^{j} (G, B) \to_{}^{\cup} H^{i + j} (G, A \otimes_{ℤ} B)

that are natural in $A$ and $B$ and satisfy the following properties:

For $i = j = 0$ , one has that the cup product
$A^{G} \otimes_{ℤ} B^{G} \to {(A \otimes_{ℤ} B)}^{G}$
is induced by the identity on $A \otimes_{ℤ} B$ .

If
$0 \to A_{1} \to A \to A_{2} \to 0$
is an exact sequence of $G$ -modules such that
$0 \to A_{1} \otimes_{ℤ} B \to A \otimes_{ℤ} B \to A_{2} \otimes_{ℤ} B \to 0$
is exact as well, then
$δ (α_{2} \cup β) = (δ α_{2}) \cup β \in H^{i + j + 1} (G, A_{1} \otimes_{ℤ} B)$
for all $α_{2} \in H^{i} (G, A_{2})$ and $β \in H^{j} (G, B)$ . (In other words, cup product on the right with a cohomology class provides a morphism of $δ$ -functors.)

If
$0 \to B_{1} \to B \to B_{2} \to 0$
is an exact sequence of $G$ -modules such that
$0 \to A \otimes_{ℤ} B_{1} \to A \otimes_{ℤ} B \to A \otimes_{ℤ} B_{2} \to 0$
is exact as well, then
$δ (α \cup β_{2}) = {(- 1)}^{i} α \cup (δ β_{2}) \in H^{i + j + 1} (G, A \otimes_{ℤ} B_{1})$
for all $α \in H^{i} (G, A)$ and $β_{2} \in H^{j} (G, B_{2})$ .

Moreover, the cup products on cohomology are the unique collection of such maps natural in $A$ and $B$ and satisfying properties (i), (ii), and (iii).

Proof.

Let $f \in C^{i} (G, A)$ and $f^{'} \in C^{j} (G, B)$ . By (1.9.3), it is easy to see that the cup product of two cocycles is a cocycle and that the cup product of a cocycle and a coboundary is a coboundary. Thus, the cup product on cochains induces cup products on cohomology. The naturality in $A$ and $B$ follows directly from the definition. Property (i) is immediate from the definition as well. Property (ii) can be seen by tracing through the definition of the connecting homomorphism. Let $f_{2}$ represent $α_{2}$ . Then $δ (α_{2})$ is obtained by lifting $f_{2}$ to a cochain $f \in C^{i} (G, A)$ , taking its boundary $𝑑𝑓 \in C^{i + 1} (G, A)$ , and then noting that $𝑑𝑓$ is the image of some cocycle $z_{1} \in Z^{i + 1} (G, A_{1})$ . Let $f^{'} \in Z^{j} (G, B)$ represent $β$ . By (1.9.3), we have $𝑑𝑓 \cup f^{'} = d (f \cup f^{'})$ . Note that $z_{1} \cup f^{'}$ has class $δ (α_{2}) \cup β$ and image $𝑑𝑓 \cup f^{'}$ in $C^{i + j + 1} (G, A \otimes_{ℤ} B)$ . On the other hand, $d (f \cup f^{'})$ is the image of a cocycle representing $δ (α_{2} \cup β)$ , as $f \cup f^{'}$ is a cocycle lifting $f_{2} \cup f^{'}$ . Since the map

C^{i + j + 1} (G, A_{1} \otimes_{ℤ} B) \to C^{i + j + 1} (G, A \otimes_{ℤ} B)

is injective, we have (ii). Property (iii) follows similarly, the sign appearing in the computation arising from (1.9.3).

The uniqueness of the maps with these properties follows from the fact that given a collection of such maps, property (i) specifies them uniquely for $i = j = 0$ , while properties (ii) and (iii) specify them uniquely for all other $i, j \geq 0$ by dimension shifting. For instance, by (ii) and Lemma 1.9.4, we have a commutative square

Dimension shifting for the cup product. A full diagram description follows.

Diagram description: Dimension shifting for the cup product

The structural squares and triangles displayed here commute.

Objects, listed by row and column:

Row 1, from left to right: column 1: H superscript (i)(G,A superscript (star)) tensor subscript (blackboard Z) H superscript (j)(G,B); column 2: H superscript (i plus j)(G,A superscript (star) tensor subscript (blackboard Z) B).
Row 2, from left to right: column 1: H superscript (i plus 1)(G,A) tensor subscript (blackboard Z) H superscript (j)(G,B); column 2: H superscript (i plus j plus 1)(G,A tensor subscript (blackboard Z) B).

Arrows and lines:

An arrow from H superscript (i)(G,A superscript (star)) tensor subscript (blackboard Z) H superscript (j)(G,B) to H superscript (i plus j)(G,A superscript (star) tensor subscript (blackboard Z) B), labelled cup.
A double-headed arrow from H superscript (i)(G,A superscript (star)) tensor subscript (blackboard Z) H superscript (j)(G,B) to H superscript (i plus 1)(G,A) tensor subscript (blackboard Z) H superscript (j)(G,B), without a label.
An arrow from H superscript (i plus j)(G,A superscript (star) tensor subscript (blackboard Z) B) to H superscript (i plus j plus 1)(G,A tensor subscript (blackboard Z) B), without a label.
An arrow from H superscript (i plus 1)(G,A) tensor subscript (blackboard Z) H superscript (j)(G,B) to H superscript (i plus j plus 1)(G,A tensor subscript (blackboard Z) B), labelled cup.

in which the lefthand vertical arrow is a surjection for all $i$ (and an isomorphism for $i \geq 1$ ). Thus, the cup products in degrees $(i, j)$ for $i \geq 1$ and $j \geq 0$ specify by the cup products in degrees $(i + 1, j)$ . Similarly, using (iii), we see that the cup products in degrees $(i, j)$ specify the cup products in degrees $(i, j + 1)$ . □

Remark 1.9.6. §

Associativity of tensor products and Lemma 1.5.10 tell us that

{Ind}^{G} (A \otimes_{ℤ} B) ≅ {Ind}^{G} (A) \otimes_{ℤ} B,

so in particular the latter modules is induced. This also implies that we have isomorphisms

{(A \otimes_{ℤ} B)}_{*} ≅ A_{*} \otimes_{ℤ} B .

If $G$ is finite, Proposition 1.5.4 tells us that we have

\begin{array}{rcl} {CoInd}^{G} (A \otimes_{ℤ} B) ≅ {CoInd}^{G} (A) \otimes_{ℤ} B & and & {(A \otimes_{ℤ} B)}^{*} ≅ A^{*} \otimes_{ℤ} B \end{array}

as well.

Corollary 1.9.7. §

Consider the natural isomorphism

s_{𝐴𝐵} : A \otimes_{ℤ} B \to B \otimes_{ℤ} A

given by $a \otimes b \mapsto b \otimes a$ , and the maps that it induces on cohomology. For all $α \in H^{i} (G, A)$ and $β \in H^{j} (G, B)$ , one has that

s_{𝐴𝐵}^{*} (α \cup β) = {(- 1)}^{𝑖𝑗} (β \cup α) .

Proof.

We first verify the result in the case $i = j = 0$ . For $a \in A^{G}$ and $b \in B^{G}$ , we have

s_{𝐴𝐵}^{*} (a \cup b) = s_{𝐴𝐵} (a \otimes b) = b \otimes a = b \cup a .

Suppose that we know the result for a given pair $(i - 1, j)$ . Let $α \in H^{i} (G, A)$ and $β \in H^{j} (G, B)$ . Recall that the maps $H^{i - 1} (G, A^{*}) \to H^{i} (G, A)$ are surjective for all $i \geq 1$ (and isomorphisms for $i \geq 2$ ), and write $α = δ (α^{*})$ for some $α^{*} \in H^{i - 1} (G, A^{*})$ . Since (1.9.4) is exact, we have by Theorem 1.9.5 that

\begin{matrix} s_{𝐴𝐵}^{*} (α \cup β) = s_{𝐴𝐵}^{*} (δ (α^{*}) \cup β) = s_{𝐴𝐵}^{*} (δ (α^{*} \cup β)) = δ (s_{𝐴𝐵}^{*} (α^{*} \cup β)) \\ = {(- 1)}^{(i - 1) j} δ (β \cup α^{*}) = {(- 1)}^{(i - 1) j} {(- 1)}^{j} β \cup δ (α^{*}) = {(- 1)}^{𝑖𝑗} β \cup α . \end{matrix}

Suppose next that we know the result for a given pair $(i, j - 1)$ . Let $α \in H^{i} (G, A)$ and $β \in H^{j - 1} (G, B)$ , and write $β = δ (β^{*})$ for some $β^{*} \in H^{j} (G, B^{*})$ . Since (1.9.4) is exact, we have by Theorem 1.9.5 that

\begin{matrix} s_{𝐴𝐵}^{*} (α \cup β) = s_{𝐴𝐵}^{*} (α \cup δ (β^{*})) = {(- 1)}^{i} s_{𝐴𝐵}^{*} (δ (α \cup β^{*})) = δ (s_{𝐴𝐵}^{*} (α \cup β^{*})) \\ = {(- 1)}^{i} {(- 1)}^{i (j - 1)} δ (β^{*} \cup α) = {(- 1)}^{𝑖𝑗} δ (β^{*}) \cup α = {(- 1)}^{𝑖𝑗} β \cup α . \end{matrix}

The result now follows by induction on $i$ and $j$ . □

Cup products also have an associative property, which can be checked directly on cochains.

Proposition 1.9.8. §

Let $A$ , $B$ , and $C$ be $G$ -modules, and let $α \in H^{i} (G, A)$ , $β \in H^{j} (G, B)$ , and $γ \in H^{k} (G, C)$ . Then

(α \cup β) \cup γ = α \cup (β \cup γ) \in H^{i + j + k} (G, A \otimes_{ℤ} B \otimes_{ℤ} C) .

Often, when we speak of cup products, we apply an auxiliary map from the tensor product of $A$ and $B$ to a third module before taking the result. For instance, if $A = B = ℤ$ , then one will typically make the identification $ℤ \otimes_{ℤ} ℤ≅ℤ$ . We codify this in the following definition.

Definition 1.9.9. §

Suppose that $A$ , $B$ , and $C$ are $G$ -modules and $𝜃 : A \otimes_{ℤ} B \to C$ is a $G$ -module homomorphism. Then the maps

H^{i} (G, A) \otimes_{ℤ} H^{j} (G, B) \to H^{i + j} (G, C), α \otimes β \mapsto 𝜃 (α \cup β)

are also referred to as cup products. When $𝜃$ is understood, we denote $𝜃 (α \cup β)$ more simply by $α \cup β$ .

Cup products behave nicely with respect to restriction, corestriction, and inflation.

Proposition 1.9.10. §

Let $A$ and $B$ be $G$ -modules. We then have the following compatibilities.

a.

Let $H$ be a subgroup of $G$ . For $α \in H^{i} (G, A)$ and $β \in H^{j} (G, B)$ , one has

Res (α \cup β) = Res (α) \cup Res (β) \in H^{i + j} (H, A \otimes_{ℤ} B),

where $Res$ denotes restriction from $G$ to $H$ .

b.

Let $N$ be a normal subgroup of $G$ . For $α \in H^{i} (G ∕ N, A^{N})$ and $β \in H^{j} (G ∕ N, B^{N})$ , one has

Inf (α \cup β) = Inf (α) \cup Inf (β) \in H^{i + j} (G, A \otimes_{ℤ} B),

where $Inf$ denotes inflation from $G ∕ N$ to $G$ . (Here, we implicitly use the canonical map $A^{N} \otimes_{ℤ} B^{N} \to {(A \otimes_{ℤ} B)}^{N}$ prior to taking inflation on the left.)

c.

Let $H$ be a subgroup of finite index in $G$ . For $α \in H^{i} (H, A)$ and $β \in H^{j} (G, B)$ , one has

Cor (α) \cup β = Cor (α \cup Res (β)) \in H^{i + j} (G, A \otimes_{ℤ} B),

where $Res$ denotes restriction from $G$ to $H$ and $Cor$ denotes corestriction from $H$ to $G$ .

Proof.

We can prove part a by direct computation on cocycles. That is, for $f \in Z^{i} (G, A)$ , $f^{'} \in Z^{i} (G, B)$ , and $h_{1}, \dots, h_{i + j} \in H$ , we have

\begin{matrix} Res (f \cup f^{'}) (h_{1}, \dots, h_{i + j}) = (f \cup f^{'}) (h_{1}, \dots, h_{i + j}) = f (h_{1}, \dots, h_{i}) \otimes h_{1} \dots h_{i} f^{'} (h_{i + 1}, \dots, h_{i + j}) \\ = Res (f) (h_{1} \dots, h_{i}) \otimes h_{1} \dots h_{i} Res (f^{'}) (h_{i + 1}, \dots, h_{i + j}) = (Res (f) \cup Res (f^{'})) (h_{1}, \dots, h_{i + j}) . \end{matrix}

Part b is similarly computed.

We now prove part c. Consider the case that $i = j = 0$ . Then $a \in A^{H}$ and $b \in B^{G}$ . By property (i) in Theorem 1.9.5 and the definition of corestriction on $H^{0}$ , we have

Cor (a) \cup b = \sum_{\bar{g} \in G ∕ H} (𝑔𝑎) \otimes b = \sum_{\bar{g} \in G ∕ H} (𝑔𝑎 \otimes 𝑔𝑏) = \sum_{\bar{g} \in G ∕ H} g (a \otimes b) = Cor (a \cup Res (b)) .

As corestriction and restriction commute with connecting homomorphisms, and as cup products behave well with respect to connecting homomorphisms on either side, we can use dimension shifting to prove the result for all $i$ and $j$ . That is, suppose we know the result for a fixed pair $(i - 1, j)$ . We prove it for $(i, j)$ . Letting $δ$ the connecting homomorphism induced by (1.7.1) for $A$ , and choose $α^{*} \in H^{i - 1} (G, A^{*})$ such that $δ (α^{*}) = α$ . We then have

\begin{matrix} Cor (α) \cup β = δ (Cor (α^{*})) \cup β = δ (Cor (α^{*}) \cup β) \\ = δ (Cor (α^{*} \cup Res (β)) = Cor (δ (α^{*} \cup Res (β))) = Cor (α \cup Res (β)) . \end{matrix}

Similarly, take $δ$ for the sequence analogous to (1.7.1) for the module $B$ and assume the result for $(i, j - 1)$ . Choosing $β^{*} \in H^{j - 1} (G, B^{*})$ with $δ (β^{*}) = β$ , we have

\begin{matrix} Cor (α) \cup β = Cor (α) \cup δ (β^{*}) = {(- 1)}^{i} δ (Cor (α) \cup β^{*}) = {(- 1)}^{i} δ (Cor (α \cup Res (β^{*}))) \\ = {(- 1)}^{i} Cor (δ (α \cup Res (β^{*}))) = Cor (α \cup δ (Res (β^{*}))) = Cor (α \cup Res (β)) . \end{matrix}

□

Notation 1.9.11. §

We may express the statement of Proposition 1.9.10a as saying that the diagram

Cup product and restriction. A full diagram description follows.

Diagram description: Cup product and restriction

This is the compatibility of restriction with cup product: Res of (alpha cup beta) equals Res(alpha) cup Res(beta). The two shifted downward Res arrows on the left act on the two tensor factors separately.

Objects, listed by row and column:

Row 1, from left to right: column 1: H superscript (i)(G,A) tensor subscript (blackboard Z) H superscript (j)(G,B); column 2: H superscript (i)(G,A tensor subscript (blackboard Z) B).
Row 2, from left to right: column 1: H superscript (i)(H,A) tensor subscript (blackboard Z) H superscript (j)(H,B); column 2: H superscript (i)(H,A tensor subscript (blackboard Z) B).

Arrows and lines:

An arrow from H superscript (i)(G,A) tensor subscript (blackboard Z) H superscript (j)(G,B) to H superscript (i)(G,A tensor subscript (blackboard Z) B), labelled cup.
A factorwise arrow labelled Res from H superscript (j)(G,B) to H superscript (j)(H,B), acting on the B tensor factor only.
A factorwise arrow labelled Res from H superscript (i)(G,A) to H superscript (i)(H,A), acting on the A tensor factor only.
An arrow from H superscript (i)(G,A tensor subscript (blackboard Z) B) to H superscript (i)(H,A tensor subscript (blackboard Z) B), labelled Res.
An arrow from H superscript (i)(H,A) tensor subscript (blackboard Z) H superscript (j)(H,B) to H superscript (i)(H,A tensor subscript (blackboard Z) B), labelled cup.

commutes (with a similar diagram for part b) and the statement of Proposition 1.9.10c as saying that the diagram

Cup product and corestriction. A full diagram description follows.

Diagram description: Cup product and corestriction

This is the projection formula for corestriction: Cor(alpha) cup beta equals Cor(alpha cup Res(beta)). Here alpha is in H superscript i of (H,A), and beta is in H superscript j of (G,B). The shifted upward Cor arrow on the left acts on the A factor; the shifted downward Res arrow acts on the B factor. The upward Cor arrow on the right acts on the cup-product value.

Objects, listed by row and column:

Row 1, from left to right: column 1: H superscript (i)(G,A) tensor subscript (blackboard Z) H superscript (j)(G,B); column 2: H superscript (i)(G,A tensor subscript (blackboard Z) B).
Row 2, from left to right: column 1: H superscript (i)(H,A) tensor subscript (blackboard Z) H superscript (j)(H,B); column 2: H superscript (i)(H,A tensor subscript (blackboard Z) B).

Arrows and lines:

An arrow from H superscript (i)(G,A) tensor subscript (blackboard Z) H superscript (j)(G,B) to H superscript (i)(G,A tensor subscript (blackboard Z) B), labelled cup.
A factorwise arrow labelled Res from H superscript (j)(G,B) to H superscript (j)(H,B), acting on the B tensor factor only.
A factorwise arrow labelled Cor from H superscript (i)(H,A) to H superscript (i)(G,A), acting on the A tensor factor only.
An arrow from H superscript (i)(H,A) tensor subscript (blackboard Z) H superscript (j)(H,B) to H superscript (i)(H,A tensor subscript (blackboard Z) B), labelled cup.
An arrow from H superscript (i)(H,A tensor subscript (blackboard Z) B) to H superscript (i)(G,A tensor subscript (blackboard Z) B), labelled Cor.

commutes.

For finite groups, we have cup products on Tate cohomology as well.

Theorem 1.9.12. §

Let $G$ be finite. There exists a unique family of maps

{\hat{H}}^{i} (G, A) \otimes_{ℤ} {\hat{H}}^{j} (G, B) \to_{}^{\cup} {\hat{H}}^{i + j} (G, A \otimes_{ℤ} B)

with $i, j \in ℤ$ that are natural in the $G$ -modules $A$ and $B$ and which satisfy the following properties:

The diagram
Diagram description: Ordinary and Tate cup products in degree zero
The structural squares and triangles displayed here commute.
Objects, listed by row and column:
- Row 1, from left to right: column 1: H superscript (0)(G,A) tensor subscript (blackboard Z) H superscript (0)(G,B); column 2: H superscript (0)(G,A tensor subscript (blackboard Z) B).
- Row 2, from left to right: column 1: hat of (H) superscript (0)(G,A) tensor subscript (blackboard Z) hat of (H) superscript (0)(G,B); column 2: hat of (H) superscript (0)(G,A tensor subscript (blackboard Z) B).
Arrows and lines:
1. An arrow from H superscript (0)(G,A) tensor subscript (blackboard Z) H superscript (0)(G,B) to H superscript (0)(G,A tensor subscript (blackboard Z) B), labelled cup.
2. An arrow from H superscript (0)(G,A) tensor subscript (blackboard Z) H superscript (0)(G,B) to hat of (H) superscript (0)(G,A) tensor subscript (blackboard Z) hat of (H) superscript (0)(G,B), without a label.
3. An arrow from H superscript (0)(G,A tensor subscript (blackboard Z) B) to hat of (H) superscript (0)(G,A tensor subscript (blackboard Z) B), without a label.
4. An arrow from hat of (H) superscript (0)(G,A) tensor subscript (blackboard Z) hat of (H) superscript (0)(G,B) to hat of (H) superscript (0)(G,A tensor subscript (blackboard Z) B), labelled cup.
commutes.
If
$0 \to A_{1} \to A \to A_{2} \to 0$
is an exact sequence of $G$ -modules such that
$0 \to A_{1} \otimes_{ℤ} B \to A \otimes_{ℤ} B \to A_{2} \otimes_{ℤ} B \to 0$
is exact as well, then
$δ (α_{2} \cup β) = (δ α_{2}) \cup β \in {\hat{H}}^{i + j + 1} (G, A_{1} \otimes_{ℤ} B)$
for all $α_{2} \in {\hat{H}}^{i} (G, A_{2})$ and $β \in {\hat{H}}^{j} (G, B)$ .
If
$0 \to B_{1} \to B \to B_{2} \to 0$
is an exact sequence of $G$ -modules such that
$0 \to A \otimes_{ℤ} B_{1} \to A \otimes_{ℤ} B \to A \otimes_{ℤ} B_{2} \to 0$
is exact as well, then
$δ (α \cup β_{2}) = {(- 1)}^{i} α \cup (δ β_{2}) \in {\hat{H}}^{i + j + 1} (G, A \otimes_{ℤ} B_{1})$
for all $α \in {\hat{H}}^{i} (G, A)$ and $β_{2} \in {\hat{H}}^{j} (G, B_{2})$ .

Proof.

Consider the complex $Q_{\cdot}$ of Theorem 1.6.10, obtained from the standard resolution $P_{\cdot}$ . The proof goes through as in Theorem 1.9.5 once we define maps $Q_{i + j} \to Q_{i} \otimes_{ℤ} Q_{j}$ satisfying the formula of Lemma 1.9.2. There are six cases to consider (the case $i, j \geq 0$ being as before), and these are omitted. □

Remark 1.9.13. §

Corollary 1.9.7, Proposition 1.9.8, and Proposition 1.9.10 all hold for cup products on Tate cohomology as well. We can also compose cup products with $G$ -module maps from the tensor product, and we again denote them using the same symbol, as in Definition 1.9.9.

1.10. Tate cohomology of cyclic groups

In this section, let $G$ be a cyclic group of finite order. We prove that the Tate cohomology groups with coefficients in a module $A$ are periodic in the degree of period $2$ , up to isomorphisms determined by a choice of generator $g$ of $G$ .

The first thing that we will observe is that for such a group $G$ , there is an even nicer projective resolution of $ℤ$ than the standard one: i.e., consider the sequence

\dots \to ℤ [G] \to_{}^{N_{G}} ℤ [G] \to_{}^{g - 1} ℤ [G] \to_{}^{N_{G}} ℤ [G] \to_{}^{g - 1} ℤ [G] \to_{}^{𝜀} ℤ \to 0,

(1.10.1)

where the boundary maps are multiplication $N_{G}$ in even degree and by $g - 1$ in odd degree. We can splice this together with its dual as in Theorem 1.6.10.

Proposition 1.10.1. §

The $G$ -cohomology groups of $A$ are the cohomology groups of the complex

\dots \to A \to_{}^{g - 1} A \to_{}^{N_{G}} A \to_{}^{g - 1} A \to \dots,

with a map $g - 1$ following the term in degree $0$ .

Proof.

Note first that for $i \in {- 1, 0}$ , the group ${\hat{H}}^{i} (G, A)$ is by definition isomorphic to the $i$ th cohomology group of the complex in question. Let $C_{\cdot}$ denote the projective resolution of $ℤ$ given by (1.10.1). The complex $C_{\cdot} \otimes_{ℤ [G]} A$ that ends

\dots \to A \to_{}^{g - 1} A \to_{}^{N_{G}} A \to_{}^{g - 1} A \to 0

computes $H_{i} (G, A)$ , yielding the result for $i \leq - 2$ .

Multiplication by $g$ induces the endomorphism

{Hom}_{ℤ [G]} (ℤ [G], A) \to {Hom}_{ℤ [G]} (ℤ [G], A), φ \mapsto (x \mapsto φ (𝑔𝑥) = 𝑔𝜑 (x)) .

Via the isomorphism of $G$ -modules ${Hom}_{ℤ [G]} (ℤ [G], A) \to_{}^{\sim} A$ given by evaluation at $1$ , the latter endomorphism is identified with multiplication by $g$ on $A$ . The complex ${Hom}_{ℤ [G]} (C_{\cdot}, A)$ that computes $H^{i} (G, A)$ is therefore isomorphic to

0 \to A \to_{}^{g - 1} A \to_{}^{N_{G}} A \to_{}^{g - 1} A \to \dots,

providing the result for $i \geq 1$ . □

Corollary 1.10.2. §

For any $i \in ℤ$ , we have

{\hat{H}}^{i} (G, A) ≅ {\begin{matrix} {\hat{H}}^{0} (G, A) & i even \\ {\hat{H}}^{- 1} (G, A) & i odd. \end{matrix}

We show that, in fact, these isomorphisms can be realized by means of a cup product. As usual, consider $ℤ$ as having a trivial $G$ -action. We remark that

{\hat{H}}^{- 2} (G, ℤ) ≅ H_{1} (G, ℤ) ≅ G^{ab} ≅𝐺

by Proposition 1.4.5. Any choice of generator $g$ of $G$ is now a generator $u_{g}$ of this Tate cohomology group. Note that $ℤ \otimes_{ℤ} 𝐴≅𝐴$ for any $G$ -module $A$ via multiplication. Here then is the result.

Proposition 1.10.3. §

Let $G$ be cyclic with generator $g$ , and let $A$ be a $G$ -module. Then the map

{\hat{H}}^{i} (G, A) \to {\hat{H}}^{i - 2} (G, A), c \mapsto u_{g} \cup c

is an isomorphism for any $i \in ℤ$ .

Proof.

Consider the two exact sequences of $G$ -modules:

\begin{array}{rcl} 0 \to I_{G} \to ℤ [G] \to_{}^{𝜀} ℤ \to 0 & and & 0 \to ℤ \to_{}^{N_{G}} ℤ [G] \to_{}^{g - 1} I_{G} \to 0 . \end{array}

As ${\hat{H}}^{i} (G, ℤ [G]) = 0$ for all $i \in ℤ$ , we then have two isomorphisms

{\hat{H}}^{- 2} (G, ℤ) \to_{}^{δ} {\hat{H}}^{- 1} (G, I_{G}) \to_{}^{δ} {\hat{H}}^{0} (G, ℤ) .

(In fact, tracing it through, one sees that the image of $u_{g}$ under this composition is $1$ modulo $| G |$ . This is not needed for the proof.)

Since we have

δ (δ (u_{g})) \cup c = δ (δ (u_{g} \cup c))

by property (ii) of Theorem 1.9.12, it suffices to show that cup product with the image of $1$ in ${\hat{H}}^{0} (G, ℤ)$ is an isomorphism. For this, using property (iii) of Theorem 1.9.12 to dimension shift, the problem reduces to the case that $i = 0$ . In this case, we know that the cup product is induced on ${\hat{H}}^{0}$ by the multiplication map on $H^{0}$ ’s:

ℤ \otimes_{ℤ} A^{G} \to A^{G}, m \otimes a \mapsto 𝑚𝑎 .

However, $1 \cdot a = a$ , so the map ${\hat{H}}^{0} (G, A) \to {\hat{H}}^{0} (G, A)$ induced by taking cup product with the image of $1$ is an isomorphism. □

Given the $2$ -periodicity of the Tate cohomology groups of a finite cyclic group, we can make the following definition.

Definition 1.10.4. §

Let $G$ be a finite cyclic group and $A$ a $G$ -module. Set $h_{0} (A) = | {\hat{H}}^{0} (G, A) |$ and $h_{1} (A) = | {\hat{H}}^{1} (G, A) |$ , taking them to be infinite when the orders of the Tate cohomology groups are infinite. If both $h_{0} (A)$ and $h_{1} (A)$ are finite, we then define the Herbrand quotient $h (A)$ by

h (A) = \frac{h_{0} (A)}{h_{1} (A)} .

Clearly, if $A$ is finitely generated, then $h (A)$ will be defined. The following explains how Herbrand quotients behave with respect to modules in short exact sequences.

Theorem 1.10.5. §

Let

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

be an exact sequence of $G$ -modules. Suppose that any two of $h (A)$ , $h (B)$ , and $h (C)$ are defined. Then the third is as well, and

h (B) = h (A) \cdot h (C) .

Proof.

It follows immediately from Proposition 1.10.3 that we have an exact hexagon

The exact Tate-cohomology hexagon. A full diagram description follows.

Diagram description: The exact Tate-cohomology hexagon

This is an exact cyclic sequence. Following the arrows gives H hat superscript 0 of (G,A), H hat superscript 0 of (G,B), H hat superscript 0 of (G,C), H hat superscript 1 of (G,A), H hat superscript 1 of (G,B), H hat superscript 1 of (G,C), and back to the first term. It is exact at every term, rather than a commutative hexagon.

Objects, listed by row and column:

Row 1, from left to right: column 2: hat of (H) superscript (0)(G,A); column 3: hat of (H) superscript (0)(G,B).
Row 2, from left to right: column 1: hat of (H) superscript (1)(G,C); column 4: hat of (H) superscript (0)(G,C).
Row 3, from left to right: column 2: hat of (H) superscript (1)(G,B); column 3: hat of (H) superscript (1)(G,A).

Arrows and lines:

An arrow from hat of (H) superscript (0)(G,A) to hat of (H) superscript (0)(G,B), without a label.
An arrow from hat of (H) superscript (0)(G,B) to hat of (H) superscript (0)(G,C), without a label.
An arrow from hat of (H) superscript (1)(G,C) to hat of (H) superscript (0)(G,A), without a label.
An arrow from hat of (H) superscript (0)(G,C) to hat of (H) superscript (1)(G,A), without a label.
An arrow from hat of (H) superscript (1)(G,B) to hat of (H) superscript (1)(G,C), without a label.
An arrow from hat of (H) superscript (1)(G,A) to hat of (H) superscript (1)(G,B), without a label.

Note that the order of any group in the hexagon is the product of the orders of the image of the map from the previous group and the order of the image of the map to the next group. Therefore, that any two of $h (A)$ , $h (B)$ , and $h (C)$ are finite implies the third is. When all three are finite, an Euler characteristic argument then tells us that

\frac{h_{0} (A) \cdot h_{0} (C) \cdot h_{1} (B)}{h_{0} (B) \cdot h_{1} (A) \cdot h_{1} (C)} = 1,

(1.10.2)

hence the result. More specifically, the order of each cohomology group is the product of the orders of the images of two adjacent maps in the hexagon, and the order of the image of each such map then appears once in each of the numerator and denominator of the left-hand side of (1.10.2). □

As an immediate consequence of Theorem 1.10.5, we have the following.

Corollary 1.10.6. §

Suppose that

0 \to A_{1} \to A_{2} \to \dots \to A_{n} \to 0

is an exact sequence of $G$ -modules with $h (A_{k})$ finite for at least one of each consecutive pair of subscripts $k$ , including at least one of $A_{n}$ and $A_{1}$ . Then all $h (A_{k})$ are finite and

\prod_{k = 1}^{n} h {(A_{k})}^{{(- 1)}^{k}} = 1 .

Next, we show that the Herbrand quotients of finite modules are trivial.

Proposition 1.10.7. §

Suppose that $A$ is a finite $G$ -module. Then $h (A) = 1$ .

Proof.

Let $g$ be a generator of $G$ , and note that the sequence

0 \to A^{G} \to A \to_{}^{g - 1} A \to A_{G} \to 0

is exact. As $A$ is finite and any alternating product of orders of finite groups in exact sequences of finite length is $1$ , we therefore have $| A^{G} | = | A_{G} |$ . On the other hand, we have the exact sequence

0 \to {\hat{H}}^{- 1} (G, A) \to A_{G} \to_{}^{Ñ_{G}} A^{G} \to {\hat{H}}^{0} (G, A) \to 0

defining ${\hat{H}}^{i} (G, A)$ for $i = 0, - 1$ . As $h_{1} (A) = | {\hat{H}}^{- 1} (G, A) |$ , we therefore have $h (A) = 1$ . □

We therefore have the following.

Proposition 1.10.8. §

Let $f : A \to B$ be a $G$ -module homomorphism with finite kernel and cokernel. Then $h (A) = h (B)$ if either one is defined.

Proof.

This follows immediately from the exact sequence

0 \to \ker f \to A \to B \to coker f \to 0,

Corollary 1.10.6, and Proposition 1.10.7. □

1.11. Cohomological triviality

In this section, we suppose that $G$ is a finite group.

Definition 1.11.1. §

A $G$ -module $A$ is said to be cohomologically trivial if ${\hat{H}}^{i} (H, A) = 0$ for all subgroups $H$ of $G$ and all $i \in ℤ$ .

In this section, we will give conditions for a $G$ -module to be cohomologically trivial.

Remark 1.11.2. §

Every free $G$ -module is also a free $H$ -module for every subgroup $H$ of $G$ and any group $G$ , not necessarily finite. In particular, $ℤ [G]$ is free over $ℤ [H]$ on any set of cosets representatives of $H ∖ G$ .

We remark that it follows from this that induced $G$ -modules are induced $H$ -modules, as direct sums commute with tensor products. We then have the following examples of cohomologically trivial modules.

Examples 1.11.3. §

a.

Induced $G$ -modules are cohomologically trivial by Proposition 1.6.7.

b.

Projective $G$ -modules are cohomologically trivial. To see this, suppose that $P$ and $Q$ are projective $G$ -modules with $P \oplus Q$ free over $ℤ [G]$ , hence over $ℤ [H]$ . Then

{\hat{H}}^{i} (H, P) ↪ {\hat{H}}^{i} (H, P) \oplus {\hat{H}}^{i} (H, Q) ≅ {\hat{H}}^{i} (H, P \oplus Q) = 0

for all $i \in ℤ$ .

We need some preliminary lemmas. Fix a prime $p$ .

Lemma 1.11.4. §

Suppose that $G$ is a $p$ -group and that $A$ is a $G$ -module of exponent dividing $p$ . Then $A = 0$ if and only if $A_{G} = 0$ and if and only if $A^{G} = 0$ .

Proof.

Suppose $A^{G} = 0$ , and let $a \in A$ . The submodule $B$ of $A$ generated by $a$ is finite, and $B^{G} = 0$ . The latter fact implies that the $G$ -orbits in $B$ are either ${0}$ or have order a multiple of $p$ . Since $B$ has $p$ -power order, this forces the order to be $1$ , so $B = 0$ . Since $a$ was arbitrary, $A = 0$ . On the other hand, if $A_{G} = 0$ , then $X = {Hom}_{ℤ} (A, 𝔽_{p})$ satisfies $𝑝𝑋 = 0$ and

X^{G} = {Hom}_{ℤ [G]} (A, 𝔽_{p}) = {Hom}_{ℤ [G]} (A_{G}, 𝔽_{p}) = 0 .

By the invariants case just proven, we know $X = 0$ , so $A = 0$ . □

Lemma 1.11.5. §

Suppose that $G$ is a $p$ -group and that $A$ is a $G$ -module of exponent dividing $p$ . If $H_{1} (G, A) = 0$ , then $A$ is free as an $𝔽_{p} [G]$ -module.

Proof.

Lift an $𝔽_{p}$ -basis of $A_{G}$ to a subset $Σ$ of $A$ . For the $G$ -submodule $B$ of $A$ generated by $Σ$ , the quotient $A ∕ B$ has trivial $G$ -coinvariant group, hence is trivial by Lemma 1.11.4. That is, $Σ$ generates $A$ as an $𝔽_{p} [G]$ -module. Letting $F$ be the free $𝔽_{p} [G]$ -module generated by $Σ$ , we then have a canonical surjection $π : F \to A$ , and we let $R$ be the kernel. Consider the exact sequence

0 \to R_{G} \to F_{G} \to_{}^{\bar{π}} A_{G} \to 0

that exists since $H_{1} (G, A) = 0$ . We have by definition that the map $\bar{π}$ induced by $π$ is an isomorphism, so we must have $R_{G} = 0$ . As $𝑝𝑅 = 0$ , we have by Lemma 1.11.4 that $R = 0$ , and so $π$ is an isomorphism. □

We are now ready to give a module-theoretic characterization of cohomologically trivial modules that are killed by $p$ .

Proposition 1.11.6. §

Suppose that $G$ is a $p$ -group and that $A$ is a $G$ -module of exponent dividing $p$ . The following are equivalent:

$A$ is cohomologically trivial
$A$ is a free $𝔽_{p} [G]$ -module.
There exists $i \in ℤ$ such that ${\hat{H}}^{i} (G, A) = 0$ .

Proof.

Immediate.
This is a special case of Lemma 1.11.5, since $H_{1} (G, A) ≅ {\hat{H}}^{- 2} (G, A)$ .
Suppose $A$ is free over $𝔽_{p} [G]$ on a generating set $I$ . Then
$𝐴≅ℤ [G] \otimes_{ℤ} ⨁_{i \in I} 𝔽_{p},$
so $A$ is induced, hence cohomologically trivial.
We note that the modules $A_{*}$ and $A^{*}$ that we use to dimension shift, as in (1.7.2) and (1.7.1) are killed by $p$ since $A$ is. In particular, it follows by dimension shifting that there exists a $G$ -module $B$ such that $𝑝𝐵 = 0$ and
${\hat{H}}^{j - 2} (H, B) ≅ {\hat{H}}^{j + i} (H, A)$
for all $H \leq G$ and $j \in ℤ$ . In particular, $H_{1} (G, B) = {\hat{H}}^{- 2} (G, B)$ is trivial. By Lemma 1.11.5, $B$ is $𝔽_{p} [G]$ -free. However, we have just shown that this implies that $B$ is cohomologically trivial, and therefore so is $A$ .

□

We next consider the case that $A$ has no elements of order $p$ .

Proposition 1.11.7. §

Suppose that $G$ is a $p$ -group and $A$ is a $G$ -module with no elements of order $p$ . The following are equivalent:

$A$ is cohomologically trivial.
There exists $i \in ℤ$ such that ${\hat{H}}^{i} (G, A) = {\hat{H}}^{i + 1} (G, A) = 0$ .
$A ∕ 𝑝𝐴$ is free over $𝔽_{p} [G]$ .

Proof.

Immediate.
Since $A$ has no $p$ -torsion,
$0 \to A \to_{}^{p} A \to A ∕ 𝑝𝐴 \to 0$
is exact. By (ii) and the long exact sequence in Tate cohomology, we have ${\hat{H}}^{i} (G, A ∕ 𝑝𝐴) = 0$ . By Proposition 1.11.6, we have therefore that $A ∕ 𝑝𝐴$ is free over $𝔽_{p} [G]$ .
By Proposition 1.11.6, we have that $A ∕ 𝑝𝐴$ is cohomologically trivial, and therefore multiplication by $p$ is an isomorphism on each ${\hat{H}}^{i} (H, A)$ for each subgroup $H$ of $G$ and every $i \in ℤ$ . However, the latter cohomology groups are annihilated by the order of $H$ , so must be trivial since $H$ is a $p$ -group.

□

We next wish to generalize to arbitrary finite groups.

Proposition 1.11.8. §

Let $G$ be a finite group and, for each $p$ , choose a Sylow $p$ -subgroup $G_{p}$ of $G$ . Let $A$ be a $G$ -module. Then $A$ is cohomologically trivial if and only if $A$ is cohomologically trivial as a $G_{p}$ -module for each $p$ .

Proof.

Suppose that $A$ is cohomologically trivial for all $G_{p}$ . Let $H$ be a subgroup of $G$ . Any Sylow $p$ -subgroup $H_{p}$ of $H$ is contained in a conjugate of $G_{p}$ , say $g G_{p} g^{- 1}$ . By the cohomological triviality of $G_{p}$ , we have that ${\hat{H}}^{i} (g^{- 1} H_{p} g, A) = 0$ . As $g^{*}$ is an isomorphism, we have that ${\hat{H}}^{i} (H_{p}, A) = 0$ . Therefore, we see that the restriction map $Res : {\hat{H}}^{i} (H, A) \to {\hat{H}}^{i} (H_{p}, A)$ is $0$ . Since this holds for each $p$ , Corollary 1.8.25 implies that ${\hat{H}}^{i} (H, A) = 0$ . □

In order to give a characterization of cohomologically trivial modules in terms of projective modules, we require the following lemma.

Lemma 1.11.9. §

Suppose that $G$ is a $p$ -group and $A$ is a $G$ -module that is free as an abelian group and cohomologically trivial. For any $G$ -module $B$ which is $p$ -torsion free, we have that ${Hom}_{ℤ} (A, B)$ is cohomologically trivial.

Proof.

Since $B$ has no $p$ -torsion and $A$ is free over $ℤ$ , we have that

0 \to {Hom}_{ℤ} (A, B) \to_{}^{p} {Hom}_{ℤ} (A, B) \to {Hom}_{ℤ} (A, B ∕ 𝑝𝐵) \to 0

is exact. In particular, ${Hom}_{ℤ} (A, B)$ has no $p$ -torsion, and

{Hom}_{ℤ} (A ∕ 𝑝𝐴, B ∕ 𝑝𝐵) ≅ {Hom}_{ℤ} (A, B ∕ 𝑝𝐵) ≅ {Hom}_{ℤ} (A, B) ∕ p {Hom}_{ℤ} (A, B) .

Since $A ∕ 𝑝𝐴$ is free over $𝔽_{p} [G]$ with some indexing set that we shall call $I$ , we have

{Hom}_{ℤ} (A ∕ 𝑝𝐴, B ∕ 𝑝𝐵) ≅ \prod_{i \in I} {Hom}_{ℤ} (𝔽_{p} [G], B ∕ 𝑝𝐵) ≅ {Hom}_{ℤ} (ℤ [G], \prod_{i \in I} B ∕ 𝑝𝐵),

so ${Hom}_{ℤ} (A, B ∕ 𝑝𝐵)$ is coinduced, and therefore $𝔽_{p} [G]$ -free. By Proposition 1.11.7, we have that ${Hom}_{ℤ} (A, B)$ is cohomologically trivial. □

Proposition 1.11.10. §

Let $G$ be a finite group and $A$ a $G$ -module that is free as an abelian group. Then $A$ is cohomologically trivial if and only if $A$ is a projective $G$ -module.

Proof.

We have already seen that projective implies cohomologically trivial. Suppose that $A$ is cohomologically trivial as a $G$ -module. Since $A$ is $ℤ$ -free, it follows that ${Ind}^{G} (A)$ is a free $G$ -module, and the sequence

0 \to {Hom}_{ℤ} (A, A_{*}) \to {Hom}_{ℤ} (A, {Ind}^{G} (A)) \to {Hom}_{ℤ} (A, A) \to 0

(1.11.1)

is exact. Moreover, $A_{*}$ is rather clearly $ℤ$ -free since $A$ is, so it follows from Lemma 1.11.9 that the module ${Hom}_{ℤ} (A, A_{*})$ is cohomologically trivial. In particular, by the long exact sequence in cohomology attached to (1.11.1), we see that

{Hom}_{ℤ [G]} (A, {Ind}^{G} (A)) \to {Hom}_{ℤ [G]} (A, A)

is surjective. In particular, the identity map lifts to a homomorphism $A \to {Ind}^{G} (A)$ , which is a splitting of the natural surjection ${Ind}^{G} (A) \to A$ . It follows that $A$ is projective as a $G$ -module. □

Finally, we consider the general case.

Theorem 1.11.11. §

Let $G$ be a finite group and $A$ a $G$ -module. The following are equivalent.

$A$ is cohomologically trivial.
For each prime $p$ , there exists some $i \in ℤ$ such that ${\hat{H}}^{i} (G_{p}, A) = {\hat{H}}^{i + 1} (G_{p}, A) = 0$ .
There is an exact sequence of $G$ -modules
$0 \to P_{1} \to P_{0} \to A \to 0$
in which $P_{0}$ and $P_{1}$ are projective.

Proof.

This follows from the definition of cohomologically trivial.
Let $F$ be a free $G$ -module that surjects onto $A$ , and let $R$ be the kernel. As $F$ is cohomologically trivial, we have ${\hat{H}}^{j - 1} (G_{p}, A) ≅ {\hat{H}}^{j} (G_{p}, R)$ for every $j \in ℤ$ . It follows that ${\hat{H}}^{j} (G_{p}, R)$ vanishes for two consecutive values of $j$ . Since $R$ is $ℤ$ -free, being a subgroup of $F$ , we have by Propositions 1.11.7, 1.11.8, and 1.11.10 that $R$ is projective.
This follows from the fact that projective modules are cohomologically trivial and the long exact sequence in Tate cohomology.

□

1.12. Tate’s theorem

We continue to assume that $G$ is a finite group, and we choose a Sylow $p$ -subgroup $G_{p}$ of $G$ for each prime $p$ . We begin with a consequence of our characterization of cohomologically trivial modules to maps on cohomology.

Proposition 1.12.1. §

Let $κ : A \to B$ be a $G$ -module homomorphism. Viewed as a $G_{p}$ -module homomorphism, let us denote it by $κ_{p}$ . Suppose that, for each prime $p$ , there exists a $j \in ℤ$ such that

κ_{p}^{*} : {\hat{H}}^{i} (G_{p}, A) \to {\hat{H}}^{i} (G_{p}, B)

is surjective for $i = j - 1$ , an isomorphism for $i = j$ , and injective for $i = j + 1$ . Then

κ^{*} : {\hat{H}}^{i} (H, A) \to {\hat{H}}^{i} (H, B)

is an isomorphism for all $i \in ℤ$ and subgroups $H$ of $G$ .

Proof.

Consider the canonical injection of $G$ -modules

κ \oplus ι : A \to B \oplus {CoInd}^{G} (A),

and let $C$ be its cokernel. As ${CoInd}^{G} (A)$ is $H$ -cohomologically trivial for all $H \leq G$ , we have

{\hat{H}}^{i} (H, B \oplus {CoInd}^{G} (A)) ≅ {\hat{H}}^{i} (H, B)

for all $i \in ℤ$ . The long exact sequence in $G_{p}$ -cohomology then reads

\dots \to {\hat{H}}^{i} (G_{p}, A) \to_{}^{κ_{p}^{*}} {\hat{H}}^{i} (G_{p}, B) \to {\hat{H}}^{i} (G_{p}, C) \to_{}^{δ} {\hat{H}}^{i + 1} (G_{p}, A) \to_{}^{κ_{p}^{*}} {\hat{H}}^{i + 1} (G_{p}, B) \to \dots .

Consider the case $i = j - 1$ . The map $κ_{p}^{*}$ being surjective on ${\hat{H}}^{j - 1}$ and injective on ${\hat{H}}^{j}$ implies that ${\hat{H}}^{j - 1} (G_{p}, C) = 0$ . Similarly, for $i = j$ , the map $κ_{p}^{*}$ being surjective on ${\hat{H}}^{j}$ and injective on ${\hat{H}}^{j + 1}$ implies that ${\hat{H}}^{j} (G_{p}, C) = 0$ . Therefore, $C$ is cohomologically trivial by Theorem 1.11.11, and so each map $κ^{*} : {\hat{H}}^{i} (H, A) \to {\hat{H}}^{i} (H, B)$ in question must be an isomorphism by the long exact sequence in Tate cohomology. □

We now prove the main theorem of Tate and Nakayama.

Theorem 1.12.2. §

Suppose that $A$ , $B$ , and $C$ are $G$ -modules and

𝜃 : A \otimes_{ℤ} B \to C

is a $G$ -module map. Let $k \in ℤ$ and $α \in {\hat{H}}^{k} (G, A)$ . For each subgroup $H$ of $G$ , define

Θ_{H, α}^{i} : {\hat{H}}^{i} (H, B) \to {\hat{H}}^{i + k} (H, C), Θ_{H, α}^{i} (β) = 𝜃^{*} (Res (α) \cup β) .

For each prime $p$ , suppose that there exists $j \in ℤ$ such that the map $Θ_{G_{p}, α}^{i}$ is surjective for $i = j - 1$ , an isomorphism for $i = j$ , and injective for $i = j + 1$ . Then for every subgroup $H$ of $G$ and $i \in ℤ$ , one has that $Θ_{H, α}^{i}$ is an isomorphism.

Proof.

First consider the case that $k = 0$ . Then the map $ψ : B \to C$ given by $ψ (b) = 𝜃 (a \otimes b)$ , where $a \in A^{G}$ represents $α$ , is a map of $G$ -modules, since

ψ (𝑔𝑏) = 𝜃 (a \otimes 𝑔𝑏) = 𝜃 (𝑔𝑎 \otimes 𝑔𝑏) = 𝑔𝜃 (a \otimes b) = 𝑔𝜓 (b) .

We claim that the induced maps on cohomology

ψ^{*} : {\hat{H}}^{i} (H, B) \to {\hat{H}}^{i} (H, C)

agree with the maps given by left cup product with $Res (α)$ . Given this, we have by Proposition 1.12.1 that the latter maps are all isomorphisms in the case $k = 0$ .

To see the claim, consider first the case that $i = 0$ , in which the map $ψ^{*}$ is induced by $ψ : B^{H} \to C^{H}$ . For $b \in B^{H}$ , we have $ψ (b) = 𝜃 (a \otimes b)$ , and the class of the latter term is $𝜃^{*} (Res (a) \cup b)$ by (i) of Theorem 1.9.12. For the case of arbitrary $i$ , we consider the commutative diagram

Induced modules and a module pairing. A full diagram description follows.

Diagram description: Induced modules and a module pairing

The two displayed rows are exact, and the squares commute.

Objects, listed by row and column:

Row 1, from left to right: column 1: 0; column 2: (A tensor subscript (blackboard Z) B) subscript (star); column 3: Ind superscript (G)(A tensor subscript (blackboard Z) B); column 4: A tensor subscript (blackboard Z) B; column 5: 0.
Row 2, from left to right: column 1: 0; column 2: C subscript (star); column 3: Ind superscript (G)(C); column 4: C; column 5: 0.

Arrows and lines:

An arrow from 0 (row 1, column 1) to (A tensor subscript (blackboard Z) B) subscript (star), without a label.
An arrow from (A tensor subscript (blackboard Z) B) subscript (star) to Ind superscript (G)(A tensor subscript (blackboard Z) B), without a label.
An arrow from (A tensor subscript (blackboard Z) B) subscript (star) to C subscript (star), labelled tilde of (theta).
An arrow from Ind superscript (G)(A tensor subscript (blackboard Z) B) to A tensor subscript (blackboard Z) B, without a label.
An arrow from Ind superscript (G)(A tensor subscript (blackboard Z) B) to Ind superscript (G)(C), labelled Ind superscript (G)( theta ).
An arrow from A tensor subscript (blackboard Z) B to 0 (row 1, column 5), without a label.
An arrow from A tensor subscript (blackboard Z) B to C, labelled theta.
An arrow from 0 (row 2, column 1) to C subscript (star), without a label.
An arrow from C subscript (star) to Ind superscript (G)(C), without a label.
An arrow from Ind superscript (G)(C) to C, without a label.
An arrow from C to 0 (row 2, column 5), without a label.

(1.12.1)

where ${Ind}^{G} (𝜃) = {id}_{ℤ [G]} \otimes 𝜃$ , and where $\tilde{𝜃}$ is both the map making the diagram commute and ${id}_{I_{G}} \otimes 𝜃$ , noting Remark 1.7.1. In fact, by Remark 1.9.6, we have an exact sequence isomorphic to the top row of (1.12.1), given by

0 \to A \otimes_{ℤ} B_{*} \to A \otimes_{ℤ} {Ind}^{G} (B) \to A \otimes_{ℤ} B \to 0

and then a map $\tilde{ψ} : B_{*} \to C_{*}$ given by $\tilde{ψ} (b^{'}) = \tilde{𝜃} (a \otimes b^{'})$ for $b^{'} \in B_{*}$ . We then have two commutative diagrams

Connecting maps and the induced pairing. A full diagram description follows.

Diagram description: Connecting maps and the induced pairing

The structural squares and triangles displayed here commute.

Objects, listed by row and column:

Row 1, from left to right: column 1: hat of (H) superscript (i minus 1)(H,B); column 2: hat of (H) superscript (i)(H,B subscript (star)).
Row 2, from left to right: column 1: hat of (H) superscript (i minus 1)(H,C); column 2: hat of (H) superscript (i)(H,C subscript (star)).

Arrows and lines:

An arrow from hat of (H) superscript (i minus 1)(H,B) to hat of (H) superscript (i)(H,B subscript (star)), labelled delta and isomorphism symbol.
An arrow from hat of (H) superscript (i minus 1)(H,B) to hat of (H) superscript (i minus 1)(H,C), without a label.
An arrow from hat of (H) superscript (i)(H,B subscript (star)) to hat of (H) superscript (i)(H,C subscript (star)), without a label.
An arrow from hat of (H) superscript (i minus 1)(H,C) to hat of (H) superscript (i)(H,C subscript (star)), labelled delta and isomorphism symbol.

In the first, the left vertical arrow is $Θ_{H, α}^{i - 1}$ and the right vertical arrow is the map ${\tilde{Θ}}_{H, α}^{i}$ given on $β^{'} \in {\hat{H}}^{i} (H, B_{*})$ by

{\tilde{Θ}}_{H, α}^{i} (β^{'}) = {\tilde{𝜃}}^{*} (Res (α) \cup β^{'}) .

In the second, the left vertical arrow is $ψ^{*}$ and the right is ${\tilde{ψ}}^{*}$ . Supposing our claim for $i$ , we have ${\tilde{ψ}}^{*} = {\tilde{Θ}}_{H, α}^{i}$ . As the connecting homomorphisms in the diagrams are isomorphisms, we then have that $ψ^{*} = Θ_{H, α}^{i}$ . I.e., if the claim holds for $i$ , it holds for $i - 1$ . The analogous argument using coinduced modules allows us to shift from $i$ to $i + 1$ , proving the claim for all $i \in ℤ$ , hence the theorem for $k = 0$ .

For any $k \in ℤ$ , the result is again proven by dimension shifting, this time for $A$ . Fix $α \in {\hat{H}}^{k - 1} (H, A)$ , and let $α^{'} = δ (α) \in {\hat{H}}^{k} (H, A_{*})$ . We note that the top row of (1.12.1) is also isomorphic to

0 \to A_{*} \otimes_{ℤ} B \to {Ind}^{G} (A) \otimes_{ℤ} B \to A \otimes_{ℤ} B \to 0 .

Much as before, define $Θ_{H, α^{'}}^{i} : {\hat{H}}^{i} (H, B) \to {\hat{H}}^{i + k} (H, C_{*})$ by $Θ_{H, α^{'}}^{i} (β) = {\tilde{𝜃}}^{*} (Res (α^{'}) \cup β)$ , where $\tilde{𝜃} : A_{*} \otimes_{ℤ} B \to C_{*}$ is the map determined by $𝜃$ . The diagram

Dimension shifting for the induced pairing. A full diagram description follows.

Diagram description: Dimension shifting for the induced pairing

The structural squares and triangles displayed here commute.

Objects, listed by row and column:

Row 1, from left to right: column 1: hat of (H) superscript (i)(H,B); column 2: hat of (H) superscript (i)(H,B).
Row 2, from left to right: column 1: hat of (H) superscript (i plus k minus 1)(H,C); column 2: hat of (H) superscript (i plus k)(H,C subscript (star)).

Arrows and lines:

Equality joins hat of (H) superscript (i)(H,B) (row 1, column 1) and hat of (H) superscript (i)(H,B) (row 1, column 2), without a label.
An arrow from hat of (H) superscript (i)(H,B) (row 1, column 1) to hat of (H) superscript (i plus k minus 1)(H,C), labelled capital Theta subscript (H, alpha) superscript (i).
An arrow from hat of (H) superscript (i)(H,B) (row 1, column 2) to hat of (H) superscript (i plus k)(H,C subscript (star)), labelled capital Theta subscript (H, alpha prime) superscript (i).
An arrow from hat of (H) superscript (i plus k minus 1)(H,C) to hat of (H) superscript (i plus k)(H,C subscript (star)), labelled delta and isomorphism symbol.

then commutes as

δ \circ Θ_{H, α}^{i} (β) = δ \circ 𝜃^{*} (Res (α) \cup β) = {\tilde{𝜃}}^{*} δ (Res (α) \cup β) = {\tilde{𝜃}}^{*} (Res (α^{'}) \cup β) = Θ_{H, α^{'}}^{i} (β) .

There exists by assumption $j \in ℤ$ such that the map $Θ_{G_{p}, α}^{i}$ is surjective for $i = j - 1$ , an isomorphism for $i = j$ , and injective for $i = j + 1$ . By the commutativity of the diagram, the same holds for $Θ_{G_{p}, α^{'}}^{i}$ . Assuming the theorem for $k$ , we then have that all of the maps $Θ_{H, α^{'}}^{i}$ are isomorphisms, and therefore again by commutativity that so are the maps $Θ_{H, α}^{i}$ . Thus, the theorem for a given $k$ implies the theorem for $k - 1$ . By the analogous argument using coinduced modules, the theorem for $k$ implies the theorem for $k + 1$ as well. □

The following special case was first due to Tate.

Theorem 1.12.3 (Tate). §

Let $A$ be a $G$ -module, and let $α \in H^{2} (G, A)$ . Suppose that, for every $p$ , the group $H^{1} (G_{p}, A)$ is trivial and $H^{2} (G_{p}, A)$ is a cyclic group of order $| G_{p} |$ generated by the restriction of $α$ . Then the maps

{\hat{H}}^{i} (H, ℤ) \to {\hat{H}}^{i + 2} (H, A), β \mapsto Res (α) \cup β

are isomorphisms for every $i \in ℤ$ and subgroup $H$ of $G$ .

Proof.

For $H = G_{p}$ , the maps in question are surjective for $i = - 1$ , as $H^{1} (G_{p}, A) = 0$ , and injective for $i = 1$ , as

H^{1} (G_{p}, ℤ) = Hom (G_{p}, ℤ) = 0 .

For $i = 0$ , we have

{\hat{H}}^{0} (G_{p}, ℤ) ≅ℤ ∕ | G_{p} | ℤ,

and the map takes the image of $n \in ℤ$ to $n Res (α)$ (which is straightforward enough to see by dimension shifting, starting with the known case of cup products of degree zero classes), hence is an isomorphism by the assumption on $H^{2} (G_{p}, A)$ . □