a tool used to compute invariants of group actions using methods from homology theory, such as invariants, coinvariants, extensions... Use with (homology-cohomology).

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2
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1answer
30 views

Showing that the Hopf fibration is a non-trivial fibre bundle

I want to show that the Hopf bundle $$ \mathbb{S}^1 \rightarrow \mathbb{S^3} \rightarrow \mathbb{S}^2$$ is non-trivial as a principal fibre bundle. I have seen hints of several different approaches: ...
3
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0answers
23 views

Explicit computation of $H^2(\mathbb{F}_p^n, \mathbb{R}/\mathbb{Z})$.

I'm interested in the computation of the second cohomology group of the elementary abelian group $\mathbb{F}_p^n$ with coefficients in $\mathbb{R}/\mathbb{Z}$: $$H^2(\mathbb{F}_p^n, ...
0
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1answer
24 views

First homology group of $\pi$ with $G$ when $(|\pi|,|G|)=1$

Let $\pi$ and $G$ be finite groups with a homomorphism (action) $\pi\rightarrow G$. If $|\pi|$ and $|G|$ are relatively prime, then it can be shown that $Z^1(\pi,G)=B^1(\pi,G)$ (group of $1$-co-cycles ...
1
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1answer
12 views

first homology group with coefficients in divisible group

I had (perhaps very elementary) doubt in the understanding of the computation of first homology group of a finite group over a divisible group. Let $\pi$ be a finite group of order $n$ and $D$ be a ...
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0answers
28 views

Crossed homomorphism from cyclic group

Let $\langle x\rangle$ be a cyclic group, and $N$ any group. It is easy to tell when a map $x\mapsto n $ can be extended to a homomorphism: if $o(x)$ is infinite then always; if $o(x)$ is finite then ...
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12 views

Equivalence of crossed homomorphisms: understanding

Let $\pi \times_{\varphi} G$ be semi-direct product with $G$ normal; $f_1,f_2\colon \pi \rightarrow G$ be crossed homomorphisms: $$f_i(\sigma\tau)=f_i(\sigma)^{\tau} f_i(\tau),\,\,\,\,\,\,\,\, \mbox{ ...
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1answer
17 views

Crossed homomorphism from semi-direct product: confusion in definition

(Ref: this) Let $\pi \times_{\varphi} G$ be semi-direct product in which $G$ is normal and $\pi$ is complement. Let $\omega$ be another complement of $G$ in above semi-direct product (so $\pi ...
2
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1answer
21 views

Finding a representative cocycle for a given group extension

Suppose we have a group extension $$0 \to N \stackrel{\iota}{\to} E \stackrel{\pi}{\to} G \to 1$$ where $N$ is abelian. How to find a representative 2-cocycle that produces this extension? Or more ...
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60 views

The cohomology of $\mathrm{GL}_n$ over an algebraically closed field

How does one go about computing the cohomology groups $H^*(\mathrm{GL}_{\kern{0.1em}{m}}(\overline{\mathbb{F}}_p),M)$? I am particularly interested in the case when $M$ is an algebraic representation. ...
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24 views

Invariant cohomology for non-compact groups

Suppose I have a compact $G$-space $M$, and a differential form $\omega$ on $M$ with the property that $$ \forall g\in G\quad g\omega = \omega + d\lambda_g, \quad(*) $$ i.e. $g\omega$ is cohomologous ...
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14 views

Group cohomology classical exercise: Short exact sequence induces long exact sequence

This is probably so easy that I didn't find any other questions asking this exact question. Suppose that $1\to A\to B\to C\to 1$ is an exact sequence of $G$-modules. I can then easily prove that ...
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1answer
32 views

Direct limit of a direct system looking like a cochain complex of objects.

I would like to ask you about a special kind of direct systems $ (A_i, f_{i}^{j} )_{ i,j \in ( I , \leq ) } $ looking like a cochaîn complex $ (A_i , f_{i}^{j} )_{ i,j \in ( \mathbb{N}^* , \leq ) } $ ...
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0answers
12 views

Question on subgroup cohomology restricting proper, simplicial actions of an algebraic group

I have a question regarding an assertion made in p. 2 of these notes on Bruhat-Tits buildings. The question concerns the group $G_p=SL_n(\mathbb{Q}_p)$ and its subgroup $\mathbb{Z}^{n-1}$ (the ...
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0answers
31 views

constructing a universal cover from K(G,1) space

Let G be a torsion-free group and G' a group of finite index. Suppose G' has finite cohomological dimension. Then it has a finite-dimensional K(G',1) complex. Its universal cover X' is also ...
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26 views

If $G$ is cyclic of order $2$ show $|H^1(G,\mathbb{Z})|=2$.

Let $G$ be the cyclic group of order $2$ acting by inversion on $\mathbb{Z}$. Show $|H^1(G,\mathbb{Z})|=2$. A hint is provided: if $E=\mathbb{Z} \rtimes G$ then every element in $E - \mathbb{Z}$ has ...
3
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44 views

Homomorphisms from a $p$-group to $\mathbb{F}_p$

I'm doing a problem on group cohomology and have reduced it to the following: if $P$ is a $p$-group then $\textrm{Hom}(P,\mathbb{F}_p) \simeq P/\Phi(P)$ where $\Phi(P)$ is the Frattini subgroup of ...
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0answers
23 views

Doubt about the module of coinvariants

Suppose that $G$ is group, $R$ is a ring (commutative, associative with $1\neq0$) and $M$ is a left $RG-$módulo such that $M$ is free as a left $R-$module. Is it true that the module of coinvariants ...
1
vote
1answer
33 views

Group cohomology for $\mathbb{Z}[G]$-modules versus $k[G]$-modules.

I am trying to get familiar with group (Tate) cohomology. I am for instance reading Brown's Cohomology of Groups. Now something seems unclear to me What information do we hope to attain from studying ...
0
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1answer
36 views

Coefficients in Tate cohomology, mod-p Tate cohomology vs integral Tate cohomology

I am new to group cohomology and Tate cohomology. I have some questions in that regard. I have not yet understood exactly what information we hope to gain from the (Tate) cohomology modules. i) What ...
3
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1answer
42 views

Where can I found an explanation of group cohomology from the point of view of invariants?

I heard once that we can view group cohomology as the right derived functor quantifying precisely (i.e. by the usual long exact sequence) how much the functor of "taking the invariants" is not right ...
3
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2answers
36 views

Does H2 depends only on abelian quotient?

Consider a finite group $G$ and an abelian group $N$. Let $G$ act trivially on $N$. Is $H^{2}(G,N)\cong H^{2}(G^{ab},N)$? ($G^{ab}=G/[G,G]$ the abelianization of $G$) I don't get group cohomology so ...
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63 views

Existence of class modules for finite groups

I'm reading Cohomology of Number Fields by Neukirch et al. Let $G$ be a finite group. A $G$-module C is a class module if, for all subgroups $H \subset G$: 1) $H^1(H,C)=0$ 2) $H^2(H,C)$ is cyclic of ...
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1answer
26 views

stuck on proof of inflation restriction exact sequence in group cohomology

The following proof of the inflation-restriction exact sequence is taken from Milne's notes on class field theory. My question is: why does $\phi':G/H \to M$ actually take values in $M^H$? In other ...
2
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1answer
52 views

perfect groups with non-trivial group homology over rational coefficients

A group $G$ is perfect if $G=[G,G]$. For perfect groups, we know that the first group homology $H_1(G, \mathbb{Z})=G/[G,G]=0$. A group $G$ is called acyclic if its group homology $H_i(G, ...
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48 views

Can a group have the same 3-cocycles as one of its quotients?

Let $G$ be a finite group and let $Z^3(G,U(1))$ be the set (indeed, group under pointwise multiplication) of all normalized 3-cocycles $G\times G\times G\to U(1)$, with trivial action on $U(1)$. For ...
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0answers
35 views

universal coefficient theorem for mod p cohomology

In the book Algebraic Topology, Allen Hatcher, p. 266, Corollary 3A.6 (b): Question: I want to rewrite the above statement into a cohomology version. If I replace all homologies with cohomologies, ...
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0answers
40 views

When does the first cohomology group commute with inverse limit?

Let $M_i,i\in\mathbb{N}$ be an inverse system of continous, discrete G-modules and let $M=\varprojlim M_i$. Under what conditions on $M$ and $M_i$ do we have $\varprojlim H^1(G, M_i) \cong H^1(G, M)$? ...
3
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1answer
61 views

Absolute Galois group $\text{Gal}(\overline{K}/K)$ of any number field $K$ has a non-open subgroup of any prime index $p$?

Let $K$ be a number field, and let $p$ be a prime number. Does $G = \text{Gal}(\overline{K}/K)$ necessarily have a subgroup of index $p$ that is not open?
3
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1answer
45 views

Easy computational example of first cohomology group: Is this how we do it?

I'm an undergraduate student learning a little bit of group cohomology on my own. I'd like to compute a few examples of the low-dimensional cohomology groups in some special cases to get some ...
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0answers
12 views

Notation in group cohomology context

Let $M$ be a $G$-module and let $\psi\colon G\to M$ be a map. What is usually meant with the symbol $\psi_{\sigma}$ where $\sigma\in G$ in the group cohomology context? I am using the appendix from ...
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31 views

For any closed form $a$ with compact support, there exists a form $b$ w.c.s. in the unit ball such that $a-b$ is exact.

Let $\alpha$ be a closed (differential) $k$-form with compact support in $\mathbb{R}^{n}$. We want to prove that there exists a $l$-form $\beta$ with compact support in the unit ball of ...
5
votes
1answer
60 views

Let $G$ be a group. Why is $ \operatorname{Ext}_{\mathbb{Z}G}^1(\mathbb{Z},\mathbb{Z})\cong \operatorname{Hom}_{Grp}(G,\mathbb{Z})$?

Let $G$ be a Group and $\mathbb{Z}G$ is the ring of the formal sums $$\sum_{g\in G}n_gg$$ with multiplication $$(\sum_{g\in G}n_gg)(\sum_{h\in G}m_hh)=\sum_{g\in G}(\sum_{h\in ...
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0answers
31 views

How to prove $M\otimes_{\,G}A≅ {Hom}_{G}({Hom}_{\,\mathbb Z}(M,{\,\mathbb Z}),A)$?

Given that G is a finite group, M is a finitely generated right free G-module and A is a left G-module, there exists a natural G-isomorphism $\phi\ : M\otimes_{\,G}A\rightarrow ...
0
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1answer
53 views

Cartesian product of compact triangulated spaces

Let $X$ and $Y$ two compact triangulated spaces, I am trying to show that $X\times Y$ is also a compact (this is obvious) triangulated space and $$\chi(X\times Y)=\chi(X)\cdot\chi(Y)$$ Any tips on ...
0
votes
1answer
28 views

How to prove $\rm{Hom}_{\,\mathbb Z}(\mathbb{Z}G,A)^{S}\cong\rm{Hom}_{\,\mathbb Z}(\mathbb{Z}(G/S),A)$?

Given that S is a normal subgroup of a finite group G and A is a G-module, I have difficulty in proving $\rm{Hom}_{\,\mathbb Z}(\mathbb{Z}G,A)^{S}\cong\rm{Hom}_{\,\mathbb Z}(\mathbb{Z}(G/S),A)$. Would ...
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1answer
9 views

The boundary formula and cohomology of finite groups

I've a very basic notational question on group cohomology. Let $G$ be a finite group and $M$ a $G$-module. For $i\geq 0$, let $P_i=\mathbb Z[G^{i+1}]$ be the free $\mathbb Z$-module on $G^{i+1}$, made ...
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1answer
31 views

An odd expression appearing in proof that kernel and image of certain map on group ring $\mathbb Z[G]$ are equal

Let $G = \langle g \rangle$ be a cyclic group of order $n$. Consider the free ring $\mathbb Z[G]$ of all formal sums of elements from $G$ with coefficients from $\mathbb Z$, with multiplication given ...
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18 views

Locally cyclic group and homology

I would like to ask about the accuracy of the following statement. If $G$ is locally cyclic. Then the $n$-th homology group of $G$ $($H_n(G) $)$ (specially for $n=2$, $n=3$) is also locally cyclic. ...
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0answers
35 views

Dimension of $\text{H}^n(G, KG)$

I was reading the paper "The second cohomology group of $G$ with $Z_2G$ coefficients" by Thomas Farrell. I came across the following question which was asked by Serre in 1974: Let $G$ be a ...
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1answer
48 views

Augmentation ideal and the abelianization of $G$

On a qual problem recently, I came across the following fact: If $G$ is a finite group, and $\mathfrak{a}$ is the augmentation ideal of the integral group ring $\mathbb{Z}G$, then ...
3
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0answers
53 views

Historical notes on the Jordan-Hölder program

I'm looking for any material (books, articles..) documenting the historical process of the formulation, partial work and/or the actual stage of the Jordan-Hölder program. I'm not sure if there is any ...
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1answer
61 views

Calculation of Group Cohomology of $\mathbb{Z}/2\mathbb{Z}$ over $\mathbb{Z}$

I am trying to learn some group cohomology and I'm starting to get my head around the theory, but I find it hard to find some explicit examples of the calculation of group cohomology of some small ...
10
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2answers
255 views

Which groups act freely on $S^n$?

When $n$ is even, it is easy to classify groups which act freely on $S^n$ using degree theory: if $G$ acts on $S^n$, then associating to each element $g \in G$ the degree of the map obtained from ...
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0answers
35 views

Group cohomology as a right derived functor

In this Wiki page, it says that group cohomology can be defined as right derived functor of $F$, where $F(M)=M^G$. There are two different equivalent definition in the page, by explicit cochains and ...
3
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2answers
155 views

I want to show the group cohomology $H^{n>1}\left(F,\,M\right)$ vanishes whenever $F$ is free.

I want to show the group cohomology $H^{n>1}\left(F,\,M\right)$ vanishes whenever $F$ is free. I tried to show $\text{pdim}_{\mathbb{Z}\left[F\right]}\left(\mathbb{Z}\right)\le 1$, but we know ...
0
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0answers
21 views

Open subgroups $V\subset U$ of profinite group such that the index $[U:V]$ is divisible by $p$.

Let $G$ be a profinite group of $p$-cohomological dimension, $cd_p(G)\neq 0,\infty$. Consider an open subgroup $U\leq G$. Does there exist an open subgroup $V\leq U$ such that its index, $[U:V]$ is ...
4
votes
1answer
82 views

Bourbaki's definition of semidirect product

This recess I'm off to learn about group extensions and the cohomological methods to characterize those extensions, but I'm a bit stuck on wraping my head around all the new concepts (I just finished ...
1
vote
1answer
33 views

Determining if a (Lie algebra) central extension is trivial.

Given a central extension for a given Lie algebra, is there any simple way to check that it is/isn't isomorphic to the trivial extension ("simple" meaning, not as tedious [and daunting, for an algebra ...
1
vote
1answer
30 views

Reference for a result

A friend of mine told me that the cohomology of $\pi_1(M)$ was isomorphic to the cohomology of the manifold $M$. Is that true (maybe there are some hypothesis) ? Does someone know a reference for this ...
0
votes
1answer
19 views

$G$ acts $\rm{Hom}_{\,\mathbb Z}(\mathbb{Z}G,A)$

Let $A$ be a $G$-module where the action $G$ on $A$ is trivial. Then also $G$ acts on $\rm{Hom}_{\,\mathbb Z}(\mathbb{Z}G,A)$ such that $g\cdot f(x):=f(xg)$. On the other hand, we know that ...