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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8
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1answer
458 views

Torsion-free virtually-Z is Z

It is well known that a torsion-free group which is virtually free must be free, by works of Serre, Stallings, Swan... Is there a simple cohomological proof of the fact that a torsion-free group ...
4
votes
2answers
184 views

What exactly is a trivial module?

Yes, this is a quite basic answer, but I have to admit to be absolutely confused about this notion. Searching on the web, I managed to found two possible definition of trivial modules, referring ...
1
vote
2answers
81 views

Show that the categories $G$-mod and $\mathbb{Z}G$-mod are equivalent.

I have another basic question inspired from reading the sixth chapter of Weibel's "An Introduction to Homological Algebra". First version of the question: a bit ambiguous At the first paragraph, ...
86
votes
1answer
2k views

Is there a characterization of groups with the property $\forall N\unlhd G,\:\exists H\leq G\text{ s.t. }H\cong G/N$?

A common mistake for beginning group theory students is the belief that a quotient of a group $G$ is necessarily isomorphic to a subgroup of $G$. Is there a characterization of the groups in which ...
35
votes
3answers
775 views

What does the group ring $\mathbb{Z}[G]$ of a finite group know about $G$?

The group algebra $k[G]$ of a finite group $G$ over a field $k$ knows little about $G$ most of the time; if $k$ has characteristic prime to $|G|$ and contains every $|G|^{th}$ root of unity, then ...
10
votes
4answers
703 views

What do higher cohomologies mean concretely (in various cohomology theories)?

Superficially I think I understand the definitions of several cohomologies: (1) de Rham cohomology on smooth manifolds (I understand this can be probably extended to algebraic settings, but I haven't ...
15
votes
1answer
646 views

Group cohomology versus deRham cohomology with twisted coefficients

Let $G$ be a simple simply-connected Lie group, let $M$ be a 3-manifold and $P \to M$ a principal $G$-bundle. Let $A$ be a flat connection in this bundle, and let $\text{Ad} P$ be the associated ...
5
votes
1answer
78 views

Show that image of $res$ lies in $H^n(H,A)^{G/H}$

Let $G$ and $G^{\prime}$ be groups, $A$ and $A^{\prime}$ be $G$-module and $G^{\prime}$-module respectively, $C^n(G,A)$ be set of all maps from $G \times \cdots \times G$ ($n$ times) to $A$, $d_n ...
6
votes
2answers
184 views

Motivation for the relations defining $H^1(G,A)$ for non-commutative cohomology

First let me review the definition of first non-commutative cohomology. Let $G$ be a group and $A$ a left $G$-group, i.e. for any $\sigma, \tau\in G$ and $a, b\in A$, one has ...
8
votes
1answer
588 views

Calculating the group co-homology of the symmetric group $S_3$ with integer coefficients.

I have been trying for a while to make sense of Ex V.3.5 & Ex III.10.1 in Brown's book 'Co-homology of Groups': Calculate the Co-homology of $S_3$ with co-efficients in $\mathbb{Z}$, possibly ...
6
votes
1answer
222 views

$H^1$ of $\Bbb Z$ as a trivial $G$-module is the abelianization of $G$ [duplicate]

Let $G$ be a group and $\mathbb{Z}$ regarded as a trivial $G$-module. As title, I'm trying to prove that $H^1(G,\mathbb{Z})$ is isomorphic to $G/[G,G]$. It is easy to see that $H^1$ is isomorphic to ...
0
votes
0answers
27 views

A clarification about the meaning of “Let $\mathbb{Z}$ be *the* trivial $G$-module”.

I have a question regarding a definition/lemma in the book from Charles A. Weibel, "An introduction to Homological Algebra". At page 161, there is a claim starting as follows: Let $A$ be any ...
12
votes
2answers
1k views

What is the intuition between 1-cocycles (group cohomology)?

This is, I'm sure, an incredibly naive question, but: is there a simple explanation for why one should be interested in 1-cocycles? Let me explain a bit. Given an action of a group $G$ on another ...
10
votes
1answer
418 views

Finite groups with periodic cohomology

I'm trying to understand Chapter 12, Section 11 in Cartan + Eilenberg's Homological Algebra, which concerns finite groups with periodic cohomology. Unfortunately I am jumping right to this section in ...
10
votes
4answers
673 views

Why is the cohomology of a $K(G,1)$ group cohomology?

Let $G$ be a (finite?) group. By definition, the Eilenberg-MacLane space $K(G,1)$ is a CW complex such that $\pi_1(K(G,1)) = G$ while the higher homotopy groups are zero. One can consider the singular ...
7
votes
2answers
293 views

What is the motivation for defining both homogeneous and inhomogeneous cochains?

In my few months of studying group cohomology, I've seen two "standard" complexes that are introduced: We let $X_r$ be the free $\mathbb{Z}[G]$-module on $G^r$ (so, it has as a $\mathbb{Z}[G]$-basis ...
7
votes
1answer
109 views

Finite generation of Tate cohomology groups

Let $G$ be a finite group, and let $F$ be a complete resolution for $G$. In other words, $F$ is an acyclic chain complex of projective $\mathbb{Z}G$-modules together with a map ...
6
votes
1answer
322 views

The Zig Zag Lemma in Cohomology

I´m reading the Zig Zag lemma in Cohomology and i want to prove the exactness of cohomology sequence at $ H^k(A)$ and $H^k(B)$ : A short exact sequence of cochain complexes $ 0 \to A \ ...
5
votes
2answers
89 views

Showing that $\operatorname {Br}(\Bbb F_q)=0$

I want to prove that $\operatorname {Br}(\Bbb F_q)=0$ using the cohomological description of the Brauer group. We have: $\operatorname {Br}(\Bbb F_q)=H^2(\operatorname {Gal}(\overline {\Bbb ...
3
votes
1answer
137 views

Finite groups such that $H^1(G,M)=0$ for any simple $G$-module $M$

I'm trying to understand for which finite groups $G$ the augmentation ideal of $\mathbb{F}_2G$ is generated by a single element over $\mathbb{F}_2G$. I'm reading a paper with a result that implies ...
3
votes
2answers
82 views

Is a group always contained in a group that surjects onto its automorphism group?

Let $G$ be a group. I am interested in embedding $G$ in a group $\tilde G$ such that there is a surjective map $\tilde G\rightarrow\operatorname{Aut}G$ whose restriction to $G$ yields the homomorphism ...
1
vote
1answer
134 views

What can we say about groups $G$ with $H_3(G)=0$?

Let $G$ be a group. What can we say about groups such that $H_3(G)=0$? If a characterization is not possible, then knowing examples of such groups would be good? Any help is appreciated. Thanks
7
votes
1answer
282 views

Why is the Herbrand quotient of the dual $\hat{A}$ equal to the inverse of the Herbrand quotient of $A$ in this situation?

I'm reading Serre's Local Fields, and I'm trying to understand the proof of Prop. 9 in $\S$5 of Chap. 8 (p.136). First, the setup: $p$ is a prime number $G$ is a cyclic group of order $p$ $A$ is a ...
3
votes
1answer
106 views

First group cohomology and composition factors

Let $G$ be a finite group. Let $k$ be a field ($\text{char}(k)=p>0$). Let $P(k)$ be the projective cover of $k$. Assume that for any nontrivial simple $kG$-module $M$ we have $H^1(G,M)=0$. Does it ...
2
votes
1answer
74 views

Embedding $G$ in a $Z(G)$ extension of $\operatorname{Aut}G$.

The present question follows up this one, in which I accidentally asked for less than I actually wanted. Given a group $G$, I would like to find an extension $\tilde G$ of its automorphism group ...