The tag has no wiki summary.

learn more… | top users | synonyms

6
votes
1answer
348 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 ...
81
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 ...
9
votes
4answers
556 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
538 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 ...
6
votes
2answers
142 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 ...
6
votes
1answer
199 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 ...
24
votes
2answers
545 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 ...
8
votes
2answers
640 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 ...
9
votes
4answers
544 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 ...
10
votes
1answer
338 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 ...
7
votes
1answer
86 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 ...
7
votes
2answers
229 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 ...
5
votes
2answers
87 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 ...
5
votes
1answer
246 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 \ ...
3
votes
1answer
134 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 ...
1
vote
1answer
131 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
415 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 ...
4
votes
1answer
61 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 ...
3
votes
1answer
95 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 ...