16 votes
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Generators for the first cohomology of free groups

They are a generating set. In fact, even more is true. Recall that for a group $G$ and a $G$-module $M$, the first cohomology group $H^1(G;M)$ is the abelian group $Der(G,M)$ of derivations $G \...
Andy Putman's user avatar
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11 votes
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Cohomology of $\operatorname{GL}_3(\mathbb{F}_2)$

As you mention in your update, you have a general answer, but if you want a concrete answer for the low-dimensional integral cohomology of $G = \operatorname{GL}(3,2)$ (or any other finite group!), ...
Carl-Fredrik Nyberg Brodda's user avatar
11 votes

Cohomology of $\operatorname{GL}_3(\mathbb{F}_2)$

For the mod $2$ cohomology, see Section 11 of Benson and Carlson, “Diagrammatic methods for modular representations and cohomology” Comm in Alg 15 (1987), 53-121 for the ring structure. If you want to ...
Dave Benson's user avatar
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11 votes
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Trivial group cohomology induces trivial cohomology of subgroups

For any abelian group $A$ we have a canonical isomorphism $\bigwedge^2A\to H_2(A,\mathbb{Z})$, given by the (anti-symmetric) Pontrjagin product $H_1(A,\mathbb{Z})\times H_1(A,\mathbb{Z}) \to H_2(A,\...
Dave Benson's user avatar
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11 votes
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For which subgroups the transfer map kills a given element of a group?

The answer to Q1 is yes, the order of $a$ might be smaller than the gcd: Let $G=\langle x,y\mid x^8=y^2=1,x^y=x^3\rangle$ be the semidihedral group of order $16$. Let $a=[x]\in G_{\text{ab}}=C_2\times ...
Kasper Andersen's user avatar
11 votes
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Do acyclic amenable groups exist?

(1) Acyclic amenable groups do exist, because binate amenable groups exists: for instance, Philipp Hall's "universal locally finite group", which is by definition the Fraïssé limit of all ...
Nicolas Monod's user avatar
10 votes

Pullbacks of classifying spaces

This is not true. Note that $BG$ is only well-defined up to homotopy equivalence, so the only question that makes sense is when the square $$\begin{array}{ccc} B\big(G \underset H\times H'\big) & \...
R. van Dobben de Bruyn's user avatar
8 votes
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Subgroups of top cohomological dimension

For the sake of completeness: Is false, just take the free product $G=H\star H$, where $H$ is a (geometrically finite) group of cohomological dimension $n$, $0<n<\infty$. This is true when $G$ ...
Moishe Kohan's user avatar
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8 votes
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Example of continuous cohomology vs cohomology

Consider the circle $G=\mathbb{R}/\mathbb{Z}$ and its trivial representation on $M=\mathbb{R}$. Since $G$ is abelian, $H^1(G,M)\cong \text{Hom}(G,M)$ and $H^1_c(G,M)\cong \text{Hom}_c(G,M)$ (these are ...
Uri Bader's user avatar
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8 votes

How is the classification of groups extensions by $H^2$ related to Yoneda Ext?

I can tell you what to do in one direction; then the problem is to prove that this is well defined and bijective. Suppose you're given an element of $\operatorname{\rm Ext}^2_{\mathbb{Z}G}(\mathbb{Z},...
Dave Benson's user avatar
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7 votes
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Ker of corestriction of Galois cohomology

(Not sure why this question comes up naturally. The more interesting question, and the one analogue to the kernel of restriction, is to ask what is the cokernel of corestriction. That turns up a lot. ...
Chris Wuthrich's user avatar
7 votes
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Pontryagin dual of a group-cohomology class

Ok, I think I worked this out based on my last comment, writing down the usual double complex for $\operatorname{Ext}(A^\vee,C^\vee)$ using a projective resolution of $A^\vee$ and an injective ...
Achim Krause's user avatar
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7 votes

Rational group homology of an infinite product of finite groups

The higher homology is not necessarily zero. For example $H_1(X,\mathbb Q)$ is $\mathbb Q \otimes \pi_1(X)^{\rm ab} $, so if one takes an infinite product of cyclic groups $\prod_{n \in \mathbb N} \...
Phil Tosteson's user avatar
7 votes
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Pullbacks of classifying spaces

On the other hand, if we define $BG$ by the usual simplicial construction, then the functor $B$ does indeed preserve pullbacks. Indeed, if we start with a pullback square $(G,H,J,K)$ then in the ...
Neil Strickland's user avatar
6 votes
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Extension of base field for modules of groups and cohomology

There is a commutative diagram $\require{AMScd}$ \begin{CD} k[G]\text{-Mod} @>{-\otimes_{k[G]}K[G]}>> K[G]\text{-Mod}\\ @V{-^G}VV @V{-^G}VV\\ k\text{-Vect} @>{-\otimes_kK}>> K\text{-...
Kenta Suzuki's user avatar
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6 votes
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The second Tate-Shafarevich group of a permutation module is trivial

We write $G_w={\rm Gal}(L_w/K_v)$. Definition. For $n\ge 1$, we denote $$Ш_\omega^n(G,M)=\ker\Big(H^n(G,M)\to\prod_C H^n(C,M)\Big)$$ where $C$ runs over the cyclic subgroups of $G$. Remark. $Ш^2(L/...
Mikhail Borovoi's user avatar
6 votes

Is there a clear pattern for the degree $2n$ cohomology group of the $n$'th Eilenberg-MacLane space?

Let me turn my comment into an answer, since I think it qualifies as an answer to the literal question you ask at the end of 2. (although you might be disappointed about the non-explicitness) Your ...
Achim Krause's user avatar
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6 votes
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Explicit formula for general group extension in terms of cartesian product set

My understanding is that this works as follows. Choose a map $\psi\colon Q\to \mathrm{Aut}(N)$ that "lifts" $\rho$. Then $\psi$ won't be a homomorphism so there is a function $f\colon Q\...
Benjamin Steinberg's user avatar
6 votes
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Reference for isomorphism between group cohomology and singular cohomology

For CW-complexes, this is roughly spelled out in Ken Brown's bible: Proposition II.4.1 (which is the combination of Proposition I.4.2 and Proposition II.2.4), assuming the algebraic-definition of ...
Chris Gerig's user avatar
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5 votes

A Tate resolution for $\Sigma_p$ - Reference request

Your resolution is the minimal Tate complex resolving the trivial module for $\mathbb{F}_p\Sigma_p$. Each of the exterior powers of the natural permutation module is indecomposable and projective. The ...
Dave Benson's user avatar
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5 votes
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A map in group cohomology from $H^n(G,G^{\vee})$ to $H^{n+1}(G,U(1))$

Let $I_G$ be the augmentation ideal of $\mathbb{Z}G$. Given that $G$ is abelian, we have $I_G/I_G^2\cong G$ by the Hurewicz isomorphism sending the coset of $g-1$ to $g$. Then your map is induced by ...
Dave Benson's user avatar
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5 votes
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Projective representations of a finite abelian group

The answer is $\ \displaystyle\bigoplus_{i<j}\ \mathbb{Z}/\!\gcd(n_i,n_j)$. The reason is that for $G$ finite, $H^2(G,U(1))$ is the dual abelian group of $H_2(G,\mathbb{Z})$. Now use the Künneth ...
Dave Benson's user avatar
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5 votes
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Group homology for a metacyclic group

The name metacyclic is normally used for a group which is cyclic-by-cyclic (ie. a group $G$ with a cyclic normal subgroup $N$ such that $G/N$ is also cyclic). I will therefore refer to a finite group $...
Kasper Andersen's user avatar
5 votes

Classifying abelian (but non-central) group extensions using homotopy theory

The short note Group Extensions and $H^3$ by P. J. Morandi works out the details of extensions by nonabelian groups. Any extension of $G$ by $A$ has an induced group homomorphism $\def\Out{{\sf Out}} ...
Dmitri Pavlov's user avatar
5 votes
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Cohomological dimension of kernel

Take $n = 2$,$r = 0$ and $M = \mathbb R^2 - \{0,1\}$ and $N = \mathbb R^2 - \{0\}$ and $M \to N$ to be the natural inclusion. Then on fundamental groups we have the surjection $F_2 \to \mathbb Z$ ...
user19232801's user avatar
5 votes
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Lifting SL2(k) to a subgroup of Witt vectors

I found the answer myself with the help of a very useful hint from user "nobody" in the comments, so I'm going to post a community-wiki answer in case anyone else finds it useful. My ...
5 votes
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Pontryagin product on the homology of cyclic groups

I recommend always looking at the canonical reference: Ken Brown's "Cohomology of Groups". Here Chapter V.5 is literally titled "The Pontraygin product" and then the very next ...
Chris Gerig's user avatar
  • 17.1k
5 votes
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Cohomology of $S$-arithmetic groups with trivial coefficients such as $H^n(\rm{PGL}_2(\mathbb{Z}[1/N]);\mathbb{Z})$

This is a hard problem in general. When $N$ is not prime, then even the first homology of $\operatorname{PSL}_2(\mathbf{Z}[\frac{1}{N}])$ is non-trivial to compute (cf. Corollary 4.4 of [1]), but I ...
Carl-Fredrik Nyberg Brodda's user avatar
4 votes

Cohomology of the adjoint representation of $\mathrm{SL}_2(k)$

(Too long for a comment.) A Magma computation shows that for $k=\mathbf{F}_p$ with $p$ prime the group $H^1(\operatorname{PSL}_2(k);k^3)$ equals $0$ for $p=3$ and $7\le p\le 17$ while the cohomology ...
Kasper Andersen's user avatar
4 votes

Cohomology of the adjoint representation of $\mathrm{SL}_2(k)$

$\DeclareMathOperator\PSL{PSL}\newcommand\Ad{\mathrm{Ad}}\newcommand\triv{\mathrm{triv}}$I believe that if the characteristic of $k$ is 2, then $H^1(\PSL_2(k),k^3)$ is nonzero, while if the ...
Peter McNamara's user avatar

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