16
votes
Accepted
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 \...
11
votes
Accepted
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!), ...
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 ...
11
votes
Accepted
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,\...
11
votes
Accepted
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 ...
11
votes
Accepted
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 ...
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) & \...
8
votes
Accepted
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$ ...
8
votes
Accepted
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 ...
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},...
7
votes
Accepted
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. ...
7
votes
Accepted
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 ...
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} \...
7
votes
Accepted
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 ...
6
votes
Accepted
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{-...
6
votes
Accepted
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/...
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 ...
6
votes
Accepted
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\...
6
votes
Accepted
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 ...
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 ...
5
votes
Accepted
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 ...
5
votes
Accepted
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 ...
5
votes
Accepted
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 $...
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}} ...
5
votes
Accepted
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 $F_2 \to \mathbb Z$ ...
5
votes
Accepted
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 ...
Community wiki
5
votes
Accepted
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 ...
5
votes
Accepted
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 ...
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 ...
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 ...
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