## New answers tagged group-cohomology

0
votes

Accepted

### Is group cohomology with the inversion action order two?

This is false.
Let's work with $H^2(G,U(1))$ instead of $H^3(G,\mathbb{Z})$.
A counterexample is given by the group $\mathbb{Z}_3 \times (\mathbb{Z}_3 \rtimes \mathbb{Z}_2)$ with the map $x(a,b,c)=c$. ...

4
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 ...

3
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### Group homology for a metacyclic group

Everything follows from the p-primary decomposition theorem, which is nicely explained in Ken Brown's group cohomology bible (Kasper's answer is the restriction-corestriction argument written circa ...

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

0
votes

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### Examples of Lie groups where $G\to G/H$ splits topologically but not as groups

A summary of some of the comments:
@YCor proposed a collection of examples given by $G\to [G, G]$ where $G$ is simply-connected nilpotent non-abelian.
Specifically, if $H$ is the Heisenberg group, we ...

4
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 $...

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 ...

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) & \...

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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