New answers tagged group-cohomology
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 \...
3
votes
Accepted
How is the classification of groups extensions by $H^2$ related to Yoneda Ext?
As I mentioned in my comment above, Gruenberg gave a direct bijection between $\operatorname{Ext}^1_{\mathbb ZG}(I_G,A)$ and group extensions $1\to A\to H\to G\to 1$. The details can be found in ...
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},...
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
votes
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 ...
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