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Results tagged with group-cohomology
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user 41644
In mathematics, group cohomology is a set of mathematical tools used to study groups using cohomology theory, a technique from algebraic topology. Analogous to group representations, group cohomology looks at the group actions of a group G in an associated G-module M to elucidate the properties of the group.
4
votes
Accepted
Shapiro's lemma in the language of group extensions
Shapiro's Lemma boils down to the following isomorphism for a subgroup of finite index: Let us write $G=\bigcup_{i=1}^t g_iH$ for left coset representatives of $H$ in $G$. For a $G$-module $P$ we have …
- 4,969
4
votes
Accepted
Identifying projective representations using "gauge-invariant" traces tr[V_g V_h V_k ... ]
Since the action on $U(1)$ is trivial, and since $U(1)$ is injective as an abelian group, the Universal Coefficients Theorem will give you an isomorphism
$$H^2(G,U(1))\cong Hom(H_2(G,\mathbb{Z}),U(1) …
- 4,969
14
votes
Accepted
The term $H^1(N,A)^{G/N}$ in the inflation-restriction exact sequence
You do not need LHS spectral sequence for this action.
The functors $H^*(N,-)$ are the derived functors of $(-)^N:G\text{-mod}\rightarrow G/N\text{-mod}$,
so they will carry a structure of $G/N$-modul …
- 4,969
2
votes
1
answer
151
views
Number of generators for the Schur multiplier of a finite group
Let $G$ be a finite group, and let $M(G)=H_2(G,\mathbb{Z})$ be its Schur multiplier. Are there any known bounds on the number of generators of $M(G)$ in terms of $G$? For example, if $G$ is abelian of …
- 4,969
2
votes
Accepted
Generalization of a lemma of Livne
The answer is yes. First, notice that if $\phi:G\rightarrow G'$ is an epimorphism of 2-groups, then $\phi(N_4(G)) = N_4(G')$. Let now $H$ be the group in your statement. Assume that $N_4(H)$ is nontri …
- 4,969
5
votes
Accepted
Computation of group homology $H_2 ((\mathbb{Z}/3\mathbb{Z}) \rtimes (\mathbb{Z}/4\mathbb{Z}...
In the spectral sequence, notice that by the remarks of YCor and Derek Holt, it is almost trivial: since the orders of $Z/4$ and $Z/3$ are prime to each other, all homology groups of the form $H_p(Z/4 …
- 4,969
6
votes
Accepted
$G$ cocycle split to a coboundary in $J$, via a group extension
In case d=1, the answer is always negative: 1-cocycles are homomorphisms, 1-coboundaries are always trivial, and inflation is injective.
If you do not restrict yourself to the case where $N$ is abeli …
- 4,969
3
votes
Accepted
$SO(3)$ 2-cocycle trivialized to a 2-coboundary in $SU(2)$?
Let $G$ be a group, and let $$1\to A\to J\to G\to 1$$ be an extension of groups with an abelian kernel. Choose a set-theoretical lifting $s:G\to J$ of the quotient map $p:J\to G$. Now define a functio …
- 4,969
16
votes
0
answers
357
views
Representation categories and homology
Let $G$ be a finite group.
Let $\mathcal{C}=Rep-G$ be the rigid $\mathbb{C}$-linear symmetric monoidal category of finite dimensional complex representations of $G$.
Can we recover some homological …
- 4,969
2
votes
Accepted
Explicit 2-cocycle from a 2nd cohomology group $H^2[Q_8 \times \mathbb{Z}/2\mathbb{Z}, U(1)]$
For any two finite groups $G$ and $H$ the Kuenneth formula will give you a homomorphism $$H^1(G,U(1))\otimes_{\mathbb{Z}} H^1(H,U(1))\to H^2(G\times H,U(1)).$$
We can describe this map explicitly. Let …
- 4,969
3
votes
Accepted
Trivialize a cup-product 3-cocycle of $G$ in a larger group $J$
The group $J$ has to be of order divisible by 16, due to the fact that if the order of the group $J$ is $8n$ where $n$ is an odd number, then the inflation of the cocycle will not be trivial (for exam …
- 4,969
2
votes
Accepted
How to claculate the $T$-stable subgroup of second cohomology group
The action is trivial, because the only action of $C_{p^2}$ on the abelian group $\mathbb{Z}/p$ is trivial. You can also see it more directly, by thinking of a two cocycle as giving a twisted group al …
- 4,969
6
votes
$G$ cocycle split and trivialized to a coboundary in $J$, given a group homomorphism $J \ove...
A small remark first: I believe that the denominator in the exponent should be 2, and not $2^2$.
In any case, write $Q_8=\langle x,y|x^2=y^2, xyx^{-1}=y^{-1},y^4=1\rangle$ so that each element in the …
- 4,969