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Questions about the branch of algebra that deals with groups.
13
votes
4
answers
982
views
Number of finite index subgroups in a free abelian group
Let $n,m\in\mathbb{N}$. Is there a formula for the number of subgroups of index $n$ in $\mathbb{Z}^m$? Perhaps in terms of the divisors of $n$?
1
vote
Free generators for the fat commutator subgroup
We can do this step by step, using the fact that any surjective map from a free group onto a free group splits. More precisely: a basis for the group $X$ is given by $\{wxw^{-1}\}_{w\in \langle y,z\ra …
1
vote
Accepted
Splitting of certain short exact sequences in context of Clifford theory
The idea is to consider your extension as a factor set $$f:Y/X\times Y/X\to 1+J(A)$$ and show that it must be equivalent to a trivial one. For this you first consider the image of $f$ in $(1+J(A))/(1+ …
10
votes
Accepted
Infinite groups of finite exponent inside of SL(2,C)
A theorem of Burnside says that a linear group of finite exponent is finite. So the answer is no.
2
votes
Accepted
embedding of finite groups into product
The answer is no, unless you restrict somehow the prime divisors of $|G_i|$. Take $G_i = \mathbb{Z}/p_i$ where $p_i$ is the $i$-th prime number. Take $K$ to be the trivial subgroup. Now, if you have s …
14
votes
2
answers
615
views
Semisimple representations of discrete groups
Let $G$ be a discrete group. Let $V$ and $W$ be two finite dimensional complex simple $G$ representations.
QUESTION. Must the tensor product $V\otimes_{\mathbb{C}} W$ with the diagonal action be …
1
vote
To calculate $Tor_1^G(\mathbb{Z},N_{ab})$ and $Tor_1^Q(\mathbb{Z},N_{ab})$
One possible solution would be to use a long exact sequence in homology arising from a short exact sequence in the following way: the group $N$ has order $p^a$ for some $a$. Therefore, $\textit{as a $ …
2
votes
1
answer
175
views
Number of homomorphism, or number of solution to equations, in finite groups
Let $G$ be a finite group, and let $P$ be a finitely generated group.
Consider the number $$n=\#Hom_{Grp}(P,G).$$
It is known (see Number of solutions to equations in finite groups) that under relativ …
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 …
2
votes
Is it possible to construct K(G, 1) out of a subgroup and its quotient?
If the total spaces $EQ$ and $EK$ are given as $Q$ and $K$ CW-complexes respectively, you can also use a construction of C.T.C. Wall (see C. T. C. Wall, Resolutions for extensions of groups, Proc. Cam …
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 …
2
votes
Orthogonal idempotents with sum equal to 1 in $k[G]$ span sub-Hopf algebra
In general no. Take for example $G=C_3, e = \frac{1}{|G|}\sum_{g\in G}g$ and $B=\{1-e,e\}$. Then $B$ spans a sub algebra, but not a sub-Hopf algebra. This was a particular example, but for most finite …
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 …
2
votes
1
answer
393
views
Finite quotients of an infinite product of finite groups
Let $G$ be a finite group.
Consider the direct product $\Gamma=\prod_{i=1}^{\infty}G$ of (countably) infinitely many copies of $G$. For every finite set of numbers $\{i_1,\ldots,i_n\}$ we have the nat …
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 …