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Questions on group theory which concern finite groups.
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 …
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
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 …
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 …
5
votes
0
answers
132
views
sums of quadratic forms over finite abelian groups
Let $A$ be a finite abelian group. Let $q:A\times A\to \mathbb{C}^{\times}$ be a non-degenerate bicharacter (that is: for every $a\in A$ $q(a,-)$ and $q(-,a)$ are characters of $A$, which are trivial …
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 …
5
votes
0
answers
150
views
lifting of idempotents in group ring
Let $G$ be a finite group, and let $\pi:G\to Q$ be a surjective group homomophism. The map $\pi:G\to Q$ does not necessarily split, but we can always find a set theoretical splitting $s:Q\to G$. In ot …
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 …
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
Involutive automorphisms of a finite abelian p-group
There is a big difference between the case $p=2$ and $p\neq 2$.
If $p\neq 2$ then $A$ will be the direct sum of $(\mathbb{Z}/p^n,1)$ and $\mathbb{Z}/p^n,-1)$ for some $n$.
In this case all the indeco …
3
votes
Jacobson radical of group algebra
Assuming that the question is the one of YCor: $J(F_{p^k}\otimes F_p G) = F_{p^k}\otimes J(F_pG)$, the answer is yes. The reason is the following: For a finite dimensional algebra $A$ over a field $K$ …
9
votes
1
answer
236
views
First order formulas for finite groups and invariant theory
Let $G$ be a finite group, and let $K$ be a field of characteristic zero.
Let $\phi(x_1,\ldots,x_n)$ be a first order formula in the language of group theory (so $\phi$ can be for example something of …
6
votes
0
answers
148
views
Subalgebra of group algebra generated by idempotents
Let $G$ be a finite group, and let $A$ and $B$ be two abelian subgroups of $G$. Let $K$ be a number field such that all characters of $A$ and of $B$ take values in $K$. Let $\mathcal{O}_K$ be the ring …
12
votes
Accepted
Which finite groups have no irreducible representations other than characters?
A group satisfies this property if and only if it is an extension of a $p$ group by an abelian group. The reason is the following: Assume that indeed all the irreducible representations of $G$ are one …