Questions tagged [gr.group-theory]
Questions about the branch of algebra that deals with groups.
1,094 questions
4
votes
1
answer
524
views
What is the outer automorphism group of the Lie group $\text{SL}_2(\mathbb{R})$ as an abstract group?
I hope to ask what the outer automorphism group of the Lie group $\text{SL}_2(\mathbb{R})$ is, just as an abstract group. It seems like Dieudonné's paper On the automorphisms of the classical groups ...
4
votes
2
answers
504
views
Unique product group which is not right orderable
(1) I am looking for an example of a u.p (unique product) group which is not right orderable (RO).
Almost any group I pick up (obviously torsion-free, as u.p. group cannot have nontrivial torsion ...
4
votes
0
answers
199
views
Generalization of the fundamental theorem of cyclic groups 2
This post is a sequel of Generalization of the fundamental theorem of cyclic groups
Let $G$ be a finite group then the fundamental theorem of cyclic groups can be formulated as follows:
Theorem: $G$ ...
4
votes
3
answers
376
views
Proof that lifts of geodesics are quasi-geodesics (relatively hyperbolic groups)
$\DeclareMathOperator\Cay{Cay}$Suppose $G$ is a relatively hyperbolic group with peripheral subgroups $P_1,P_2,\dots, P_n$, and suppose $\mathcal{S}$ is a finite generating set for $G$. Let $X=\Cay(G,\...
4
votes
1
answer
195
views
Finite p-groups and their fibered products
Is every finite $p$-group an epimorphic image of a fibered product of two finite $p$-groups which can be generated by $2$ elements?
4
votes
1
answer
204
views
Totally right preorderable groups
Are there any known non-trivial sufficient conditions, or full characterizations, of a totally right-preorderable group?
More precisely:
totally right-preorderable: has a non-trivial total right-...
4
votes
5
answers
1k
views
Groups of order $p(p^2+1)/2$
It seems that when $p>3$ is a prime, then each group of order $p(p^2+1)/2$ is abelian as I checked by Gap for small $p$. Is it true for each $p$?
Thanks for your answers
4
votes
2
answers
433
views
What is the current status of representation theory of $n$-ary groups in terms of hypermatrices?
An $n$-ary group is a generalization of the usual concept of a group where the binary operation (2-argument operation) is instead an $n$-ary ($n$-argument) operation. More info here on Wikipedia.
I ...
4
votes
1
answer
515
views
Generating infinite index subgroups of a free group
Let $F$ be nonabelian finitely generated free group, let $H \leq F$ be a finitely generated subgroup of infinite index and let $x,y \in F \setminus H$. Must there be some $a \in F$ such that $[F : \...
4
votes
2
answers
308
views
Another quotient of Hurwitz group
The paper An update on Hurwitz groups by Marston Conder seems to suggest
that the Chevalley group $G(2,5)$ of order $5859000000$ is a quotient of
$G := \langle a, b \ | \ a^2, b^3, (ab)^7, [a,b]^{10} \...
4
votes
1
answer
237
views
Minimal number of generators of satellite knot groups
In light of Knot groups with big number of generators, I was wondering...
Question 1 What is the minimal number of generators of the fundamental group of a satellite knot?
Another more specific ...
4
votes
1
answer
142
views
Can the degree of $k$-nilpotence of a simple simply connected compact Lie group be in $(0,1)$?
Let $G$ be a simple (i.e. every proper normal subgroup is discrete) simply connected compact Lie group. Define the degree of $k$-nilpotence of $G$ to be the Haar measure of the set
$$\{(x_1,\dotsc,x_{...
4
votes
2
answers
388
views
Bound on the index of an abelian subgroup in discrete subgroup of the euclidean group?
I asked the following question on math.SE a couple of days ago. Dietrich Burde gave an answer for the case that the subgroup is not only discrete but also acts cocompactly.
What about the general ...
4
votes
1
answer
246
views
How do "Kummer closures" of fields look?
Let $F$ be a field and $A$ a finite abelian group. You can ask: does the regular representation $F[A]$ of $A$ split as a direct sum of 1-dimensional representations? This is equivalent to the ...
4
votes
0
answers
238
views
Non-abelian representations and their induced abelian representations
$\DeclareMathOperator\Aut{Aut}\DeclareMathOperator\Out{Out}\newcommand\ab{^{\text{ab}}}$Let $G$ and $H$ be a groups and assume that $H$ is non-abelian. Then we have a morphism
$$\Out(H)\rightarrow \...
4
votes
1
answer
458
views
Does there exist an order in a number field of deg>1 with a map to F_p for all p?
This question is motivated by a computational issue. Suppose $R$ is a product of orders in numberfields such that there is no ring homomorphism $R \to \mathbb Z$, then can one write an algorithm that ...
4
votes
1
answer
238
views
No lifts in an exact sequence of profinite groups?
In pg. 24 of his book on Galois cohomology, Serre gives the following exercise:
"Give an example of an extension $1 \to P \to E \to G \to 1$ of profinite groups with the following properties:
(i) $...
4
votes
0
answers
159
views
Can the numbers of elements of distinct prime orders of a finite simple group coincide? [duplicate]
Does there exist a finite simple group $G$ and distinct prime numbers
$p$ and $q$ dividing the order of $G$ such that the numbers of elements
of $G$ of order $p$ and $q$ are the same?
Remark 1: It ...
4
votes
1
answer
918
views
Heisenberg group in characteristic two
I have seen two constructions called the Heisenberg group. If $k$ is a field of characteristic not equal to $2$ and $V$ is a $2n$-dimensional vector space over $k$ with symplectic form $\omega : V \...
4
votes
1
answer
348
views
Is there a topologizable group admitting only Raikov-complete group topologies?
Definition. A group $G$ is called complete (resp. non-topologizable) if each Hausdorff group topology on $G$ is Raikov-complete (resp. discrete). It is clear that each non-topologizable group is ...
4
votes
1
answer
218
views
Inverse limits of perfect groups
Is every group isomorphic to an inverse limit (that is, projective limit) of perfect groups?
I guess, the answer is no. In that case: Is there a characterization of the groups that are isomorphic to ...
4
votes
0
answers
458
views
Mackey theory application - semidirect product abelian-by-finite
In order to advance my research I'm supposed to understand this fact:
Let $A$ be an abelian group and $S$ a finite group acting on $A$.
This defines the semi-direct product $A\rtimes S$:
Let $\chi$ ...
4
votes
1
answer
625
views
Is there an algorithm for computing Schur multiplier?
Suppose we are given group $G=\langle a_1,\ldots,a_n \mid R_1=1,\ldots R_m=1 \rangle$. Is there an algorithm which computes a finite presentation for the Schur multiplier, i.e. second homology group $...
4
votes
2
answers
356
views
Defining parity on the set of all bijections of ${\mathbb{N}}$
Let $\phi:{\mathbb{N}}\rightarrow {\mathbb{N}}$ be a bijection. Can we extend the notion of parity (of a finite permutation) to $\phi$ ?
In other words, Can we define a group homomorphism $\Lambda $...
4
votes
0
answers
210
views
Profinite closure of characteristic subgroup
Let $F$ be a free group of finite rank, and $K\subset F$ a finite index characteristic subgroup.
Let $\hat{F}$ be the profinite completion of $F$ (i.e. a free profinite group of same rank), and $\bar{...
4
votes
1
answer
383
views
Finite dimensional compact abelian group that is not a product of connected and a totally disconnected
Let $G$ be a compact abelian group. A compact abelian group is said to have dimension $n$ if $\dim_\mathbb{Q} \mathbb{Q}\otimes \hat G = n$. Equivalently one can show that this holds if $G$ is ...
4
votes
1
answer
221
views
On groups without word growth $\succeq\exp(n^{1/2})$
I just asked today the following question:
On groups with finite pro-$p$ completion for all primes $p$. However, I can actually simplified what I am interested in, but as it is somewhat a more general ...
4
votes
2
answers
448
views
Does $\text{Sym}(\omega)$ have $2^{\aleph_0}$ pairwise non-isomorphic subgroups?
Let $\text{Sym}(\omega)$ denote the set of all bijections $f:\omega\to\omega$ together with composition as group operation. Does $\text{Sym}(\omega)$ have $2^{\aleph_0}$ pairwise non-isomorphic ...
4
votes
0
answers
640
views
Closed subgroups of a connected Lie group
Is it true that for any closed subgroup $H$ of a connected Lie group $G$, the group of connected components $\pi_0(H)$ is finite or countable? (inspired by the comment of nfdc23 to this question ).
4
votes
0
answers
226
views
Possible dimensions for triples of unitary irreducible representations whose tensor product contains the identity
For which triples $\{A,B,C\}$ of positive integers does there exist a (finite or compact) group $G$ with unitary irreducible representations of dimensions $A$,$B$, and $ C$ whose tensor product ...
4
votes
0
answers
732
views
Properties of Higman's group
The infinite group of Higman which has no finite quotient is given by the presentation (with 4 generators and 4 relations):
$$
G = \langle a_i, i \in \mathbb{Z}/4\mathbb{Z} \mid a_ia_{i+1 \,(\text{mod ...
4
votes
1
answer
344
views
Extensions of compact Lie groups
Let $G$, $H$ be two compact Lie groups (possibly disconnected). Two short exact sequences of compact Lie groups
$$
0\rightarrow G\rightarrow M_1 \rightarrow H\rightarrow 0,
$$
$$
0\rightarrow G\...
4
votes
0
answers
593
views
Galois classes of L-functions
Around one month ago, I posted on math.stackexchange a draft I wrote in which I define the notion of Galois class of L-functions: see https://math.stackexchange.com/questions/280876/definition-of-a-...
4
votes
1
answer
518
views
Finite groups with the same character table *including* class types, and square-free order
There are non-isomorphic finite groups with the same (complex) character table, as $D_4$ and $Q_8$.
$$\scriptsize\begin{array}{c|c}
\text{class}&1&2A&2B&2C&4 \newline
\text{...
4
votes
3
answers
717
views
Finite, abelian, yet "fugitive" orthogonal subgroups
Update July 29, 2013.
I have still not found a good textbook for this topic, if you point one to me I will be grateful :) I have accepted BS's answer anyway, since their explanation was useful to me ...
4
votes
2
answers
704
views
Behaviour of cohomology groups under extension of scalars
Let $\hat{R}\to R$ be a homomorphism of commutative unital rings and let
$\hat{M}$ be an $\hat{R}G$-module for a group $G$. Does the $R$-module isomorphism $$H^n(G,\hat{M}\otimes R)\cong H^n(G,\hat{M}...
4
votes
1
answer
373
views
Classification of finite abelian hypergroups and table algebras
Update: Originally, I formulated this question for finite abelian hypergroups, but in a discussion with Geoff Robinson below I realized that the abelian hypergroups defined below are equivalent to ...
4
votes
1
answer
158
views
Counting "deflected" permutations: Part I
Let $\mathfrak{S}_n$ denote the group of permutations on $\{1,2,\dots,n\}$. Now, introduce the sets
$$\mathcal{A}_n^{(k)}:=\{\pi\in\mathfrak{S}_n: -1\leq \pi(j)-j\leq k,\,\forall j\}.$$
I would like ...
4
votes
2
answers
393
views
Is $(G\rtimes H,H)$ a Gelfand pair iff $G$ is abelian?
Let $G$ be a finite group and $K \subset G$ a subgroup. Then $(G,K)$ is a Gelfand pair if the double coset Hecke algebra $\mathbb{C}(K \backslash G / K)$ is commutative.
Let $H$ be a subgroup of $...
4
votes
1
answer
242
views
Which 3-manifolds have positive rank gradient?
For which $3$-manifolds $M$ is the fundamental group $\pi_1(M)$
finitely generated and has positive rank gradient?
Recall that the rank gradient of a finitely generated group $G$ is defined to be $$...
4
votes
0
answers
248
views
Generalization of a theorem of Øystein Ore in group theory: the infinite case
This post is the infinite version of this one, and is motivated by an exchange with Carmela Musella and Maria De Falco. We are interested in relative versions of the following Ore's theorem and ...
4
votes
2
answers
448
views
Number of torsion-free abelian groups
Let $\mathfrak{c}$ be the cardinality of the continuum. How much Choice, if any, is needed to prove that there are $2^{\mathfrak{c}}$ distinct (mutually nonisomorphic) torsion-free abelian groups of ...
4
votes
1
answer
416
views
Fricke involution on GL(3)
Define $\Gamma_0(N)=\{\begin{pmatrix}
a&b&c\\
d&e&f\\
g&h&i
\end{pmatrix}
\in SL(3,\mathbb{Z})|g\equiv h\equiv 0(\mod N)\}$ be the $N$-level congruence subgroup on GL(3).
...
3
votes
0
answers
217
views
References on a certain generalization of Dedekind groups
Recall that a group is called a Dedekind group if all of its subgroups
are normal. Also recall that a weaker property of a subgroup than normality
is that of being a TI-subgroup: a subgroup $H$ of a ...
3
votes
1
answer
214
views
What is the growth of the rank of a power of a finite simple group?
Which asymptotic bounds (upper and lower) are known for $s_n$ - the minimal number of generators of $S^n$ where $S$ is a nonabelian finite simple group?
3
votes
0
answers
233
views
A bridge between the algebraic / differential geometry of $\frak{sl}_2(\mathbb{C})$ and the Sheffer-Appell calculus and combinatorics
In "Four examples of Beilinson-Bernstein localization", Anna Romanov introduces the basis
$m_k = \frac{(-1)^k}{k!} \partial^k \delta $
on p. 9, where $\partial$ is a partial derivative and $\...
3
votes
4
answers
1k
views
Automorphisms of the Selberg class
Hello,
assuming Selberg's orthonormality conjecture, let's consider bijective maps $f$ from Selberg's class $\mathcal{S}$ to itself such as:
1) $f$ maps a primitive function of $\mathcal{S}$ to a ...
3
votes
1
answer
2k
views
Automorphisms of quotient groups
I have a general question about automorphism groups. Sorry in advance if I'm talking about well known facts, but I didn't find much in the literature.
Let $G$ be a group and let $N$ be a ...
3
votes
1
answer
384
views
Using the Lehmer quintic to solve $11$-degree equations and higher?
(This is a natural continuation of a previous post.)
I. Quintic method
Given the Lehmer quintic,
$$x^5 + n^2x^4 - (2n^3 + 6n^2 + 10n + 10)x^3 + (n^4 + 5n^3 + 11n^2 + 15n + 5)x^2 + (n^3 + 4n^2 + 10n + ...
3
votes
1
answer
740
views
Bases of free groups
Let $F$ be a free group on a finite set $X$. Let $A \subseteq X$ be a subset of $X$ contained in some $H \leq F$, a subgroup of finite index in $F$. Must there be a basis (free generating set) for $H$ ...