Search Results
Search type | Search syntax |
---|---|
Tags | [tag] |
Exact | "words here" |
Author |
user:1234 user:me (yours) |
Score |
score:3 (3+) score:0 (none) |
Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
Views | views:250 |
Code | code:"if (foo != bar)" |
Sections |
title:apples body:"apples oranges" |
URL | url:"*.example.com" |
Saves | in:saves |
Status |
closed:yes duplicate:no migrated:no wiki:no |
Types |
is:question is:answer |
Exclude |
-[tag] -apples |
For more details on advanced search visit our help page |
Questions about the branch of algebra that deals with groups.
2
votes
Generating sets for $\mathrm{SU}(1,1;\mathcal{O}_K)$ or $\mathrm{PU}(1,1;\mathcal{O}_K)$?
To expand on my comment above : as noted by john mangual $SU(1,1)$ is isomorphic to $SL(2, \mathbb R)$. If $h$ is an Hermitian form defined over a ring of integers $\mathcal O_K$, where $K$ is imagina …
2
votes
1
answer
182
views
Fixed directions and Zariski density of hyperbolic groups
It is a fact that if $\Lambda$ is a nonelementary subgroup of ${\rm PSL_2}(\mathbb{C})$ which contains an hyperbolic transformation and moreover ${\rm tr}(g)\in\mathbb{R}/\pm 1$ for all $g\in\Lambda$ …
8
votes
Accepted
Morphisms $\mathrm{SL}_n(\mathbb Z) \to \mathrm{SL}_m(\mathbb Z)$
I'll try to flesh out the first comment to give an answer: for all $n\ge 4$ there is a representation $\rho:A_{n+1}\to \rm SL_n(\bf Z)$ whose image normally generates $\rm SL_n(\bf Z)$ (because it inj …
1
vote
Generators for a certain congruence subgroup of SL(n,Z)
I guess the folllowing is a short explicit proof of Tits' result for this very special case, assuming CSP for $\rm{SL}_n(\bf{Z})$ (so this should be seen as a lenghty comment on the above answer).
Th …
7
votes
Accepted
Unique product group which is not right orderable
Such a group has been found by N. Dunfield, see the appendix to
the paper
Steffen Kionke, Jean Raimbault, Nathan Dunfield, On geometric aspects of diffuse groups, Documenta Mathematica, Vol. 21 (2016 …
8
votes
0
answers
340
views
Polynomial growth without Gromov's theorem
It is a consequence of the theorem of M. Gromov on groups with polynomial growth and a result of P. Pansu(*) that if a finitely generated group $\Gamma = \langle S\rangle,\, S=S^{-1}$ satisfies $(\dag …
23
votes
1
answer
1k
views
Kaplansky's unit conjecture and unique products
There are three conjectures on group rings that bear the name of Kaplansky (see for example this question). The 'unit conjecture' in the title of the present question is the strongest of them, and sta …
7
votes
2
answers
912
views
Subgroups of $SL_2(F)$ generated by unipotent elements
I am interested in the following problem : given a finite field $F$ and two unipotent elements $g_1,g_2\in\mathrm{SL}_2(F)$ which do not commute, what can we say about the subgroup they generate? More …
14
votes
1
answer
932
views
The number of involutions in a permutation group
If $G$ is a group let $I(G)$ be the number of involutions (elements of order 2) in $G$. My question is then easily stated: does there exists a constant $C > 1$ such that for every $n \ge 1$ and every …
7
votes
1
answer
435
views
A criterion for loxodromicity in Gromov-hyperbolic spaces
Recall that an isometry of a Gromov-hyperbolic space $X$ is called loxodromic if it has exactly two fixed points on the Gromov boundary $\partial X$, one being "attracting" and the other "repelling". …
1
vote
About normalizers of infinite cyclic subgroups of Hilbert modular group
This is not a complete answer but it's too long to fit in a comment, so here it goes. In case $g \in \mathrm{SL}_2(\mathcal O_k)$ is semisimple there are two possibilities :
As you observed, if $g$ …
9
votes
Accepted
Equivalence of surjections from a surface group to a free group
This is true, and it is written up in lemma 2.2 of "The co-rank conjecture for 3--manifold groups" by C. Leininger and A. Reid https://arxiv.org/abs/math/0202261. They state the result in slightly dif …
10
votes
Accepted
Random pairs of commuting permutations
This is not the case, in fact in both cases the limit is 1/2. This is because with probability going to 1 as $n \to +\infty$ a uniformly random morphism $\mathbb F_2 \times \mathbb F_2 \to \mathrm{Sym …
3
votes
A question about congruence subgroups
This is true in general, by an argument similar to the one you used for the coprime case. The subgroup $\Gamma$ generated by $\Gamma(N_1) \cup \Gamma(N_2)$ contains the matrices
$$
a = \begin{pmatrix …
4
votes
1
answer
248
views
Subgroups and representations of finite groups of Lie type
Is there a usable bound for the minimal index of a proper subgroup in a finite simple group of Lie type in terms of its rank and the characteristic (or even cardinality) of its field of definition?
On …