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
$2$-cohomology group of semi-direct products
From the spectral sequence mentioned by Anton one can derive an exact sequence
in low degrees
$$ 0\to E_2^{1,0}\to E^1\to E_2^{0,1}\to E_2^{2,0}
\to \mathrm{ker}\left[E^2\to E_2^{0,2}\right]\to E …
3
votes
1
answer
421
views
Conjugation of group extensions
Let $H$ be a finite group. We write ${{\mathbb{C}}}^{*n}$ for the $n$-dimensional complex torus $({{\mathbb{C}}}^*)^n$.
We have a short exact sequence
$$ 0\to {{\mathbb{Z}}}^n\to {{\mathbb{C}}}^n\to{{ …
5
votes
Projective arrows
In addition to Fernando's answer, note that projective arrows were introduced in 1966 by A. V. Roiter (under the name of projective morphisms)
in the paper On integral representations belonging to one …
16
votes
Accepted
In a compact lie group, can two closed connected subgroups generate a non-closed subgroup?
The abstract subgroup generated by $H$ and $K$ is closed.
We may assume that $G$ is connected.
The groups $G$, $H$, $K$ are the groups of real points of real algebraic groups $\mathbf{G}$, $\mathbf{H …
2
votes
1
answer
615
views
Does any finite group of order $2m$ with odd $m$ have a subgroup of index 2? [closed]
Let $G$ be a finite group of order $2m$ where $m>1$ is an odd natural number.
Question. Is it true that any such $G$ has a subgroup $H$ of index 2?
If yes, I would be grateful for a reference or a …
9
votes
Accepted
About the conjugation of semi-simple subgroups
The answer is YES. It suffices to assume that $H_1$ and $H_2$ are conjugate over $\mathbb{C}$ or, what is the same, that they are conjugate over $\overline{\mathbb{Q}}$.
Theorem 1. Let $G$ be a co …
6
votes
2
answers
350
views
The Tits classes of simply connected simple real groups
Let $G$ be a simply connected semisimple group over a perfect field $k$ (at the moment I am interested in the case $k=\mathbb R$).
Then $G$ is an inner form of a quasi-split $k$-group $G_{\rm qs}$:
th …
3
votes
1
answer
219
views
Real automorphisms of the "quaternionic" real group ${\rm SO}^*(4m)$
Let $m\ge 2$, and let $G={\rm SO}^*(4m)$ denote the "quaternionic" real form of the special orthogonal group ${\rm SO}(4m,\mathbb C)$ of type ${\sf D}_{2m}$.
Let $\tau\in{\rm Aut}_{\Bbb R}(G)$ be a re …
3
votes
2
answers
395
views
Indecomposable integral representations of a group of order 2 "by hand"
This question is a duplicate of
that 2010 MO question.
I am interested in classifying isomorphism classes of $n$-dimensional integral representations of the cyclic group $C_2$ of order $2$.
Clearly, a …
4
votes
1
answer
171
views
A small rank linear combination of a small number of elements of a group
This is a version of
this question of Klim Efremenko.
Let $r>2$ be a natural number, say $r=3$ or $r=10$. Let $G$ be a finite group and $\rho$
be an irreducible complex representation of $G$. We con …
2
votes
0
answers
318
views
Surjective homomorphisms of non-connected Lie groups
Let $\psi\colon B\to C$
be a homomorphism of real Lie groups, where the group $C$ is connected.
Let $B^0$ denote the identity component of $B$, and we set $\pi_0(B)=B/B^0$, then $\pi_0(B)$ is a discr …
1
vote
on z-extensions
No, if the center $Z(G)$ is not connected, we cannot construct $G'$ with connected $Z(G')$.
Indeed, let $\pi\colon G'\to G$ be an epimorphism of connected reductive $F$-groups with central kernel. Ch …
4
votes
Homomorphism from noncompact semisimple Lie group to compact Lie group
See below a detailed version of the comment of @LSpice. (Edited taking into account a comment of @YCor.) This is an answer to the question on homomorpisms of real algebraic groups.
Proposition.
Let $ …
6
votes
1
answer
143
views
Finite $\Gamma$-modules with trivial $H^2$, where $\Gamma$ is a group of order 2
Let $\Gamma=\{1,\gamma\}$ be a group of order 2. Let $A$ be a finite $\Gamma$-module,
that is, a finite abelian group on which $\Gamma$ acts.
It is a hopeless problem to classify finite $\Gamma$-modul …
1
vote
Indecomposable integral representations of a group of order 2 "by hand"
See Appendix A in M. Borovoi and D. A. Timashev, Galois cohomology and component group of a real reductive group.