Skip to main content
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
Results tagged with
Search options not deleted user 4149

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 …
Mikhail Borovoi's user avatar
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{{ …
Mikhail Borovoi's user avatar
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 …
Mikhail Borovoi's user avatar
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 …
Mikhail Borovoi's user avatar
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 …
Mikhail Borovoi's user avatar
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 …
Mikhail Borovoi's user avatar
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 …
Mikhail Borovoi's user avatar
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 …
Mikhail Borovoi's user avatar
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 …
Mikhail Borovoi's user avatar
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 …
Mikhail Borovoi's user avatar
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 …
Mikhail Borovoi's user avatar
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 …
Mikhail Borovoi's user avatar
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 $ …
Mikhail Borovoi's user avatar
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 …
Mikhail Borovoi's user avatar
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.
Mikhail Borovoi's user avatar

1
2 3 4 5 6
15 30 50 per page