bio | website | halgebra.math.msu.su/staff/… |
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location | Moscow | |
age | ||
visits | member for | 2 years, 10 months |
seen | 7 hours ago | |
stats | profile views | 1,565 |
Nov 3 |
comment |
Is there an infinite group with exactly two conjugacy classes?
@Gerry, I do not think this is a homework. An example can be easily constructed using iterated HNN-extensions. A finitely generated example also exists but this is a highly non-trivial result of D. Osin. |
Nov 2 |
accepted | Are compact simple groups homotopically non-abelian? |
Oct 31 |
comment |
Are compact simple groups homotopically non-abelian?
Here is the link: ams.org/journals/bull/1960-66-04/S0002-9904-1960-10487-9/… |
Oct 31 |
comment |
Are compact simple groups homotopically non-abelian?
Thank you, Ramiro, this is what I sought! |
Oct 31 |
asked | Are compact simple groups homotopically non-abelian? |
Oct 8 |
comment |
About sets with two mutually associative group structures
In particular, this means that the two groups must be isomorphic. |
Oct 1 |
awarded | Caucus |
Aug 31 |
comment |
A generalization of an old group problem
@nadal, (2) is not true (without additional assumptions); see Yves's comment. |
Aug 31 |
comment |
Basis removal gives a basis
Oh, thank you, @domotorp! |
Aug 31 |
comment |
Basis removal gives a basis
I do not understand about 28. Can you clarify? |
Aug 31 |
comment |
Basis removal gives a basis
@Will and domotorp: you are right of course; I edited this part. That was an incorrect generalisation of the 3-dimensional case. |
Aug 31 |
revised |
Basis removal gives a basis
deleted 36 characters in body |
Aug 30 |
comment |
Basis removal gives a basis
@fedja, if there is a pair of twins, then the other vectors must lie in some hyperplane $U$, because otherwise we would have a basis avoiding these twins and the harmony would give a basis containing the twins; this is a contradiction. So, all vectors except this pair of twins is a harmonic subset of $U$. |
Aug 30 |
answered | Embedding of a “quotient graph” |
Aug 28 |
awarded | Nice Question |
Aug 28 |
comment |
Basis removal gives a basis
@Gjergji: honestly, there are no research-level motivations. It was intended to be a problem for students (with the integer 10 probably) but --- I discovered that my question seems to be not so easy (for me). So, if there is a simple solution for 10, I shall be glad. |
Aug 28 |
comment |
Basis removal gives a basis
@Gerry, it does not matter whether we are speaking about bases of $X$ (= bases of $\langle X\rangle$) or about bases of the whole space $V$. The answers will be obviously the same. |
Aug 28 |
comment |
Basis removal gives a basis
@Gerry, A basis of a set of vectors is its maximal linearly independent subset. So, Joël and Tom are right. Thank you, Tom. |
Aug 27 |
asked | Basis removal gives a basis |
Aug 25 |
revised |
Is there a better description of this class of discrete groups?
added 100 characters in body |