bio | website | halgebra.math.msu.su/staff/… |
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location | Moscow | |
age | ||
visits | member for | 3 years, 1 month |
seen | 2 hours ago | |
stats | profile views | 1,653 |
Dec 11 |
comment |
Applications of the Chinese remainder theorem
Excuse me, KConrad (and @Zack), I do not understand the problem about squares. Suppose that we have integers $a$ and $b$ such that $f(a)\ne0$ modulo $p^2$ and $f(b)\ne0$ modulo $q^2$, then CRT provides us with an integer $c$ equal to $a$ modulo $p^2$ and equal to $b$ modulo $q^2$. Thus, $f(c)=f(a)\ne0$ modulo $p^2$ and $f(c)=f(b)\ne0$ modulo $q^2$, which is a contradiction. Or I miss something? |
Nov 5 |
comment |
Minimal number of generators of subgroups of Noetherian groups
@Ian, indeed, "Are all finitely presented Noetherian groups virtually polycyclic? is (an open) Question 11.38 (due to S.V.Ivanov, 1990) from the Kourovka Notebook. |
Nov 4 |
awarded | Enlightened |
Nov 4 |
awarded | Nice Answer |
Nov 3 |
comment |
The rank of the intersection of subgroups of a free group
Igor, it was asked about the case, where one of the subgroups is of finite index. This is indeed an exercise on Schreier formula. |
Nov 3 |
revised |
Minimal number of generators of subgroups of Noetherian groups
added 7 characters in body |
Nov 3 |
answered | Minimal number of generators of subgroups of Noetherian groups |
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 |
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Basis removal gives a basis
Oh, thank you, @domotorp! |
Aug 31 |
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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 $U$. |