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bio website halgebra.math.msu.su/staff/…
location Moscow
age
visits member for 2 years, 6 months
seen 2 hours ago

Dec
15
comment Large abelian characteristic subgroups in abelian-by-countable groups
Excuse me, Yves. I do not understand your comment. The finitely supported matrices form a normal subgroup of $G$, right? If they do, then the commutator must be finitely supported.
Dec
14
comment Large abelian characteristic subgroups in abelian-by-countable groups
Thank you, Yves!! Surely, you mean $g(e_q)\in e_q+Vect(\dots)$ in the first paragraph. Also, $G_K$ in the second paragraph is the same as $G_K^Q$. Furthermore, I guess that the "easy play with commutators" is something like the following. The commutator of any matrix and an elementary matrix is finitely supported, i.e. this commutator lies in a finite-dimensional unitriangular subgroup $UT_n(K)$ which is nilpotent and, hence, any its non-trivial normal subgroup non-trivially intersects the centre, which consists of elementary matrices.
Dec
14
asked Large abelian characteristic subgroups in abelian-by-countable groups
Dec
12
comment Commutator Width of a direct limit of hyperbolic groups
I do not understand. Any group having only finitely many conjugacy classes must have finite commutator width, because conjugate elements have the same commutator length.
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
comment Basis removal gives a basis
Oh, thank you, @domotorp!