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bio website halgebra.math.msu.su/staff/…
location Moscow
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visits member for 2 years, 6 months
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Jan
14
comment Minimal generating sets of groups
Actually, any (inclusion-)minimal generating set of a Tarski monster consists of two elements (if $p$ is prime).
Jan
9
comment relatively free groups in $Var(S_3)$
Yes, exactly. By the way ,the group of polynomial-with-coefficients functions is also free but in a variety of different universal algebras --- a variety of groups with marked elements (i.e., groups with additional 0-ary operations.
Jan
9
comment relatively free groups in $Var(S_3)$
@M.Shahryari, see the edit. I guess that 324 is the number of one -variable polynomial functions over $S_3$ in a different sense of the word polynomial; propably they mean polynomials with coefficients from the group --- such as $x(12)x^2(123)$
Jan
9
revised relatively free groups in $Var(S_3)$
added 1235 characters in body
Jan
8
answered relatively free groups in $Var(S_3)$
Dec
17
comment Large abelian characteristic subgroups in abelian-by-countable groups
Dear Yves, if you do not mind, we shall include your example in our preprint (op. cit.) with a proper reference of course. As for the Podoski--Szegedy example, it is maybe simpler but we do not know how to convert it to the normal-but-not-characteristic settings. Their example is (in essence) the subgroup $\hbox{Alt}(\mathbb Z)A$ of the uncountable symmetric group $\hbox{Sym}(\mathbb Z)$, where $\hbox{Alt}(\mathbb Z)$ is the alternating group (consisting of even finitary permutations) and $A$ is the elementary abelian 2-group consisting of permutations preserving absolute values of integers.
Dec
17
accepted Large abelian characteristic subgroups in abelian-by-countable groups
Dec
15
comment Large abelian characteristic subgroups in abelian-by-countable groups
Oh, yes. I mixed up finitely supported matrices and matrices $f$ with finite $\hbox{rank}(f-1)$ (i.e operators $f$ such that $\hbox{ker}(f-1)$ is of finite codimension). The latter form a normal subgroup but the former do not. Inside $H$, these two subgroups coincides, right? (I mean, their intersections with $H$ coincides.)
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.