bio | website | halgebra.math.msu.su/staff/… |
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location | Moscow | |
age | ||
visits | member for | 3 years, 1 month |
seen | 4 hours ago | |
stats | profile views | 1,641 |
Jan 16 |
comment |
Almost uniquely generated groups
This condition is indeed weaker. See the (wrong) answer of M. Shahryari. |
Jan 16 |
comment |
Almost uniquely generated groups
No. Even $F_1$ does not satisfy. The question is about inclusion-minimal sets. |
Jan 16 |
comment |
Almost uniquely generated groups
It seems that you use the letter $n$ in two different senses. You are talking about $m$-generated free groups in the varieties of nilpotent groups of exponent $p^n$, right? |
Jan 16 |
asked | Almost uniquely generated groups |
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 |