bio  website  halgebra.math.msu.su/staff/… 

location  Moscow  
age  
visits  member for  2 years, 2 months 
seen  15 mins ago  
stats  profile views  1,036 
13h

comment 
Are infinite groups “locally topologizable”?
It is on page 339 of English edition. Unfortunately, Google preview rejects to show this page to me (but I may read the table of contents). 
2d

comment 
Are infinite groups “locally topologizable”?
I have a Russian edition. This is Chapter 10, Section 31, Subsection 3. 
Aug 25 
comment 
Invariant planes of a nilpotent matrix with two Jordan blocks of size two
This is the description of all of them, because any invariant subspace containing a vector $v$ shoud contain $Nv$. 
Aug 25 
comment 
Invariant planes of a nilpotent matrix with two Jordan blocks of size two
Take any vector $v$ not belonging to the kernel of $N$ and consider the space $\langle v,Nv\rangle$. This family and the kernel of $N$ are all spaces you ask about. 
Aug 23 
comment 
A sequence of subsets of an infinite group
A related question: mathoverflow.net/q/179177/24165 . 
Apr 24 
comment 
An element $g$ in a group such that neither $g=1$ nor $g\ne 1$ can be proved.
@Dan, surely. What you said is basically Metatheorem 3.5.2 from my answer. 
Jan 20 
comment 
Why the axiomatic rank of the variety of groups is equal to three?
Your octonion argument is much simpler and works perfectly. The algebra of all unit octonions or all nonzero octonions (with operations $\cdot$, ${\ }^{1}$, and $1$) is nonassociative but satisfies all twovariable laws that are consequences of the group axioms. 
Jan 17 
comment 
Almost uniquely generated groups
Oh yes, thank you! 
Jan 17 
comment 
Almost uniquely generated groups
"Any finite $p$group which is relatively free in some variety has this property by the same argument. In light of Geoff's answer perhaps these are the only ones?"  No. The quaternion group of order 8 is not relatively free but satisfies the conditions. 
Jan 17 
comment 
Almost uniquely generated groups
Why $\{a_{1},b_{1},g_{2},g_{3}, \ldots, g_{n} \}$ is irredundant? 
Jan 17 
comment 
Almost uniquely generated groups
No. $S$ is unique up to automorphisms; so, the complement of $S$ is not necessary the Frattini subgroup.Take a cyclic group of prime order, for example. 
Jan 16 
comment 
Almost uniquely generated groups
Yes. And probably you mean Question 2 as you use that the orders are finite. 
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 inclusionminimal 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 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 polynomialwithcoefficients functions is also free but in a variety of different universal algebras  a variety of groups with marked elements (i.e., groups with additional 0ary 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 $x(12)x^2(123)$ 
Dec 17 
comment 
Large abelian characteristic subgroups in abelianbycountable 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 PodoskiSzegedy example, it is maybe simpler but we do not know how to convert it to the normalbutnotcharacteristic 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 2group consisting of permutations preserving absolute values of integers. 
Dec 15 
comment 
Large abelian characteristic subgroups in abelianbycountable groups
Oh, yes. I mixed up finitely supported matrices and matrices $f$ with finite $\hbox{rank}(f1)$ (i.e operators $f$ such that $\hbox{ker}(f1)$ 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.) 