1,687 reputation
418
bio website halgebra.math.msu.su/staff/…
location Moscow
age
visits member for 1 year, 10 months
seen Apr 1 at 17:06

Apr
1
comment Intersection of conjugates of subgroups in free groups
I do not know any references, but I think Fact 1 can be strengthened: $A$ and $B^f$ generate their free product for some $f$.
Apr
1
revised An element $g$ in a group such that neither $g=1$ nor $g\ne 1$ can be proved.
deleted 4 characters in body
Mar
31
answered Number of relations and free subgroups
Mar
31
answered Groups where every two generator subgroup is free
Mar
30
answered An element $g$ in a group such that neither $g=1$ nor $g\ne 1$ can be proved.
Feb
9
comment Ring-theoretic version of a matrix problem
Felix, it is a mistake. A sum of four orthogonal matrices cannot be arbitrary large. Clearly, the matrix $100I$ is NOT a sum of four orthogonal matrices.
Dec
15
comment Polynomial maps between noncommutative groups
They have a link to the English translation: dx.doi.org/10.1023%2FA%3A1025001013073 (full text is freely available).
Dec
15
revised Polynomial maps between noncommutative groups
added 2 characters in body
Dec
15
comment Given a rational number a/b does there exist a finite group G and an automorphism f s.t. f maps exactly a/b elements of G to their own inverses?
Yes, this is a well-known chestnut: mathoverflow.net/questions/48 .
Dec
14
answered Polynomial maps between noncommutative groups
Dec
11
comment Ascending chain condition on ideals of free products
And my is an amalgamated free product of two free groups.
Dec
11
answered Ascending chain condition on ideals of free products
Nov
14
comment Kernel of linear representation of Baumslag-Solitar group
Actually, if $|m|=|n|$, then $f$ is not njective and its kernel is not generated by commutators (1).
Nov
13
comment Kernel of linear representation of Baumslag-Solitar group
If $|m|=|n|=1$, then $f$ is not injective and the kernel is not generated by the commutators (1).
Nov
13
comment Kernel of linear representation of Baumslag-Solitar group
The answer is No, if you take $|n|=1=|m|$.
Nov
7
comment Applications of Frobenius theorem and conjecture
@Nick, just to clarify. The theory of groups was written by Marshall Hall, and the paper I cited is authored by Philip Hall. The theorem you mentioned about the number of solution to $x^n=c$ belongs to Frobenius (according to Philip Hall, see the same paper). P.Hall's generalisation is much more complicated.
Oct
23
answered Sylow theorems for infinite groups
Oct
12
awarded  Good Answer
Oct
11
comment Does $C'\left(\frac{5}{11}\right)$ imply exponential growth?
I would ask a more specific question. Does $\mathbb Z^2$ have a $C'(1/6+\varepsilon)$-presentation?
Oct
11
awarded  Commentator