bio | website | halgebra.math.msu.su/staff/… |
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location | Moscow | |
age | ||
visits | member for | 1 year, 10 months |
seen | Apr 1 at 17:06 | |
stats | profile views | 997 |
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 |