1,702 reputation
519
bio website halgebra.math.msu.su/staff/…
location Moscow
age
visits member for 2 years, 1 month
seen Apr 25 at 1:16

Apr
1
comment Intersection of conjugates of subgroups in free groups
Ashot, you are right. But this is the only problem. Hence, Fact 1 is reduced to the case where $A=\langle x\rangle$ is cyclic (generated by a letter). Now, once again recall that $B$ is a free factor in a finite-index subgroup. This reduces the situation to the case where $B$ is a free factor of $F$ and $A$ is cyclic (arbitrary). In this case, we have no problems, right?
Apr
1
comment Intersection of conjugates of subgroups in free groups
If $A$ is a free factor, i.e. $A$ is generated by some letters (after a change of basis), then it is easy to conjugate $B$ so that all elements of $B^f$ would start and end with a letter not belonging to $A$. (You can take $f=x^n$, where $x$ is a letter not lying in $A$ and $n$ is an integer large enough with respect to the Schreier basis of $B$.)
Apr
1
comment Intersection of conjugates of subgroups in free groups
If you like to avoid mentioning graphs, you may recall that any f.g. subgroup is just a free factor of a finite-index subgroup. This reduces the problem to the case where $A$ is a free factor of $F$...
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