bio | website | halgebra.math.msu.su/staff/… |
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location | Moscow | |
age | ||
visits | member for | 2 years, 1 month |
seen | Apr 25 at 1:16 | |
stats | profile views | 1,012 |
Apr 1 |
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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 |
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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 |
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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 |
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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.
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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 |
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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 |
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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
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Dec 15 |
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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 |
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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 |
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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 |
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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 |
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Kernel of linear representation of Baumslag-Solitar group
The answer is No, if you take $|n|=1=|m|$. |
Nov 7 |
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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 |