Anton Klyachko
|
Registered User
|
|
|
Jun 3 |
awarded | ● Yearling |
|
May 12 |
comment |
Magic trick based on deep mathematics The sum of all elements of a finite abelian group is non-zero if and only if the 2-Sylow subgroup of this group is a nontrivial cyclic. |
|
May 12 |
comment |
Example of a group with unsolvable word problem The Russian original of Borisov's paper is freely available here: mi.mathnet.ru/eng/mz6959 |
|
Apr 11 |
comment |
Elements of minimal length in normal closures of elements in free groups Thank you, Alexey. I did not know that. |
|
Apr 11 |
comment |
Elements of minimal length in normal closures of elements in free groups I think there are no good answers. Sometimes there are no non-trivial normal roots, e.g., when $w$ is a proper power (by Newman's theorem); sometimes such normal roots exist, e.g., $[x,y]\in \langle\langle xy^{2013}\rangle\rangle$. |
|
Apr 11 |
awarded | ● Scholar |
|
Apr 11 |
comment |
The sum of same powers of all matrices modulo p Matthieu, the solution goes as follows: Ilya showed that the sum in question vanishes for sufficiently small $k$; then ya-tayr showed that if the sum vanishes for all sufficiently large $k$, than it vanishes always; when these bounds have met, "we're done!". This is all right, but honestly I hoped for a simpler solution. ---- Thanks, ya-tayr, Ilya, Will, and everybody involved! |
|
Apr 9 |
comment |
The sum of same powers of all matrices modulo p ... and it is easy to find the LCM of all unitary polynomials of degree $p$. |
|
Apr 9 |
comment |
The sum of same powers of all matrices modulo p ya-tayr, Ilya, I am not sure I understand. Should "determinant of $A$" read "determinant of $I-Ax$"? If so, there are exactly $p^{p-1}$ different denominators: they are just polynonomials reciprocal to characteristic polynomials of all matrices (= all unitary polynomials of degree $p$). |
|
Apr 8 |
comment |
The sum of same powers of all matrices modulo p Thanks, Ilya!!! |
|
Apr 8 |
comment |
The sum of same powers of all matrices modulo p Thanks, Sergey! |
|
Apr 8 |
comment |
The sum of same powers of all matrices modulo p Will, if the size is not a multiple of the characteristic, it suffice to evaluate the sum of traces. because the sum in question is, clearly, a scalar matrix. But I argee that the problem seems non-trivial for any sizes. The case where size $=p$ is just an additional difficulty. |
|
Apr 7 |
awarded | ● Nice Question |
|
Apr 7 |
asked | The sum of same powers of all matrices modulo p |
|
Apr 4 |
comment |
Embedding a semigroup into a divisible semigroup The Russian original of Shutov's paper is freely available: mathnet.ru/php/… . |
|
Apr 1 |
comment |
Intersection of conjugates of subgroups in free groups Ashot, if you find $g$ such no power of $x$ belongs to $B^g$, then the problem would be solved (just conjugate $B$ by $g$ and then by $x^n$ and the resulting group $B^{gx^n}$ intersects $A$ trivially). So, the problem is as follow: you have two subgroups: cyclic $\langle x\rangle$ and infinite−index f.g. $B$ and have to find an element $g$ such that $\langle x\rangle\cap B^g=1$. This is the same as Fact 1 but with cyclic subgroup instead of $A$. |
|
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. |
