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Anton Klyachko

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 Name Anton Klyachko Member for 1 year Seen May 16 at 2:45 Website Location Moscow Age
 Jun3 awarded ● Yearling May12 comment Magic trick based on deep mathematicsThe 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. May12 comment Example of a group with unsolvable word problemThe Russian original of Borisov's paper is freely available here: mi.mathnet.ru/eng/mz6959 Apr11 comment Elements of minimal length in normal closures of elements in free groupsThank you, Alexey. I did not know that. Apr11 comment Elements of minimal length in normal closures of elements in free groupsI 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., $[x,y]\in \langle\langle xy^{2013}\rangle\rangle$. Apr11 awarded ● Scholar Apr11 comment The sum of same powers of all matrices modulo pMatthieu, 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! Apr9 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 $p$. Apr9 comment The sum of same powers of all matrices modulo pya-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 $p$). Apr8 comment The sum of same powers of all matrices modulo pThanks, Ilya!!! Apr8 comment The sum of same powers of all matrices modulo pThanks, Sergey! Apr8 comment The sum of same powers of all matrices modulo pWill, 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. Apr7 awarded ● Nice Question Apr7 asked The sum of same powers of all matrices modulo p Apr4 comment Embedding a semigroup into a divisible semigroupThe Russian original of Shutov's paper is freely available: mathnet.ru/php/… . Apr1 comment Intersection of conjugates of subgroups in free groupsAshot, 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 infiniteâˆ’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 $A$. Apr1 comment Intersection of conjugates of subgroups in free groupsAshot, 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? Apr1 comment Intersection of conjugates of subgroups in free groupsIf $A$ 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 $B$.) Apr1 comment Intersection of conjugates of subgroups in free groupsIf 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 $F$... Apr1 comment Intersection of conjugates of subgroups in free groupsI do not know any references, but I think Fact 1 can be strengthened: $A$ and $B^f$ generate their free product for some $f$. Apr1 revised An element $g$ in a group such that neither $g=1$ nor $g\ne 1$ can be proved.deleted 4 characters in body Mar31 answered Number of relations and free subgroups Mar31 answered Groups where every two generator subgroup is free Mar30 answered An element $g$ in a group such that neither $g=1$ nor $g\ne 1$ can be proved. Feb9 comment Ring-theoretic version of a matrix problemFelix, 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.