bio | website | halgebra.math.msu.su/staff/… |
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location | Moscow | |
age | ||
visits | member for | 2 years, 3 months |
seen | 9 hours ago | |
stats | profile views | 1,037 |
Apr 9 |
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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 |
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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 |
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The sum of same powers of all matrices modulo p
Thanks, Ilya!!! |
Apr 8 |
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The sum of same powers of all matrices modulo p
Thanks, Sergey! |
Apr 8 |
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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 |
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Embedding a semigroup into a divisible semigroup
The Russian original of Shutov's paper is freely available: mathnet.ru/php/… . |
Apr 1 |
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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 |
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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 |
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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 |
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Polynomial maps between noncommutative groups
added 2 characters in body |