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seen Apr 25 at 1:16

Jul
27
comment Generalized free product of semigroups with amalgamated subsemigroups
@BorisNovikov, yes, I do not fully understand the question. Note that even for groups, if we have 3 groups $G_i$ with 3 subgroups $H_{ij}=G_i\cap G_j$, this amalgam may be non-embeddable in any common group. There is, however, the notion (due to Bass and Serr) of the fundamental group of a graph of groups, but this is not an amalgamated product, this is a composition of amalgamated products and HNN-extensions (see, Misha's comment). What is your question, actually?
Jul
27
comment Generalized free product of semigroups with amalgamated subsemigroups
@AndreasBlass, surely, you are right.
Jul
27
answered Generalized free product of semigroups with amalgamated subsemigroups
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
accepted The sum of same powers of all matrices modulo p
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 $A$.