bio | website | halgebra.math.msu.su/staff/… |
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location | Moscow | |
age | ||
visits | member for | 2 years, 3 months |
seen | 12 hours ago | |
stats | profile views | 1,042 |
Aug 5 |
comment |
Sets which are not fixed by any non-identity isomorphism
@A.B., yes, this is nontrivial. But I understand the question as: find all fields such that, for any f.d. space over this field, blah-blah-blah. |
Aug 5 |
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Sets which are not fixed by any non-identity isomorphism
Anton, in two-dimensional space, there are only three non-zero vectors and we can choose nothing (singletons are obviously bad and two-elements subsets are no better as their complements are singletons). |
Aug 5 |
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Is there a better description of this class of discrete groups?
Now, the answer is all countable groups, because the trivial group $H$ belongs to the class by (1); so, all countable groups $G$ lie in the class by (2). |
Aug 5 |
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Sets which are not fixed by any non-identity isomorphism
So, Anton answered the question positively for any fields except two. For the two-element field, the answer is obviously negative. What about the only remaining case? |
Aug 5 |
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Sets which are not fixed by any non-identity isomorphism
What if the characteristic is 2? |
Aug 4 |
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Sets which are not fixed by any non-identity isomorphism
"we must therefore have either $\alpha_iv_i$ or $\alpha_jv_j$ in $S$". Why? We may take some combination, e.g., $v_i+\alpha v_j$... |
Jul 30 |
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Can closed compacts in a topological group behave “paradoxically” with respect to unions, intersections, and one-sided translations?
Compacts are always closed, or I miss something? |
Jul 30 |
awarded | Informed |
Jul 28 |
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Generalized free product of semigroups with amalgamated subsemigroups
@BorisNovikov, This is all right but my point is that it is much easier to answer more exact questions, something like "Is it true that...? Note that a similar statement for groups is true [HN48]." |
Jul 27 |
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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 |
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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 |
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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 |
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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 |
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Elements of minimal length in normal closures of elements in free groups
Thank you, Alexey. I did not know that. |
Apr 11 |
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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! |