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Aug
8
answered Subgroups of p-groups
Aug
5
comment Sets which are not fixed by any non-identity isomorphism
You are right about $F_3$. So, to answer the question "Find all fields..." it remains to analyse the two-dimensional space over the three-element field. This can be done by brute force surely. But to answer the question "Find all spaces..." we have to deal with almost all spaces over the two-element field...
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
comment 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
comment 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
comment 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
comment Sets which are not fixed by any non-identity isomorphism
What if the characteristic is 2?
Aug
4
comment 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
comment 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
comment 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
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