bio | website | halgebra.math.msu.su/staff/… |
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location | Moscow | |
age | ||
visits | member for | 2 years, 3 months |
seen | Sep 6 at 13:50 | |
stats | profile views | 1,085 |
Aug 27 |
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Are infinite groups “locally topologizable”?
It is on page 339 of English edition. Unfortunately, Google preview rejects to show this page to me (but I may read the table of contents). |
Aug 26 |
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Are infinite groups “locally topologizable”?
I have a Russian edition. This is Chapter 10, Section 31, Subsection 3. |
Aug 25 |
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Invariant planes of a nilpotent matrix with two Jordan blocks of size two
This is the description of all of them, because any invariant subspace containing a vector $v$ shoud contain $Nv$. |
Aug 25 |
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Invariant planes of a nilpotent matrix with two Jordan blocks of size two
Take any vector $v$ not belonging to the kernel of $N$ and consider the space $\langle v,Nv\rangle$. This family and the kernel of $N$ are all spaces you ask about. |
Aug 23 |
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A sequence of subsets of an infinite group
A related question: mathoverflow.net/q/179177/24165 . |
Apr 24 |
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An element $g$ in a group such that neither $g=1$ nor $g\ne 1$ can be proved.
@Dan, surely. What you said is basically Metatheorem 3.5.2 from my answer. |
Jan 20 |
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Why the axiomatic rank of the variety of groups is equal to three?
Your octonion argument is much simpler and works perfectly. The algebra of all unit octonions or all non-zero octonions (with operations $\cdot$, ${\ }^{-1}$, and $1$) is non-associative but satisfies all two-variable laws that are consequences of the group axioms. |
Jan 17 |
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Almost uniquely generated groups
Oh yes, thank you! |
Jan 17 |
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Almost uniquely generated groups
"Any finite $p$-group which is relatively free in some variety has this property by the same argument. In light of Geoff's answer perhaps these are the only ones?" -- No. The quaternion group of order 8 is not relatively free but satisfies the conditions. |
Jan 17 |
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Almost uniquely generated groups
Why $\{a_{1},b_{1},g_{2},g_{3}, \ldots, g_{n} \}$ is irredundant? |
Jan 17 |
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Almost uniquely generated groups
No. $S$ is unique up to automorphisms; so, the complement of $S$ is not necessary the Frattini subgroup.Take a cyclic group of prime order, for example. |
Jan 16 |
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Almost uniquely generated groups
Yes. And probably you mean Question 2 as you use that the orders are finite. |
Jan 16 |
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Almost uniquely generated groups
This condition is indeed weaker. See the (wrong) answer of M. Shahryari. |
Jan 16 |
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Almost uniquely generated groups
No. Even $F_1$ does not satisfy. The question is about inclusion-minimal sets. |
Jan 16 |
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Almost uniquely generated groups
It seems that you use the letter $n$ in two different senses. You are talking about $m$-generated free groups in the varieties of nilpotent groups of exponent $p^n$, right? |
Jan 14 |
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Minimal generating sets of groups
Actually, any (inclusion-)minimal generating set of a Tarski monster consists of two elements (if $p$ is prime). |
Jan 9 |
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relatively free groups in $Var(S_3)$
Yes, exactly. By the way ,the group of polynomial-with-coefficients functions is also free but in a variety of different universal algebras --- a variety of groups with marked elements (i.e., groups with additional 0-ary operations. |
Jan 9 |
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relatively free groups in $Var(S_3)$
@M.Shahryari, see the edit. I guess that 324 is the number of one -variable polynomial functions over $S_3$ in a different sense of the word polynomial; propably they mean polynomials with coefficients from the group --- such as $x(12)x^2(123)$ |
Dec 17 |
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Large abelian characteristic subgroups in abelian-by-countable groups
Dear Yves, if you do not mind, we shall include your example in our preprint (op. cit.) with a proper reference of course. As for the Podoski--Szegedy example, it is maybe simpler but we do not know how to convert it to the normal-but-not-characteristic settings. Their example is (in essence) the subgroup $\hbox{Alt}(\mathbb Z)A$ of the uncountable symmetric group $\hbox{Sym}(\mathbb Z)$, where $\hbox{Alt}(\mathbb Z)$ is the alternating group (consisting of even finitary permutations) and $A$ is the elementary abelian 2-group consisting of permutations preserving absolute values of integers. |
Dec 15 |
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Large abelian characteristic subgroups in abelian-by-countable groups
Oh, yes. I mixed up finitely supported matrices and matrices $f$ with finite $\hbox{rank}(f-1)$ (i.e operators $f$ such that $\hbox{ker}(f-1)$ is of finite codimension). The latter form a normal subgroup but the former do not. Inside $H$, these two subgroups coincides, right? (I mean, their intersections with $H$ coincides.) |