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seen Apr 25 at 1:16

Apr
24
comment 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
comment 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
comment Almost uniquely generated groups
Oh yes, thank you!
Jan
17
comment 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
comment Almost uniquely generated groups
Why $\{a_{1},b_{1},g_{2},g_{3}, \ldots, g_{n} \}$ is irredundant?
Jan
17
comment 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
comment Almost uniquely generated groups
Yes. And probably you mean Question 2 as you use that the orders are finite.
Jan
16
comment Almost uniquely generated groups
This condition is indeed weaker. See the (wrong) answer of M. Shahryari.
Jan
16
comment Almost uniquely generated groups
No. Even $F_1$ does not satisfy. The question is about inclusion-minimal sets.
Jan
16
comment 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
comment 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
comment 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
comment 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
comment 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
comment 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.)
Dec
15
comment Large abelian characteristic subgroups in abelian-by-countable groups
Excuse me, Yves. I do not understand your comment. The finitely supported matrices form a normal subgroup of $G$, right? If they do, then the commutator must be finitely supported.
Dec
14
comment Large abelian characteristic subgroups in abelian-by-countable groups
Thank you, Yves!! Surely, you mean $g(e_q)\in e_q+Vect(\dots)$ in the first paragraph. Also, $G_K$ in the second paragraph is the same as $G_K^Q$. Furthermore, I guess that the "easy play with commutators" is something like the following. The commutator of any matrix and an elementary matrix is finitely supported, i.e. this commutator lies in a finite-dimensional unitriangular subgroup $UT_n(K)$ which is nilpotent and, hence, any its non-trivial normal subgroup non-trivially intersects the centre, which consists of elementary matrices.
Dec
12
comment Commutator Width of a direct limit of hyperbolic groups
I do not understand. Any group having only finitely many conjugacy classes must have finite commutator width, because conjugate elements have the same commutator length.
Dec
11
comment Applications of the Chinese remainder theorem
Excuse me, KConrad (and @Zack), I do not understand the problem about squares. Suppose that we have integers $a$ and $b$ such that $f(a)\ne0$ modulo $p^2$ and $f(b)\ne0$ modulo $q^2$, then CRT provides us with an integer $c$ equal to $a$ modulo $p^2$ and equal to $b$ modulo $q^2$. Thus, $f(c)=f(a)\ne0$ modulo $p^2$ and $f(c)=f(b)\ne0$ modulo $q^2$, which is a contradiction. Or I miss something?
Nov
5
comment Minimal number of generators of subgroups of Noetherian groups
@Ian, indeed, "Are all finitely presented Noetherian groups virtually polycyclic? is (an open) Question 11.38 (due to S.V.Ivanov, 1990) from the Kourovka Notebook.