445 reputation
212
bio website
location Université Kasdi Merbah Ouargla, Ouargla, Algeria
age 29
visits member for 2 years, 6 months
seen Sep 2 at 9:51

"Aujourd’hui, j’exprimerais la leçon que j’ai apprise alors ainsi : "les scientifiques", des plus illustres aux plus obscurs, sont des gens exactement comme tous les autres ! Je m’étais complu à m’imaginer que "nous" étions quelque chose de mieux, que nous avions quelque chose en sus - il m’a fallu bien un an ou deux pour me débarrasser de cette illusion-là, décidément tenace !"

Alexandar Grothendieck.


Aug
28
comment Generalized identities of (soluble) groups
$x^a=a^{-1}xa$, is this sufficient? In your comment, you mean $x^{a_1}\dots x^{a_n}=1$,...
Aug
27
comment Nilpotent pro-$p$ groups
It can be embedded in $GL_n(\mathbb{Z}_p)$, for some $n$ (such a group is $p$-adic analytic).
Aug
27
asked Generalized identities of (soluble) groups
Aug
8
awarded  Self-Learner
Aug
8
awarded  Yearling
Jul
25
comment A question on direct limits of finite $p$-groups
Artuto, I'm really thankful (sorry for the delay).
Jul
25
accepted A question on direct limits of finite $p$-groups
Jul
25
revised Torsion in profinite groups
deleted 43 characters in body
Jul
25
revised Torsion in profinite groups
edited body
Jul
25
revised Torsion in profinite groups
added 707 characters in body
Jul
22
comment Torsion in profinite groups
You have mentioned that Zelmanov's solution of RBP, implies that every finitely generated torsion profinite group is finite. I'm not sure about this; it is more safe to say that it implies that every finitely generated profinite group of finite exponent is finite. The result that you mentioned follows from a more general result of Zelmanov (which I'm not sure that is equivalent to the positive solution of the RBP).
Jul
22
comment Torsion in profinite groups
-containing $\overline{G^n}$. Pick such a $N$ such that $G/N$ has the maximal possible order. If $M$ is another normal open subgroup containing $\overline{G^n}$, then so is $M\cap N$, if $N$ does not lie in $M$, then $G/(M\cap N)$ has order greater than $|G/N|$, a contradiction. Thus $N$ is contained in every normal open subgroup containing $\overline{G^n}$; so $N=\overline{G^n}$.
Jul
22
comment Torsion in profinite groups
@Pablo: If there is a bound on the n's, we may assume that $x^n\in K$, for some $n$ and for all $x$. The subgroup $G^n$ generated by all the $x^n$'s lies in $K$, and so is $\overline{G^n}$. We have only to show that $G/\overline{G^n}$ is finite. First, note that $\overline{G^n}$ is the intersection of open normal subgroups containing it. By the solution of RBP, there is a bound $f(d,n)$ on the orders of $d$-generated finite $p$-groups satisfying the identity $x^n$. If we assume that $G$ is $d$-generated, then $|G/N|$ is bounded by $f(d,n)$, for every normal open subgroup $N$
Jul
22
revised Torsion in profinite groups
added 269 characters in body
Jul
21
comment Torsion in profinite groups
As usual, I done a stupid mistake. I'm using trivially to mean "empty". The answer is false. I have to delete it.
Jul
21
comment Torsion in profinite groups
Ok, in a compact space, from every family of closed subsets which intersect trivially, one can extract a finite subfamily which intersects trivially. (Please check if I done a mistake in the answer; usually, this is the case)
Jul
21
awarded  Revival
Jul
21
revised Torsion in profinite groups
added 86 characters in body
Jul
21
answered Torsion in profinite groups
Jul
21
comment Is there a Noetherian profinite group of infinite rank?
The question is well-known for pro-$p$ groups. You can find it under the form : is every noetherian pro-$p$ group analytique $p$-adic? (See for instance "New Horizons in pro-$p$ Groups", by du Sautoy et Al; Appendix, Problem 1.