Yassine Guerboussa
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 Sep 19 comment Generalized identities of (soluble) groups Thank you so much and sorry for the delay. Quite elementary and ingenious argument. You mean that $k$ is the minimal degree of a possible identity, as $k$ can have arbitray large values (for instance consider any multiple of $qp$). Sep 19 accepted Generalized identities of (soluble) groups 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