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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 |