Is there a finitely generated profinite group $G$ with a closed subgroup of infinite index $K \leq G$ such that for every $g \in G$ there exists some $n \in \mathbb{N}$ for which $g^n \in K$ ?
Can $G$ be pro-$p$, for some prime number $p$?
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$\begingroup$ Do you want to exclude groups like $\mathbb{Z}/2^\mathbb{N}$. $\endgroup$– HenrikRüpingFeb 11, 2015 at 0:45
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1$\begingroup$ @HenrikRüping I am not sure which group are you talking about, but if it is $(\mathbb{Z}/2\mathbb{Z})^\mathbb{N}$ then it is excluded since I am asking about finitely generated groups. $\endgroup$– PabloFeb 11, 2015 at 9:29
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2$\begingroup$ @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 $N$ $\endgroup$– Yassine GuerboussaJul 22, 2015 at 5:52
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2$\begingroup$ -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 $N=\overline{G^n}$. $\endgroup$– Yassine GuerboussaJul 22, 2015 at 5:59
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2$\begingroup$ 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). $\endgroup$– Yassine GuerboussaJul 22, 2015 at 6:05
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1 Answer
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Too long for a comment.
I note first that I made an attempt to reduce the problem to the case where $K$ is normal, but it turned out to be false; I'm thankful to Ian Agol for his discussion. The case where $K$ is normal follows at once from a theorem of Zelmanov stating that every periodic torsion group is locally finite.
This second attempt is not a complete answer; however, it reduces the problem considerably:
Let $G$ be as above, then, virtiually, $L_p(G)$ satisfies a PI, for every prime $p$; with $L_p(G)$ denotes the Lie algebra associated to the dimension subgroups (over the field of $p$ elements).
Indeed, let $K_n$ denote the set of elements of $G$ satisfying $x^n\in K$. As $K$ is closed, each $K_n$ is closed. By assumption, $\cup_{n\geq1} K_n=G$, so by the standard Baire category theorem, there exists $n$ such that $K_n$ contains an open subset, so there is an open normal subgroup $N$ and $t\in G$ such that $tN \subseteq K_n$.
Let $H=\langle t \rangle N$, then $H$ is open. Consider the subgroup generated by $X$, the set of the elements $(tx)^n$, with $x\in N$. Then $X$ is a normal subset of $M$, and $X \subseteq K$ by the above paragraph. It follows that the closed subgroup $L$ generated by $X$ lies $K$.
Let us work now in the finitely generated profinite group $M/L$ ($K$ may be identified with $K/L$. We have $M/L$ satisfies the coset identity $X^n=1$ with respect to $N/L$ (see Wilson and Zelmanov's http://www.sciencedirect.com/science/article/pii/0022404992901386), by the main result in the previous paper, the Lie algebra $L_p(M/L)$ satisfies a polynomial identity. This proves the claim.
Remark. If $G$ is a pro-$p$ group, then we can find a finite generating set of $M/L$ in which every element satisfies the identity $X^n=1$ (take $t$ together with $tx_1,..,tx_s$, where $x_1,..,x_s$ generate $N$; or actually thier images in $M/L$). I wished to deduce from this (using the remark by Professor Yiftach Barnea in his answer Elements of infinite order in a profinite group) that $M/L$ is finite, from which it follows that $K$ has a finite index in $G$. Unfortunately, it seems that this remark is incorrect.
answered Jul 21, 2015 at 15:35
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$\begingroup$ Maybe you can expand the argument in your fourth sentence - I don't understand why one has only finitely many conjugates to intersect in the identity. $\endgroup$– Ian AgolJul 21, 2015 at 16:27
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$\begingroup$ 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) $\endgroup$ Jul 21, 2015 at 18:40
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$\begingroup$ By intersect trivial, you mean intersect in the identity element? I don't see why this follows from point-set topology. $\endgroup$– Ian AgolJul 21, 2015 at 18:56
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$\begingroup$ As usual, I done a stupid mistake. I'm using trivially to mean "empty". The answer is false. I have to delete it. $\endgroup$ Jul 21, 2015 at 19:14