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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Sign up to join this communityIs 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 $p$?
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.