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Let $F$ be a finitely generated free group and let $S \subseteq F$ be a subset for which there is some $\epsilon > 0$ such that for any epimorphism to a finite group $\phi \colon F \to G$ we have that $\frac{|\phi(S)|}{|G|} \geq \epsilon$ (that is, the closure of $S$ in the profinite completion of $F$ has positive Haar measure).

Is it possible that the closure of $S$ with respect to the profinite topology on $F$ does not contain a coset of any finite index subgroup of $F$ ?

This is the same as asking whether it is possible that $S$ is nowhere dense (the interior of the closure is empty) with respect to the profinite topology.

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  • $\begingroup$ Is this true in $\mathbb{Z}$? $\endgroup$ – HJRW Dec 21 '14 at 16:09
  • $\begingroup$ @HJRW, After thinking of it a bit I was not able to figure it out. It can have a different answer in the nonabelian case in which I am interested most. $\endgroup$ – Pablo Dec 21 '14 at 16:12
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Yes, $S$ can be nowhere dense. The idea is to build a 'fat Cantor set' inside $F$.

Let $H_0=F,H_1,H_2,\dots$ be finite-index subgroups of $F$ such that $H_{i+1}\leq H_i$ for each $i$ and such that $\bigcap_{i=0}^\infty H_i = \{1\}$. Start with $i_0=0$ and $C_0=\{H_0\}$, and then iteratively do the following. Having defined $C_n$ as a certain set of more than half the cosets of $H_{i_n}$, let $S_n\subset F$ consist of one member from each coset $gH_{i_n}\in C_n$, and then choose $i_{n+1}$ so large that $$\frac{|C_n|}{|F/H_{i_n}|} - \frac{|C_n|}{|F/H_{i_{n+1}}|} > 1/2,$$ and so that when we decompose each $gH_{i_n} \in C_n$ into a union of cosets of $H_{i_{n+1}}$ we can remove at least one of these cosets without uncovering any member of $S_0\cup \dots \cup S_n$. Let $C_{n+1}$ be the resulting set of cosets of $H_{i_{n+1}}$.

In the end let $S=\bigcap_{n=0}^\infty \bigcup C_n$, and let $\bar{S}$ be the closure of $S$ in $\hat{F} = \lim F/H_i$. Then $S$ is closed in the profinite topology of $F$ and contains no coset of any $H_i$, so $\bar{S}$ is nowhere dense. On the other hand $S$ contains $\bigcup_{i=0}^\infty S_i$, so the image of $S$ in $F/H_i$ always has size at least $|F/H_i|/2$, so $\bar{S}$ has measure at least $1/2$.

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  • $\begingroup$ It looks like the this works for $F=\mathbb Z$, or indeed any infinite, residually finite $F$. $\endgroup$ – HJRW Dec 22 '14 at 17:25
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For the special case of $\mathbb{Z}$ we can construct a set $S=\{n_1, \dotsc, n_k, \dotsc\}$ such that $$\sum_{i=1}^\infty \frac{1}{n_i} < \infty,$$ and such that $S$ contains every residue modulo every $n>1$ (this is essentially a triviality - for each $n$ construct a complete residue set which starts at $n^3,$ or thereabouts, then take the union of all these). This will project onto every quotient of $\mathbb{Z}$ and have natural density $0.$

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  • $\begingroup$ Isn't $S$ dense in $\mathbb{Z}$? $\endgroup$ – HJRW Dec 21 '14 at 20:35
  • $\begingroup$ @HJRW It is indeed. So the closure contains a coset and this is not an answer. $\endgroup$ – Pablo Dec 21 '14 at 20:36
  • $\begingroup$ @HRJW how is this dense in $\mathbb{Z}?$ It has natural density $0,$ so does not contain any arithmetic progression... $\endgroup$ – Igor Rivin Dec 21 '14 at 21:58
  • $\begingroup$ Igor, nevertheless, it's dense in the profinite topology. $\endgroup$ – HJRW Dec 21 '14 at 22:10
  • $\begingroup$ @HJRW Sorry, I am quite dense: how so? $\endgroup$ – Igor Rivin Dec 21 '14 at 22:11

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