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Definition: Let $G$ be a group. $G$ is said to be double coset separable if given any finitely generated subgroups $H$ and $K$ in $G$, given any $g\in G$ and $h\not\in HgK$, there exists a finite index normal subgroup $G_0$ in $G$ such that if $\pi$ is the projection of $G$ onto $G/G_0$, then $$\pi(h)\not\in\pi (HgK).$$

Here G. Niblo proved that

Theorem: A surface gorup is double coset separable.

Definition: Let $\{\Gamma_m\}_{m\in\mathbb{N}}$ be a sequence of nested finite index normal subgroups of $\Gamma_0=\Gamma$. We say that $\{\Gamma_m\}_{m\in\mathbb{N}}$ is a vanishing sequence if for all $\gamma ,\eta \in \Gamma$ and for any set $H$ invariant by left multiplication by $\gamma$ and right multiplication by $\eta$ whose projection in $<\eta> \setminus \Gamma /<\gamma>$ is finite, there exists $n_0$ such that for all $n>n_0$, $H\subset \Gamma_n\cap<\eta>.<\gamma>$.

Here Corollary 11.0.5 F. Labourie stated that

As surface subgroups are double coset separable and countable, Vanishing sequence exists.

Q) What is the proof of the above statement?

PS: I can show double coset separability inpleis subgroup separability and residually finiteness, but I can't prove the above statement.

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This is not correct (the stated conclusion implies $H \subset \langle \eta \rangle \cdot \langle \gamma \rangle$, which need not be true), so there must be a typographical error. Probably the conclusion should be $H \cap \Gamma_n \subset \langle \eta \rangle \cdot \langle \gamma \rangle$. Since $\Gamma$ is countable, there are only countably many possibilities for the triple $(\eta, \gamma, H)$. Given $\eta_m$, $\gamma_m$ and $H_m$ as described, double coset separability implies there is a finite-index normal subgroup $\Gamma_m'$, such that $H_m \cap \Gamma_m' \subset \langle \eta_m \rangle \cdot \langle \gamma_m \rangle$. Let $\Gamma_n = \bigcap_{m=1}^n \Gamma_m'$.

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  • $\begingroup$ Why double coset separability implies that there is a finite-index normal subgroup $\Gamma_m'$, such that $H_m \cap \Gamma_m' \subset \langle \eta_m \rangle \cdot \langle \gamma_m \rangle$? From my point of view double coset separability tells you what is not in $\Gamma_n$. How does it implies something is in $\Gamma_n.$ $\endgroup$
    – Cusp
    Commented Mar 27, 2015 at 8:47
  • $\begingroup$ You are absolutely right that double coset separability only tells us that something is not in $\Gamma_n$. (So the original conclusion cannot possibly be correct.) However, telling us something is not in $\Gamma_n$ is the same as putting a restriction on the things that are in $\Gamma_n$. $\endgroup$
    – Dave Witte Morris
    Commented Mar 27, 2015 at 16:12
  • $\begingroup$ To simplify, suppose the image of $H_m$ in $\langle\eta\rangle \backslash \Gamma /\langle\gamma\rangle$ has only one double coset $\langle\eta\rangle h \langle\gamma\rangle$ that is not in $\langle\eta\rangle \cdot \langle\gamma\rangle$. By double coset separability, there exists $\Gamma_m'$, such that $h \notin \langle\eta\rangle \Gamma_m' \langle\gamma\rangle$. Since $\langle\eta\rangle h \langle\gamma\rangle$ is the only one double coset that is not in $\langle\eta\rangle \cdot \langle\gamma\rangle$, this implies $H_m \cap \Gamma_m' \subseteq \langle\eta\rangle \cdot \langle\gamma\rangle$. $\endgroup$
    – Dave Witte Morris
    Commented Mar 27, 2015 at 16:14
  • $\begingroup$ The key point in my answer is that I moved the assertion "$h \in \Gamma_n$" from the conclusion to the hypotheses. $\endgroup$
    – Dave Witte Morris
    Commented Mar 27, 2015 at 18:47
  • $\begingroup$ Indeed, there was a typo here and the correct (and needed) conclusion is the one stated by Dave. $\endgroup$
    – François Labourie
    Commented Nov 15, 2016 at 11:35

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