Let $\Gamma_g$ be a surface group of genus $g \geq 2$, that is we have a presentation: $$\Gamma_g = \langle x_1,y_1 \dots, x_g,y_g \vert \prod_{i = 1}^g [x_i,y_i] = 1\rangle$$

Let $H \leq \Gamma_g$ be a finitely generated subgroup.

Must $H$ be closed in the profinite topology on $\Gamma_g$?

That is, must there be a family of finite index subgroups of $\Gamma_g$ intersecting in $H$?

If the answer is no, then must there be for any $M \in \mathbb{N}$ a subgroup $H \leq K \leq \Gamma_g$ with $\infty > [\Gamma_g : K] \geq M$?


closed as off-topic by Benjamin Steinberg, abx, YCor, Stefan Kohl, Ryan Budney Oct 24 '14 at 17:08

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    $\begingroup$ Yes, surface group are LERF, it's Peter Scott's theorem. $\endgroup$ – YCor Oct 24 '14 at 8:49
  • $\begingroup$ Great! Do you have a reference? You can post this as an answer. $\endgroup$ – Pablo Oct 24 '14 at 9:24
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    $\begingroup$ See the references here: arxiv.org/abs/1204.5135. It includes the reference to Scott original paper, and an erratum. $\endgroup$ – YCor Oct 24 '14 at 9:29
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    $\begingroup$ voting to close as no longer relevant because it is answered in comments. $\endgroup$ – Benjamin Steinberg Oct 24 '14 at 14:23