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Let $K \leq H$ be two subgroups of a topological group $G$ and suppose that $K$ has finite index in $H$. Does it follow that $\bar{K}$ has finite index in $\bar{H}$ ?

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Yes, if your group has a countable basis (so the closure of a set is the set of limits of sequences of points in that set). Let $\{h_1,...,h_n\}$ be representatives of the left cosets of $K$ in $H$. Then for any point $h \in \bar{H}$, express it as a limit of a sequence $x_1, x_2, x_3, ...$ of points $x_i \in H$. These points can then be represented using the coset representatives as $h_{i_1}y_1, h_{i_2}y_2, h_{i_e}y_3,...$, with $i_k \in \{1,...,n\}$ and $y_n \in K$.

Now some $h_k$ must appear infinitely often, thus by passing to a subsequence we have that $h$ is the limit of a sequence $h_ky_1, h_ky_2, h_ky_3,...$ and so by the continuity of group multiplication we have $h = h_ky$, where $y \in \bar{K}$ is the limit of the $y_i$.

This shows that any point $h \in \bar{H}$ lies in $h_k\bar{K}$ for some $k \in \{1,...,n\}$, thus the same coset representatives of $K$ in $H$ also form a set of (possibly redundant) coset representatives of $\bar{K}$ in $\bar{H}$, and in particular $\bar{K}$ has finite index in $\bar{H}$.

I suspect the sequences in this argument could be adapted into more of a "for any open set around $h$, there is ..." sort of language to make the proof work for general topological groups, but I prefer the intuition you get with sequences.

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Probably you could replace "sequence" with "net" and get the result without the first countability (I guess that's what you really want, instead of second countability). With nets you can show that the closure of $h_iK$ is $h_i\bar{K}$, and then, since closure commutes with finite unions, $\bar{H}=\bigcup_{i=1}^n h_i\bar{K}$. –  Keenan Kidwell Sep 27 '11 at 23:39
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You don't even need nets. We have $\overline{h_iK}\subset h_i\overline{K}$ because $h_i\overline{K}$ is a closed set containing $h_iK$ (translation is a homeomorphism). Conversely, the statement $\overline{h_iK}\supset h_i\overline{K}$ is equivalent to $h_i^{-1}\overline{h_iK}\supset \overline{K}$, and again this is clear because $h_i^{-1}\overline{h_iK}$ is a closed set containing $K$. –  Kevin Sep 28 '11 at 0:30
    
@Kevin Oh yeah, there you go. –  Keenan Kidwell Sep 28 '11 at 1:33
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The answer is yes in general, and here is a proof, which is an adaptation of MartianInvader's:

Let $K$ have finite index in $H$, with coset reps. $h_1,\ldots,h_n$. Since multiplication by any element of $G$ is a homeomorphism from $G$ to itself (since $G$ is a topological group), we see that each coset $h_i \overline{K}$ is a closed subset of $G$, and hence so is their union $h_1\overline{K} \cup \cdots \cup h_n \overline{K}$. Now this closed set contains $H$, and hence contains $\overline{H}$. Thus $\overline{H}$ is contained in the union of finitely many $\overline{K}$ cosets, and hence contains $\overline{K}$ with finite index.

[Note: I hadn't paid proper attention to Keenan and Kevin's comments on MartianInvader's answer when I wrote this, and this answer essentially replicates the content of their comments.]

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