At page 79 of Basic Number Theory by André Weil, there is an argument showing that a subgroup of a topological group is discrete,

because there is a compact neighbourhood of $1$ with finite intersection with that subgroup $\Gamma$.

So I am wondering how one proves the following:

If a sub-group of a topological Hausdorff group has finite intersection with a compact neighbourhood of $1$, then it is discrete.

Since one has earlier spotted the statement that a discrete subset of a compact set is finite, I think, in this book, by discrete one understands closed discrete.
Now, we know that this subset $\Gamma$ has a discrete intersection with a compact neighbourhood. But how could this fact help us showing the discreteness of the whole subset $\Gamma$? This is where I am stuck.
Any hint is well-appreciated.

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    $\begingroup$ Hello awllower! Since the intersection is finite, we know that the subspace described by the intersection is finite. So, in particular, there exists a neighborhood $U$ of $1$ in $G$ such that $U\cap \Gamma=\{1\}$. Then, for any point $g\in\Gamma$ you have that $\{g\}=gU\cap\Gamma$. Doesn't this show discreteness? Perhaps I am misunderstanding something. $\endgroup$ – Alex Youcis Aug 6 '13 at 7:25
  • $\begingroup$ @AlexYoucis Thanks for the response; might I ask how do you deduce the existence of such $U$? This is indeed a crucial step. $\endgroup$ – awllower Aug 6 '13 at 7:28
  • $\begingroup$ Perhaps I am misunderstanding your use of "compact neighborhood"--I took it to mean in the Bourbaki sense. If so, then you have some open set $V$ of $1$ such that $V\cap\Gamma$ is finite. Take open sets separating $1$ from all the other elements of $V\cap\Gamma$ and then intersect them all as well as with $V$. $\endgroup$ – Alex Youcis Aug 6 '13 at 7:30
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    $\begingroup$ Oh! So the point is the Hausdorff-ness of the space, right? Thanks for this approach! $\endgroup$ – awllower Aug 6 '13 at 7:32
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    $\begingroup$ Yes, for sure! Groups that aren't Hausdorff aren't even $T_0$! Don't mess with them. $\endgroup$ – Alex Youcis Aug 6 '13 at 7:33

Let me put the comments of Alex Youcis together to see if I understand rightly.
Firstly, we need only to find one neighbourhood of $1$ that does not contain any other elements of $\Gamma$, since we are working with a topological group.
Then we use the Hausdorff-ness of the group to find neighbourhoods $U_i$ of $1$ such that $x_i\not\in U_i \forall x_i\in\Gamma\cap V, x_i\not=1,$ where $V$ is the neighbourhood in question. And the desired neighbourhood is just $\bigcap U_i\cap V,$ a finite (number of ) intersection(s) by assumption.
Indications of errors are mostly welcomed.

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    $\begingroup$ Unsurprisingly, it looks right to me. $\endgroup$ – Alex Youcis Aug 6 '13 at 7:43

Alex's comments, as turned into an answer by the OP, are indeed correct.

Let me just point out that the Hausdorff condition really is the key here. For instance, let $G$ be any group admitting a nontrivial finite subgroup $H$. Endow $G$ with the indiscrete topology (in which the only open subsets are $\varnothing$ and $G$): this makes $G$ into a non-Hausdorff topological group. But the desired conclusion fails: $G$ is a compact neighborhood of the identity which has finite intersection with the subgroup $H$, and $H$ is not discrete (rather it is indiscrete).

  • $\begingroup$ Thanks for this illusstrating example! $\endgroup$ – awllower Aug 7 '13 at 4:28

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