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For some reason I cannot figure out the following.

Suppose $G$ is a compact, Hausdorff and totally disconnected (but I think compact should suffice) group and $U$ an open subgroup. Why is the normalizer of $U$ open?

This was mentioned in a book I read some time ago but I don't see why.

Edit: As said below $U \subseteq N(U)$ and therefore $\infty > [G:U] > [G:N(U)]$.

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migrated from Aug 12 '13 at 16:52

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because it contains $U$ –  user35353 Aug 12 '13 at 16:04
Ah yes of course, I get it. –  Horstenson Aug 12 '13 at 16:30
The finiteness of indices is a red herring. A supergroup of an open group is always open, being a union of open sets (viz. cosets of the smaller group). This works in any topological group, you don’t need it to be compact or Hausdorff. –  Emil Jeřábek Aug 12 '13 at 16:51
See also "Subgroups of finite index in profinite groups" –  Dietrich Burde Aug 12 '13 at 18:32

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