Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\infty > [G:U] > [G:N(U)]$.

share|improve this question

migrated from mathoverflow.net Aug 12 '13 at 16:52

This question came from our site for professional mathematicians.

3  
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
1  
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" math.wisc.edu/~boston/FLTJensen.pdf. –  Dietrich Burde Aug 12 '13 at 18:32

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.