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.

A profinite group is by defination a topological group $G$ which is Hausdorff , compact and totally disconnected. How to prove the following equivalent defination:

A compact Hausdorff group is profinite if and only if its neutral element admits a basis of neighbourhoods consisting only of nomal subgroups.

Besides, for proving the "$\Leftarrow$" direction, I only use the fact: its neutral element admits a basis of neighbourhoods consisting only of $subgroups$. So is it also true that "A compact, Hausdorff group with the neutral element admits a basis of neighbourhoods consisting only of $subgroups$ is a profinite group.

share|improve this question
    
You can see a proof of the equivalence of the definitions in Neurkirch, Cohomology of number field (the very first pages). (books.google.com/…). The $\Rightarrow$ part is quite tricky. –  user10676 Aug 4 '11 at 12:37
    
Great, thank you! –  Li Zhan Aug 4 '11 at 13:16

1 Answer 1

up vote 7 down vote accepted

The existence of a basis of neighborhoods made up of subgroups is equivalent to the existence of a basis of neighborhoods made up of normal subgroups, given that you are assuming compactness.

If $H$ is an open subgroup of a compact topological group, then since the cosets of $H$ cover $G$ and $G$ is compact, it follows that $H$ is of finite index and in particular is clopen. The intersection of all conjugates of $H$ is therefore also of finite index and also clopen (only finitely many distinct conjugates of $H$, since its normalizer is of finite index), and of course normal. So if you replace each subgroup in a basis by the intersection of its conjugates, you obtain a basis made up of normal subgroups. There is, thus, no more generality in requiring the subgroups to be normal.

share|improve this answer
    
Thank you for your answer to my second question! –  Li Zhan Aug 4 '11 at 13:09

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.