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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.

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    $\begingroup$ 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. $\endgroup$ – user10676 Aug 4 '11 at 12:37
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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.

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