TITLE: Strongly complete profinite group
QUESTION [6 upvotes]: Let $G$ be a profinite group (or equivalently a  compact and totally disconnected topological group ) with the property that all of its normal subgroups of finite index are open sets. 
Does this imply that all of its subgroups of finite index are open sets ? 
(if all subgroups of finite index from $G$ are open sets, than $G$ is called strongly complete ; this motivates the title of this post)

REPLY [7 votes]: Yes.
Lemma: Let $H$ be a subgroup of finite index in a group $G$. Then $H$ contains a normal subgroup of finite index, namely $\bigcap_{g \in G} gHg^{-1}$.
Proof. $G$ acts on the left cosets $G/H$ by translation. Since $|G/H|$ is finite, the kernel of this action has finite index (dividing $|G/H|!$), and it is precisely the above intersection. $\Box$
So every subgroup of finite index is a union of cosets of a normal subgroup of finite index. Hence if the latter are open, then so are the former.