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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)

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up vote 5 down vote accepted

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.

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