# Strongly complete profinite group

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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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. $\Box$
The kernel of this action is contained in $H$ (because $g \in G$ fixing all cosets implies $g$ fixing the coset $H$), so it should have bigger index than $|G/H|$... right? still, it should have finite index because $G/H$ is a finite set, so we can take representatives and write $\bigcap_{g \in G} gHg^{-1}$ as a finite intersection of subgroups of $G$ of finite index $H_i \overset{def}= g_i H g_i^{-1}$, from which we deduce $$|G/(H_1 \cap \cdots \cap H_n)| \le |G/H_1| \cdots |G/H_n| = |G/H|^n < \infty.$$ I always apply the Orbit-Stabilizer theorem upside down, too... – Patrick Da Silva Nov 9 '14 at 21:08
@Patrick: did you not see the factorial symbol? The kernel of the action has index dividing $|G/H|!$ because that's the size of the group of permutations of $|G/H|$. – Qiaochu Yuan Nov 9 '14 at 21:37