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Let G be a compact group, H be an open subgroup of G. Show that if H is profinite, then G is also profinite.

Lemma to use as a hint is this:

Let G be a compact group and $ {N_i | i \in I}$ be directed family of closed normal subgroups of G of finite index such that $\cap N_i=1.$(i.e. intersection of them is 1). Then G is profinite.

I know that if H is open subgroup, then H is closed of finite index and since it is profinite, H is inverse limit of inverse limit systen of finite groups. Somehow I have to construct these $N_i$‘s from the closed subgroups that construct H. I also know that intersection of all open normal sungroups is 1. But I cant see the way to combine al of these. Any hint is welcomed.

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An application of the Van-Dantzig theorem is that a topological group (Hausdorff) is profinite if and only if it is compact and totally disconnected. Now, the proof follows easily: H is clopen and totally disconnected, hence G is as well. But G is compact and so it is profinite.

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