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Let $G$ a compact Hausdorff and totally disconnected topological group. Then every neighborhood of 1 contains an open normal subgroup of finite index in $G$. I need this lemma to prove that every compact Hausdorff and totally disconnected topological group is a profinite group. I am trying but I can not prove it. Any suggestions?

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Lemma 3.9 in here.

If and when I have more time, I can copy the proof or you can DIY.

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  • $\begingroup$ I was using the book "Introduction to Profinite Groups and Galois Cohomology" by Ribes, and the argument is similarly, but he picked up another symmetric neighborhood. $\endgroup$ – Kaique Matias Oct 9 '14 at 11:58

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