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I've been reading about profinite groups and have encountered the notion of strong completeness. I.e. that a profinite group $G$ is strongly complete if it is isomorphic to it's profinite completion or equivalently, if every subgroup of finite index is open. My problem is that I am not understanding why these conditions are equivalent. I cannot find a reference for this fact; every mention of it I find states that this equivalence is "obvious." I believe the equivalence stems from the fact that if all subgroups of finite index of $G$ are open then the set of subgroups of finite index forms a fundamental system of open neighborhoods of $1$ in $G$ which allows one to reconstruct the topology. I would appreciate any help understanding this, or any references on this fact.

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1 Answer 1

The profinite completion is the inverse limit of all quotients by finite index normal subgroups. Any profinite group is the inverse limit of its quotients by open normal subgroups. Since open normal subgroups have finite index a profinite group is strongly complete iff the open normal subgroups are cofinal among finite index notmal subgroups. But this is equivalent to all finite index subgroups are open.

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