0
$\begingroup$

This question already has an answer here:

Let $G$ be à profinite group and $\widehat{G}$ his profinite completion. I want to show (I hope it's true) that $\widehat{G}$ is strongly complete that is all normal subgroups of finite index are open sets.

For that I take $\widehat{N}$ a finite index normal subgroup of $\widehat{G}$ and try to find $N$ normal subgroup of finite index of $G$ such that $\widehat{N}=\ker\left(\varphi_N:\widehat{G}\to G/N\right)$. I tried $N=\theta^{-1}\left(\widehat{N}\right)$ where $\theta:G\to\widehat{G}$ but could not conclude.

Not rely with this question I think: I don't want a different definition of strongly complete but I want to check that somewath ($\widehat{G}$ here) is strongly complete.

$\endgroup$

marked as duplicate by Dietrich Burde, Claude Leibovici, Parcly Taxel, Namaste group-theory Oct 17 '16 at 11:13

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ For an answer see also here. It has been shown on MSE, too. $\endgroup$ – Dietrich Burde Oct 15 '16 at 20:12
  • $\begingroup$ Thanks for the links: so I have to check that $\widehat{\widehat{G}}=\widehat{G}$ for $G$ profinite... any help? $\endgroup$ – Macadam Oct 15 '16 at 21:23
  • $\begingroup$ Yes, I have help - see here. $\endgroup$ – Dietrich Burde Oct 16 '16 at 8:04
  • $\begingroup$ So it seems to be false: take $V$ an $\mathbb{F}_p$-vector space of infinite dimension and let $G=\widehat{V}$ (for the additive structure). Then $G$ is profinite but $\widehat{\widehat{G}}\neq\widehat{G}$. Right? $\endgroup$ – Macadam Oct 16 '16 at 8:47