# Strongly complete completion of profinite group [duplicate]

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.

## marked as duplicate by Dietrich Burde, Claude Leibovici, Parcly Taxel, Namaste group-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Oct 17 '16 at 11:13

• Thanks for the links: so I have to check that $\widehat{\widehat{G}}=\widehat{G}$ for $G$ profinite... any help? – Macadam Oct 15 '16 at 21:23
• 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? – Macadam Oct 16 '16 at 8:47