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Let $G$ be a finitely generated, residually finite group and $\widehat{G}$ its profinite completion. Must the natural image of $G$ inside $\widehat{G}$ be a normal subgroup of $\widehat{G}$? Must it be characteristic?

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No. Suppose $G$ is residually finite and is normal in $\hat{G}$.

Any element of $\hat{G}$ that commutes with all elements of $G$ commutes with all elements of $\hat{G}$, the closure of $G$. So $C_{\hat{G}}(G)=Z(\hat{G})$. Also an element $g\in G$ is in the centre of $\hat{G}$ if and only if it's in the centre of $G$, so we have injective homomorphisms $$G/Z(G)\hookrightarrow\hat{G}/Z(\hat{G})\hookrightarrow\operatorname{Aut}(G),$$ the second one induced by the action of $\hat{G}$ on $G$ by conjugation.

If $G$ is finitely generated then $\operatorname{Aut}(G)$ is countable. But $\hat{G}/Z(\hat{G})$ is a profinite group and so must be finite or uncountable.

So if $G$ is a finitely generated residually finite group with $|G:Z(G)|$ infinite (for example, a free group on two generators) then $G$ is never normal in $\hat{G}$.

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  • $\begingroup$ If $G/Z(G)$ is infinite, how do we know that $[\widehat{G}:Z(\widehat{G})] = [G:Z(G)]$? Is that a standard fact? $\endgroup$ – oxeimon Apr 21 '15 at 23:41
  • $\begingroup$ @oxeimon I've edited my answer to add some details. $\endgroup$ – Jeremy Rickard Apr 22 '15 at 11:38

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