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If $G$ and $H$ are finitely generated residually finite groups such that $G\leq H\leq \hat G$, where $\hat G$ denotes the profinite completion of $G$, does it follow that $$\hat H \cong \hat G$$

Thanks!

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By the universal property of profinite completions, we have to show that every homomorphism of groups $G \to K$, where $K$ is a profinite group, extends uniquely to a homomorphism of groups $H \to K$.

Well, we have only one choice: $G \to K$ corresponds to a homomorphism of profinite groups $\hat{G} \to K$, and this can be restricted to $H \leq \hat{G}$. This is clearly an extension of $G \to K$.

Now, assume that two homomorphism $f,g : H \to K$ agree on $G$. Then $\hat{f},\hat{g} : \hat{H} \to K$ are homomorphisms of profinite groups which agree on $\hat{G}$. Then they also agree on $H \leq \hat{G}$, i.e. $f=g$.

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  • $\begingroup$ How is $\hat H$ inside $\hat G$? $\endgroup$ – Zeppelin Mar 19 '13 at 13:45
  • $\begingroup$ You need the inclusion map from $G$ to $\hat H$ be continuous? $\endgroup$ – Zeppelin Mar 19 '13 at 13:52
  • $\begingroup$ I am sorry I do not understand your proof. $\endgroup$ – Zeppelin Mar 19 '13 at 13:57
  • $\begingroup$ $G$ is discrete. And I've corrected the proof. Is it OK now? $\endgroup$ – Martin Brandenburg Mar 19 '13 at 14:20
  • $\begingroup$ G is not discrete, it has the profinite topology. By definition of profinite completion, you need the inclusion map from $G$ to $\hat H$ to be continuous. But I do not think it is the case always. $\endgroup$ – Zeppelin Mar 19 '13 at 14:32

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