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Let $G$ be finitely generated residually finite group and $\hat{G}$ its profinite completion. Let $H \leq \hat{G}$ be a dense subgroup. Does it follow that $\hat{H}$ is isomorphic to $\hat{G}$?

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I think for non-finitely generated groups, the profinite completion of the profinite completion can grow. You might be able to take H = G-hat for a counterexample. Maybe none of the examples are residually finite, but I kind of thought that was a different sort of finiteness condition. – Jack Schmidt Oct 18 '11 at 22:08
I forgot to add finitely generated condition. Thanks for pointing out. – Mustafa Gokhan Benli Oct 18 '11 at 22:12

I believe this is correct.

If $\hat{G}$ is finitely generated (as a topological group) and if $\hat{H}$ is another profinite group with the same isomorphism classes of finite groups as continuous finite images then $\hat{G}\cong\hat{H}$.

In the case above, it remains to show that $\hat{G}$ and the dense subgroup $H$ have the same finite images. This can be achieved using the second isomorphism theorem.

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