Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\hat{G}$?

share|improve this question
    
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
add comment

1 Answer 1

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.