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Suppose $G$ is a profinite group and $H$ and $K$ are subgroups with conjugate closures. Does it follow that $H$ and $K$ themselves are conjugate in $G$?

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I guess if $H$ is not closed in the profinite topology, and $K$ is the closure of $H$, they would have identical closures, but $H$ and $K$ would not actually be conjugate. – John M Jul 22 '11 at 17:51
Thanks this works as a counter example. I am still curious whether this is the only case. – Mustafa Gokhan Benli Jul 22 '11 at 19:39
I now see that my question was not a good question. Non isomorphic groups can have the same profinite completion so no hope for a satisfactory answer to my question. – Mustafa Gokhan Benli Jul 22 '11 at 19:40
up vote 1 down vote accepted

The question has been decided in the comments, but let me just record a specific counterexample.

Let $G = \hat{\mathbb{Z}}$ be the profinite completion of $\mathbb{Z}$. Let $H = \hat{\mathbb{Z}}$ and let $K$ be any subgroup with $\mathbb{Z} \subseteq K \subsetneq \hat{\mathbb{Z}}$. Then $H$ and $K$ have conjugate (indeed equal) closures, but are not conjugate (equivalently equal, since $G$ is abelian) in $G$. It is easy to see that there are uncountably many choices for $K$.

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