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The normal definition of subgroup separability is:

A group $G$ is said to be subgroup separable if for every finitely generated subgroup $H\leq G$ and $g\in G\setminus H$ there exists a subgroup of finite index $K_g\leq G$ such that $H\leq K_g$ and $g\notin K_g$.

Why is this equivalent to the following?

$G$ is subgroup separable if every finitely generated subgroup of $G$ is closed in the profinite topology of $G$.

The profinte topology is the topology generated by normal subgroups of finite index and there cosets as open sets.

It is easy to show, that the first definition implies the second:

$$H=\bigcap_{g\notin H}K_g$$ We can assume that $K_g$ is normal for all $g\notin H$. So these are open normal subgroups of finite index. Hence they are closed. So the intersection is an intersection of closed sets in the profinite topology and hence closed. So it follows that $H$ is closed.

But how does the second implies the first?

Thanks for help.

I asked this question on mathstackexchange.com, too.

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  • $\begingroup$ If H is closed there is a finite index subgroup K so that Kg does not intersect H. Then KH is a finite index subgroup containing H and not g. $\endgroup$ – Benjamin Steinberg Nov 6 '13 at 13:10
  • $\begingroup$ That said I vote to send this back to MSE. $\endgroup$ – Benjamin Steinberg Nov 6 '13 at 13:11
  • $\begingroup$ K should also be normal in my previous comment. Forgot to say it. $\endgroup$ – Benjamin Steinberg Nov 6 '13 at 13:20
  • $\begingroup$ Thanks. I don't know why this subgroup $K$ exists, if $H$ is closed. Since $H$ is closed, $gH$ is closed. It follows that the complement of $gH$ is the union of basiselements of the toplogy. Hence the union of normal subgroups of finite index. But how can I proof the existence of $K$? $\endgroup$ – Timo Nov 6 '13 at 13:57
  • $\begingroup$ @Timo, basic nbhds of 1 are finite index normal subgroups so basic nbhds of g are cosets gK with K of finite index. $\endgroup$ – Benjamin Steinberg Nov 6 '13 at 14:26
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I will give some explaination following Benjamin: let $H$ closed in $G$, and $g\in H^c$, where $H^c=G-H$, then complement of $H$ in $G$. Now $Hg^{-1}$ is closed for $H$ is closed, and $(Hg^{-1})^c$ is open. Note $1 \in (Hg^{-1})^c$, so $(Hg^{-1})^c=\cup N$ where $N$ is normal with finite index in $G$ (because the normal subgroups of finite index forms the basis of identity). This means $Hg^{-1} \cap N$ is empty. Now $g \not \in HN$, as required.

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