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How can I prove that a residually finite group $G$ is totally disconnected? I considered the topology generatad by $\{Ng\}_{N\in\eta,\;g\in G}$ where $\eta=\{N\unlhd G \;, |G:N|<\infty\}$ and I considered $$ \alpha:G\longrightarrow\prod_{N\in\eta} G/N\;. $$ defined by $g\mapsto\{Ng\}_{N\in\eta}$ which are all the nhbrd of $g$.

Clearly $\ker\alpha=\bigcap_{N\in\eta}N=1$ by definition of residually finite.

Then I took $Y\subseteq G,\;Y\neq\emptyset$. If by contradiction $Y$ is connected and $|Y|\ge2$ then we can take $g,h\in Y$; being connected we see that exists a nhbrd of $g$ which contains $h$, i.e. $h\in Ng$, $\exists N\in\eta$. Hence $g^{-1}h\in N$.

How can I proceed from here? If this last condition could hold for all $N\in\eta$ I would finish... but I can't go on.

Thanks!

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    $\begingroup$ Are you interested in a more abstract proof? I don't see a nice way to finish the way you started. $\endgroup$ Jun 16 '14 at 17:44
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    $\begingroup$ Inverse limit of finite Hausdorff groups is always TD. $\endgroup$ Jun 16 '14 at 17:53
  • $\begingroup$ @DanielFischer: ok thanks! Propose your argument, although the tools we suppose to use should be really minimal: some general topology. We have only introduced the profinite topology for topological groups! But if you have a proof... you're welcome! $\endgroup$
    – Joe
    Jun 16 '14 at 18:05
  • $\begingroup$ Take $g,h\in Y$ with $g\neq h$. Then $g^{-1}h\neq 1$, so there is a normal subgroup $N$ of finite index in $G$ for which $g^{-1}h\not\in N$. Then $gN\neq hN$, so in fact, $gN\cap hN = \emptyset$. Of course, $gN, hN\in\eta$, and you're done. $\endgroup$
    – James
    Jun 17 '14 at 3:01
  • $\begingroup$ @James: thank you but... can you explain your argument? Where is the contradiction? The fact that $gN\cap hN=\emptyset$ doesen't imply that $Y$ is disconnected (if was that the point you looked for). Finally $gN,hN\in\eta$ is not true ($gN,hN$ in general are not even subgroup of $G$). $\endgroup$
    – Joe
    Jun 17 '14 at 6:16
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Each $G/N$ is a finite group. We endow it with the discrete topology. Then the group

$$P = \prod_{N\in\eta} G/N,$$

endowed with the product topology, is a topological group. Since each factor is totally disconnected, $P$ is totally disconnected (if $\varnothing \neq C\subset P$ is connected, so is $\pi_N(C)$ for each $N$, so it consists of a single point, and therefore $C$ also consists of a single point).

The homomorphism $\alpha$ embeds $G$ as a subgroup into $P$. But it also is continuous (it is even a topological embedding), since $\pi_N \circ\alpha$ is the canonical factor map $G\to G/N$, and that is continuous since the preimage of any subset of $G/N$ is a union of cosets of $N$, hence open. Thus $G$ must be totally disconnected, since the continuous image of any connected set is again connected - hence empty or a singleton.

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