Let $G$ be a (countable) residually finite group whose profinite completion is topologically finitely generated. Must $G$ be finitely generated?
$\begingroup$
$\endgroup$
4
-
1$\begingroup$ Wouldn't $\hat{\mathbb{Z}}$ be a counterexample? Or am I missing something? $\endgroup$– Donu ArapuraCommented May 10, 2015 at 16:58
-
$\begingroup$ @DonuArapura my fault. Edited. $\endgroup$– PabloCommented May 10, 2015 at 17:01
-
1$\begingroup$ Isn't $\mathbb{Z}_{(p)}$ still a counterexample? $\endgroup$– Johannes HahnCommented May 10, 2015 at 17:39
-
1$\begingroup$ @DonuArapura: yes it works (that subgroups of finite index in $\hat{\mathbf{Z}}$ can be checked by hand without any fancy theorem. By Nikolov-Segal (and Serre in the pro-$p$-case), any infinite topologically finitely generated profinite group also works. Of course it provides no countable example, but very simple countable examples have been pointed out as well. $\endgroup$– YCorCommented May 10, 2015 at 20:55
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
0
No. Take $G=\oplus_{p} C_p$, where the sum is over all primes $p$ and $C_p$ is the cyclic groups of order $p$. Then $G$ is clearly locally finite and infinite and therefore not finitely generated. However, it is easy to see that the profinite completion of $G$ is $\Pi_p C_p$ which is a cyclic profinite group.
answered May 10, 2015 at 18:52