most recent 30 from http://mathoverflow.net 2013-05-23T08:16:21Z http://mathoverflow.net/feeds/question/59167 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/59167/profinite-completion Paulo 2011-03-22T11:24:25Z 2011-03-22T12:09:21Z

<p>What is the profinite completion of the group $S^1$? </p> <p>where $S^1= \{ z\in\mathbb{C}: |z|=1 \}$ is a compact and abelian group.</p>

http://mathoverflow.net/questions/59167/profinite-completion/59175#59175 quid 2011-03-22T11:57:55Z 2011-03-22T11:57:55Z

<p>It is trivial, as $S^1$ has no non-trivial (normal) finite index subgroups. </p> <p>Recall that the profinite completion is the inverse limit of the $G/N$ where $N$ is a normal subgroup of finite index. So, if the only normal finite index subgroup is the full group, the profinit completion is trivial. And, for your group this is the case, for example as it is divisible, or by a direct argument. </p>

http://mathoverflow.net/questions/59167/profinite-completion/59176#59176 Pete L. Clark 2011-03-22T12:00:41Z 2011-03-22T12:09:21Z

<p>Recall that a group $G$ is <strong>divisible</strong> if for every positive integer $n$, the map $ x \mapsto x^n$ is surjective on $G$.</p> <p>It is easy to show that any quotient of a divisible commutative group is again divisible. Moreover, no nontrivial finite group can be divisible, since if $|G| = n$ then $x \mapsto x^n$ sends every element to the identity.</p> <p>It follows that a divisible commutative group has no proper finite index subgroups, so it has trivial profinite completion. </p> <p>Note that the group $(\mathbb{R},+)$ is divisible, hence so is its quotient $S^1 = (\mathbb{R},+)/(\mathbb{Z},+)$. Therefore the profinite completion of $S^1$ is the trivial group.</p> <p><b>Added</b>: Having said this much, I might as well add a little more to show that even a divisible non-commutative group has trivial profinite completion. To see this, note first that as above a divisible group can have no proper finite index <em>normal</em> subgroups, and second that if a group $G$ has a subgroup $H$ with $1 &lt; [G:H] &lt; \infty$, then it also has a normal subgroup $H'$ with $1 &lt; [G:H'] &lt; \infty$: indeed by orbit-stabilizer considerations $H$ has only finitely many distinct conjugates $g H g^{-1}$ in $G$, and we may take $H'$ to be the intersection of all these conjugates. </p>