Dense cyclic subgroup - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T01:25:22Z http://mathoverflow.net/feeds/question/37239 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/37239/dense-cyclic-subgroup Dense cyclic subgroup Bad English 2010-08-31T09:24:27Z 2010-08-31T10:38:49Z <p>Does anyone know a continuous group (not necessarily locally compact) with dense cyclic subgroup other than a torus? </p> http://mathoverflow.net/questions/37239/dense-cyclic-subgroup/37241#37241 Answer by Charles Matthews for Dense cyclic subgroup Charles Matthews 2010-08-31T09:56:02Z 2010-08-31T09:56:02Z <p>How about the Bohr compactification of the infinite cyclic group? </p> http://mathoverflow.net/questions/37239/dense-cyclic-subgroup/37242#37242 Answer by Xandi Tuni for Dense cyclic subgroup Xandi Tuni 2010-08-31T09:58:07Z 2010-08-31T09:58:07Z <p>How about the infinite cyclic group itself with the discrete topology? Or p-adic integers?</p> http://mathoverflow.net/questions/37239/dense-cyclic-subgroup/37245#37245 Answer by Gjergji Zaimi for Dense cyclic subgroup Gjergji Zaimi 2010-08-31T10:10:02Z 2010-08-31T10:10:02Z <p>You already have some examples in the other answers. Groups which have a dense cyclic subgroup are called <em>Monothetic</em> groups. In the article "On monothetic groups" by P.R. Halmos and H. Samelson, you can find many of their properties, such as</p> <blockquote> <p>Every compact connected separable (abelian) group is monothetic.</p> </blockquote> http://mathoverflow.net/questions/37239/dense-cyclic-subgroup/37247#37247 Answer by Keivan Karai for Dense cyclic subgroup Keivan Karai 2010-08-31T10:29:40Z 2010-08-31T10:29:40Z <p>First, it is clear the group has to be abelian. Now, if you assume that $G$ is locally compact, then by the classification you can decompose $G$ as $G={\mathbb R}^n \times H$ where $H$ has a compact open subgroup. Clearly, there can be no ${\mathbb R}^n$ factor, so $G$ has a compact open subgroup. Now, suppose $G$ is itself compact and topologically generated by $g$. Then any character $\chi$ in the dual of $G$ vanishing on $g$ will be identically zero. So, the map $\chi \mapsto \chi (g)$ is injective, hence the dual is a subgroup of $U(1)$. Conversely, you can also see that if $\Gamma$ is a subgroup of $U(1)$ (considered with the discrete topology) then the dual of $\Gamma$ has a dense cyclic subgroup. By taking various subgroups you can, for instance, get the $p$-adic integers, or the n-torus. </p>