Does anyone know a continuous group (not necessarily locally compact) with dense cyclic subgroup other than a torus?
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$\begingroup$ Anything procyclic. Someone below already listed the $p$-adic integers, although I'm not sure why this example wasn't one in which you're interested. Another is the absolute Galois group of a finite field of size $q$, which is topologically generated ($=$ equal to the closure of the abstract subgroup generated) by the Frobenius automorphism $x\mapsto x^q$. This is just $\hat{\mathbb{Z}}$, the product over all primes $p$ of $\mathbb{Z}_p$. $\endgroup$– Keenan KidwellCommented Aug 31, 2010 at 12:11
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$\begingroup$ Take your favorite group, and give it the indiscrete topology. Then the identity element comprises a dense cyclic subgroup. Presumably you don't want this. $\endgroup$– Theo Johnson-FreydCommented Aug 31, 2010 at 15:36
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4 Answers
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You already have some examples in the other answers. Groups which have a dense cyclic subgroup are called Monothetic groups. In the article "On monothetic groups" by P.R. Halmos and H. Samelson, you can find many of their properties, such as
Every compact connected separable (abelian) group is monothetic.
answered Aug 31, 2010 at 10:10
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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.
answered Aug 31, 2010 at 10:29
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How about the Bohr compactification of the infinite cyclic group?
answered Aug 31, 2010 at 9:56
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How about the infinite cyclic group itself with the discrete topology? Or p-adic integers?
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$\begingroup$ I care about is not a question of existence obvious examples like this. In some sense and for some reasons p-adic integers for me are not much better than infinite dimensional torus (IDT). It is easy to understand, that forgetting about topology, p-adic integers realized subgroup of IDT. Need something unlike torus. Sorry for my english. $\endgroup$ Commented Aug 31, 2010 at 11:09