If $\chi(g)$ generates a dense subgroup of $\chi(G)$ for all $\chi$ then $g$ generates a dense subgroup of $G$.

Question

This question arised from something in Ergodic theory, however this is not necessary to state or answer the question.

Suppose that $G$ is a compact abelian group and $g\in G$.

Are the following two equivalent:

1. For every continuous character $\chi:G\rightarrow S^1$, $\chi(g)$ generates a dense subgroup of $\chi(G)$.

2. $g$ generates a dense subgroup of $G$.

Clearly $2$ implies $1$, but I can't figure out whether the opposite holds.

For those who are interested in the connection with ergodic theory. Property $2$ means that $(G,R_g)$ is an ergodic system, and I ask whether this is equivalent to property 1 which means that all the factors $(\chi(G),R_{\chi(g)})$ are ergodic.

Answer 1

Short proof: if 2 fails, the group $G/\overline{\langle g\rangle}$ is a nontrivial compact abelian group and hence has a nontrivial continuous character $\chi$. Then $\chi(g)=1$ generates the trivial subgroup, which is not dense in $\chi(G)$, so 1 fails.

Answer 2

The answer is Yes!.

Suppose by contradiction that $1$ holds but $2$ doesn't. Let $H=\overline{\left<g\right>}$, then by assumption $H\leq G$ is a proper subgroup.

Look at $G/H$. This is a non-trivial group and so admits a non-trivial character $\chi:G/H\rightarrow S^1$. We may compose with $G\rightarrow G/H$ to obtain a character $\tilde{\chi}:G\rightarrow S^1$ which is trivial on $H$. In particular $\tilde{\chi}(g)$ is trivial, however $\tilde{\chi}(G) = \chi(G/H)$ is a non-trivial subgroup of $S^1$ - contradiction.

This completes the proof.

Edit: I just figured even another answer for that!

Let $s\in G$, for all $\chi\in\hat G$ there exists a net $n_\alpha$ such that $\chi(g^{n_\alpha})\rightarrow \chi(s)$.

Using a diagonal argument we can choose the same sub-net $n_\beta$ for all characters, also by moving into a sub-sub-net if necessary we may assume that $g^{n_\beta}\rightarrow h\in G$. Since all the characters are continuous we have that $\chi(h)=\chi(s)$ for all $\chi$. Using the fact that characters separates points we have that $h=s$, which means that $g^{n_\beta}$ converges to $s$. In particular $s\in \overline{\left<g\right>}$