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


*

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

*$g$ 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. 

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 $s\in \overline{\left<g\right>}$
