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. 

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.
