[An extended comment, in community-wiki mode.] There is a closely related fact: Take $G$ to be an algebraic torus such as the multiplicative group, defined over a field $k$ which is not an algebraic extension of a finite field. (The characteristic doesn't matter, but this condition does.) Then there exists an element of infinite order in $G_k$ which generates a dense (cyclic) subgroup of $G$.

For a direct proof when $G$ is $k$-split, see Proposition 8.8 in Borel's book Linear Algebraic Groups (Springer, GTM 126). He remarks also that the split assumption can be dropped, using a more delicate argument from Tits' lectures at Yale.

What I've just quoted does not exactly fit your question. since it doesn't state that $G_k$ itself is cyclic, but it does illustrate a sort of algebraic parallel to the existence of a topological generator for a compact torus.