Let $G$ be a linear algebraic group defined over some algebraically closed field $\mathbb{K}$ and also over some subfield $k\subset \mathbb{K}$. There is thus a natural group structure on the set of $k$-points of $G$. It seems impossible to me that this group is cyclic and infinite. Is this true?
EDIT: user48841 explained that this happens with elliptic curves. I am in fact interested in linear algebraic groups.
There are elliptic curves $E/\mathbb{Q}$ with $E(\mathbb{Q}) \simeq \mathbb{Z}$.
We may assume, by replacing $G$ by the Zariski closure of $G(k)$, that $G(k)$ is Zariski dense. By assumption, $G(k)$ is $\mathbb Z$; hence $G$ is abelian and after replacing $G$ by the connected component of identity of $G$, we may assume that $G$ is connected. \vskip 5mm
Now a conneced abelian algebraic group is a product of ${\mathbb G}_a$ and ${\mathbb G}_m$. Hence the assumptions mean that $k$ or $k^*$ must be $\mathbb Z$; this is impossible for a subfield of $\mathbb C$.
In detail, $G$ is a product of copies of ${\mathbb G}_a$ and copies of ${\mathbb G}_m$. Hence $G(K)$ is the product of (copies of) $K$ and $K^*$. Assume that a factor like a product of ${\mathbb G}_a$ occurs. Then $G(k)$ contains the subgroup $k\supset {\mathbb Q}$. However, $G(k)$ being $\mathbb Z$, any infinite subgroup must be $\mathbb Z$ but $\mathbb Q$ is not $\mathbb Z$.
A similar but more involved reasoning may be used to tackle the case when $G$ is a product of the multiplicative group ${\mathbb G}_m$ over $K$.
Since the questioner has asked for details, if we assume that $G$ over $K$ does not have the additive group components, then over the smaller field $k$, $G$ is isogenous to a product of ${\mathbb G}_m$ and the groups $R^1_{l/k}({\mathbb G}_m$ where $R_1$ refers to the group of norm one elements. This is a theorem due to Ono (for reference, one may see Borel-Tits). The ${\mathbb G}_m$ factor may easily be taken care of. As to the other group $R^1$, one has to work a little to say that the group of norm one elements in $l$ cannot be isomorphic to $\mathbb Z$, for any char zero $l/k$.
[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.
