Let $G$ be a linear algebraic group over a field $K$. (Say $K=\mathbb{F}_q$ or $K=\mathbb{C}$; do not assume $K$ is algebraically closed or of characteristic $0$.) Let $H_1$, $H_2$ be algebraic subgroups of $G$. Consider the multiplication map

$\phi:H_1\times H_2\to G$.

The image of $\phi$ is a constructible set, i.e., a variety with perhaps a few varieties of lower dimension deleted from it. (This is a special case of a result of Chevalley's.)

Question: when is $H_1(K) H_2(K)$ equal to the set of $K$-points of an algebraic subgroup $H$ of $G$?

There are two issues here: closure (i.e., really getting a variety rather than a constructible set as the image) and rationality.

Getting more specific, since the question above may be too hairy in general:

(a) Assume that $G$ is solvable. Does that help? Can we then answer the question in the affirmative?

(b) Say, furthermore, that both $H_1$ and $H_2$ are in the same unipotent subgroup of $G$, or that $H_1$ is unipotent and $H_2$ is a subgroup of a corresponding maximal torus. Does that help?