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The following is an answer to a previous version of the question, which asked whether there exists an algebraic subgroup $H$ of $G$ such that $H(K)=H_1(K)H_2(K)$:

There are two necessary conditions on $H_1, H_2$:

First, the set $\Gamma:=H_1(K)H_2(K)$ has to be a subgroup of $G(K)$.

Also, since any algebraic subgroup of an algebraic group over a field is closed, the set $\Gamma$ has to closed in $G$.

These conditions are also sufficient. This follows from from the following fact: Let $K$ be a field and $\Gamma$ a subgroup of $\operatorname{GL}_n(K)$ which is closed (for the Zariski toplogy on $\operatorname{GL}_n(K)$). Then there exists an algebraic subgroup $G$ of $\operatorname{GL}_n$ such that $G(K)=\Gamma$. This is (part of) Theorem 4.8 of these notes of Milne: http://www.jmilne.org/math/CourseNotes/aag.html

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There are two necessary conditions on $H_1, H_2$:

First, the set $\Gamma:=H_1(K)H_2(K)$ has to be a subgroup of $G(K)$.

Also, since any algebraic subgroup of an algebraic group over a field is closed, the set $\Gamma$ has to closed in $G$.

These conditions are also sufficient. This follows from from the following fact: Let $K$ be a field and $\Gamma$ a subgroup of $\operatorname{GL}_n(K)$ which is closed (for the Zariski toplogy on $\operatorname{GL}_n(K)$). Then there exists an algebraic subgroup $G$ of $\operatorname{GL}_n$ such that $G(K)=\Gamma$. This is (part of) Theorem 4.8 of these notes of Milne: http://www.jmilne.org/math/CourseNotes/aag.html