Let $G$ be an algebraic group, i.e., a topological group homeomorphic to an algebraic variety. In particular, $G$ is a topological manifold. Let $H\subset G$ be a subgroup which is a closed submanifold in the manifold topology, not Zariski topology.
How far is $H$ from being Zariski-closed, i.e., an algebraic subgroup?
How far is the homogeneous space $G/H$ from being an algebraic variety? Maybe projective variety?
It is known that a Lie group can be defined as a smooth group, but then the group structure forces it to be necessarily analytic. Because the multiplicative law in $H$ is still algebraic (that of $G$), I thought maybe this forces $H$ to be Zariski closed.
Note that I am neither an algebraist nor a geometer, so a very basic language would be appreciated. Thank you.