A $C$-closed but non-Zariski-closed subgroup of an algebraic group 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. **Questions:** 1. How far is $H$ from being Zariski-closed, i.e., an algebraic subgroup? 2. 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.

The easiest example I can think of where both fail miserably is $S^1\subseteq\mathbb C-\\{0\\}$ under multiplication, where we treat $\mathbb C-\\{0\\}$ as a complex variety. $S^1$ is uncountably infinite, whereas the proper Zariski closed subsets are finite. The quotient is the group of positive reals, which is not algebraic over the complex numbers.