I have read in many textbooks proofs that any Lie group has no small subgroups, that is, there is an open neighborhood of the unity element containing no nontrivial subgroups. In particular, $GL_n(\mathbb{C})$ with the usual topology has no small subgroups.
My question is, does $GL_n(\mathbb{C})$ have no small subgroups also as an algebraic group, that is, does it exist a Zariski-open neighborhood of 1 containing no nontrivial subgroups?
The group $GL_n(\mathbb{C})$ does have small subgroups for the Zariski topology. Indeed, let $U$ be an open subset of $GL_n(\mathbb{C})$ containing the identity matrix. Consider the diagonal morphism $$ \Delta:\mathbb{C}^\times \to GL_n(\mathbb{C}): \lambda\to diag(\lambda, \ldots, \lambda).$$
Then $\Delta^{-1}(U)$ is a non-empty Zariski-open subset of $\mathbb{C}^\times$; in other words, it is a cofinite subset of $\mathbb{C}^\times$. In particular, there exists a natural number $m>1$ such that all $m$-th roots of unity are in $\Delta^{-1}(U)$. Let $1, \zeta, \zeta^2, \ldots, \zeta^{m-1}$ be these roots of unity. Then the matrices $diag(\zeta^i, \ldots, \zeta^i)$ for $i=0, \ldots, m-1$ form a non-trivial finite subgroup of $GL_n(\mathbb{C})$ contained in $U$.