Closed finite-index subgroups of rational points of smooth connected groups over algebraically closed fields are the whole point group

By [F1] the space $\mathrm{\mid }G\mathrm{\mid }$ is irreducible, and by [F2] the subset $G(k)$ is dense in $\mathrm{\mid }G\mathrm{\mid }$; a dense subset of an irreducible space is irreducible by [F3], so $G(k)$ is irreducible in the Zariski topology.

For $g\in G(k)$ let ${\lambda }_g:G\to G$, $x\mapsto gx$, be left translation; it is an automorphism of $k$-schemes with inverse ${\lambda }_{g^{-1}}$, hence induces a homeomorphism of $G(k)$ onto itself. The cosets of $H$ in $G(k)$ are the images ${\lambda }_g(H)$ of $H$, and because $H$ is closed in $G(k)$, every coset is closed in $G(k)$; distinct cosets are disjoint and nonempty.

Suppose $H\mathrm{\ne }G(k)$. Since $H$ has finite index, $G(k)$ is the disjoint union of the finitely many distinct cosets $g_{1}H,\dots ,g_{r}H$ with $r\ge 2$, each closed and nonempty by [step 1.2]. Then $H$ and the union $g_{2}H\cup \cdots \cup g_{r}H$ are two disjoint nonempty closed subsets whose union is $G(k)$, contradicting the irreducibility of $G(k)$ from [step 1.1] by [F3]. Hence $H=G(k)$.