Let $G$ be a linear algebraic group over an algebraically closed field $K$. For a subgroup $H \subset G$, we denote with $\overline{H}$ the closure of $H$ in $G$ with respect to the Zariski topology. Note, that then $\overline{H} \subset G$ is a subgroup. I want to know, if then $\overline{H}$ is again linear algebraic, or in other words:
My question is, if a closed subgroup of an linear algebraic group, is again a linear algebraic group.
A little context, as for why I want to know this. I have a subgroup $H \subset G$ as above and consider its identity component $H^0$. Then $H^0$ is a normal subgroup of $H$. So I have a chain of subgroups $$H^0 \subset H \subset G$$ and I want to verify, if $$\overline{H^0} \subset \overline{H} \subset{G}$$ holds true. If $\overline{H}$ were again linear algebraic, then I could use the same statement above for $G_1 := \overline{H}$ with $H^0 \subset G_1$.