I am going through the proof of a basic lemma from affine algebraic groups. Since I am not following it I would appreciate any clarification.
Lemma: Let $G$ is an algebraic groupIf $H$ in $G$ is an abstract subgroup, then $\overline{H}$ is a closed subgroup. Proof: Multiplication by $g$ is an isomorphism of varieties $G \rightarrow G$. Thus $g \bar{H} = \overline{gH}$ and $\overline{H} g = \overline{Hg}$. Therefore $\overline{H} \cdot \overline{H} \subset \overline{H}$. Similarly for inversion we have $(\overline{H})^{-1} = \overline{H}$.
1) First I am struggling with how to show $g \overline{H} = \overline{gH}$. I can see that $\overline{gH} \subset g \overline{H} $ but how can I show the other direction?
2) How does $\overline{H} \cdot \overline{H} \subset \overline{H}$ follow?
Thank you!