Since $i_G (H) = 2$, $H \unlhd G$ as you correctly pointed out. The result is a lemma to the following proof.
Consider $h \in H$. Then $h^2 \in H$ by closure in $H$. Now consider some $g \in G$ s.t. $g \notin H$. Since $H \unlhd G$, then $G$ is partitioned by the cosets $He$ and $Hg$ i.e. $G = He \cup Hg$ and $He \cap Hg = \emptyset$.
Now $Hg Hg = Hgg = Hg^2$, once again since $H \unlhd G$, and thus $g^2 \in Hg^2$.
It follows that $Hg^2 \cap Hg = \emptyset$ (since $g \neq e$, otherwise $g \in H$), and there are only two cosets of $H$ in $G$. Thus it follows that $Hg^2 = He$ and hence $g^2 \in H$.