Subset of a subgroup is not closed under group actions In the following image (from "Field Arithmetic by Fried & Jarden" Page 6, Lemma 1.2.2(b)), red rectangle,
 
I'm trying to figure out why it's right to claim $h^{-1} \in H$.
I thought the following solved it:
$g=k_ih_i$ and $g=kh^{-1}$ $\Rightarrow k_ih_i=kh^{-1} \Rightarrow h_ih=k_i^{-1}k \in H$ because right hand side is in H.
So $h^{-1}=k^{-1}g=k^{-1}k_ih_i$, hence, because $k_i^{-1}k \in H$ we get $h^{-1} \in H \Rightarrow g=kh^{-1} \in KH$.
But then I realized there's no logic in assuming $h_ih \in H$ because we're talking subsets here, not subgroups. So is it ok to assume $h_ih \in H$? if so, why?
If not, I'd appreciate an explanation for the red part..

H is as an intersection of subgroups a subgroup by itself. As you can see almost immediatly there has to exist an invert in H for every element of H. It follows the existence of $h^{-1}$ in H.
