Subset of a subgroup is not closed under group actions

Question

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..

Answer 1

The proof is not correct.

You have $g = k_ih_i$ with $k_i \in K, h_i \in H_i$. But then $h_i = k_i^{-1}g \in K^{-1}g$ and we conclude $H_i \cap K^{-1}g \ne \emptyset$ and not $H_i \cap g^{-1}K \ne \emptyset$. But then we can correctly show that there exists $h \in \bigcap_{i \in I} (H_i \cap K^{-1}g) = (\bigcap_{i \in I} H_i) \cap K^{-1}g = H \cap K^{-1}g$. Hence $h \in H$ and $h = k^{-1}g$ for some $k \in K$. This means $g = kh \in KH$.

Answer 2

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.