Galois theory proving $\mathrm{Gal}(E/K_1 \cap K_2)= \left<\mathrm{Gal}(E/K_1),\mathrm{Gal}(E/K_2)\right>$ I found this question in an exercise serie:
Given E/F a galois extension and $F\subseteq K_1,K_2 \subseteq E$ two intermediate fields. Let's denote $H_1= \mathrm{Gal}(E/K_1)$ and $H_2=\mathrm{Gal}(E/K_2)$.
We must show that
$$\mathrm{Gal}(E/K_1\cap K_2)= \langle H_1,H_2 \rangle $$
where $\langle H_1,H_2 \rangle$ is the group generated by $H_1$ et $H_2$.
I know this must be true, intuitively, but I have difficulties proving it properly. The inclusion $\mathrm{Gal}(E/K_1\cap K_2) \supseteq \langle H_1,H_2 \rangle$ is clear, but the other side isn't...
Thanks a lot for your help :)
 A: Let $P = \{ E' : E'/F  \}$ be the poset of intermediate field extensions $E/E'$ and let $Q =   \{H: H\leqslant G\}$ be the poset of subgroups of $G = \operatorname{Gal}(E/F)$. 
There are canonical arrows
$$ \psi : P \longrightarrow Q, \quad \phi :Q\longrightarrow P$$
$$ E\longmapsto \operatorname{Gal}(E/F), \quad H\longmapsto E/E^H$$
that are order reversing, and are bijections since your extension is Galois.
It suffices you show that such maps between posets (in general, forget the field theory context for a while) send join to meets and meets to joins, which is really easy! Since $\langle H_1,H_2\rangle$ is the join of $H_1$ and $H_2$ in the poset of subgroups and $K_1\cap K_2$ is the meet in the poset of extensions, you are done. 
Concretely, you want to prove the following: let $f,g$ be inverse order reversing bijections between posets $P,Q$, and let $x,y\in P$. Then
$$ f(x\wedge y) = f(x) \vee f(y).$$
Now note that since $f$ is order reversing, and since $x\wedge y\leqslant x,y$ we certainly have $$f(x)\vee f(y)\leqslant f(x\wedge y).$$ Since $f$ is invertible, we can apply the same reasoning to its inverse $g$ to get 
$$g(u\vee v) \leqslant g(u)\wedge g(v).$$
Now set $u=f(x)$ and $v = f(y)$ and apply $g$ to get the reverse equality
$$f(x) \vee  f(y) \geqslant f(x\wedge y).$$
