Let $K_1/K$ and $K_2/K$ be Galois field extensions, and consider the compositum $K_1K_2$. Then $K_1K_2/K$ is Galois. We have then a natural map
$$\varphi: \operatorname{Gal}(K_1K_2/K_1) \to \operatorname{Gal}(K_2/K)$$
given by restricting to $K_2$. It is easy to see that the image of $\varphi$ lies inside the subgroup $\operatorname{Gal}(K_2/(K_1\cap K_2))$, and that the morphism is injective, as an element in the kernel is an automorphism fixing $K_1$ and $K_2$, hence in must fix $K_1K_2$.
Now, I guess that the image of $\varphi$ should be $\operatorname{Gal}(K_2/(K_1\cap K_2))$, and this could be deduced proving that the groups $\operatorname{Gal}(K_2/(K_1\cap K_2))$ and $\operatorname{Gal}(K_1K_2/K_1)$ have the same cardinal.
My question is then: how can one prove that $$|\operatorname{Gal}(K_2/(K_1\cap K_2))|=|\operatorname{Gal}(K_1K_2/K_1)|?$$
Best Answer
I will rephrase your question to a more general case. Let $K \subset \Omega$ be an arbitrary field extension and assume that there exist two intermediate fields $M$ and $L$, such that $K \subset L$ is a finite Galois extension. There is a different way you can prove that $\text{Gal}(ML/M) \cong \text{Gal}(L/L\cap M)$.
Convince yourself that the extension $M \subset ML$ is a finite Galois extension and that $M \cap L$ is a field. Consider the restriction map $\text{res}:\text{Gal}(ML/M) \to \text{Gal}(L/K)$ that sends $\tau \mapsto \tau|_L$ and notice that this is an injective homomorphism. Therefore the image of $\text{Gal}(ML/M)$ is isomorphic to some subgroup of $\text{Gal}(L/K)$ and because of the Galois correspondence this image will be of the form $\text{Gal}(L/F)$ for some intermediate field $K \subset F \subset L$. Now, this $F$ is exactly the field that is fixed by the automorphisms $\tau|_L$ with $\tau \in \text{Gal}(ML/M)$, where $\tau$ fixes $M$ as an automorphism of $ML$ (you should reread this sentence a couple of times, because nothing really exceptional happens). Therefore, $F = L \cap M$. And you conclude that $\text{Gal}(ML/M) \cong \text{Gal}(L/L\cap M)$.