Galois group of $K(X)/K$ Let $K$ be an infinite field, if $K(X)$ is the field of rational function I want to find the Galois group of the extension $K(X)/K$.

Lemma 1: If $L$ is a field such that $K\subsetneq L\subseteq K(X)$ then $[K(X):L]$ is finite.
Proof. It is easy to show that $X$ is algebraic over $L$, so $K(X)/L$ is a finite extension.

Lemma 2: $\operatorname{Gal}(K(X)/K)$ contains only finite (proper) subgroups.
Proof. Suppose that $H<G$ is infinite;  for the lemma 1 we have  that $[K(X):\operatorname{Fix}(H)]=n$, and so $|\operatorname{Gal}(K(X)/\operatorname{Fix}(H))|\le n$. But  $\operatorname{Gal}(K(X)/\operatorname{Fix}(H))\supseteq H$ and this is a contraddiction.

Now I know that the group $\operatorname{Gal}(K(X)/K)$ is a group with only finite subgroups, but I can't find other informations about its structure. Maybe this group depends strongly from the field $K$.
Thanks in advance.
 A: The methods of Galois theory are not so good for studying transcendental extensions – instead you should consider the methods of birational geometry. For simplicity, assume $K$ is algebraically closed. Then $K (X)$ is the function field of the projective curve $\mathbb{P}^1_K$, and the group $K$-automorphisms of $K (X)$ is canonically isomorphic to (the opposite of) the group of birational automorphisms of $\mathbb{P}^1_K$. This is called the Cremona group of order 1.
Now, notice that $K (X)$ is generated over $K$ by $X$, so $K$-automorphisms of $K (X)$ are uniquely determined by the image of $X$. By geometrical considerations, we see that the image of $X$ must be of the form
$$\frac{a X + b}{c X + d}$$
because otherwise the induced rational map $\mathbb{P}^1_K \to \mathbb{P}^1_K$ would send more than one point to $0$ or more than one point to $\infty$. Moreover, we must have $a d - b c \ne 0$, so that 
$$\frac{1}{a d - b c} \frac{d X - b}{- c X + a}$$
corresponds to the inverse automorphism. Thus the automorphism group of $K (X)$ over $K$ is none other than the Möbius group $\mathrm{PGL}_2 (K)$.
