Artin's lemma and some morphisms. Let $E:=Q(t)$ the field of rational function over $Q$. Consider the following morphisms:
$$ \begin{split} 
   & a_1: f(x) \to f(x) \\
   & a_2: f(x) \to f(1/x) \\
   & a_3: f(x) \to f(1-x) \\
   & a_4: f(x) \to f(1/(1-x)) \\
   & a_5: f(x) \to f(x/(x-1)) \\
   & a_6: f(x) \to f((x-1)/x)
\end{split}$$
I have shown theese are automorphisms of $E$ and they form a subgroup of Aut($E$) with the usual composition product.

Let's consider $F_0$ the set of elements of $E$ that are fixed by every $a_i$. Show  that $[E:F_0] \ge 6$

What I thought is the opposite. The six automorphisms form a group $G$ of automorphisms of $E$. $F_0$ is by definition Inv($G$), so thanks to Artin's lemma shoudn't $[E:F_0]$ be less or equal to $|G| = 6$?
 A: In fact, the general result is the following:

Let $E$ be a field, and let $G\subseteq \operatorname{Aut(E)}$ be a finite subgroup. Let $F = E^G$ be the elements of $E$ which are fixed by all elements of $G$, i.e. $F = \{x\in E: g(x) = x, \forall g \in G\}$. Then the extension $E/F$ is Galois, with Galois group $G$.

As to the proof, the difficult part is Artin's lemma, and the question you have is actually the easy part.
By Artin's lemma, we know that the extension $E/F$ is finite. Now for any finite extension $E/F$, there are always inequalities:
$$\#\operatorname{Aut}(E/F) \leq \#\operatorname{Hom}_F(E, \overline F) \leq \deg(E/F),$$
where:


*

*$\operatorname{Aut}(E/F)$ is the set of automorphisms of $E$ fixing elements in $F$;

*$\operatorname{Hom}_F(E, \overline F)$ is the set of embeddings of $E$ into $\overline F$, an algebraic closure of $F$, which are identity on $F$;

*$\deg(E/F)$ is the degree of the extension $E/F$.


The first inequality becomes an equality if and only if $E/F$ is normal, and the second inequality becomes an equality if and only if $E/F$ is separable. Thus both equalities hold if and only if $E/F$ is Galois.
In our case, we know that $G$ is a subgroup of $\operatorname{Aut}(E/F)$, hence we have $\deg(E/F) \geq \#\operatorname{Aut}(E/F) \geq \#G$; but Artin's lemma says that $\#G \geq \deg(E/F)$, so all terms are equal.
