Use Artin's Theorem to show that $[E:k] \ge n$

Question

Artin's Theorem- Let $E$ be a field and $G$ be a group of automorphisms of $E$ and $k$ be the set of elements of $E$ fixed by $G$. Then $k$ is a subfield of $E$ and $E$ has finite degree over $k$ iff $G$ is finite. In that case, $[E:k]=|G|.$

Question- Use Artin's Theorem to show that for every field $E$ with $n$ distinct automorphisms, if $k$ is the fixed field of this set of automorphisms, then $[E:k] \ge n$.

Attempt- If $[E:k]$ is infinite we are done.

So we take the case where $[E:k]$ is finite. Let $A= \{\sigma_{1}, \ldots\ ,\sigma_{n}\}$ be the set of distinct automorphisms of $E$. Then $k=\{a\in E\ | \sigma_{i}(a)=a\ , 1\le i \le n \}$.

Now the hint given in class says "consider the group $G$ generated by $\sigma_{i}'s$ and $F=\{a\in E \ |\ \sigma(a)=a\ \forall\ \sigma \in G\}$. (Here $G$ is the subgroup of all automorphisms of $E$)"

But isn't $A$ itself a group so $G$ must be equal to $A$ and thus $F=k.$ And thus $[E:k]=[E:F]$. What did we achieve this way?

Answer 1

$A$ is not itself a group, think about the set of transpositions which generated $S_n$, (a permutation whose signature is even is not a transposition). The Artin theoem's allows to say that and $[E:F]=|G|\geq |A|$.

Answer 2

The group $G$ generated by $n$ automorphisms has atleast $n$ elements, and the fixed field of $G$ is the same as the fixed field of the set of $n$ automorphims. So by Artin $$[ E : F] = |G| \geq n$$