Field extension of fixed field has degree greater than the size of the group Let $K$ be a field, $G\leqslant\mathrm{Aut}(K)$ a (finite) group, and $K^G$ the fixed subfield of $K$.
How exactly would you go about proving the following?$$[K:K^G]\geqslant|G|$$
For some reason I have come to a complete blank trying to even start on this!
I would much prefer some hints to a complete solution, but either is appreciated.
Edit:
Sorry, just saw that this hint is at the bottom of the question:
You may assume that if $\sigma_1,\ldots,\sigma_n$ are distinct field homomorphisms from $K$ to $L$ then $\{\sigma_1,\ldots,\sigma_n\}$ is linearly independent over $L$.
 A: I abbreviate $F=K^G$. Let $n=[K:F]$. Consider the set $V$ of $F$-linear mappings from $K$ to itself. The $V$ has the usual structure of a vector space over $F$, and clearly has dimension $n^2$, after all $V$ is isomorphic to the space of $n\times n$ matrices over $F$.
But $V$ also has a structure of a vector space over $K$. Just multiply the values by scalars from $K$ instead of restricting yourself to $F$ (this works only because the range, here $K$, already has multiplication by elements of $K$ defined.
What is the dimension of $V$ as a vector space over $K$? Call it $m$. The same argument that you have seen when proving the product formula in a tower of field extensions shows that
$$
n^2=\dim_FV=[K:F]\dim_KV=nm.
$$
This implies that $m=n$.
It sounds like you want to think about the rest (great!), so I spoilerize the rest (mouseover the pale pink block to see the text).

 The Lemma in your hint exhibits a set of $|G|$ elements of $V$ that are linearly independent over $K$. Basic linear algebra tells us that the size of a linearly independent set is bounded from above by the dimension of the space. Therefore 

$$
|G|\le \dim_KV=n=[K:F].
$$
Q.E.D.
