Proving the order of a Galois group is equal to the dimension of $F$ over its fixed field w Suppose $F/K$ is a finite dimensional field extension and $G = Aut_KF$.  Let $G'$ be the fixed field of $F/K$, i.e. the set of members of $F$ which are fixed by every element of $G$.  Before the Fundamental Theorem of Galois Theory is established, through a lengthy proof that $[Aut_LF : Aut_MF] \leq [M : L]$ for any intermediate fields $L,M$ of $F/K$, and another proof that $[J' : H'] \leq [H : J]$ for any subgroups $H, J \subseteq G$ (where $J' = \{ c \in F : \phi(c) = c $ for all $\phi \in J \}$), one can prove indirectly that $[Aut_LF : Aut_MF] = [M : L]$, which implies that $|G| = [F : G']$ as a corollary.
My question is, are there any shorter, more direct proofs that $|G| = [F : G']$?  I'm trying to develop a different approach to field theory than the one from my textbook, and it was going quite well until I realized I didn't have a nice proof of this fact.
 A: The proof that I know of may not be shorter, but takes a different approach, using only linear algebra to prove this. I don't know whether or not the additional results I'm using are known to you, so I've proved those as well, which has made this answer much longer than I would've liked it to be. I know some people don't define the Galois group for non-Galois extensions, so here:
$$\text{Gal}(K/F)=\lbrace\sigma\in\text{Aut}(K):\sigma(a)=a,\text{ for all }a\in F\rbrace=\lbrace\sigma\in\text{Aut}(K):F\le\text{Fix}(\sigma)\rbrace$$


Lemma. Let $\sigma_1,\dots,\sigma_n\in\text{Gal}(K/F)$ be distinct. Then $\sigma_1,\dots,\sigma_n$ are linearly independent over $K$.

Proof. If $n=1$, then $a_1\sigma_1=0$, for all $x\in K$, for some $a_1\in F\Rightarrow 
a_1\sigma_1(1)=a_1=0$.
Assume false, and choose $n$ minimal such that it is false. So $n>1$, and:
\begin{equation}
  \tag{1}
  a_1\sigma_1(x)+\cdots+a_n\sigma_n(x)=0
\end{equation}
for all $x\in K$. Since $n$ is minimal, $a_1,\dots,a_n\ne 0$. So multiplying 
through by $a_n^{-1}$, we can assume that $a_n=1$.
Since $\sigma_1\ne\sigma_n$ (they are distinct), there exists an $\alpha\in K$ 
such that $\sigma_1(\alpha)\ne\sigma_n(\alpha)$. Replace $x$ in $(1) $
by $\alpha x$:
\begin{equation}
  \tag{2}
  a_1\sigma_1(x)\sigma_1(\alpha)+\cdots+a_{n-1}\sigma_{n-1}(x)\sigma_{n-1}(\alpha)
  +\sigma_n(x)\sigma_n(\alpha)=0
\end{equation}
Then:
\begin{equation*}
  (2)-\sigma_n(\alpha)(1)
  =a_1\sigma_1(x)(\sigma_1(\alpha)-\sigma_n(\alpha))+\cdots
  +a_{n-1}\sigma_{n-1}(x)(\sigma_{n-1}(\alpha)-\sigma_n(\alpha))=0
\end{equation*}
This eliminates the final term, and the coefficient of $\sigma_1(x)$ is $a_1(\sigma_1(\alpha)-\sigma_n(\alpha))\ne 0$ 
by choice of $\alpha$. So we have linear dependence between 
$\sigma_1,\dots,\sigma_{n-1}$, contradicting the minimality of $n$. QED


Proposition. If $|K:F|<\infty$, then $|\text{Gal}(K/F)|\le|K:F|$.

Proof. Let $|K:F|=n$, with basis $\omega_1,\dots,\omega_n$ of $K$ over $F$. Let $\sigma_1,\dots,\sigma_m\in\text{Gal}(K/F)$ 
be distinct. Assume for a contradiction that $m>n$.
Consider the following system of $n$ homogeneous linear equations in $m$ 
unknowns:
\begin{align}
  \tag{1}
  \sigma_1(\omega_1)x_1+\cdots+\sigma_m(\omega_1)x_m&=0 \\
  &~~\vdots \notag\\
  \tag{$n$}
  \sigma_1(\omega_n)x_1+\cdots+\sigma_m(\omega_n)x_n&=0
\end{align}
By Linear Algebra, there is a non-zero solution $x_i=a_i\in K$, for 
$i=1,\dots,m$, with $a_1,\dots,a_m$ not all zero.
We claim that $a_1\sigma_1(x)+\cdots+a_m\sigma_n(x)=0$, for all $x\in K$, 
which would contradict the Lemma. To see this, let $x\in K$, and 
since $\omega_1,\dots,\omega_n$ are a basis for $K$ over $F$, 
$x=b_1\omega_1+\cdots+b_n\omega_n$, for some $b_1,\dots,b_n\in F$. Consider 
$b_1(1)+\cdots+b_n(n)$. Since $b_1,\dots,b_n\in 
F$, they are fixed by $\text{Gal}(K/F)$, and hence $a_1\sigma_1(x)+\cdots+a_m\sigma_m(x)=0$ 
as claimed. QED


Theorem. Let $G\le\text{Aut}(K)$, $|G|<\infty$, and let $F=\text{Fix}(G)$. Then $|G|=|K:F|$.

Proof. Let $G=\lbrace\sigma_1=1,\sigma_2,\dots,\sigma_m\rbrace$. By the Proposition, 
$m\le|K:F|$ since $G\subset\text{Gal}(K/F)$.
Suppose for a contradiction that $m<|K:F|$. That is, there exists $\omega_1,\dots,\omega_n\in K$ 
linearly independent over $F$ with $m<n$.
This time, we set up $m$ 
linear homogeneous equations in $n$ unkowns:
\begin{align*}
  \sigma_1(\omega_1)x_1+\cdots+\sigma_1(\omega_n)x_n&=0 \\
  &~~\vdots \\
  \sigma_m(\omega_1)x_1+\cdots+\sigma_m(\omega_n)x_n&=0
\end{align*}
Again, there is a non-zero solution $x_1=a_1,\dots,x_n=a_n$.
We choose a solution with $k$ non-zero $a_i$'s such that $k$ is minimal. We 
may assume that $a_k=1$, and $a_1,\dots,a_k$ are non-zero, and:
\begin{equation}
  \tag{1}
  \sigma_i(\omega_1)a_1+\cdots+\sigma_i(\omega_{k-1})a_{k-1}
  +\sigma_i(\omega_k)=0
\end{equation}
for $i=1,\dots,m$. Since $\omega_k\ne 0$, we must have $k>1$. In particular, 
for $i=1$ we get $\omega_1a_1+\cdots+\omega_{k-1}a_{k-1}+\omega_k=0$.
So $a_1,\dots,a_k$ cannot all be in $F$ since $\omega_1,\dots,\omega_k$ are 
linearly independent over $F$. So assume that $a_1\notin 
F=\operatorname{Fix}(G)$.
So for some $j$, $\sigma_j(a_1)\ne a_1$. We apply $\sigma_j$ 
to the $m$ equations in $(1)$. Since $G$ is a group, we have $\lbrace \sigma_j\sigma_1,\dots,\sigma_j\sigma_m\rbrace
=\lbrace\sigma_1,\dots,\sigma_m\rbrace$, just reordering the permutations. 
So (after reordering the equations) we get:
\begin{equation}
  \tag{2}
  \sigma_i(\omega_1)\sigma_j(a_1)+\cdots+\sigma_i(\omega_{k-1})
  \sigma_j(a_{k-1})+\sigma_i(\omega_k)=0
\end{equation}
for $i=1,\dots,m$. And:
\begin{align*}
  (2)-(1)
  &=\sigma_i(\omega_1)(\sigma_j(a_1)-a_1)+\cdots
  +\sigma_i(\omega_{k-1})(\sigma_j(a_{k-1})-a_{k-1})
  +\sigma_i(\omega_k)(\underbrace{\sigma_j(a_k)-a_k}_{=0}) \\
  &=\sigma_i(\omega_1)(\sigma_j(a_1)-a_1)+\cdots
  +\sigma_i(\omega_{k-1})(\sigma_j(a_{k-1})-a_{k-1})
\end{align*}
Since $\sigma_j(a_1)-a_1\ne 0$, this is a solution with less than $k$ 
non-zero terms, contradicting the minimality of $k$. QED
