Is every subgroup of automorphisms a Galois group? 
Let $G=\{\sigma_1=1, \sigma_2, \dots,\sigma_n\}$ be a subgroup of automorphisms of a field $K$ and let $F$ be the fixed field. Then $[K:F]=|G|$

Why is this always true? I thought this would be true only if $K$ was the Galois extension of $F$ and $G$ was $\text{Aut}(K/F)$
 A: Let $E$ be a field, $G$ be a finite subgroup of $\textrm{Aut}(E)$ and $F:=\textrm{Fix}(G)$. The result you are trying to prove is usually called fixed subfield formula and relies on the Artin's lemma, that is:

Lemma 1. $[E:F]\leqslant |G|$.

Proof. A proof can be found in the Jacobson's Basic Algebra I:
 $\Box$
To establish the fixed subfield formula, one has now to show that:

Lemma 2. $[E:F]\geqslant|G|$.

Proof. Let $m:=\left[E:F\right]$, $n:=|G|$ and assume by contradiction that $m<n$. Let $(x_i)_{i\in\{1,\ldots,m\}}\in E^m$ be a $F$-basis of $E$ and $\sigma_1,\ldots,\sigma_n$ be the distinct elements of $G$. Since $m<n$ there exists $(y_i)_{i\in\{1,\ldots,n\}}\in E^n$ a nonzero solution to the following homogeneous system: $$\forall i\in\{1,\ldots,m\},\sum_{i=1}^n\sigma_j(x_i)Y_i=0_E.$$
Let $x\in E$, there exists $(\lambda_i)_{i\in\{1,\ldots,m\}}\in(E^G)^m$ such that: $$x=\sum_{i=1}^m\lambda_ix_i.$$
One derives from the construction of $(y_i)_{i\in\{1,\ldots,n\}}$: $$\sum_{i=1}^n\sigma(x)y_i=\sum_{i=1}^m\lambda_i\sum_{j=1}^n\sigma_i(x_j)y_i=0_L.$$
Therefore, one has: $$\sum_{i=1}^n\sigma y_i=0_{\textrm{Map}(G,E)}.$$
Since $(y_i)_{i\in\{1,\ldots,n\}}$ is nonzero and $\sigma_1,\ldots,\sigma_n$ are pairwise distinct, here is a contradiction to Dedekind's lemma (see Lemma $1.1.$ at p$.1$). $\Box$
