Order of the Galois Group is $|G(E/F)| = [E:F_G]$ where $F_G$ is the field fixed by $G$

Question

I was reading Lisl Gaal' book on Galois Theory and, on page 84, the author uses the result in the title without proving it, leaving the proof to the reader (see below):

I want to prove the following: $|G(E/F)| = [E:F_G]$ where $F_G$ is the field fixed by $G = G(E/F)$ and $E/F$ is a finite extension.

I could find a proof of this on Michael Artin's book but, the way the problem was presented in Gaal's book, it seems there should be simpler proof using the results presented so far there, but I am not able to prove it.

If you could provide any hint I would appreciate a lot.

Answer 1

$G=Aut(E/F)=Aut(E/F_G)$.

Every $a\in E$ is a root of $$f_a=\prod_{b\in Ga}(x-b)$$ which is separable, in $F_G[x]$, and of degree $\le |G|$.

So $E/F_G$ is separable, from which $$[E:F_G] = |Hom_{F_G}(E,\overline{E})|$$

Any element of $Hom_{F_G}(E,\overline{E})$ sends $a$ to some root of $f_a$, which is in $E$. So $Hom_{F_G}(E,\overline{E}) = Aut(E/F_G)$ and hence $$[E:F_G]=|G|$$

Answer 2

Since $[E:F_G] < \infty$ in this problem, wlog $E=F_G(\alpha_1,....,\alpha_m)$. Hence the problem reduces to $[F_G(\alpha):F_G] = |Aut(F_G(\alpha)/F_G)|$ by induction on $[E:F_G]$. This is because restriction map is a homomorphism between $Aut(F_G(\alpha,\beta)/F_G)$ and $Hom_{F_G}(F_G(\alpha),F_G(\alpha,\beta))$. Kernel of this map is $Aut(F_G(\alpha,\beta)/F_G(\alpha))$. Hence $$Aut(F_G(\alpha,\beta)/F_G)/Aut(F_G(\alpha,\beta)/F_G(\alpha)) \tilde{=} Hom_{F_G}(F_G(\alpha),F_G(\alpha,\beta)) \supseteq Aut(F_G(\alpha)/F_G).$$ Hence we can restrict ourselves to proving:$$[F_G(\alpha):F_G] = |Aut(F_G(\alpha)/F_G)|.$$ Following is an attempt to analyze if this equality holds. This can be done by observing that $\alpha$ is mapped to a root of minimal polynomial of $\alpha$ by an Automorphism and if there are distinct roots of minimal polynomial of $\alpha$ inside $F_G(\alpha)$, we have the equality which is to be proved. We have distinct roots as $F_G(\alpha)/F_G$ is a separable extension.

Editing based on feedback comments below (inspired by proof given by @reuns ) : Assumption in the proof of $$[F_G(\alpha):F_G] = |Aut(F_G(\alpha)/F_G)|.$$ The proof assume that minimal polynomial of $\alpha$ is separable (distinct factors) and $F_G(\alpha)$ is a splitting field of minimial polynomial of $\alpha$. This can be partially proved as follows: Since $E$ has all roots of minimial polynomial of every element present in it including $\alpha$, we have : $$[F_G(\alpha):F_G] = |Hom_{F_G}(F_G(\alpha),E)| = \frac{|Aut(E/F_G)|}{|Aut(E/F_G(\alpha))|} \geq |Aut(F_G(\alpha)/F_G)|$$.

The second equality in the above equation is because every homomorphism in $Hom_{F_G}(F_G(\alpha),E)$ can be extended to an automorphism in $Aut(E/F_G)$ and this is an injective embedding and further restriction map for automorphisms in $Aut(E/F_G)$ and quotienting on the kernel of this restriction map gives the equality. The third inequality if equality completes the proof. As @reuns pointed out, this need not be true but i am leaving the answer as it is for clarity to others.