Some exercises about Galois Groups and Automorphisms of a finite field extension

Question

Hey I have these two exercises where I am having some problems solving them:

Let $E/K$ be a finite field extension, let $G$ be a subgroup of $Aut(E/K)$. Show:

a) $G$ is finite and $|G|$ is a divisor of $[E : K]$. b) Exactly then $|G| = [E : K]$ if $E/K$ is Galois with $G =Gal(E/K)$.

For a) I have:

Since $|G|\leq |Aut(E/K)|\leq[E:K]$, $G$ is finite. $G$ is a subgroup of $Aut(E/K)$, therefore $|G|$ divides $|Aut(E/K)|$. Since for every $\tau \in Aut(E/K), \tau(K)=K$ every $\tau$ is just a permutation on the elements of $E\K$ therefore the number of the elements divides $[E:K]$. From this follows our assumption.

Here I am not sure if I have done any mistake.

For b) I have only one direction of the implication: $|G| = [E : K]$ $\Leftrightarrow$ $E/K$ is Galois with $G =Gal(E/K)$

$\Rightarrow$: Since $|G|\leq |Aut(E/K)|\leq[E:K]$ so we have $|Aut(E/K)|=[E:K]$ and therefore is $E/K$ Galois with $G =Gal(E/K)$.

For $\Leftarrow$ I really dont know what to do. Can someone help me?

Answer 1

I don't understand your proof of (a) past the part where you show inequality.

For (a), we have $E^G$, the fixed field of G, gives a Galois extension $E/E^G$ with Galois group $G$. Thus, $|G| = [E : E^G]$. Note that $K \subseteq E^G$. I'll leave it to you to complete the argument from here. It's a simple degree argument.

For the converse of (b), you should review the several equivalent definitions of a Galois extension. It follows rather directly from the extension being Galois.