Let $K/E$ and $E/F$ be Galois extensions. If every $σ ∈ \text{Aut}(E/F)$ is the restriction of an element of $\text{Aut}(K/F)$, then $K/F$ is Galois.

Question

$\newcommand{\Q}{\mathbb{Q}}$$\DeclareMathOperator{\Aut}{Aut}$Let $K/E$ and $E /F$ be Galois extensions. I would like to show that, if every $\sigma \in \Aut(E/F)$ is the restriction of an element of $\Aut(K/F)$, then $K \supset F$ is Galois.

I know that this is not necessarily true when only the information "Let $K /E$ and $E/F$ be Galois extensions." is given, for instance when $\Q \subset \Q(\sqrt2) \subset \Q(\sqrt[4]2)$, $K/F$ is not Galois.

I am not sure how to use the restriction aspect of the question, how to proceed?

Thank you.

Answer 1

$\DeclareMathOperator{\Aut}{Aut}$ Hint 1

You want to prove that if $a \in K \setminus F$, then there is $g \in \Aut(K/F)$ such that $a g \ne a$.

Hint 2

If $a \notin E$, then you even achieve that with an element of $g \in \Aut(K/E)$.

Hint 3

So you are left with $a \in E \setminus F$. You know that there is $h \in \Aut(E/F)$ such that $a h \ne a$.

Hint 4

And now use your hypothesis about the restriction...