# 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.

$$\newcommand{\Q}{\mathbb{Q}}\DeclareMathOperator{\Aut}{Aut}$$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.

$$\DeclareMathOperator{\Aut}{Aut}$$ 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