Galois extentions 

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*If $F \subset K \subset E$, with $E/K$ and $K/F$ galois, then $E/F$ is galois.


*If $F \subset K \subset E$, with $E/F$ galois, then $E/K$ and $K/F$ are galois.
$\def\less{\smallsetminus}$

For the first one: $E/F$ is galois because $E/K$ is galois you can extend an automorphism $\delta$ on $F$ to an automorphism $δ'$ on $K$ such that $δ'(α) \neq α$  for $α \in K \less F$. And in a similar way for elements in $E \less K$. Therefore there is an automorphism on $E$ which permutes all elements not in $F$.
How would I prove the second one is right or wrong?
 A: If $Gal(E/F)$ is abelian, then the second statement is true.
In general case, $E/K$ is Galois, given $E/F$ is. This is the statement of Fundamental Theorem of Galois theory.
However $K/F$ needn't be Galois if $Gal(E/F)$ is not abelian.
Let $E$ be the splitting field of $x^{3} - 2$, an irreducible polynomial, over $\mathbb{Q} = F$ and let $K =  \mathbb{Q} (2^{1/3})$. Then $E$ is Galois over $F$ , because it's the splitting field of a polynomial,hence over $K$ as well,  However $K/F$ is not a normal extension (Why?)  , hence not Galois. 
A: You don’t make clear just what your question is. I presume, though, that you’ve been asked to say whether your two statements are true or false; and if false, show a counterexample.
Galoisness, in fact, normality, is not a tower property, that is, a normal extension of a normal extension is not necessarily normal, and the standard example is $\Bbb Q\subset\Bbb Q(\sqrt2\,)\subset\Bbb Q(\sqrt[4]2\,)$. The big field isn’t normal over $\Bbb Q$ ’cause it doesn’t contain the root $i\sqrt[4]2$ of $X^4-2$. The two separate layers are normal because they’re quadratic, and every quadratic extension is normal.
But you’ve made a statement in your attempted proof of the first proposition that might be a common misapprehension. You seem to say that an automorphism of the subfield $F$ can be extended to an automorphism of the Galois extension $K$ of $F$. This is emphatically not true. Take the automorphism of $F=\Bbb C(z)$ that sends $z$ to $z+1$, so takes a rational function $q(z)$ to $q(z+1)$. And take $K=\Bbb C(z^{1/2})$, a quadratic extension, so Galois. But $(z+1)^{1/2}\notin\Bbb C(z^{1/2})$, that is, the automorphism I’ve specified can’t be extended to the extension field.
