Is normal extension of normal extension always normal? Let F be a char 0 field,
K be a normal extension of F
and L be a normal extension of K.
Can it be proved or disproved that L is normal extension of F ?
 A: Here I try to explain better of zcn's example. 
Let $E=\mathbb{Q}, F=\mathbb{Q}(\sqrt{2}), G=\mathbb{Q}(2^{1/4}), H=\mathbb{Q}(2^{1/4},i)$. Then we see that
$$
\textrm{Gal}(F|E)=\mathbb{Z}/2\mathbb{Z}, \textrm{Gal}(G|F)=\mathbb{Z}/2\mathbb{Z},\textrm{Gal}(H|E)=\mathbb{D}_{4}
$$
where the generators for $\textrm{Gal}(H|E)$ are $g:i\rightarrow -i$ (complex conjugation) and $h:2^{1/4}\rightarrow i*2^{1/4}$ (permutation of the roots). The way we think about it is $F$ is the fixed field of $H$ under $\langle g,h^2\rangle$, and $G$ is the fixed subfield of $H$ under $\langle g\rangle$, whereas $E$ is the fixed subfield of the whole group. If $G$ is a normal extension over $E$, then by Galois correspondence $\langle g\rangle$ will be a normal subgroup of $\langle g,h\rangle$. But we know that
$$
h^{-1}gh=gh^3\not\in \langle g\rangle
$$
Therefore it cannot be normal. But we do have $\langle g\rangle \unlhd \langle g,h^2\rangle$ and $\langle g,h^2\rangle \unlhd \langle g, h\rangle$. 
I try to do it by hand mainly because the question has showed up again and again:
Are normal subgroups transitive?
Example of composition of two normal field extensions which is not normal.
K⊂M⊂L tower of fields. Find counterexample for statement "if L normal over K, then M normal over K"
Does $K/E$ and $E/F$ being normal mean $K/F$ is normal?
and probably in other places. But somehow an explicit calculation is missing. Not sure if this helps. This is essentially the same as the Wikipedia example with $\mathbb{D}_{4}$ instead of $\mathbb{S}_{3}$. I remember this was a well-known freshmen Galois theory class execrise, and for unknown reason I never really did it. 
A: Let $F = \mathbb{Q}$, $K = \mathbb{Q}(\sqrt{2})$, $L = \mathbb{Q}(\sqrt[4]{2})$. Then $K/F, L/K$ are degree $2$ extensions, hence normal (and Galois) but $L/F$ is not normal (the splitting field of $x^4 - 2$ has degree $8$ over $\mathbb{Q}$).
