Property of Galois extension Let $F\subseteq K \subseteq L$ be fields. If $L$ is Galois over $F$, then $K$ is Galois over $F$. Is this true?
 A: No. Consider $F=\Bbb Q, K=\Bbb Q(\sqrt[3]{2}), L=\Bbb Q(\sqrt[3]{2},\zeta_3)$, where $\zeta_3$ is a third root of unity.
A: We have a notion of Galois closure: if $K/F$ is a finite separable extension, its Galois closure is a minimal extension $L/K$ such that $L/F$ is Galois. Seeing as we have such a notion, then it is clear that $L/F$ 
being Galois does not entail $K/F$ being Galois. For a concrete example, pick any finite separable non-Galois extension, say $\Bbb Q(\sqrt[3]2)/\Bbb Q$,
and work out its Galois closure.
A: The other answers give counterexamples, but here is the "real reason" why $K/F$ need not be Galois:
If $G=\operatorname{Gal}(L/F)$ is the Galois group of $L/F$, then Galois theory tells us that intermediate fields $F \subseteq K \subseteq L$ correspond 1-to-1 with subgroups $H < G$. Furthermore, $K/F$ is Galois if and only if $H \trianglelefteq G$, i.e. the corresponding subgroup is normal in the Galois group.
Therefore your statement is not true in general because there exist Galois groups $G$ with subgroups $H$ which are not normal. In the other answers (by MatheinBoulomenos and Lord Shark the Unknown) there are counterexamples where $G \cong S_3$ and $H \cong C_2 \cong \langle (1\,2) \rangle \subseteq S_3$ and $H \not\trianglelefteq G$.
On the other hand, if every subgroup of $G$ is normal (such a group is called a Dedekind group) then every intermediate field is normal. In particular if $G$ is abelian then this is true.
