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Let $F\subseteq E$ be a finite degree extension and $E$ be separable over $F$. Let $L\supseteq E$ be a splitting field over $F$ for some polynomial $g\in F[X]$ with the property that every irreducible factor of $g$ in $F[x]$ has a root in $E$. First, I want to prove that $L$ is Galois over $F$ and second $|L:F|$ divides $|E:F|!$.

In order to prove that $L$ is Galois over $F$, I have a following theorem for Galois extension:

If $|L:F|$ is finite and $L$ is a splitting field over $F$ for some "separable" polynomial over $F$, then $L$ is Galois over $F$.

Since $E$ is separable over $F$ and $L\supseteq E$ be a splitting field over $F$ for some polynomial $g\in F[X]$ with the property that every irreducible factor of $g$ in $F[x]$ has a root in $E$, an irreducible factor of $g$ is a unit multiple of a minimal polynomial $f=min_F(\alpha)$, where $\alpha \in E$. Hence, $g$ is separable over $F$ implying that $L$ is Galois over $F$ by the theorem above.

Now, I have that $F\subseteq E \subseteq L$ such that $|L:F|=|L:E|.|E:F|$ and by Fundamental theorem of Galois Theory $|L:F|=|Gal(L/F)|$.

My question is that how can I show that $|L:F|$ divides $|E:F|!$ using my works above.

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By Galois Correspondence, this can be translated in group theory: let $G=Gal(L/F)$ and $H=Gal(E/F)$, then $|G:H|=[E:F]$. Observe that, because the extensions are normal and separable, $G$ acts transitively and this implies that the intersection of all the conjugates of $H$ in $G$ is trivial, that is $core_G(H)=1$. Then the so-called $n!$-Theorem of group theory now gives the required result: let $G$ act on the left cosets by left multiplication. This group action has trivial kernel ($=core_G(H)$) and hence embeds $G$ homomorphically in $S_n$. Hence $|G|$ divides $n!$.

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