Given a tower of field extensions, does this equality involving Galois group orders hold in general?

Suppose we have a tower of field extensions:

$\overline{F} \subset K \subset E \subset F$

Is it true in general that $|G(K/F)| = |G(K/E)| \cdot |G(E/F)|$?

I was able to verify some specific examples, like $\mathbb{Q}(\sqrt[3]{2}, \omega)$ for $x^3-2$ and another extension, but how could I show that this holds in general for all such towers of extensions?

• Are you assuming that $K$ and $E$ are Galois extensions of $F$?. If so then this is equivalent to the standard identity $|K:F| = |K:E| \cdot |E:F|$. – Rolf Hoyer May 17 '16 at 19:37
• All I know is that they are extensions. – Ninosław Brzostowiecki May 17 '16 at 19:38
• Then does $G(K/F)$ just mean $Aut(K/F)$, the automorphisms of $K$ fixing $F$? Or is it the Galois group of the normal closure of $K$ over $F$? – Rolf Hoyer May 17 '16 at 19:42
• It would have to be the latter,the Galois group of the normal closer of K over F. – Ninosław Brzostowiecki May 17 '16 at 19:52

This is generally not the case. For instance, take $K = \Bbb Q(\sqrt[4]{2}), E = \Bbb Q(\sqrt{2}), F = \Bbb Q$. Then $K/E$ and $E/F$ are both Galois extensions with $|G(K/E)| = |G(E/F)| = 2$. However, the extensions $K/F$ is not Galois, with there being only two automorphisms of $K$ fixing $F$, since such an automorphism is determined by its action on $\sqrt[4]{2}$, which can only be sent to $\pm \sqrt[4]{2}$, the only roots of $x^4-2$ lying in $K$. Thus, $|Aut(K/F)| = 2$.
If you interpret $|G(K/F)|$ to be the order of the Galois group of the Galois closure of $K$ over $F$, then the order is 8 instead, as the Galois closure, $\Bbb Q(\sqrt[4]{2}, i)$ is a degree 8 extension.
• @NinosławCiszewski If all three extensions $K/F, E/F, K/E$ are Galois, then the equality would hold, yes. – Rolf Hoyer May 18 '16 at 20:25