# Proof subextension of Abelian extension is also abelian

This is a property listed on MathWorld:

One nice property of an Abelian extension $K$ of a field $F$ is that any intermediate subfield $E$, with $F \subset E \subset K$, must be a Galois extension field of $F$ and, by the fundamental theorem of Galois theory, also an Abelian extension

What specifically about the fundamental theorem of Galois Theory shows that? To my knowledge, it only shows a one-to-one correspondence from the Galois subgroups and the intermediate fields. How does that make it abelian?

Thanks!

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At the moment this has a tag "finite-fields", which isn't really relevant to the question. –  KCd Apr 1 '12 at 5:03

The fundamental theorem tells you that $E/F$ is Galois and that $\text{Gal}(E/F)$ is a quotient of $\text{Gal}(K/F)$--and quotients of abelian groups are abelian.
How do you know that $E \subset F$ is normal? –  Larry X Apr 1 '12 at 4:44
@Larry: the fundamental theorem tells you that $E/F$ is Galois iff $\text{Gal}(K/E)$ is normal in $\text{Gal}(K/F)$, which is automatically true if $\text{Gal}(K/F)$ is abelian. –  Qiaochu Yuan Apr 1 '12 at 4:46
Because you know that $E/F$ will be normal if and only if the subgroup of $\text{Gal}(K/F)$ corresponding to $E$ is normal. But, $\text{Gal}(K/F)$ is abelian so every subgroup is normal. –  Alex Youcis Apr 1 '12 at 4:46