I keep seeing this theorem used in many textbooks but none of them provide proof (or there is no text layer so I can't find it!). Here is the statement in Algebra (Artin, pg. 540):
(1.6) Theorem. For any finite extension $K/F$, the order $|G(K/F)|$ of the Galois group divides the degree $[K:F]$ of the extension.
I am brand new to Field Theory and Galois Theory, but this seems rather nontrivial to prove and so I don't know why I am having such trouble finding a book which proves it.
Just to be clear, here is my understanding of a few of the definitions first. If I have misunderstood, please correct me.
The Galois group $G(K/F)$ is the set of all automorphisms of $K$ which fix every element of $F$.
The degree $[K:F]$ is the dimension of $K$ as an $F$-vector space, that is, the number of vectors in any basis of $K$ where linear combinations are taken with coefficients in $F$.
Could someone please show me a proof of this theorem?
Best Answer
(This answer follows the comment by Brandon Carter.)
Consider $K^G$, the fixed field of $G$. This is a subfield of $K$ such that $[K:K^G]=|G|$. This is the main non-trivial point. Since $[K:F]=[K:K^G]\,[K^G:F]$, we conclude that $|G|$ divides $[K:F]$.