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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?

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It's a consequence of Theorem 4.6 on pg. 554, for which a proof is given. – Brandon Carter Nov 15 '11 at 22:35
up vote 6 down vote accepted

(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 $[K:F]$.

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Isn't the order of the Galois group also equal to the degree of extension? – user23238 Mar 17 '14 at 5:24
In the case of a Galois extension, yes. – grantfgates May 22 '14 at 22:21

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