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This question (of the title) rose from the following theorem and its proof:

Theorem: Let $K/F $ be a finite extension of fields. Then rhe Galois group $G (K/F) $ is a finite group and satisfies $[K : F] \geq [G : (K/F)] $ Proof: As $F $ is a subfield of $K^S $ for any nonempty subset $S $ of $G (K/F) $ we have $[K:F] \geq [K:K^S] \geq |S|$ for any finite $S $. It follows that $G (K/F) $ is finite and satisfies $[K : F] \geq [G (K/F)] $

I cannot comprehend why "it follows that $G (K/F) $ is finite". Since $K/F $ is finite $K=F (a_1,...,a_n) $ with $a_i $'s algebraic, and as any automorphism is isomorhism the $a_i $'s cannot be mapped to any element in $F $. My undersranding of the (untold) part of the proof is that the $a_i$'s permute under action by $G (K/F) $. But why must the $a_i $'s get mapped to another $a_j $? It seems that it may get mapped to some $f \in F (a_1,...,a_n) - (F \cup \{a_1,...,a_n\} $. Am I missing something here?

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If $G(K/F)$ were infinite, you could choose a finite subset $S\subset G(K/F)$ with more than $[K:F]$ elements (since $[K:F]$ is finite). The inequality $[K:F]\geq[K:K^S]\geq|S|$ then gives a contradiction.

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Each $a_i$ is the root of some polynomial $f_i(X)\in F[X]$. For every $\sigma\in G(K/F)$, it is then clear that also $f_i(\sigma (a_i))=0$. So while $G$ may not induce a permutation on the finite set $\{a_,\ldots,a_n\}$, it certainly does on the - still finite - set obtained by adding all other roots of all $f_i$ to the set. Note that this answers "Why is $G$ finite?" and the main question "Where do the $a_i$ get mapped to?". To see that the order of $G$ is in fact at most the degree of the field extension, follow the original argument.

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