I am self studying Fields and Galois Theory from Algebra by Thomas Hungerford and I have a question in this theorem s proof.

Its imageenter image description here

I have question in line 1 where author writes "In any case $G$ is a subgroup of $\operatorname{Aut}_K F $ .

Definition of $\operatorname{Aut}_K F $ is group of all $K$-automorphisms of FF , where $G$ is a group of automorphisms of $F$.

So, I think since conditions are on $\operatorname{Aut}_K F $ that it must be a $K$-module homomorphism , so $G$ must be a larger set $\operatorname{Aut}_K F$ , but the opposite is given. So, can anyone please explain why opposite is given.

  • $\begingroup$ No, $G$ need not be "larger" because it is assumed that $K$ is the fixed field of $G$ in $F$. $\endgroup$ Apr 1, 2020 at 19:03

1 Answer 1


By definition $\operatorname{Aut}_KF$ is the group of all automorphisms of $F$ that fix $K$ pointwise. So if $G$ is a group of automorphisms of $F$ that fix $K$ pointwise, then $G$ is a subgroup of $\operatorname{Aut}_KF$.

  • $\begingroup$ sorry I should have paid more attention. $\endgroup$
    – user775699
    Apr 1, 2020 at 19:12

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