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Question essentially in the title.

I know that if $ K \subseteq L $ is a finite extension, then $L/K$ is Galois if and only if the fixed field of $ \mbox{Aut}(L/K)$ is $K$.

But is it possible for there to be a finite subgroup $G < \mbox{Aut}(L/K)$ that also has $K$ as a fixed field?

Thanks

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    I think I've just solved the problem myself. If $G$ is a finite group of automorphisms acting on $L$, with $K$ the fixed field, then $ |G| \geq [L:K] $. But we have that $L$ is Galois, so $ |\mbox{Aut}(L/K)| = [L:K] $. So we must have that $G = \mbox{Aut}(L/K) $.2011-12-06
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    You can post your own answer. Your comment looks like a good answer, in any case.2013-01-22

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