Let $K$ be a fixed field in $\mathbb C$ (complex numbers) of an automorphism of $\mathbb C$. Prove that every finite extension of $K$ in $\mathbb C$ is cyclic.
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As Jyrki Lahtonen kindly pointed out in a comment below, there was a gap in the previous version of this answer. After having read Jyrki's comment, I also noticed that an answer, which (unlike mine) was correct, had been given by Jiangwei Xue in comment of May $3$, $2011$. I'll write Jiangwei Xue's solution below, and turn this answer into a community wiki.
Recall the following.
Let $G$ be a finite group of automorphisms of a field $L$, and let $K$ be the fixed field. Then $L/K$ is Galois with Galois group $G$.
If $G$ is abelian, any sub-extension $S/K$ of $L/K$ is Galois with Galois group a quotient of $G$. Thus $S/K$ is abelian.
If $G$ is cyclic, then so is any sub-extension $S/K$ of $L/K$.
Now here is Jiangwei Xue's argument.
Let $\sigma$ be an automorphism of a field $M$, let $K$ be the fixed field, and let $L/K$ be a finite degree sub-extension of $M/K$. Then the restriction of $\sigma$ to $M$ generates a finite group $G$ of automorphisms of $M$ whose fixed field is $K$, and the above observations show that $L/K$ is cyclic.
Aside. If $G$ be a finite group of automorphisms of an algebraically closed field $L$, then $G$ has order $1$ or $2$. Here are three references for this:
$\bullet$ These notes of our friend Pete L. Clark on Field Theory (see Theorem $124$ p. $75$, called "Grand Artin-Schreier Theorem"),
$\bullet$ This nLab entry.