I'm working through some practice problems on Galois Theory and really trying to understand the concepts. I've come across the following question:
Is there a finite Galois extension $F$ of $K$ such that there are exactly two fields $E$ such that $K \leq E \leq F$ and $[F:E] = 2$?
I'm pretty sure the answer is no, but I'm not sure how to rigorously prove it. I know $F$ must be finitely generated and algebraic over both $E_x$ and $E_{y}$, but once I have the bases written out, I don't know how to use this to show a third $E_{z}$ must exist. Any help is appreciated.
If you translate your question in terms of group theory, you get the following question: does there exists a finite group $G$ with exactly two subgroups of order two, that is exactly two elements of order $2$ ?
The answer is NO: if $x,y$ have order $2$, so is $yxy^{-1}$. By assumption $yxy^{-1}=y$ or $x$. The first possibility is not possible since it will give $y=x$. Then we get the other possibility, and $xy=yx$. But $xy\neq 1$, since otherwise $y=x^{-1}=x$, and$(xy)^2=x^2y^2=1$. Hence $xy$ has order $2$. So $xy=x$ or $xy=y$, and we get the contradiction $x=1$ or $y=1$.