What is the Conjugacy Criterion for Galois Subgroups?

[masterslave]

Homework Statement

Let K be a Galois extension of F. Two intermediate fields E and L of field F are said to be conjugate if there exists
\sigma\in\text Gal_F K such that \sigma (E) = L.
Prove that E and L are conjugates of F if and only if \text Gal_E K and \text Gal_L K are conjugate subgroups of \text Gal_F K.

The Attempt at a Solution

From left to right, I have it already. I can't figure out how to get anywhere going from the right to left part of the proof.
I want to show that \sigma (E)=L for some \sigma\in Gal_F K.
Let \alpha\in Gal_L K , \beta\in Gal_E K.
Since Gal_L K, Gal_E Kare conjugates, then Gal_E K=\{\sigma\alpha\sigma^{-1} | \alpha\in Gal_L K\} or Gal_L K=\{\sigma^{-1}\beta\sigma | \beta\in Gal_E K\}.

\alpha=\sigma\beta\sigma^{-1}
\beta=\sigma^{-1}\alpha\sigma

I know that \beta fixes E and that \alpha fixes L.

\alpha (L)=\sigma\beta\sigma^{-1} (L)
\beta (E)=\sigma^{-1}\alpha\sigma (E)

L=\sigma\beta\sigma^{-1} (L)
E=\sigma^{-1}\alpha\sigma (E)

\sigma^{-1} (L)=\beta (\sigma^{-1} (L))
\sigma (E)=\alpha (\sigma (E))

\beta fixes \sigma^{-1} (L) and \alpha fixes \sigma (E)

\beta\in Gal_{\sigma^{-1} (L)} K and \alpha\in Gal_{\sigma (E)} K

From earlier: \beta\in Gal_E K and \alpha\in Gal_L K

So the only conclusion that I have is that the Gal_{\sigma (E)} K \cap Gal_L K and Gal_{\sigma^{-1} (L)} K \cap Gal_E K are nontrivial.

I feel like I should be using the Galois extension, i.e. Galois correspondence, to my advantage here, but I just don't see how it is applicable. It gives me that there exists a isomorphism from the intermediate fields to their respective Galois subgroup, i.e. \tau (E)=Gal_E K. And since isomorphisms are order preserving, I get \left| {Gal_E K} \right|=\left| {E} \right| and \left| {Gal_L K} \right|=\left| {L} \right|.

Any direction on the problem is appreciated.

Thanks in advance.

 

[Billy Bob]

I know that \beta fixes E and that \alpha fixes L.

\alpha (L)=\sigma\beta\sigma^{-1} (L)
\beta (E)=\sigma^{-1}\alpha\sigma (E)

L=\sigma\beta\sigma^{-1} (L)
E=\sigma^{-1}\alpha\sigma (E)

\sigma^{-1} (L)=\beta (\sigma^{-1} (L))
\sigma (E)=\alpha (\sigma (E))

\beta fixes \sigma^{-1} (L) and \alpha fixes \sigma (E)

Shouldn't that argument be carried out on elements of E (and elements of L) rather than on the set E itself (resp. L)? Still get the same conclusion though.

So the only conclusion that I have is that the Gal_{\sigma (E)} K \cap Gal_L K and Gal_{\sigma^{-1} (L)} K \cap Gal_E K are nontrivial.

No, better than that, didn't you show

Gal_L K\subset Gal_{\sigma (E)} K and Gal_E K\subset Gal_{\sigma^{-1} (L)} K ?


Would it help now to use G_1 \subset G_2 implies K_{G_1} \supset K_{G_2} where K_G denotes the subfield of K fixed by G?

 

[VKint]

Let G = \textrm{Gal}_L K and H = \textrm{Gal}_E K. You have G = \sigma H \sigma^{-1}. Why is it true that if a \in E and \tau \in G, then \tau \circ \sigma (a) = \sigma(a)? Why does this tell you that \sigma(E) \subseteq L (hint: fundamental theorem of Galois theory)? (The reverse inclusion is basically the same.)

 

[masterslave]

Let G = \textrm{Gal}_L K and H = \textrm{Gal}_E K. You have G = \sigma H \sigma^{-1}. Why is it true that if a \in E and \tau \in G, then \tau \circ \sigma (a) = \sigma(a)? Why does this tell you that \sigma(E) \subseteq L (hint: fundamental theorem of Galois theory)? (The reverse inclusion is basically the same.)

Thank you both for your kind help. I see how <br /> \sigma(E) \subseteq L <br /> , but i don't see how to get <br /> L \subseteq \sigma(E)<br />

Thanks again!

 

[Billy Bob]

Thank you both for your kind help. I see how <br /> \sigma(E) \subseteq L <br /> , but i don't see how to get <br /> L \subseteq \sigma(E)<br />

Thanks again!

You can get \sigma^{-1}(L) \subseteq E so then you have it.

 

[masterslave]

Wow, yeah, it's so obvious now. Thank you all for your time. I get it now.