# Conjugacy of Galois subgroups

## 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 $$Gal_L K, Gal_E K$$are 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.

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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?

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.)

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 $$\sigma(E) \subseteq L$$ , but i dont see how to get $$L \subseteq \sigma(E)$$

Thanks again!

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Thank you both for your kind help. I see how $$\sigma(E) \subseteq L$$ , but i dont see how to get $$L \subseteq \sigma(E)$$

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

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