Thread: Galois conjugates

Let K/F be extension field. Let E and L be two intermediate fields. E and L are called conjugate if there is f in Gal(K/E) s.t. f(E)=L .

Show that E and L two conjugate iff Gal(K/E) and Gal(K/L) are conjugate subgroups
of Gal(K/F).

For simplicity let A=Gal(K/E) and B=Gal(K/L) . Since E and L are conjugates
then there is f in Gal(K/F) s.t. f(E)=L .

Pleas help on this question.

Actually the first direction is easy. I think the answer is:
We show that f.A.f^(-1)= B. i.e., for every g in A, f.A.f^-1 is in B ( i.e. f.A.f^-1 fixes L ) and for every h in B, f.B.f^-1 is in A (i.e. fixes E).
Doing the first part, let x be an element in L then f^-1(x)= y in E , and g fixes y , and f again send y to L, so f.A.f^-1 fixes L and similarly for the other assertion, hence the first direction is proved . I am having trouble in the reverse direction.

I think I got it. there is f in Gal(K/F) s.t. f.A.f^-1=B we show that f(E) is subset of L.
we have f.A.f^-1 is a subset of B, i.e. f.A.f^-1 fixes L.
consider f(e) where e is an element in E. then f^-1(f(e))=e (since f is bijection) and for g be in A g(e)= d where d is another element in E (since g fixes E)
so f(d) is in L hence we have f.A.f^-1 , hence f(e)) is in L . similarly i can show the other direction and hence f(E)=L . Is this right ?