Proving Isomorphism and Galois Group Existence in Abstract Algebra Homework

[Mystic998]

Homework Statement

Show that G is isomorphic to the Galois group of an irreducible polynomial of degree d iff is has a subgroup H of index d such that \bigcap_{\sigma \in G} \sigma H \sigma^{-1} = {1}.

Homework Equations

The Attempt at a Solution

I know that if G acts transitively as a permutation group on a set (and it does in this case being isomorphic to the Galois group of an irreducible), then the stabilizer of an element has the given property. And since the stabilizer is functionally the same for any root, it really shouldn't matter which one I choose. But I don't know how to show that it's of index d. I've tried counting permutations that fix an element and permutations that don't (should be |G|/d or d respectively, assuming I'm doing this right), but I can't find any obvious way to do so in general since obviously the Galois group can be lots and lots of weird things (all subgroups of a symmetric group of course).

As for the other direction, I think it's probably easy to show that if all the conjugates of a subgroup intersect in the trivial subgroup, then G has to act transitively on some set, and so it can be the Galois group of an irreducible with the same number of roots as the set it's acting on. But again I'm not sure why the index being d implies that it has to act transitively on a set of d elements.

Thanks for any help. Also, I might post some more stuff in this topic if I don't have any luck on some other problems.

 

[morphism]

Did you mean to say "such that \bigcap_{\sigma \in G} \sigma H \sigma^{-1} = H"?

For the index bit, can't you invoke the fundamental theorem of Galois theory? Remember the connection between indices of subgroups and degrees of field extensions.

 

[Mystic998]

Oh, I meant for the intersection to be trivial. I'll think about what you said though.