Prove that if the Galois group of the splitting field of a cubic over is the cyclic group of order 3 then all the roots of the cubic are real.
So here is what I know
since the group is cyclic it is Abelian( and so are all of its subgroups)
The Galois group has only the trivial subgroups and they are normal.
This implies that each field extention is Galois and that there is a 1-1 corrispondance to the subgroups of G and the subfields of ie Since there are no non trivial sub groups there are not any non trivial subfields???
I think I need to argue the degree of the extentions, but I don't see how to make the connection.
Thanks for any help