Since it tells us that where . If is an automorphism of then it is determined by its value on . Remember that automorphisms permute zeros of polynomials therefore is a zero of and is a zero of . Thus, . We see that there are at most automorphisms but since it means we must have an automorphism for each one of these possibilities. Define by , define by . We see that is another automorphism of . Therefore, with . The non-trivial proper subgroups of are: (just like ).(b) Determine the Galois Group of over .
The proper non-trivial intermediate are: . Notice that is fixed by therefore (fill in the minor details in this argument, do you see how or you want me to?) that likewise . For notice that is fixed because the two minus signs go away, therefore (again you need to fill in a small detail) .(c) Find all intermediate fields between and , by refering to the Fundamental Theorem of Galois Theory [FTG] to show you have them all.