from the base field by adjoining , then , each element of can be written as: Its Galois group comprises the automorphisms of which fix . Such automorphisms must send to or , and send to or , since they permute the roots of any irreducible polynomial. Suppose that exchanges and , so and exchanges and , so These are clearly automorphisms of , respecting its addition and multiplication. There is also the identity automorphism which fixes each element, and the composition of and which changes the signs on both radicals: Since the order of the Galois group is equal to the degree of the field extension, , there can be no further automorphisms: which is isomorphic to the Klein four-group. Its five subgroups correspond to the fields intermediate between the base and the extension . The trivial subgroup corresponds to the entire extension field . The entire group corresponds to the base field The subgroup corresponds to the subfield since fixes . The subgroup corresponds to the subfield since fixes . The subgroup corresponds to the subfield since fixes . Example 2 The following is the simplest case where the Galois group is not abelian. Consider the splitting field K of the irreducible polynomial over ; that is, where θ is a cube root of 2, and ω is a cube root of 1 (but not 1 itself). If we consider K inside the complex numbers, we may take , the real cube root of 2, and Since ω has minimal polynomial , the extension has degree:, with -basis as in the previous example. Therefore the Galois group has six elements, determined by permutations of the three roots of :Since there are only 3! = 6 such permutations, G must be isomorphic to the symmetric group of all permutations of three objects. The group can be generated by two automorphisms f and g defined by: and , obeying the relations . Their effect as permutations of is (in cycle notation): . Also, g can be considered as the complex conjugation mapping. The subgroups of G and corresponding subfields are as follows: As always, the trivial group {1} corresponds to the whole field K, while the entire group G to the base field . The unique subgroup of order 3, , corresponds to the subfield of degree two, since the subgroup has index two in G: i.e. . Also, this