Consider the splitting field "K" of the irreducible polynomial formula_17 over formula_18; that is, formula_19 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 formula_20 Since "ω" has minimal polynomial formula_21"," the extension formula_22 has degree:formula_23, with formula_18-basis formula_25 as in the previous example. Therefore the Galois group formula_26 has six elements, determined by permutations of the three roots of formula_17:formula_28Since 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: