subgroup is normal, so the subfield is normal over , being the splitting field of . Its Galois group over the base field is the quotient group , where [g] denotes the coset of g modulo H; that is, its only non-trivial automorphism is the complex conjugation g. There are three subgroups of order 2, and corresponding respectively to the subfields These subfields have degree 3 over since the subgroups have index 3 in G. The subgroups are not normal in G, so the subfields are not Galois or normal over . In fact, each subfield contains only a single one of the roots , so none has any non-trivial automorphisms. Example 3 Let be the field of rational functions in the indeterminate λ, and consider the group of automorphisms: here we denote an automorphism by its value , so that . This group is isomorphic to (see: six cross-ratios). Let be the fixed field of , so that . If is a subgroup of , then the coefficients of the polynomial generate the fixed field of . The Galois correspondence implies that every subfield of can be constructed this way. For example, for , the fixed field is and if then the fixed field is . The fixed field of is the base field where is the -invariant written in terms of the modular lambda function:Similar examples can be constructed for each of the symmetry groups of the platonic solids as these also have faithful actions on the projective line and hence on . Applications The theorem classifies the intermediate fields of E/F in terms of group theory. This translation between intermediate fields and subgroups is key to showing that the general quintic equation is not solvable by radicals (see Abel–Ruffini theorem). One first determines the Galois groups of radical extensions (extensions of the form F(α) where α is an n-th root of some element of F), and then uses the fundamental theorem to show that solvable extensions correspond to solvable groups. Theories such as Kummer theory and class field theory are predicated on the fundamental theorem. Infinite case Given an infinite algebraic extension we can still define it to be Galois if it is normal and separable. The problem that one encounters in the infinite case is that the bijection in the fundamental theorem does not hold as we get too many subgroups generally. More precisely if we