9.4. MORE ON ALGEBRAIC FIELD EXTENSIONS 231

and α moves exactly four numbers. Then since B is normal, α1 ≡

(i5, i4, i3)(i1, i2)(i3, i4)(i3, i4, i5) ∈ B

Then α1α−1 =(i5, i4, i3)(i1, i2)(i3, i4)(i3, i4, i5)(i3, i4)(i1, i2) ∈ B

Then i1 → i1, i2 → i2, i3 → i4, i4 → i5, i5 → i3. Thus this permutation moves only threenumbers and so α cannot be of the form given in 9.30. It follows that case 9.29 does notoccur. ■

Definition 9.4.38 A group G is said to be simple if its only normal subgroups are itself andthe identity.

The following major result is due to Galois [26].

Proposition 9.4.39 Let n≥ 5. Then An is simple.

Proof: From Lemma 9.4.37, if B is a normal subgroup of An, B ̸= {ι} , then it containsa 3 cycle α = (i1, i2, i3), (

i1 i2 i3i2 i3 i1

)Now let ( j1, j2, j3) be another 3 cycle.(

j1 j2 j3j2 j3 j1

)Let σ be a permutation which satisfies

σ (ik) = jk

Then

σασ−1 ( j1) = σα (i1) = σ (i2) = j2

σασ−1 ( j2) = σα (i2) = σ (i3) = j3

σασ−1 ( j3) = σα (i3) = σ (i1) = j1

while σασ−1 leaves all other numbers fixed. Thus σασ−1 is the given 3 cycle. It followsthat B contains every 3 cycle not just a particular one. By Proposition 9.4.36, this impliesB = An. The only problem is that it is not known whether σ is in An a product of an evennumber of transpositions. This is where n ≥ 5 is used. If necessary, you can modify σ ontwo numbers not equal to any of the {i1, i2, i3} by multiplying by a transposition so that thepossibly modified σ is expressed as an even number of transpositions. ■

9.4.7 Solvable GroupsRecall the fundamental theorem of Galois theory which established a correspondence be-tween the normal subgroups of G(K,F) and normal field extensions whenever K is thesplitting field of a separable polynomial p(x). Also recall that if H is one of these nor-mal subgroups, then there was an isomorphism between G(KH ,F) and the quotient groupG(K,F)/H. The general idea of a solvable group is given next.