234 CHAPTER 9. SOME ITEMS WHICH RESEMBLE LINEAR ALGEBRA

Proof: By the above theorem, it suffices to show that H(k) = {ι} for some k. Letf be the homomorphism. Then H ′ = f (G′). To see this, consider a commutator ofH, f (a) f (b) f (a)−1 f (b)−1 = f

(aba−1b−1

). It follows that H(1) = f

(G(1)

). Now con-

tinue this way, letting G(1) play the role of G and H(1) the role of H. Thus, since G issolvable, some G(k) = {ι} and so H(k) = {ι} also. ■

Now as an important example, of a group which is not solvable, here is a theorem.

Theorem 9.4.46 For n≥ 5,Sn is not solvable.

Proof: It is clear that An is a normal subgroup of Sn because if σ is a permutation,then it has the same sign as σ−1. Thus σασ−1 ∈ An if α ∈ An because both α and σασ−1

are a product of an even number of transpositions. If H is a normal subgroup of Sn, forwhich Sn/H is Abelian, then H contains the commutator S′n. However, ασα−1σ−1 ∈ Anobviously so An ⊇ S′n. By Proposition 9.4.39 (An is simple), this forces S′n = An. So what isS′′n? If it is Sn, then S(k)n ̸= {ι} for any k and it follows that Sn is not solvable. If S′′n = {ι} ,the only other possibility, then An/{ι} is Abelian and so An is Abelian, but this is obviouslyfalse because the cycles (1,2,3) ,(2,1,4) are both in An. However, (1,2,3)(2,1,4) is(

1 2 3 44 2 1 3

)

while (2,1,4)(1,2,3) is (1 2 3 41 3 4 2

)Alternatively, by Theorem 9.4.43, if Sn is solvable, then so is An. However, An is simple

so there is no normal subgroup other than An and ι . Now An/{ι}= An is not commutativefor n≥ 4. ■

Note that the above shows that An is not Abelian for n = 4 also.

9.4.8 Solvability by RadicalsThe idea here is to begin with a big field F and show that there is no way to solve thepolynomial in terms of radicals of things in the big field. It will then follow that there isno way to get a solution in terms of radicals of things in a smaller field like the rationalnumbers. The most interesting conclusion is what will be presented here, that you can’t doit. This amazing conclusion is due to Abel and Galois and dates from the 1820’s. It put astop to the search for formulas which would solve polynomial equations.

First of all, in the case where all fields are contained in C, there exists a field which hasall the nth roots of 1. You could simply define it to be the smallest sub field of C such thatit contains these roots. You could also enlarge it by including some other numbers. Forexample, you could include Q. Observe that if ξ ≡ ei2π/n, then ξ

n = 1 but ξk ̸= 1 if k < n

and that if k < l < n, ξk ̸= ξ

l . The following is from Herstein [20]. This is the kind of fieldconsidered here.

Lemma 9.4.47 Suppose a field F has all the nth roots of 1 for a particular n and supposethere exists ξ such that the nth roots of 1 are of the form ξ

k for k = 1, · · · ,n, the ξk being

distinct, as is the case when all fields are in C. Let a ∈ F be nonzero. Let K denote the