Solubility and Simplicity
14.3 Cauchy’s Theorem
We next prove Cauchy’s Theorem: if a prime p divides the order of a finite group, then the group has an element of order p. We begin by recalling several ideas from group theory.
Definition 14.9. Elements a and b of a group G are conjugate in G if there exists g∈ G such that a = g−1bg.
Conjugacy is an equivalence relation; the equivalence classes are the conjugacy classesof G.
If the conjugacy classes of G are C1, . . . ,Cr, then one of them, say C1, contains only the identity element of G. Therefore |C1| = 1. Since the conjugacy classes form a partition of G we have
|G| = 1 + |C2| + · · · + |Cr| (14.2) which is the class equation for G.
Cauchy’s Theorem 167 Definition 14.10. If G is a group and x ∈ G, then the centraliser CG(x) of x in G is the set of all g ∈ G for which xg = gx. It is always a subgroup of G.
There is a useful connection between centralisers and conjugacy classes.
Lemma 14.11. If G is a group and x ∈ G, then the number of elements in the conju-gacy class of x is the index of CG(x) in G.
Proof. The equation g−1xg= h−1xh holds if and only if hg−1x= xhg−1, which means that hg−1∈ CG(x), that is, h and g lie in the same coset of CG(x) in G. The number of these cosets is the index of CG(x) in G, so the lemma is proved.
Corollary 14.12. The number of elements in any conjugacy class of a finite group G divides the order of G.
Definition 14.13. The centre Z(G) of a group G is the set of all elements x ∈ G such that xg = gx for all g ∈ G.
The centre of G is a normal subgroup of G. Many groups have trivial centre, for example Z(S3) = 1. Abelian groups go to the other extreme and have Z(G) = G.
Lemma 14.14. If A is a finite abelian group whose order is divisible by a prime p, then A has an element of order p.
Proof. Use induction on |A|. If |A| is prime the result follows. Otherwise take a proper subgroup M of A whose order m is maximal. If p divides m we are home by induction, so we may assume that p does not divide m. Let b be in A but not in M, and let B be the cyclic subgroup generated by b. Then MB is a subgroup of A, larger than M, so by maximality A = MB. From the First Isomorphism Theorem, Lemma 14.3(1),
|MB| = |M||B|/|M ∩ B|
so p divides the order r of B. Since B is cyclic, the element br/phas order p.
From this result we can derive a more general theorem of Cauchy in which the group need not be abelian:
Theorem 14.15 (Cauchy’s Theorem). If a prime p divides the order of a finite group G, then G has an element of order p.
Proof. We prove the theorem by induction on the order |G|. The first few cases |G| = 1, 2, 3 are obvious. For the induction step, start with the class equation
|G| = 1 + |C2| + · · · + |Cr|
Since p||G|, we must have p-|Cj| for some j ≥ 2. If x ∈ Cjit follows that p||CG(x)|, since |Cj| = |G|/|CG(x)|.
If CG(x) 6= G then by induction CG(x) contains an element of order p, and this element also belongs to G.
Otherwise CG(x) = G, which implies that x ∈ Z(G), and by choice x 6= 1, so Z(G) 6= 1.
Either p||Z(G)| or p-|Z(G)|. In the first case the proof reduces to the abelian case, Lemma 14.14. In the second case, by induction there exists x ∈ G such that the image ¯x∈ G/Z(G) has order p. That is, xp∈ Z(G) but x 6∈ Z(G). Let X be the cyclic group generated by x. Now X Z(G) is abelian and has order divisible by p, so by Lemma 14.14 it has an element of order p, and again this element also belongs to G.
This completes the induction step, and with it the proof.
Cauchy’s Theorem does not work for composite divisors of |G|. See Exercise 14.6.
EXERCISES
14.1 Show that the general dihedral group
Dn= ha, b : an= b2= 1, b−1ab= a−1i
is a soluble group. Here a, b are generators and the equalities are relations between them.
14.2 Prove that Snis not soluble for n ≥ 5, using only the simplicity of A5. 14.3 Prove that a normal subgroup of a group is a union of conjugacy classes. Find
the conjugacy classes of A5, using the cycle type of the permutations, and hence show that A5is simple.
14.4 Prove that Snis generated by the 2-cycles (12), . . . , (1n).
14.5 If the point α ∈ C is constructible by ruler and compasses, show that the Galois group of Q(α) : Q is soluble.
14.6 Show that A5has no subgroup of order 15, even though 15 divides its order.
14.7 Show that Snhas trivial centre if n ≥ 3.
14.8 Find the conjugacy classes of the dihedral group Dndefined in Exercise 14.1.
Work out the centralisers of selected elements, one from each conjugacy class, and check Lemma 13.7.
14.9 If G is a group and x, g ∈ G, show that CG(g−1xg) = g−1CG(x)g.
14.10 Show that the relation ‘normal subgroup of’ is not transitive. (Hint: Consider the subgroup G ⊆ V ⊆ S4generated by the element (12)(34).)
Exercises 169 14.11 There are (at least) two distinct ways to think about a group homomorphism.
One is the definition as a structure-preserving mapping, the other is in terms of a quotient group by a normal subgroup. The relation between these is as follows. If φ : G → H is a homomorphism then
ker(φ )C G and G/ker(φ ) ∼= im(φ ) If NC G then there is a natural surjective homomorphism
φ : G → G/N with ker(φ ) = N
Show that the first and second isomorphism theorems are the translations into
‘quotient group’ language of two facts that are trivial in ‘structure-preserving mapping’ language:
(1) The restriction of a homomorphism to a subgroup is a homomorphism.
(2) The composition of two homomorphisms is a homomorphism.
14.12* By counting the sizes of conjugacy classes, prove that the group of rotational symmetries of a regular icosahedron is simple. Show that it is isomorphic to A5.
14.13 Mark the following true or false.
(a) The direct product of two soluble groups is soluble.
(b) Every simple soluble group is cyclic.
(c) Every cyclic group is simple.
(d) The symmetric group Snis simple if n ≥ 5.
(e) Every conjugacy class of a group G is a subgroup of G.
Chapter 15
Solution by Radicals
The historical aspects of the problem of solving polynomial equations by radicals have been discussed in the introduction. Early in his career, Galois briefly thought that he had solved the quintic equation by radicals, Figure 22. However, he found a mistake when it was suggested that he should try some numerical examples. This motivated his work on solubility by radicals.
The object of this chapter is to use the Galois correspondence to derive a con-dition that must be satisfied by any polynomial equation that is soluble by radicals, namely: the associated Galois group must be a soluble group. We then construct a quintic polynomial equation whose Galois group is not soluble, namely the disarm-ingly straightforward-looking t5− 6t + 3 = 0, which shows that the quintic equation cannot be solved by radicals.
Solubility of the Galois group is also a sufficient condition for an equation to be soluble by radicals, but we defer this result to Chapter 18.