UNIT 3 THE SYLOW THEOREMS The Sylow Theorems Structure Page Nos. 3.1 Introduction 57 Objectives 3.2 Direct Product of Groups 58 External Direct Product Internal Direct Product 3.3 Existence of Subgroups of Given Orders 62 3.4 Applications of Sylow’s Theorems 70 3.5 Summary 73 3.6 Solutions / Answers 74 3.1 INTRODUCTION In this unit, we use the notions and results about group actions developed in the previous unit to derive some basic theorems about finite groups and their structures. To be able to describe, and analyse, the structure of a finite group, we need the concept of a direct product of groups. In Sec.3.2, you will study this concept. In Sec.3.3, we look at when the converse of Lagrange’s theorem holds, namely, given a factor k of ),G(o where G is finite, does there exist a subgroup of G of order ?k As you will see, the answer is “no” in general. Further, in Sec.3.3 and Sec.3.4, you will see that there are several special cases when the answer is “yes”. To reach this stage, you will be banking on very important theorems of group theory, which are due to the mathematician Sylow. In Sec.3.3, you will study the proofs of three theorems due to him. In Sec.3.4, you will be studying several applications of these theorems. As noted above, you will be closely looking at several theorems and techniques for analysing the structure of finite groups. Therefore, after studying this unit, come back to this point and make sure you have achieved the following objectives of studying this unit. Objectives After studying this unit, you should be able to: • construct the direct product of a finite number of groups; • check if a group is the direct product of its subgroups; • give examples to show when the converse of Lagrange’s theorem holds, and that it doesn’t hold in general; • state, and prove, all three Sylow theorems; and • apply Sylow’s theorems to prove the existence of subgroups of a specified order, and to analyse the structure of some finite groups. 57 Group Theory 3.2 DIRECT PRODUCT OF GROUPS In this section you will study a very important method of constructing new groups by using given groups as building blocks. We will first look at one way in which two groups can be combined to form a third group. Then we will discuss a similar way in which two subgroups of a group can be combined to form another subgroup. 3.2.1 External Direct Product In this sub-section, you will study one way of constructing a new group from two or more groups that are given. Let (G1 ,∗ 1 ) and (G2 ,∗ 2 ) be two groups. Consider their Cartesian product G= G1 × G 2 = {(x,y)|x∈ G,y1 ∈ G}.2 Can we define a binary operation on G by using the operations on G1 and G2 ? Let us try the obvious method, namely, componentwise multiplication. That is, we define the operation ∗ on G by (a,b)∗ (c,d) = (a ∗1 c,b ∗2 d) ∀ a,c ∈ G,b1 , d ∈ G2 . The way we have defined ∗ ensures that it is a binary operation. To check that (G,∗ ) is a group, you need to solve the following exercise. E1) Show that the binary operation ∗ on G is associative. Find its identity element and the inverse of any element (x, y) in .G So, by solving E1, you have proved that GGG=1 × 2 is a group with respect to ∗ . We call G the external direct product of (G1 ,∗ 1 ) and (G2 ,∗ 2 ). For example, R2 is the external direct product of R with itself. Another example is the direct product (,)(,)Z + ×R∗ ⋅ in which the operation is given by (m,x)∗ (n,y)= (m+ n,xy) for m, n∈Z , x, y ∈ R∗ . Remark 1: The groups forming the direct product do not need to have the Z same properties or algebraic structure. For instance, S3 × is a well-defined Z Z direct product of (S3 ,o ) and ( ,+ ), where S3 is non-abelian and is abelian, in fact, cyclic. We can define the external direct product of 3, 4 or more groups along the same lines. Definition: Let (G,1∗ 1 ),(G, 2 ∗ 2 ),K ,(G,n∗ n ) be n groups. Their external direct product is the group (G,∗ ), where GGGG=1 × 2 ×K × n and (x,x,1 2 K ,x)(y,y,n∗ 1 2 K ,y)n = (x1 ∗ 1 y,x 1 2 ∗ 2 y, 2 K ,xn∗ n y) n ∀ x,y i i ∈ G. i Thus, Rn is the external direct product of n copies of R,∀ n∈ N . We would like to make a remark about notation now. 58 The Sylow Theorems Remark 2: Henceforth, we will assume that all the operations ∗,,, ∗1 K ∗n are multiplication, unless mentioned otherwise. Thus, the operation on GGGG=1 × 2 ×K × n will be given by (a,1 K ,a).(b,n 1 K ,b)n= (ab,ab, 1 1 2 2 K ,ab)n n∀ a,b i i ∈ G. i Now, you know that given two sets A and B, A× B≠ B× A. So, you may Z Z wonder if, for example, ×S3 and S3 × are different groups. It turns out they are essentially the same, which is what the following exercise says. ~ E2) Show that GG1× 2GG, 2× 1 for any two groups G1 and G2 . Because of E2, we can speak of the direct product of 2 (or n ) groups without bothering about the order in which their products are taken. Now, let G be the external direct product G1× G 2 . Consider the projection map π1:G 1 × G 2 → G: 1 π 1 (x,y) = x. You should check that π 1 is a surjective group homomorphism. Also, Kerπ1 = {(x,y) ∈ G1 × G| 2 π 1 (x,y) = e}1 ={(e,y)|y1 ∈ G}2 = {e} 1 × G. 2 ∴{e1 } × G 2 G1× G 2 . So, by the Fundamental Theorem of Homomorphism ~ (G1× G 2 ) ({e 1 }× G 2 ) G. 1 You can, similarly, prove that G1× {e 2 } GG1× 2 and ~ (G1× G 2 ) (G 1 × {e 2 }) G.2 In the following exercises you can prove some general facts about external direct products of groups. E3) Show that GG1× 2 is the product of its normal subgroups H= G1 × {e 2 } and K= {e1 }× G 2 . Also show that (G1× {e}) 2 ∩ ({e}1 × G) 2 = {(e,e)}.1 2 E4) Prove that Z(G1× G) 2 = Z(G)1 × Z(G),2 where Z(G) denotes the centre of G. E5) Let A and B be cyclic groups of order m and n, respectively, where (m, n)= 1. Prove that A× B is cyclic of order mn. Z Z Z Z Z [Hint: Define f : →m × n :f(r) = (r + m,r + n). Then apply the Z Z ~ Z Fundamental Theorem of Homomorphism to show that m× n mn . ] E6) Show that if HG1≤ 1 and H2≤ G 2 , then HHGG.1× 2 ≤ 1× 2 Further, if H1 G1 and H2 G2 , is HH1× 2 G1× G? 2 Give reasons for your answer. E7) Give an example, with justification, to show that not every subgroup of the direct product GG1× 2 of the groups G1 and G2 is of the form 59 Group Theory HH,1× 2 where H1 is a subgroup of G1 and H2 is a subgroup of G2 . So, far we have discussed the construction of GG1× 2 from two groups G1 and G2 . Let us now see under which conditions we can express a group as a direct product of its subgroups. 3.2.2 Internal Direct Product Let us begin by recalling that if H and K are normal subgroups of a group G, then HK is a normal subgroup of .G We are interested in the case when HK is the whole of .G Related to this, consider the following definition. Definition: Let H and K be normal subgroups of a group .G We call G the internal direct product of H and K if G= HK and H∩ K= {e}. This fact is denoted by GHK= × . Let us consider some examples. Example 1: Show that the Klein 4-group, K4 , is an internal direct product of its subgroups. 2 2 Solution: K4 = {e, a, b, ab}, where a= e, b= e and ab= ba. Let H= a and K= b . Then, H and K are normal in K4 (as K4 is abelian), and H∩ K= {e}. Also, K4 = HK. ∴KHK4 = × . ~ Z ~ Z ~ Z Z Here, note that H 2 and K 2 . ∴K4 2 × 2. *** Z Example 2: Show that 10 is the internal direct product of its subgroups H= {0, 5} and K= {0,2,4,6,8}. Z Z Solution: Note that 10 is abelian, so that H and K are normal in 10 . Further, Z Z i) 10 =H + K, since any element of 10 is the sum of an element of H and an element of K (note that 1= 5 − 4 ∈ H + K ), and ii) H∩ K = {0}. Z Hence 10 =H × K. *** Now, can an external direct product also be an internal direct product? Well, go back to E3. What does it say? It says that the external product of GG1× 2 is the internal product (G1× {e}) 2 × ({e}1 × G). 2 Consider the following remark now. Remark 3: Let H and K be normal subgroups of a group .G Then the internal direct product of H and K is isomorphic to the external direct product of H and .K Therefore, when we talk of an internal direct product of 60 subgroups we can drop the word internal, and just say ‘direct product of The Sylow Theorems subgroups’.