sign in sign up
<<
Home , Direct product of groups, Sylow theorems

UNIT 3 THE SYLOW The

Structure Page Nos.

3.1 Introduction 57 Objectives 3.2 Direct of Groups 58 External Internal Direct Product 3.3 Existence of 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 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 , we need the concept of a . In Sec.3.2, you will study this concept.

In Sec.3.3, we look at when the converse of Lagrange’s holds, namely, given a factor k of ),G(o where G is finite, does there exist a of G of ?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 , 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 ∗11 ),G( and ∗22 ),G( be two groups. Consider their

21 =×= 1 ∈∈ 2}.Gy,Gx|)y,x{(GGG Can we define a on

G by using the operations on G1 and 2 ?G Let us try the obvious method, namely, componentwise . That is, we define the operation ∗ on

G by 1 2 1 b,Gc,a)db,ca()d,c()b,a( ∈∈∀∗∗=∗ 2.Gd, 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 and the inverse of any element )y,x( in .G

So, by solving E1, you have proved that ×= GGG 21 is a group with respect to

∗ . We call G the external direct product of ∗11 ),G( and ∗22 .),G(

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 ∗ = + )xy,nm()y,n()x,m( for Z ∈∈ R∗.y,x,n,m

Remark 1: The groups forming the direct product do not need to have the Z same properties or . For instance, S3 × is a well-defined Z Z direct product of 3 o ),S( 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 ∗∗ 2211 K ∗nn ),G(,),,G(),,G( be n groups. Their external

direct product is the group ∗ ),,G( where 21 K×××= GGGG n and

21 K ∗ 21n K n ∗∗= 222111 K ∈∀∗ iiinnn .Gy,x)yx,,yx,yx()y,,y,y()x,,x,x(

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

21 K×××= GGGG n will be given by

1 K 1n K = 2211n K ∈∀ iiinn .Gb,a)ba,,ba,ba()b,,b).(a,,a(

Now, you know that given two sets A and × ≠ × .ABBA,B 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 GG ×× 1221 ,GG for any two groups G1 and 2 .G

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 × 21 .GG Consider the map 11211 =π→×π .x)y,x(:GGG:

You should check that π1 is a surjective . Also,

1 121 =π×∈=π 1}e)y,x(|GG)y,x{(Ker

= 1 ×=∈ 212 .G}e{}Gy|)y,e{(

×∴ G}e { 21 × 21 .GG So, by the Fundamental Theorem of Homomorphism ~ × × )G}e({)GG( 12121 .G

You can, similarly, prove that × 21 }e{G ×GG 21 and ~ ×× 2121 })e{G()GG( 2 .G

In the following exercises you can prove some general facts about external direct products of groups.

E3) Show that × GG 21 is the product of its normal subgroups ×= 21 }e{GH

and ×= 21 .G}e{K Also show that 21 21 =×∩× 21 )}.e,e{()G}e({})e{G(

E4) Prove that 21 1 ×=× 2 ),G(Z)G(Z)GG(Z where )G(Z denotes the centre of .G

E5) Let A and B be cyclic groups of order m and n, respectively, where = .1)n,m( Prove that A× B is cyclic of order mn. Z Z Z Z Z [Hint: Define :f nm ++=×→ ).nr,mr()r(f: Then apply the Z Z ~ Z Fundamental Theorem of Homomorphism to show that × mnnm . ]

E6) Show that if ≤ GH 11 and ≤ 22 ,GH then ×≤× 2121 .GGHH Further, if

H1 G1 and H2 2 ,G is × HH 21 × 21 ?GG Give reasons for your answer.

E7) Give an example, with justification, to show that not every subgroup of

the direct product × GG 21 of the groups G1 and is of the form 59

Group Theory × 21 ,HH where H1 is a subgroup of G1 and H2 is a subgroup of 2 .G

So, far we have discussed the construction of × GG 21 from two groups G1

and 2 .G 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 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 = HKG and ∩ = }.e{KH This fact is denoted by = ×KHG .

Let us consider some examples.

Example 1: Show that the Klein 4-group, 4 ,K is an internal direct product of its subgroups.

2 2 Solution: 4 = },ab,b,a,e{K where == eb,ea and = .baab

Let = aH and = .bK Then, H and K are normal in K4 (as K4 is

abelian), and ∩ = }.e{KH Also, 4 = .HKK

4 ×=∴ KHK . ~ Z ~ Z ~ Z Z Here, note that H 2 and K 2 . K ×∴ 224 . ***

Z Example 2: Show that 10 is the internal direct product of its subgroups = }5,0{H and = }.8,6,4,2,0{K Z Z Solution: Note that 10 is abelian, so that H and K are normal in 10 . Further, Z Z i) 10 += ,KH since any element of 10 is the sum of an element of H and an element of K (note that +∈−= KH451 ), and ii) =∩ }.0{KH Z Hence 10 ×= .KH ***

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 × GG 21 is

the internal product 21 ××× 21 ).G}e({})e{G(

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 internal, and just say ‘direct product of The Sylow Theorems subgroups’.

Let us now extend the definition of the internal direct product of two subgroups to that of several subgroups.

Definition: A group G is the internal direct product of its normal subgroups

21 K H,,H,H n if i) = K n21 ,HHHG and ii) ∩ K +− K n1i1i1i =∀= K .n,,1i}e{HHHHH

For example, look at the group G generated by },c,b,a{ where 2 === cbea 22 and = = = .cbbc,caac,baab This is the internal direct ~ Z Z Z product of b,a and .c That is, G ×× 222 .

Now you may ask: can every group be written as an internal direct product of two or more of its proper normal subgroups? To answer this, consider Z . Suppose Z = × ,KH where K,H are non-zero subgroups of Z. So, = mH and = nK for some non-zero ∈Z .n,m Then ∈ ∩ .KHmn But if H× K is a direct product, ∩ = }.0{KH So, we reach a contradiction. Therefore, Z can’t be written as an internal direct product of two subgroups.

By the same reasoning, we can say that Z can’t be expressed as Z 21 K××× HHH n , where i =∀≤ K n,,2,1iH for any > .1n

Also, you know that there are some groups that do not have any proper normal subgroup except }.e{ So, they can certainly not be written as such a direct product!

So, how do we find out whether a group can be written as a direct product of its subgroups? The answer is given by the following theorem.

Theorem 1: Let a group G be the internal direct product of its subgroups H and .K Then i) each ∈Gx can be uniquely expressed as = ,hkx where ∈ ∈ ;Kk,Hh and ii) = ∀ ∈ ∈ .Kk,Hhkhhk

Proof: i) We know that = HKG and ∩ = }.e{KH Therefore, if ∈ ,Gx then

= ,hkx for some ∈ ∈ .Kk,Hh Now suppose = khx 11 also, where −1 −1 1 ∈ Hh and 1 ∈ .Kk Then = 11 .khhk 1 =∴ 1 .kkhh −1 −1 −1 −1 Now 1 ∈ .Hhh Also, since 1 1 1 ∈∈= .Khh,Kkkhh −1 −1 1 =∩∈∴ }.e{KHhh 1 =∴ ,ehh which implies that = 1 .hh Similarly,

1 = .kk Thus, the representation of x as the product of an element of H and an element of K is unique. ii) The best way to show that two elements x and y commute is to show 61

Group Theory that their −− 11 xyyx is identity. So, let h ∈ H and k ∈ K and consider −− 11 .hkkh Since K −− 11 ∈ .Khkh,G ∴ −− 11 ∈ .Khkkh By similar reasoning, −− 11 ∈ .Hhkkh ∴ −− 11 =∩∈ }.e{KHhkkh ∴ −− 11 = ,ehkkh that is, = .khhk

What Theorem 1 says is that if a group does not have normal subgroups for which (i) and (ii) are satisfied, then G cannot be an internal direct product of its subgroups.

Try the following exercise now.

E8) Let H and K be normal subgroups of G which satisfy (i) of Theorem 1. Then show that = × .KHG

Now let us look at the relationship between internal direct products and quotient groups.

Theorem 2: Let H and K be normal subgroups of a group G such that = × .KHG Then HG ~ K and KG ~ .H

Proof: We will use the second theorem to prove this result. Now = HKG and ∩ = }.e{KH Therefore, = HHKHG ~ =∩ }e{K)KH(K ~ .K You can similarly prove that KG ~ .H

We now give a result which immediately follows from Theorem 2, and which will be used later in this unit.

Corollary 1: Let G be a finite group and H and K be its subgroups such that = × .KHG Then = ).K(o)H(o)G(o

We leave the proof of Corollary 1 to you, and give you some related exercises.

E9) Let G be a finite group such that 1 K××= n ,HHG where

Hi ∀ = K .n,,1iG Prove, by mathematical induction, that n = ∏ i ,)H(o)G(o for ≥ .1n =1i

And now let us discuss some basic results about the structure of a finite group.

3.3 EXISTENCE OF SUBGROUPS OF GIVEN ORDERS

As already mentioned in Sec.3.1, this section is centred around the following question: Is the converse of Lagrange’s theorem true? That is, given a factor k of the order of a finite group ,Gdoes there exist a subgroup in G of order 62

?k Let us begin with an example to show that the answer is “no” in general. The Sylow Theorems

Example 3: Show that the A4 has no subgroup of order 6. Solution: You will see how this can be proved by contradiction. Now

( 4 ) = .12Ao Suppose H is a subgroup of A4 of order 6. Being a subgroup of index 2 in4 ,A H is normal in 4 .A Also ∈∀= 4 .H\Ag2)gH(o Thus, 2 2 2 ∈∀∈ 4.AgHg So ∈= H)231()321( and ∈= .H)321()231( In the same way all the 3-cycles in A4 are in .H So ≥ ,8)H(o a contradiction.

Hence, A4 has no subgroup of order 6. ***

In the example above, the group A4 is not abelian. So, your next question may be that if G is a finite , does the converse of Lagrange’s theorem hold true? In some cases it does, as the following theorem tells us.

Theorem 3: If G is a finite abelian group, and p is a prime dividing ),G(o then G has a subgroup of order p.

Proof: We first observe that it suffices to prove the following claim (under the given hypothesis): There exists in G an element g whose order m is divisible by p. In fact, if this claim is proved, then for this g, the element gp/m has order p, and the group generated by gp/m has order p, thus proving Theorem 3.

So let us prove the claim, which we shall call a lemma.

Lemma 1: If G is a finite abelian group such that ,|G|p then ∃ ∈ Gg s.t. ).g(op

Proof: Let us prove this by induction on |G| . If = ,1|G| then the statement is true vacuously since there is no p that divides ).G(o

Now suppose that > .1|G| Let h be any element of G other than identity and let n be its order. Note that n > 1 and that n divides .|G| If p divides n, then the lemma holds – just choose g to be h. If p does not divide n, consider the .hG Its order is ,n|G| which is divisible by p since |G|p and /| .np Now since < ,|G|n|G| we may apply the induction hypothesis to hG to conclude that there exists an element in hG whose order k is divisible by p. Let g be the pre- in G of such an element in hG (under the natural → hGG ). Let m be the order of g. Denoting hg by g , we have ( )m m == egg , where e denotes the identity element of .G Since e is the identity element of the quotient ,hG it follows that the order k of g divides m, and hence | .mp

Thus, Lemma 1 is proved, and so is Theorem 3. 63

Group Theory Let us now consider the next step in showing that the converse of Lagrange’s theorem holds for abelian groups.

Proposition 1: Let N be a finite normal subgroup of order n of a finite group G and let →π NGG: be the natural epimorphism. If K is a subgroup of order k of NGthen π−1(K) is a subgroup of G of order kn, that is, −1 =π .|N||K||)K(|

Proof: You know that the inverse image of a subgroup under a −1 homomorphism is a subgroup. So ≤π G)K( . Let g,...,gk1 be elements of G −1 such that = { 1 k }.Ng,...,NgK Then 1 ∪⋅⋅⋅∪=π k .NgNg)K( Since each −1 i Ng has n elements, and any two of them are disjoint, it follows that π )K( consists of kn elements.

Now you will see how useful Proposition 1 is for proving that the converse of Lagrange’s theorem holds for abelian groups.

Theorem 4: If G is a finite abelian group and k divides ,G then G has a subgroup of order .k

Proof: This proof is by induction on .k For ∈N ,n let )n(P be the statement, ‘if G is a finite abelian group and |G|n , then G has a subgroup of order n ’. If = ,1n then you know that }e{ is a subgroup of G of order 1, so that )1(P is true. Now suppose that k > 1 and that )n(P is true ∀ < .kn Now we shall show that )k(P is true. Let p be a prime divisor of .k ByTheorem 3, there exists a subgroup N of G of order p. Note that, as G is abelian, N .G Now, the quotient group NG has order ,p|G| which is divisible by .pk Since < ,kpk the induction hypothesis applies. Therefore, there exists a subgroup of NGof order .pk Its pre-image in G (under the natural epimorphism → NGG ) is a subgroup of G of order = kppk (by Proposition 1). So, )k(P is true.

Hence, )n(P is true ∀ ≥ .1n

So far we have looked at abelian groups. We now move to the first of the three famous theorems proved by the Norwegian mathematician Ludwig Sylow in 1872. This is one of the most important theorems in finite group theory. It tells us about the p-subgroups of any finite group, abelian or not.

Theorem 5 (Sylow’s first theorem): Let G be a finite group and p be a prime such that pr divides |G| for some ≥ .1r Then there exists a Fig. 1: L. Sylow r (1832-1918) subgroup of order p in .G

Proof: Let = r .mp|G| We shall proceed by induction on .|G| Let = r .pq If = ,1m then = q|G| so that the result clearly holds – G itself is a subgroup of the desired order. 64

So assume that > .1m Now there are two cases. The Sylow Theorems

Case 1: Suppose there exists a proper subgroup H of G such that q divides |H| . Write = ′ .mq|H| Then, < .|G||H| So, by the induction hypothesis, H (and so also G ) has a subgroup of order = r .pq

Case 2: Suppose there is no proper subgroup H such that = pq r divides .|H|

Now, since p divides ,|G| p divides ,|H||G| where ≤/ .GH So, |H:G|p for every proper subgroup H of .G So, by the Structure Theorem for Orbits, every orbit of G (and, in particular, every of G ) that is not a singleton has divisible by p. So, it follows from the Class Equation (see Sec.2.4, Unit 2) that .|)G(Z|p So, by Theorem 3, there exists a subgroup N of order p in ).G(Z Being contained in N),G(Z is normal in .G Consider the quotient group .NG Its order |N||G| is −1r .mp So, by the induction hypothesis, there exists a subgroup K of order p −1r in .NG It follows, from Proposition 1, that the pre-image of K under the natural epimorphism → NGG is a subgroup in G of order = r .pq

This important theorem has an immediate corollary, a result that was actually proved much earlier by the mathematician Cauchy, and is a generalisation of Theorem 3.

Corollary 2 (Cauchy’s theorem): If G is a finite group and p is a prime dividing ,|G| then G has an element of order p.

Proof: By Sylow’s theorem, G has a subgroup H of order p. Since |H| is prime, H is cyclic, say = .xH Then ∈ Gx s.t. = .p)x(o

Here are some examples of the use of Sylow’s first theorem and Cauchy’s theorem.

Example 4: Does a group of order 750 have at least 5 distinct proper non- trivial subgroups? Give reasons for your answer. Solution: Yes. To give the reason, let G be a group of order ××= 3 .532750 By Sylow’s first theorem, G has subgroups of order .125,25,5,3,2 All these must necessarily be distinct since they are of different orders. ***

Example 5: If G is a group of order ,4 what can its structure be? Solution: Firstly, being of order 2 G,2 is abelian. Next, since = ,4|G| by Sylow’s first theorem it has subgroups of order 2 and ,4 and at least one element of order 2 (by Cauchy’s theorem). ~ Z If G has an element of order ,4 then G is cyclic and G 4. If G has no element of order ,4 then it has three elements of order ,2 say, .c,b,a Since ∈ ∈ .Gab,Gb,a Also ≠ ≠ .bab,aab So = .cab As G is abelian, a b,G .G Also =∩ }.e{ba 65

Group Theory So ×= baG is the Klein 4-group. ***

Here is a related exercise for you to try now. While solving it, consider all the information you have about the group.

N E10) For which values of n ∈ must S5 have subgroups of order n, and why?

Sylow’s first theorem tells us about subgroups of G of order 2 K r ,p,,p,p where r .|G|p This leads us to the following definition.

Definition: Let G be a finite group and p be a prime such that = rmp|G| , with r ≥ 1 and = .1)m,p( A subgroup H of G such that = p|H| r is called a Sylow p-subgroup of .G

For instance, if = ,1400|G| then G will have a Sylow 2-subgroup (of order 8), a Sylow 5-subgroup (of order 25), and a Sylow 7-subgroup (of order 7). Also, G will have no Sylow 3-subgroup, since /| .|G|3

Remark 4: It follows from Sylow’s first theorem, as a special case, that G has Sylow p-subgroups.

Let us consider some examples of Sylow p-subgroups in some groups of small order. You can return to these examples later, and look at them again in the light of Sylow’s second and third theorems, which you will study later in this section.

Example 6: Exhibit a Sylow p-subgroup of S3 for each prime p dividing

3 .|S| Solution: The group has order = × .326 So it has a Sylow 2-subgroup of order ,2for instance, { ( )}.21,e Note that )}31(,e{ and )}32(,e{ are also

Sylow 2-subgroup of 3 .S It also has a Sylow 3-subgroup of order ,3 for instance, { ( ) ( 231,321,e )}. ***

Example 7: Obtain a Sylow p-subgroup of 4 ,S for each prime p dividing

4 ).S(o 3 Solution: ( 4 ) ×== .3224So So S4 has a Sylow 3-subgroup of order 3, for instance, { ( ) ( 231,321,e )}. It also has a Sylow 2-subgroup of order 8, for instance, { }.)3241(),4231(),32()41(),42()31(,)43()21(),43(),21(,e ***

Example 8: Does A4 have a Sylow 2-subgroup? A Sylow 3-subgroup? A Sylow 5-subgroup? Give reasons for your answers. 66

2 The Sylow Theorems Solution: ( 4 ) ×== .3212Ao So A4 has a Sylow 2-subgroup and a Sylow

3-subgroup. However, since /| A),A(o5 44 has no Sylow 5-subgroup. ***

Try an exercise now.

E11) Is every Sylow p-subgroup of Sn also a Sylow p-subgroup of +1n ?S Give reasons for your answer.

E12) Consider the direct product ×= GGG 21 of finite groups G1 and 2 ,G

where p divides 1 )G(o as well as 2 ).G(o Prove that the Sylow

p-subgroups of G are precisely those of the form × 21 ,PP where P1 and

P2 are Sylow p-subgroups of G1 and 2 ,G respectively.

In Example 6 you have seen that a group can have several Sylow p-subgroups for a given p. Are these subgroups related, apart from the fact that they have the same order? Let’s see.

Let X denote the of all Sylow p-subgroups of a finite group .G There is a “conjugation action” of G on ,X defined as follows: For g in G and P in ,X define ⋅ Pg to be the “conjugate of P by g ”, namely, −1 .gPg Note that this is a well-defined action since the conjugate of a Sylow p-subgroup is also a Sylow p-subgroup. (Why?) This leads us to the next theorem of Sylow.

Theorem 6 (Sylow’s second theorem): Let G be a finite group and p a prime. Let P be a p-subgroup of G and P0 be a Sylow p-subgroup of G.

Then P is contained in a conjugate of 0 .P In particular, any two Sylow p-subgroups are conjugates.

Proof: Let C denote the set of , 0 .P/G Restrict to P the natural action of G on .C Since P0 is a Sylow p-subgroup, it follows that (= 0 |P||G||C| ) is coprime to p. Since P is a p-group, we may invoke the result in E17 of Unit 2 to obtain ≡ P )p(mod|C||C| , where CP is the set of points of C fixed by P. Since ≡/ ),p(mod0|C| P ≠ .0|C| This means that P fixes some coset −1 0.gP So, ⊆ 0G )gP(StabP , that is, P is contained in 0 .ggP

For the proof of the second assertion, let Q be a Sylow p-subgroup of .G By −1 −1 the first part, there exists g in G such that ⊆ 0 .ggPQ But Q and 0ggP −1 have the same order. So, = 0 .ggPQ

Theorem 6 has several important consequences. Consider three of them, which we give as corollaries to the theorem.

Corollary 3: A Sylow p-subgroup of a finite group G is unique if and only if it is normal. 67

Group Theory Proof: First, assume that the Sylow p-subgroup of G is unique, and is 0 .P −1 −1 Then 0ggP is another subgroup of the same order. So 0 0 ∈∀= .GgPggP

Thus, P0 .G The converse holds by Sylow’s second theorem.

Corollary 4: Sylow p-subgroups of abelian groups are unique.

Proof: Since every subgroup is normal in an abelian group, Corollary 3 gives the result.

For a subgroup ,H let G (HN ) denote the normaliser of ,H that is, ∈= −1 = }.HgHg|Gg{)H(N Observe that (HN ) is a subgroup of G and H NG(H) G G

that H is a normal subgroup of G (HN ). The next corollary to Theorem 6 is related to this.

Corollary 5: Let P and Q be Sylow p-subgroups of .G If P normalises ,Q

that is, ⊆ G ( ),QNP then = .QP

Proof: Note that P is a Sylow p-subgroup of G ( ),QN and Q is a normal

Sylow p-subgroup of G ( ),QN and hence is unique. Thus, = .QP

Try these exercises now.

E13) Show that ( GG ( )) = G (PNPNN ) for any Sylow p-subgroup P of .G

E14) Let G be a finite group and pd divide ,|G| where p is a prime. (We are not assuming that pd is the highest power of p that divides .|G| ) Do any two subgroups of G of order pd have to be conjugates in G? Give reasons for your answers.

We now turn to the question of how many Sylow p-subgroups a group has.

In Example 6, you saw that S3 has at least 3 Sylow-2 subgroups. Can it have more? Let’s see.

Theorem 7 (Sylow’s third theorem): Let G be a group, p a prime and = r ,mp|G| with r ≥1 and m coprime to p. Let n be the number of Sylow p-subgroups of .G Then n divides m and ≡ ).p(mod1n

Proof: Let X denote the set of all Sylow p-subgroups of ,G so that = .|X|n Consider the action of G by conjugation on .X By Sylow’s second theorem, this action is transitive. It follows, from the Structure theorem for orbits, that ~ X − G )P(StabG as G-sets, where P is an arbitrary point in .X ~ Note that the stabiliser of P is G ( ).PN Thus X − G .)P(NG

In particular, the cardinality of X is G = G .|)P(N||G||)P(NG| And, since

⊆ G ( ),PNP it follows that G |)P(N||G| divides = .m|P||G| We conclude that |X| divides m.

68

For the second assertion, we restrict the action of G on X to a Sylow The Sylow Theorems p-subgroup P of .G Since P is a p-group, we may invoke the result in E17 of Unit 2 to conclude that ≡ p ),p(mod|X||X| where XP denotes the set of points of X fixed by all elements of .P Now, P −1 ∈∀=∈= },XQQpQp|Pp{X that is, XP is precisely the of Sylow p-subgroups normalised by .P By Corollary 5, there is precisely one such subgroup, namely, P itself. Thus, P =1|X| , and hence ≡ ).p(mod1|X|

Let us look at the tremendous utility of this result now.

Example 9: Obtain all the Sylow p-subgroups of the 10 .D

Solution: The dihedral group 10 ,D of all symmetries of the regular pentagon, has order = × .5210 It has Sylow 2-subgroups and Sylow 5-subgroups. By Theorem 7, the number of Sylow 5-subgroups is )5(mod1 and divides .2

Thus, D10 has a unique Sylow 5-subgroup, namely, the subgroup consisting of all its rotations (by angles that are integer multiples of π 5/2 ), which is normal. The number of Sylow 2-subgroups is )2(mod1 and divides .5 Hence, it is 1 or 5 . If it were ,1 this subgroup would have been normal, and then D10 would Z Z have been isomorphic to × 52 , and hence abelian, which it is not. So D10 has five Sylow 2-subgroups, each of which is generated by an element of order 2, that is, a reflection. ***

Example 10: Check whether or not a group of order 300 is simple. Solution: Let G be a group with 2 ××== 2.532300|G| The number of Sylow 5-subgroups is )5(mod1 and divides 12, i.e., it is 1 or 6. If it is 1, then the Sylow 5-subgroup is normal in .G

If the number is 6, then all 6 are conjugates in .G Call them 21 K 6.P,,P,P

Consider the action of G on = 1 K 6}P,,P{X by conjugation. Using E9 of

Unit 2, you can get KerG ~ ≤ϕϕ 6.SIm So KerG ϕ divides = .720!6 Now since /| ≠ϕ .}e{Ker,720300 Hence, Kerϕ is a non-trivial normal subgroup of .G Thus, in either case, G is not simple. ***

Try some related exercises now.

E15) Show that a group of order 28 is not simple. Further, how many elements of order 7 does such a group have?

E16) Obtain all the Sylow p-subgroups of the dihedral group 12 .D

Z Z E17) Obtain all the Sylow p-subgroups of the group × 3.S)2(

E18) [This exercise outlines another proof of the existence of Sylow 69

Group Theory p-subgroups, using group actions. The proof can be modified to obtain the more general result in Theorem 5. This proof does not rely on induction.] Let G be a finite group and p a prime such that = r mp|G| , with r ≥ 1 and m coprime to p. Prove the following:  r   mp   n  i) is coprime to p.   n  r    r )r,n(C,C,  p   r  ii) The regular action of G on itself induces an action of G on the denote the same number. power set of .G The set ,Xof subgroups of cardinality pr of G, form a G-invariant subset for this action.  r mp  iii) |X|=   and there is a G-orbit in X of cardinality not  r   p  divisible by p. iv) Let Y be a G-orbit of X such that |Y| is coprime to p. Let H be the stabiliser of an element in .Y Then r .p|H| [Hint: see (ii) of Unit 2.] v) |H| is divisible by r .p vi) H is a Sylow p-subgroup of .G

There are two reasons for E19) Let G be a subgroup of a group ,K and let p be a prime dividing insistence on first principles. The first is to encourage you ).G(o Let Q be a Sylow p-subgroup of .K Prove the following from to review the technique by first principles, that is, using only the results about group actions in the which the results on group previous unit and none of the results of this unit: actions are applied in this unit. The second is that the (i) G has a Sylow p-subgroup. results of this exercise form a basis for yet another (ii) Any p-subgroup of G is contained in a conjugate of Q in .K independent proof of Sylow’s theorems. (iii) Any Sylow p-subgroup is of the form ∩ ′ ,QG for some conjugate Q′ of Q in .K

Let us now look at some uses of these theorems for analysing the structure of finite groups.

3.4 APPLICATIONS OF SYLOW’S THEOREMS

Let us now use Sylow’s three theorems to analyse the structure of some groups of small orders. The following proposition will be used crucially.

Proposition 2: Let H and N be subgroups of a finite group G. Suppose that H normalises ,N that is, −1 = NhNh for all h in .H Then: i) The set = ∈ ∈ }Nn,Hh|hn{HN is a subgroup of G. In particular, if N ,G then ≤ .GH N ⋅ |N||H| ii) |HN| = . ∩ |NH| iii) If N also normalises ,H and ∩ NH is trivial, then NH is isomorphic 70

to the direct product × .NH This happens, in particular, when H and N The Sylow Theorems iv) are both normal subgroups of ,G and their intersection is trivial. v) If both H and N are normal in ,G then so is .NH

The proof of this proposition is left to you. Let us now see how to use this, and the Sylow theorems.

Example 11: Going beyond Example 7, how many Sylow 2-subgroups and

Sylow 3-subgroups does S4 have? Solution: Applying Sylow’s third theorem, you know that the number of Sylow 2-subgroups is )2(mod1 and divides .3 Hence it is 1 or ,3 each of order .8 The number of Sylow 3-subgroups is )3(mod1 and divides .8 Thus, it is 1 or .4 From Example 7, you know that one is ( ) .321 You can see that ( ) ( 431,421 ) and ( 432 ) are also Sylow 3-subgroups. So, the number is .4 Call them T,T,T 321 , and 4 .T

From Example 7, you also know at least one Sylow 2-subgroup, call it 1 .H Is 4 it unique? Let’s see. Now, UTi has 9 elements, of which one is e and the =1i other 8 are all of order .3 Also H1 has 7 elements of order 2 or .4 So, we have covered 16 elements of 4 .S There are still 8 others left, of order a power of .2 Thus, there are two other Sylow 2-subgroups, H 2 and 3 .H ***

Example 12: Construct the Sylow 2-subgroups of 5 .S

3 3 Solution: Since 5 == ,5.3.2! 5)S(o each Sylow 2-subgroup has order 2 = 8. By Sylow’s third theorem, their number is 1(mod 2) and divides 15. Thus, the number can be 1, 3, 5 or 15. In fact, you will see that it is 15. Consider the subgroup consisting of those permutations that leave one of the digits fixed (say, 5). This subgroup is isomorphic to 4 ,S and thus, it has three Sylow 2-subgroups (see Example 11). These subgroups are of order 8, and hence are Sylow 2-subgroups of S5 as well. There are 5 different subgroups of 5 ,S each isomorphic to S4 (by the choice of the digit to be fixed). An inspection shows that no two of these subgroups share a Sylow 2-subgroup. We have, thus, produced 15 distinct Sylow 2-subgroups. ***

Example 13: Apply the Sylow theorems to analyse the structure of groups of order 15. Solution: Let G be a group of order = × .531 5 Let H be a Sylow 3-subgroup and N a Sylow 5-subgroup. They are both unique by Sylow’s third theorem, and hence normal. Since the order of H and N are coprime to each other, it follows that ∩ NH is trivial. So, G is isomorphic to × .NH It follows that G is cyclic, as both H and N are cyclic (being of prime order) and their orders are coprime. *** 71

Group Theory This leads us to an important point to note.

Remark 5: The argument in Example 13 applies more generally to groups of order pq, , where p and q are primes such that p < q and ≡/ 1q (mod p). For instance, to groups of order 51, 77 and = × 317391 .2 The conclusion is that there is only one group ( isomorphism) of order pq , where p and q are prime, < ≡/ 1qq,p p),(mod namely, the cyclic one.

Try some exercises now.

a a 21 E20) Let G be a group of order 1 2 ,pp where p1 and p2 are two distinct N primes and 21 ∈ .a,a Let P1 and 2 ,P respectively, be a Sylow

p1-subgroup and a Sylow p2-subgroup of .G Suppose that both P1 and P2

are normal. Show that ×= 21 .PPG Generalise the result to the case when G has arbitrary order and for all p, its Sylow p-subgroups are normal.

A famous theorem by Burnside tells us that if G E21) How many Sylow 5- and Sylow 3-subgroups does S5 have, and why? is a finite non-abelian , then at least E22) Prove that no group of order pq is simple, where p and q are distinct three distinct primes divide )G(o . primes. What about a group of order 2 qp ?

E23) Prove Proposition 2.

Now, we shall use the Sylow theorems to understand the structure of groups of order 34, or 38, or 62, for example.

Theorem 8 (Groups of order 2p for a prime p ): There are precisely two groups of order ,p2up to isomorphism, where p is a prime.

Proof: If = ,2p then you know from Example 5 that the group can either be the or the Klein 4-group. Now, let p be odd and G be a group of order .p2 The number of Sylow p-subgroups is of the form 1(mod p) and divides 2. Hence, the Sylow p-subgroup of G is unique, and hence normal. Denote it by .N Since ( ) = N,pNo is cyclic of order p , say = ,rN where r(o = .p) Let = { s,eH }, where 2 = ,es be a Sylow 2-subgroup of .G Since ∩ NH is trivial (by considerations of order), Proposition 2 (ii) applies. Hence,= ,NHG so that G is generated by s and r. Now define =ϕ→ϕ −1 .srs)r(:NN: Since N is normal, ϕ is an automorphism of .N This automorphism is determined by what it does to r, since r generates .N − m1 In Unit 6 you will study Choose m to be an integer such that = .rsrs We have, on the one hand, 2 about generators of −1m − === mmmm1 .r)r ()srs(ssr And on the other hand, −1m = −− 11 = ,rs)srs(sssr groups, and the relations 2 they satisfy, in detail. since 2 = .es Thus, m = rr . So, 2 − ),1m(|)r(o that is, 2 − ⇒ |p)1m(|p − )1m( or |p + ).1m( 72

If − ),1m(|p then −1 = ,rsrs which means that sr = rs, that is, elements of H The Sylow Theorems commute with those of .N In this case, = × ,NHG and G is cyclic. If + ),1m(|p then we have = −− 11 .rsrs This means that G is generated by elements r and s subject to the relations 2 2 == er,es and rs = −− 11 .rs Thus, G is isomorphic to the dihedral group of order p2 .

We shall now state a result, without proof, regarding the types of groups of order 12. This, as well as many similar results, will be proved in Unit 6, while studying generators and relations.

Theorem 9: There are five isomorphism classes of groups of order 12. These are represented by: i) Z Z × Z Z;34 ii) Z Z Z Z×× Z Z;322 iii) A4 ; iv) D12 ; v) the group generated by x, y such that 4 3 === 2 .xyxy,yex

From this theorem, you can see, for example, that two types of groups of order 12 are abelian and 3 types are not.

Try this exercise now, looking at all the results you have studied in this unit.

E24) What are the possible structures of groups of order n, where = K ≠ 8n,10,,1n ? [You will study groups of order 8 in Unit 6.]

Let us now end our discussion on the analysis of the structure of some finite groups, with a summary of what you have studied in this unit.

3.5 SUMMARY

In this unit, we have discussed the following points.

1. The definition, and examples, of external direct products of groups.

2. The definition, and examples, of internal direct products of normal subgroups.

Z Z Z 3. If = 1)n,m( , then × ~ mnnm .

4. × = ).K(o)H(o)KH(o

5. The converse of Lagrange’s theorem is not true in general.

6. Cauchy’s theorem: If G is a finite group and p is a prime such that | )G(op , then G has an element of order p.

73

Group Theory 7. Sylow’s theorems: If G is a group such that o = r ,mp)G( where p is a prime, r ≥ 1 and = ,1)m,p( then

i) G has p-subgroups of orders p 2 K r .p,,p, A subgroup of order pr is called a Sylow p-subgroup. ii) Any two Sylow p-subgroups are conjugates in .G iii) The number of distinct Sylow p-subgroups of G is mod1 )p( and divides m.

8. If = q|G| ,p where p and q are primes such that p < q and q ≡/ ,)p(mod1 then G is cyclic.

9. If G| = ,p2| where p is an odd prime, then G is cyclic or G~ p2 .D

10. If = 2|G| ,1 then G has 5 possible structures.

3.6 SOLUTIONS/ANSWERS

E1) ∗ is associative: Let 332211 ∈ .G)b,a(),b,a(),b,a(

Use the fact that ∗1 and ∗2 are associative to show that

2211 1133 ∗∗=∗∗ 3322 )).b,a()b,a(()b,a()b,a())b,a()b,a((

The identity element of G is e( 21 ),e, where e1 and e2 are the identities

in G1 and 2 ,G respectively. The inverse of ∈ G)y,x( is −− 11 ),y,x( where −1 −1 1 21 =∗=∗ 2.eyy,exx

E2) Define 21 ×→× 12 = b()b,a(f:GGGG:f )a, . Then check that f is well-defined, 1-1, surjective and a homomorphism, that is, f is an isomorphism.

GG ~ ××∴ GG 1221 .

E3) We need to show that any element of ×GG 21 is of the form k ,h where

h ∈ H and ∈ .Kk For this, consider any )y,x( in ×GG 21 .

= 12 )y,e()e,x()y,x( , with 2 ∈ 1 y,e(,H)e,x( ∈ .K)

21 =×∴ .HKGG Now, let us look at H ∩ K . Let ∈ ∩ .KH)y,x(

Since =∈ 2.ey,H)y,x( Since y,x( =∈ 1.ex,K)

=∴ 21 )e,e()y,x( . =∩∴ e{(KH 21 )}.e,

E4) Now, ×∈ 21 )GG(Z)y,x(

⇔ = ×∈∀ GG) b,a()y,x()b,a()b,a()y,x( 21

⇔ y,ax()yb,xa( 1 ∈∈∀= Gb,Ga)b 2

∈∀=⇔ xxa Gaa 1 and ∈∀= Gbb yyb 2

∈⇔ 1 )G(Zx and ∈ 2 )G(Zy

1 ×∈⇔ 2 ).G(Z)G(Z)y,x( ×=×∴ ).G(Z)G(Z)GG(Z 74 21 1 2

E5) Let = xA and = ,yB where = y(o,m)x(o = .n) The Sylow Theorems Z ~ Z Then A ~ m and B n . Z Z ~ Z If we prove that × mnnm , then we will have proved that Z × BA ~ mn , that is, A × B is cyclic of order mn. Z ~ Z Z So, let us prove that if = 1)n,m( , then mn × nm . Z Z Z Z Z Define :f nm ++=×→ ).nr,mr()r(f: Now, you can check that f is well-defined and f is a homomorphism. = ∈Z ∈ Z ∩ Z}nmr|r{fKer = ∈Z ∈ Z }mnr|r{ , since = 1)n,m( = Zn.m Finally, to show that f is surjective, take any element Z Z Z Z Z ( )nv,mu ×∈++ nm . Since = ∃ t,s,1)n,m( ∈ such that + =tms 1n . So, − + − )nt1(v)ms1(u ∈Z such that − + − = + Z + Z).nv,mu())nt1(v)ms1(u(f Thus, f is surjective. Now, we apply the Fundamental Theorem of Homomorphism to find Z Z Z Z Z Z ~ Z Z that fKer ~ fIm , that is, mn ~ × nm , that is, mn × nm . ∴A × B is cyclic of order mn.

E6) Since H1 and H2 are non-empty, so is × 21 .HH

Also, for 21 )h,h( ( and 43 )h,h in −1 −1 −1 × 432121 = ()h,h()h,h(,HH 31 42 ×∈ 21 .HH)hh,hh So,

×≤× 2121 .GGHH

Next, for ×∈ GG) y,x( 21 (and ×∈ 2121 ,HH)h,h −1 21 x()h,h()y,x( )y, −1 −1 = 1 2 ×∈ 21 .HH)yyh,xxh(

Thus, if H1 G1 , H2 G2 , then × 21 )HH( × 21 ).GG(

Z Z Z Z ) Take 21 == .2GG Note that 2 has two subgroups: the trivial

one and the whole group. So the subgroups of type × HH 21 are four in

number. But ×GG 21 is the Klein 4-group (non-cyclic of order 4) which has five subgroups: the , the whole group, and three of order 2.

E8) We know that each ∈Gx can be expressed as hk , where h ∈H and k ∈K . ∴ = .H KG We need to show that ∩ = }.e{KH For this, take ∈ ∩ .KHx Then x ∈H and ∈ .Kx ∴ ∈HKx e and ∈ .H Kex So, x has two representations, xe and ex , as a product of an element of H and an element of .K But we have assumed that each element must have only one such representation. So the two representations xe and ex must coincide, that is x = e . ∴ ∩ = }.e{KH ∴ = × .KHG

E9) Let )n(P be the statement that if 1 K××= n ,HHG then n = ∏ i .)H(o)G(o =1i 75

Group Theory )1(P is trivially true, as you can see. Now, assume that )k(P is true for some ∈N.k

Consider 1 K××= HHG +1k now.   k Since H ,G G ~ K×× .HH Thus, oG  = ,)H(o so +1k H 1 k H ∏ i +1k  +1k  =1i +1k that = ∏ i o)G(o )H( , i.e., + )1k(P is true. =1i Hence )n(P is true ∀ ≥ .1n

3 E10) 5 ××= .235|S| Now, }e{ and S5 are subgroups of 5 ,S of orders 1 and

120, respectively. By Sylow’s first theorem, S5 must have subgroups of Z Z ~Z order 4,2,3,5 and .8 Further, since × mnnm for = S, 1)n,m( 5 must also have subgroups of order × × ×23,25,35 , i.e., 10,15 and 6.

We also know that A,S 44 and A5 are subgroups of 5 ,S of orders 12,24 and ,60 respectively. Thus, the required values of n are .120,60,24,15,12,10,8,6,5,4,3,2,1

E11) No. For example, consider S3 and 4 .S The order of a Sylow 2-subgroup

in S3 is 2, and in S4 is 8. So they cannot be the same.

e1 e1 E12) Let 1 = mp|G| 1 and 2 = 2 ,mp|G| with m1 and m2 coprime to p.

+ee 21 Since G,|G|p has a Sylow p-subgroup say ,P where = .p|P| Let π1

denote the projection of G onto 1 ,G that is, →×π GGG: 1211 given by

a 121 .g)g,g( Similarly, let π 2 denote the projection of G onto 2 ,G

that is, 212 →×π GGG: 2 given by ( 21 ) a 2.gg,g The projections π1

and π2 , being homomorphisms, their images π 1(P) and π 2 )P( are

d1 p-subgroups of G1 and 2 ,G respectively. Thus, 2 =π p| )P(| and

d 2 2 =π ,p|)P(| with ≤ ed 11 and ≤ 22 .ed But, since P is contained in the

+ee 21 +dd 21 subgroup 1( ) π×π ( PP 2 ) , 1 2 =π⋅π≤= .p|)P(||)P(||P|p

Hence == 2211 ,ed,ed π 1 )P( and π2 )P( are Sylow p-subgroups of P1

and P2 , respectively, and 1 P(P π×π= 2 .)P()

Conversely, if P1 and P2 are Sylow p-subgroups of G1 and 2 ,G

+ee 21 respectively, then 21 =× ,p|PP| so that × PP 21 is a Sylow p-subgroup of .G

−1 E13) Let ∈ G ).P(Nx Then = .PxPx −1 Now, for any ∈ G ),P(Ny ∈ G )P(Nxyx . −1 So, G = G )P(Nx)P(xN , that is, ∈ GG .))P(N(Nx

To prove the other inclusion, let g be an element in GG .))P(N(N Then −1 −1 −1 G = G P(Ng)P(gN ) and ⊆ G .)P(NP So, ⊆ G )P(NgPg . So gPg is

a Sylow p-subgroup of G .)P(N

But ,Pbeing normal in G ),P(N is the only Sylow p-subgroup of G .)P(N 76

−1 The Sylow Theorems So = PgPg , that is, ∈ G )P(Ng .

E14) No. Take, for instance, = = ,1d,2p and G to be the Klein 4-group. There are three subgroups of order 2 of .G Since G is abelian, each of these is normal, so no two of them are conjugate.

E15) Let 2 ×== .7228|G| By Sylow’s third theorem, the number of Sylow 7-subgroups is )7(mod1 and divides 4. Thus, it is 1. Call this unique subgroup .H Since H is normal in ,G G is not simple.

Now, any element of order 7 generates a subgroup of order 7, which must be .H Also every element of ,H apart from e, is of order 7. Thus, G has 6 elements of order 7.

2 E16) D12 has order 2 ×= .321 The number of Sylow 3-subgroups is 1 or 4.

If it is 4, then D12 has 8 elements of order 3. However, by considering its elements, you know that it has only two elements of order 3, namely, the rotations by π 3/2 and π .3/4 So there is a unique Sylow 3-subgroup. As to the Sylow 2-subgroups, there are either 1 or 3 of them. The

number cannot be 1 since D12 is not abelian. So the number is 3. The rotation by an angle π belongs to the centre of the group. It is contained in all 3 Sylow 2-subgroups. Each of these subgroups consists of e, the rotation by angle π, and a pair of reflections. To describe such a pair, consider the line through a pair of opposite vertices. The reflection in this line, along with that in the line perpendicular to it, forms a pair of reflections. Note that the perpendicular line passes through the midpoints of the two edges that are not incident on the chosen pair of opposite vertices. There being three such pairs (of opposite vertices and therefore of reflections), the Sylow 2-subgroups are all described.

E17) This group has order 2 ×=× .3262 This has Sylow 2- and Sylow 3-subgroups of orders 4 and 3, respectively. Along the lines of E12, you

can show that the Sylow 3-subgroups are of the form × P} e{ 1 where P1 Z is a Sylow 3-subgroup of 3 .S Since P1 is unique, ×S32 has a unique Sylow 3-subgroup. Using E12, there are three Sylow 2-subgroups, each of which is isomorphic to the Klein 4-group.

E18) i) Note that if a term in the numerator is divisible by p, it is of the form − sr ,npmp where = .1)n,p( Then in the denominator there is a corresponding term − sr ).npp( So ps divides both terms and p +1s does not divide either. So the powers of p in the numerator and denominator are the same and cancel out. Hence the result.

ii) G ×℘(G)→℘(G)a gS)S,g(: is an action. If =⊆= r ,}p|S|GS{X then × ⊆ .XXG

So X is a G-invariant subset of ℘(G). 77

Group Theory  rmp  iii) The number of distinct in X is   .  r   p  Now /| ,|X|p by (i) above. Also X is a of its G-orbits. So there is at least one G-orbit in ,X say ,Y such that /| .|Y|p

r iv) = G )S(StabH for ∈ ,YS where /| |Y|p and = .p|S| Now, by E6 (ii) of Unit 2, as S is H-invariant, it is the union of right cosets of .H So = r .p|S||H|

v) Y~ HG as G-sets. Since /| r Hp ,,|Y|p .||

vi) From (iv) and (v), = r ,p|H| so that H is a Sylow p-subgroup of .G

E19) i) Let X denote the set of cosets ,QK with the natural action of K on it. Restrict to G the action of K on ,X and consider the G-orbits in .X Since Q is a Sylow p-subgroup of ,K it follows that |X| is coprime to p. So there exists a G-orbit Y of X such that |Y| is coprime to p. Let y be a point of Y and consider the

stabiliser, G ).y(Stab Call it H. Now, if y represents the coset kQ of Q in ,Kthen the stabiliser of y in K is −1 ,kQk so ∩= kQkGH −1 is a p-group (being a subgroup of the p-group kQk−1 ). And, on the order hand, Y −~ ,HG as G-sets. So the index of H in G (being equal to |Y| ) is coprime to p. So H is a Sylow p-subgroup of .G

ii) Let P be a p-subgroup of G and let X be as in (i) above. Restrict to P the action of K on .X Since P is a p-group, we may invoke E17 of Unit 2 to conclude that P ≠ .0|X| Letting kQ be the coset corresponding to a fixed point of P in ,X we obtain ⊆ −1.kQkP

iii) Let P be a Sylow p-subgroup of .G By (ii) above, there exists a conjugate Q′ of Q in K such that ⊆ ′ ,QP so that ⊆ ∩ ′ .QGP But ∩ QG ′ is a p-subgroup of G (since Q′ is a p-group), so its cardinality cannot exceed that of .P Hence, = ∩ ′ .QGP

E20) We apply Proposition 2, with 1 = HP and 2 = .NP For reasons of order,

∩ PP 21 is trivial. So = 21 ,|PP||G| so that = 21 .PPG Thus, G is

isomorphic to × 21 .PP

For the general case, let P, ...,P,Pk21 be Sylow subgroups of ,G corresponding one each to the various prime divisors of .|G| Suppose that they are all normal. The argument of the previous paragraph shows that PP is × .PP . It follows, from Proposition 2(iv), that PP is 78 21 21 21

normal in .G We now apply Proposition 2 again, this time with The Sylow Theorems

= PPH 21 and = 3 ,PN to conclude that PPP 321 is normal and isomorphic ~ to PPP ××−× 321321 .PPP Proceeding thus, we conclude that G is

isomorphic to the direct product 1 ×⋅⋅⋅× k .PP

3 E21) The order of 5S5 is ⋅⋅= .532! The number of Sylow 5-subgroups can be 1 or 6. The number of Sylow 3-subgroups can be 4,1 or 10. Now, the order of a Sylow 5-subgroup is 5. Every subgroup of order 5 consists of four 5-cycles in to the identity. No two distinct subgroups of order 5 intersect non-trivially – the identity is the only

element common to them both. The number of 5-cycles in S5 being 24,

we conclude that there are 6 Sylow 5-subgroups in 5 .S

The analysis for Sylow 3-subgroups is similar. The 3-cycles in S5 are × × 345 = 20 in number. They are distributed over 10 Sylow 3 3-subgroups.

E22) Let G be a group of order pq. Assume p < q. (The same argument works if p > q. ) Then G has a Sylow q-subgroup and the number of such subgroups is )q(mod1 and divides p. Hence the number is 1. So the Sylow q-subgroup has order q and is normal in .G Hence G is not simple.

Now consider a group G of order 2 .qp If p > q , then, as above, G will not be simple. Now assume q > p. The number of Sylow q-subgroups can be p,1 or 2.p If the number is 1, then G is not simple. Since q > p, the number cannot be p. If the number of Sylow q-subgroups is p2 , then 2 − ,)1p(q that is +− .)1p()1p(q Since /| +− .)1p(q),1p(q This is only possible if = .2p Then = ,3q so that = 12|G| and G has

4 Sylow 3-subgroups. Let the Sylow 3-subgroups be 4321 .T,T,T,T 4

Then UTi consists of the identity and 8 elements of order 3. Any =1i Sylow 2-subgroup of G intersects this union trivially because any such

group intersects each Ti trivially. So a Sylow 2-subgroup must contain     the 3 remaining elements in UT\G i  and the identity. Thus, the  i  Sylow 2-subgroup is normal, and G is not simple.

Hence, in all the cases, G is not simple.

E23) i) Firstly, check that ≠ « .HN Next, use the fact that H normalises N to show that if −1 2211 ∈ ,HNnh,nh then 2211 ∈ .HN)nh()nh( Hence ≤ GHN . 79

Group Theory ii) You have seen this in Unit 1.

iii) Since ∩ = HN},e{NH is a direct product.

iv) From (i) you know that ≤ .GHN Further, for any ∈ ∈ ∈ ,Nn,Hh,Gg ghng −1 = −− 11 ∈ .HNgngghg Hence HN .G

E24) If n = 1, then = }.e{G • If = ,7,5,3,2n then G ~ Z Z.n • If = ,4n then G is cyclic or the Klein 4-group (see Example 5). • If = 6n or 10, then by Theorem 8, G is either cyclic or isomorphic to the dihedral group. • If = ,9n then G is abelian. If G has an element of order 9, it is cyclic. If not, consider its subgroup = ,xH where 3 = .ex As HG ~ Z 3Z , G ~ Z Z × Z Z.33

80