Group Theory Anupam Singh IISER, central tower, Sai Trinity building, Pashan circle, Pune 411021 INDIA E-mail address: [email protected]

Large portion of this notes is contributed by students M. E. Lakshmipriya, Avi Prasanna, A. S. Arvind, Kapil Gupta, Navi Prasad and Rajesh Yadav

CHAPTER 1

Introduction This is class notes for the course on representation theory of finite groups. We study character theory of finite groups and illustrate some properties of groups using them. The Burnside’s theorem is one of the very good applications. It states that every group of order pa q b , where p, q are distinct primes, is solvable. We will always consider finite groups unless stated otherwise. All vector spaces will be considered over fields unless mentioned otherwise.

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CHAPTER 2

Direct and Semidirect product of Groups There are some simple ways to construct new groups out of known ones. In this chapter we are also going to see how these construction help in understanding groups of larger size in terms of their subgroups. 2.1. Direct Product of Groups Let G1 and G2 be two groups. Definition 2.1 (Direct Product). Let G = G1 ×G2 . On the set G we define an operation as follows: (g1 , g2 )(g1 0 , g2 0 ) = (g1 g10 , g2 g20 ). This operation makes G into a group and is called the Direct Product of G1 and G2 . Exercise 2.2. Verify that the operation defined on G = G1 × G2 as above makes it into a group. This definition can be extended in the similar way to finitely many groups G1 , G2 , . . . , Gn to constuct the direct product group G = G1 × G2 × · · · × Gn . We list some of the properties of the direct product below as an exercise. Exercise 2.3. Let G = G1 × G2 be the direct product of two groups G1 and G2 . (1) The map G1 → G1 × G2 given by g1 7→ (g1 , e) defines an embedding of G1 into G1 × G2 . We define the image subgroup as G01 = {(g, e) | g ∈ G1 } (2) Consider the group homomorphism π : G1 × G2 → G2 defined as π(g1 , g2 ) = g2 . Clearly Ker(π) = G01 . Hence G1 is embedded as a normal subgroup inside G1 ×G2 . (3) The above statements hold for G2 as well. We call the embedding of G2 as G02 defined in the same way as G01 . (4) Any element (g1 , g2 ) of G1 × G2 can be written as a product of elements of G01 and G02 as (g1 , g2 ) = (g1 , e)(e, g2 ). (5) Clearly G01 ∩ G02 = (e, e) and hence the above representation of an element of G1 × G2 as the product of elements from G01 and G02 is unique. In case we have G1 × G2 × · · · × Gn , then G01 ∩ (G02 · · · G0n ) = {(e, e, . . . , e)}. 5

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2. DIRECT AND SEMIDIRECT PRODUCT OF GROUPS

Given groups G1 and G2 , we have constructed a group G1 × G2 of larger size having a copy of G1 and G2 as normal subgroups. The group G1 × G2 is usually called the External Direct Product of groups G1 and G2 . Now we turn our attentation to a group G and its subgroups. All the observations made above give a cue as to when a group is the direct product of its subgroups. The following theorem gives a necessary and sufficient condition as to when a group is isomorphic to a direct product of its subgroups. Theorem 2.4. Let G be a group and let H and K be subgroups of G such that (1) H ∩ K = {e}. (2) H and K are normal in G. (3) HK = G. Then G ∼ = H × K. Proof. Since H and K are normal in G and H ∩ K = {e} we observe that hk = kh for all h ∈ H and k ∈ K. As G = HK, for every g ∈ G there exists h ∈ H and k ∈ K such that g = hk. In fact h and k are unique for the given g. We define a map φ : G → H × K give by φ(g) = (h, k) where g = hk. Now we claim that φ is an isomorphism. Let g = hk and g 0 = h0 k 0 then φ(gg 0 ) = φ(hkh0 k 0 ) = φ(hh0 kk 0 ) = φ(h, k)φ(h0 , k 0 ) = φ(g)φ(g 0 ). This proves φ is a group homomorphism. Clearly φ is onto, since φ−1 (h, k) = hk. Now to prove injectivity let φ(hk) = (e, e). Then h = k = e and hence ker(φ) = {e}. Thus φ is injective and hence an isomorphism.  In this case G is called the Internal Direct Product of H and K. Exercise 2.5. Let G be a group with two subgroups H and K. Let HK = {hk | h ∈ H, k ∈ K}. Prove the following: (1) HK need not be a subgroup. Give an example. (2) If one of the H or K is normal then HK is a subgroup of G. (3) If both H and K are normal then HK is a normal subgroup of G. Proposition 2.6. Let G = H ×K be direct product of its subgroups. Suppose G is finite. If (|H|, |K|) = 1 then any subgroup L 6 G is a direct product given by L = (H ∩L)×(K ∩L). Proof. We have L ∩ H E L and L ∩ K E L and also (L ∩ H) ∩ (L ∩ K) = {e} . Thus we consider the direct product of (L ∩ H) and (L ∩ K) inside L. (L ∩ H) × (L ∩ K) ⊆ L. Hence we need to prove that when g ∈ L with g = hk, h ∈ H and k ∈ K then h ∈ H ∩ L and hence h ∈ L and similarly k ∈ L. We have hk ∈ L. (hk)|H| = k |H| ∈ L. But since |H| and |K| are coprime some power of k |H| will give k and hence k ∈ L. Similarly we can prove that h ∈ L. Thus L = (H ∩ L) × (K ∩ L). 

2.2. SEMI DIRECT PRODUCT OF GROUPS

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Example 2.7. In general (i.e., when (|H|, |K|) 6= 1) the above theorem does not hold. For example consider the group G = Z4 × Z4 , L = {(x, x) | x ∈ Z4 }. We claim that the above theorem does not hold for this L, that is there are no subgroups of Z4 , H and K such that L = H × K. This is because the projection π : L → Z4 onto the first (and also second) coordinate is the whole of Z4 . Hence if at all L = H × K, then H = K = Z4 . But clearly L is a proper subgroup of G. 2.2. Semi Direct Product of Groups The notion of Semi Direct Product is more general than Direct Product as we will see in this section. Definition 2.8 (Semi Direct Product). Let G be a group with subgroups H and N where N is normal. Suppose H and N satisfy the following conditions: (1) N H = G (2) H ∩ N = {e} Then G is called the Internal Semi Direct Product of the subgroups H and N and is usually denoted by G = N o H. We write down some properties of the semi direct product defined above as an exercise. Exercise 2.9. With the notation as above, a. Useing G = N H and H ∩ N = {e} prove that every element of g ∈ G can be written uniquely as g = nh for some h ∈ H and n ∈ N . Hence one can think of the underlying set as N × H. b. Let g = nh and g 0 = n0 h0 . Verify that gg 0 = nhn0 h0 = n(hn0 h−1 )hh0 . Hence the multiplication of G can be written in terms of elements of N and H as follows: (n, h)(n0 , h0 ) = (nhn0 h−1 , hh0 ). As N is a normal subgroup, conjugation by an element of H defines an isomorphism of N . The map φ : H → Aut(N ) given by φ(h)(n) = hnh−1 is a group homomorphism. We can rewrite the multiplication as gg 0 = nhn0 h0 = n(φ(h)(n0 ))hh0 . c. We note that if H is also a normal subgroup then G is actually the direct product of subgroups H and N . Warning! However whenever we say a group is internal semidirect product we mean one of the subgroup is normal and other one is not, so that it is not a direct product. We make use of the above observations to define the External Semi Direct product. Definition 2.10 (External Semi Direct Product). Let H and N be two groups. Suppose we have a map φ : H → Aut(N ) a group homomorphism. Then on G = N × H we define the following operation: (n, h)(n0 , h0 ) = (nφ(h)(n0 ), hh0 )

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2. DIRECT AND SEMIDIRECT PRODUCT OF GROUPS

This makes G into a group and is usually denoted as N oφ H, the semidirect product of N and H. Exercise 2.11. Prove that the above operation makes G = N × H into a group. Next we will write some properties of the semidirect product as an exercise. Exercise 2.12. With the notation as above, (1) If the map φ is trivial then the resulting group G is direct product N × H. (2) If φ is non-trivial, the group G = N oφ H is non Abelian. (Hint: Find the elements of the form (n, eH ) and (eN , h) which do not commute.) (3) Different φ may give rise to different groups. (4) The group G = N oφ H contains an isomorphic copy of N as N 0 = {(n, eH ) | n ∈ N } which is a normal subgroup of G and also a copy of H as H 0 = {(eN , h) | h ∈ H}. (5) Verify that N 0 ∩ H 0 = e ∈ G. (6) Prove that the internal semidirect product N 0 o H 0 ∼ = G. Now suppose H and N are two groups. Suppose we have homomorphisms φ, ψ : H → Aut(N ). The following theorem gives a sufficient condition when the groups N oφ H and N oψ H are isomorphic. Theorem 2.13. Let H be a cyclic group. Also assume φ and ψ are monomorphisms such that φ(H) = ψ(H). Then N oφ H ∼ = N oψ H. Proof. Let H =< x >. We also have φ(H) = ψ(H), hence φ(x) and ψ(x) generate the same cyclic subgroup of Aut(N ). Hence ∃ a, b ∈ Z such that φ(x)a = ψ(x) and ψ(x)b = φ(x). Since H is cyclic, any h ∈ H is of the form h = xk . Thus for any h ∈ H also, φ(h)a = ψ(h) and ψ(h)b = φ(h) hold. Consider the map τ : N oψ H → N oφ H given by τ (nh) = nha . We first prove that τ is a homomorphism. τ (nhn0 h0 ) = τ (nψ(h)(n0 )hh0 ) = nψ(h)(n0 )(hh0 )a = nφ(ha )(n0 )ha h0a = nha n0 h−a ha h0a = nha n0 h0a = τ (nh)τ (n0 h0 ) We define a similar map λ : N oφ H → N oψ H by λ(nh) = nhb . Consider the map τ ◦ λ : N oφ H → N oφ H, τ ◦ λ(nh) = nhab = nh. (This is because φ(x) = ψ(x)b = (φ(x)a )b = φ(xab ) and since by assumption ψ and φ are injective and hence x = xab ). Hence τ ◦ λ is the identity map on N oφ H and similarly λ on N oψ H. Hence τ and λ are actually isomorphisms. 

2.3. EXAMPLES OF DIRECT AND SEMI DIRECT PRODUCT

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2.3. Examples of direct and semi direct product Example 2.14. Any finitely generated Abelian group is isomorphic to a direct product of cyclic groups. Example 2.15. Let N = Z/nZ and H = Z/2Z. Define φ : H → N by φ(1)(x) = x−1 for all x ∈ Z/nZ. Then the Dihedral group D2n ∼ = N oφ H. Example 2.16. Let S3 be the symmetric group. Let N =< (123) > and H =< (12) >. Then S3 = N o H. Example 2.17. All groups of order p2 and p3 can be classified in terms of just direct and semi direct product of their subgroups. See next chapter. Example 2.18. Here we give an example of semidirect product where N oφ H  N oψ H where φ and ψ are different. Let us take H = Z/4Z and N = Z/5Z. Let us denote Aut(Z/5Z) = {fi | fi (1) = i, i ∈ {1, 2, 3, 4}. Here f4 is an element of order 4, f2 and f3 are elements of order 2 and f1 is the identity element. We define φ, ψ : Z/4Z → Aut(Z/5Z) as follows: φ(1) = f4 and ψ(1) = f2 . We count the number of order 2 elements in each of N oφ H and N oψ H and observe that they are not equal and hence proving that the groups are not isomorphic. In N oφ H, if (n, h) is an element of order 2, then (n, h)(n, h) = (n + φ(h)n, 2h) = (0, 0). Which implies either h = 0 or 2. In case h = 0 we get n = 0 which is (0, 0), the identity element. For h = 2, we have (n, h)(n, h) = (n + φ(2)n, 2h) = (0, 0). So we want n such that n + φ(2)n = 2n = 0 (in Z/5Z, hence n = 0). Thus the only element of order 2 is (0, 2). In N oψ H, if (n, h) is an element of order 2, then h = 2 and n+ψ(2)n = n+4n = 5n = 0. But this holds for all n ∈ Z/5Z. Hence every element (n, 2), where n ∈ Z/5Z is of order 2. Thus the number of elements of order 2 is 5. Hence N oφ H  N oψ H. Exercise 2.19. Let G be the set of n × n invertible upper triangular matrices over C (also called the standard Borel Group). Let U be the subgroup consisting of those upper triangular matrices which have 1 on its diagonal. Prove that G ∼ = D n U where D is the diagonal subgroup.

CHAPTER 3

p-Groups Let p be a prime. A group of order pr , for some r, is called a p-group. We list some of the properties of these groups. Theorem 3.1 (p-Groups). Let G be a p group of order pr . Then, (1) The center of G is nontrivial. (2) G has a subgroup of order ps for all 0 ≤ s ≤ r. (3) Moreover if G is acting on a finite set X. Then, |X| ≡ |X G | (mod p) where X G = {x ∈ X | gx = x∀g ∈ G}. (4) In particular, |Z(G)| ≡ 0 (mod p). Proof. Look at X − X G which is a union of nontrivial orbits. (We can also use class equation here).  Lemma 3.2. Let G be a finite group with center Z(G). If G/Z(G) is a cyclic group then G is Abelian. Proof. Let us denote G/Z(G) ∼ = hx.Z(G)i. Then G = hx, z | z ∈ Z(G)i. But since x commutes with the elements of the center the group G is Abelian.  Note that G/Z(G) being Abelian need not imply G is Abelian. For example take Q8 . Proposition 3.3. Let G be a group of order p, a prime. Then G is cyclic (and hence Abelian) and isomorphic to Z/pZ. Proposition 3.4. Let G be a group of order p2 where p is a prime. Then G is Abelian and is isomorphic to one of the following: Z/pZ × Z/pZ, Z/p2 Z. Proof. Let Z be the center of G. Then either |Z| = p or p2 . In the case |Z| = p2 the group G is Abelian and is isomorphic to Z/p2 Z. Let |Z| = p. Then |G/Z| = p hence G/Z is a cyclic group. Using previous Lemma we conclude that G is Abelian in this case also and is isomorphic to Z/pZ × Z/pZ.  11

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3.1. Groups of order p3 Proposition 3.5. Let P be a group of order p3 where p is a prime. Then, (1) If P is Abelian then it is isomorphic to one of the following: Z/pZ × Z/pZ × Z/pZ, Z/p2 Z × Z/pZ, Z/p3 Z. (2) If P is not Abeilan and p 6= 2 it is isomorphic to one of the following: (Z/pZ × Z/pZ) o Z/pZ, (Z/p2 Z) o Z/pZ. (3) If P is not Abelian and p = 2 then the groups are Q8 and D8 . This classification can be broadly divided into two parts. One being the set of Abelian groups and the other being Non-Abelian groups. Now, let P be a group of order p3 , where p is an odd prime. Let us first discuss the Abelian groups without going into too much detail. We know that the order of the subgroup should divide the order of the group. Hence, the subgroups of the group P can have order 1, p, p2 , p3 . The order 1 subgroup we know is the trivial group. The group of order p would be clearly isomorphic to Z/pZ. Next subgroup is the group of order p2 , and this would be isomorphic to either Z/p2 or Z/p × Z/p. And then last is the case when the order is p3 , in this case we can easily figure out that the groups will be isomorphic to one of these, namely Z/p × Z/p × Z/p, Z/p2 × Z/p, Z/p3 . Now we will discuss the Non-Abelian groups of order p3 . Let us first assume that the group has an element of order p2 and H is the subgroup generated by that element. Then, Proposition 3.6. If p is an odd prime, then there is an exactly one isomorphism class of non-Abelian group of order p3 having elements of order p2 . In fact, this group is isomorphic to Z/p2 oZ/p. We will break the proof in following Lemma. ∼ Z/n for n ∈ N, and for each 0 ≤ m < n let σm be the Lemma 3.7. Let G =< x >= endomorphism of G sending x to xm . Then Aut(G) consists precisely of those σm for which m 6= 0 and gcd(m, n) = 1. Furthermore, Aut(G) is Abelian and is isomorphic to the group (Z/nZ)× the of units of the ring Z/nZ. From this Lemma it immediately follows that Aut(Z/p2 ) ∼ = Z/p(p − 1). Hence, |Aut(Z/p2 )| = p(p − 1), which is divisible by |Z/p| = p. This proves by taking appropriate automorphism we can form the group Z/p2 o Z/p. Definition 3.8. A proper subgroup H of a group G is said to be maximal if there is no proper subgroup of G that properly contains H. Lemma 3.9. If P is a finite p-group, then every maximal subgroup of P is normal in P.

3.1. GROUPS OF ORDER p3

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It now follows from the above Lemma that the subgroup H of P of order p2 is a normal subgroup. Now we wish to show that we can find elements x and y of P such that x has order p, y has order p2 , and such that x 6∈< y >. Hence, in this case, we would have P =< y > o < x > i.e. Zp2 o Zp . So, we choose an element y of order p2 , and choose some x 6∈ < y >. Assume that xp 6= 1, for if not then we have found the desired elements x and y. Theorem 3.10. If P is a non-trivial finite p-group, then Z(P) is also non-trivial. Moreover, if 1 6= N  P , then N ∩ Z(P ) 6= 1. Lemma 3.11. If G is a non-abelian p-group, then G/Z(G) cannot be cyclic. Proof. (of Lemma 2.5) We prove this lemma by contradiction, suppose that G/Z(G) = < xZ(G) > for some x ∈ G. Then G would be generated by the set S = Z(G) ∪ {x}. In this case every pair of elements of S commutes, and S being the generator implies that < S >= G is abelian, which is a contradiction. If P is a finite p − group and |P : Z(P )| = p, then P/Z(P ) is necessarily cyclic, contradicting the above lemma.  Using above theorem and Lemma we prove, Lemma 3.12. |Z(P )| = p. Proof. The order of Z(P ) is either p or p2 . Now suppose |Z(P )| = p2 then |P/Z(P )| = p. Hence P is Abelian, a contradiction.  From Proposition 2.3, we have seen that < y >6= 1 is a normal subgroup of P and then it follows from Theorem 2.4, that N ∩ Z(P ) 6= 1, and this implies that Z(P )  N . Also since the order of the center is p, safely we can write Z(P ) = < y p >. Now, for x ∈ P , xZ(P ) ∈ P/Z(P ), which implies (xZ(P )p ) = Z(P ), hence either xp = 1 or xp = y kp , for some k, such that 1 ≤ k < p. Our next step now would be to replace y −k by y. As a result of this replacement we obtain xp = y −p . Note here the fact that we have taken into consideration the fact that y ∈ Zp2 . Now let us introduce the Commutator subgroup. Let G be a group, then G0 is called the derived group or the commutator subgroup of G. So, G0 : = subgroup of G generated by the set of all the commutators in G. i.e. G0 =< [x, y]|x, y ∈ G >, where [x, y] = xyx−1 y −1 . Proposition 3.13. Let G be a group, and let N G. Then G/N is abelian iff G0 ≤ N



If x, y ∈ P then it follows from the proposition that [x, y] ∈ Z(P ). Another important lemma related to this is the following Lemma 3.14. Let G be a group and x,y ∈ G. If [x, y] ∈ Z(G), then [xn , y] = [x, y]n and xn y n = (xy)n [x, y]n(n−1)/2 for any n ∈ N.

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Proof. (of Lemma 2.7) We will prove this lemma by the Principle of Mathematical Induction, let us consider the first statement, clearly this is true for n = 1, so assuming it to be true for n, we try to prove it for n + 1 [xn+1 , y] = xxn yx−n (y −1 y)x−1 y −1 = x[xn , y]yx−1 y −1 = x[x, y]n yx−1 y −1 = [x, y]n+1 via the induction hypothesis. Now consider the second statement; again we shall use induction on n. Assume that the statement holds true n ∈ N. Then xn+1 y n+1 = xxn y(yxn )−1 (yxn )y n = x[xn , y]yxn y n = x[x, y]n y(xy)n [x, y]n(n−1)/2 = (xy)n+1 [x, y]n(n+1)/2 via the first statement and the induction hypothesis.



Using this above lemma and replacing (xy) by x we can easily obtain xp = 1. Hence, we proved the existence of x. And also note that since p is odd, (p − 1) is even and hence divisible by 2. We have so far shown that any non-abelian group of order p3 having an element of order 2 p can be written as a semidirect product of Zp2 by Zp . We have seen earlier that Aut(Zp2 ) is an abelian group of order p(p − 1). Sylow’s theorem now implies that Aut(Zp2 ) has a unique subgroup of order p. Thus there exists a monomorphism from Zp to Aut(Zp2 ), and any two maps must have the same image. The existence and uniqueness of a semidirect product follows from the following proposition. Proposition 3.15. Let H be a cyclic group and let N be an arbitrary group. If ϕ and ψ are monomorphisms from H to Aut(N ) such that ϕ(H) = ψ(H), then we have N oϕ H ∼ = N oψ H  Hence, we have proved that there is exactly one isomorphism class of non-abelian groups of order p3 having elements of order p2 . Another important Proposition related to this is as follows Proposition 3.16. Let N and H be groups, let ψ : H −→ Aut(N ) be a homomorphism, and let f ∈ Aut(N ). If fˆ is a linear automorphism of Aut(N ) induced by f , then N ofˆøψ H ∼ = N oψ H 

3.1. GROUPS OF ORDER p3

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The second and the last part of this classification can be stated as the following proposition Proposition 3.17. If p is an odd prime, then there is exactly one isomorphism class of non-abelian groups of order p3 having no elements of order p2 . Proof. We easily see that any non-abelian group of order p3 having no elements of order p2 can be expressed as the semidirect product of a maximal subgroup generated by an element not lying in that maximal subgroup. Therefore, it suffices to consider the semidirect products of Zp × Zp by Zp . Now, we want to find the cardinality of Aut(Zp × Zp ), which is given by the following propositions Proposition 3.18. Let E be a finite abelian group of exponent p, where p is prime i.e.∼ E = Zp × Zp × . . . × Zp (n times). Then Aut(E) ∼ = GL(n,p) where n ∈ N is such that |E| = pn  Proposition 3.19. Let n ∈ N, and let q be a prime power. Then n Y |GL(n, q)| = (q n − q k−1 )  k=1

So, from the above two propositions we get Aut(Zp × Zp ) ∼ = GL(2, p), where if Zp ×   a b Zp = < u, v >, then ∈ GL(2, p) corresponds to an automorphism sending u to c d ua v c and v to ub v d . We also have from the above proposition, |Aut(Zp × Zp )| = p(p − 1)2 (p + 1), and hence all subgroups of order p of Aut(Zp × Zp ) are conjugate by Sylow’s theorem. Let ψ and τ be monomorphisms from Zp to Aut(Zp × Zp ). Then there is some f ∈ Aut(Zp ×Zp ) such that τ (Zp ) = f ψ(Zp )f −1 = (fˆøψ)(Zp ), where fˆ is the inner automorphism of Aut(Zp × Zp ) induced by f . And the final result follows from the proposition 2.8 and 2.9.  Example 3.20 (Hiesenberg Group). Example 3.21 (Infinite p-group).

CHAPTER 4

Sylow’s Theorem Let G be a group. Let p | |G|. Suppose pn is the largest power of p dividing |G|. Then a subgroup of order pn is called Sylow p-subgroup of G. We now prove Sylow’s Theorem one by one making use of group actions and the above results on p-groups. Theorem 4.1 (Sylow (1)). Let G be a group with |G| = pn m. Then Sylow p-subgroups exist. Proof. We use induction on |G|. We may assume n ≥ 1. Let us denote C for the center of G. Let us first assume p | |C|. In this case C is Abelian and it has a subgroup of order p, say D. Consider the group G/D. It has order pn−1 m. Using induction we have a sylow subgroup P/D of G/D. Hence the subgroup P is a sylow subgroup of G. Let us now assume p - |C|. Consider X = G−C. The group G acts on X by conjugation. Say X is the union of conjugacy classes Cl1 , Cl2 , . . . , Clr . Since p - |X| (else it would divide |C|) there exists i such that p - |Cli |. Say, Cli is the conjugacy class of g ∈ G. Consider H = CG (g). Then H is a proper subgroup of G (because g is not central) such that |G| pn | |H|. Since |H| = |Cl along with p - |Cli | we get the required result. Now we use i| induction hypothesis on H. Moreover any Sylow p-subgroup of H will be Sylow p-subgroup of G.  Theorem 4.2 (Sylow (2)). With the notation as above, let P be a Sylow p-subgroup. Let Q be any p-subgroup of G. There exists g ∈ G such that Q ⊂ gP g −1 . In particular, Sylow p-subgroups are conjugate. Proof. Consider the set X = G/P = {aP | a ∈ G}. The group Q acts on X by left translations as follows: q.aP = (qa)P. We use the statement from p-group action to deduce that |X| ≡ |X Q | (mod p). Since G |X| = |G| |P | 6≡ 0 (mod p) hence so is |X |. Hence there exists g ∈ G such that Q.gP = gP , i.e., Q ⊂ gP g −1 .  Theorem 4.3 (Sylow (3)). The number of Sylow p-subgroup, np ≡ 1 (mod p). 17

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4. SYLOW’S THEOREM

Proof. Let P be a Sylow p-subgroup. Let X = {P1 , P2 , . . . , Pnp } be the collection of all Sylow p-subgroups. The group P acts on X by conjugation. Hence 1 = |X P | ≡ |X| = np (mod p).  4.0.1. Classification of groups of order pq. Theorem 4.4. Let p and q be distinct primes with p > q. If p 6≡ 1 (mod q), then any group of order pq is isomorphic to Z/pqZ. If p ≡ 1 (mod q), then any Abelian group of order pq is isomorphic to Z/pqZ and there is exactly one isomorphism class of non-Abelian group of order pq which is isomorphic to Z/pZ o Z/qZ. Proof. Let G be a group of order pq. Let P be a Sylow p-subgroup of G and Q be a Sylow q-subgroup. Since |P | = p and |Q| = q, both prime, P ∼ = Z/pZ and Q ∼ = Z/qZ. |P ||Q| Using Lagrange’s theorem we conclude that P ∩ Q = 1. Hence |P Q| = |P ∩Q| = |P ||Q| and G = P Q. By Sylow’s theorem the number of conjugates of P in G divides |G : P | = q and is congruent to 1 (mod p). Since p > q this number is equal to 1 hence P has only one conjugate in G. This implies P is a normal subgroup of G. Similarly using Sylow’s theorem we see that Q has only one conjugate when p 6≡ 1 (mod q) and either 1 or p conjugates when p ≡ 1 (mod q). In the first case when Q has only one conjugate the subgroup Q is also normal in G and G = P × Q ∼ = Z/pZ × Z/qZ ∼ = Z/pqZ. In the second case when p ≡ 1 (mod q) the subgroup Q has possibly 1 or p conjugates. When it has 1 conjugate we are in the previous case and the group G is Abelian. Now suppose Q has p conjugates in G, in that case G is non-Abelian. Note that in this case G = P o Q since P  G, G = P Q and P ∩ Q = 1. However we still need to prove that there is exactly one class of non-Abelian group up to isomorphism. Let φ : Q → Aut(P ) be the conjugation homomorphism, if kernel(φ) 6= 1, kernel(φ) = Q. So it’s trivial homomorphism, hence G is Abelian. It will be direct product only. We conclude that if p ≡ 1 (mod q) and there is a non-Abelian group of order pq, then there is a monomorphism from Z/q to Aut(Z/p). Conversely, given a monomorphism φ : Z/q → Aut(Z/p) we can construct a non-Abelian group of order pq, namely Z/p oφ Z/q. To complete the proof we must exhibit a monomorphism φ : Z/q → Aut(Z/p), which shows the existence of a non-Abelian group of order pq. By proposition 1.3 and 1.4, we see that Aut(Zp ) has a unique subgroup K of order q. By proposition 1.5 we can see that there is some 1 < r < p such that K is generated by the automorphisms σr sending every element to it’s rth power. Define φ : Zq → Aut(Zp ) which sends x to σr , where Zq =< x >. Then φ is a monomorphism and φ(Zq ) = K, which proves existence.

4.2. SIMPLE GROUPS

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We must show that if ψ : Z/q → Aut(Z/p) is another such monomorphism, then the groups Z/p oφ Z/q ∼ = Z/p oψ Z/q. Now if ψ : Zq → Aut(Zp ) is another monomorphism, then by the uniqueness of K we have φ(Zq ) = K = ψ(Zq ), we now have by proposition 1.6  Zp oφ Zq ∼ = Zp oψ Zq . 4.1. Solvable Groups Let G be a group. We denote by G0 the subgroup generated by the elements xyx−1 y −1 . This subgroup is normal and called Commutator subgroup of G. Sometimes it is also denoted as [G, G]. Let us denote G1 = G0 and Gi = G0i−1 . This gives us a series: G ⊃ G 1 ⊃ G 2 ⊃ . . . Gi ⊃ . . . . Definition 4.5. A group G is called solvable if Gn = 1 for some n, i.e., the series G ⊃ G1 ⊃ G2 ⊃ . . . Gi ⊃ . . . . terminates at 1. Example 4.6. Every Abelian group is solvable. Example 4.7. Let T = Tn (k), the set of n × n invertible upper triangular matrices. This group T is solvable. Most (??) of the solvable groups are subgroup of this group. Example 4.8. The group Sn is not solvable for n ≥ 5 since An = Sn0 and An = A0n . Theorem 4.9. Let G be a finite group. (1) p-groups : Every p-group is solvable. (2) Burnside : Every group of order pa q b where p and q are prime is solvable. (3) Feit-Thompson : Every group of odd order is solvable. Proof. The proof of the first one follows from part 2 of the Theorem 3.1. We will prove part two as an application of representation theory later in this book. However the proof of part three runs in several hundred pages and one can refer to the original journal article.  4.2. Simple Groups A group G is called simple if it has no proper normal subgroup.

CHAPTER 5

Linear Groups

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Bibliography [AB] Alperin, J. L.; Bell R. B., “Groups and Representations”, Graduate Texts in Mathematics, 162, Springer-Verlag, New York, 1995. [CR1] Curtis C. W.; Reiner I., “Methods of representation theory vol I, With applications to finite groups and orders”, Pure and Applied Mathematics, A Wiley-Interscience Publication. John Wiley & Sons, Inc., New York, 1981. [CR2] Curtis C. W.; Reiner I., “Methods of representation theory vol II, With applications to finite groups and orders”, Pure and Applied Mathematics, A Wiley-Interscience Publication. John Wiley & Sons, Inc., New York, 1987. [CR3] Curtis C. W.; Reiner I., “Representation theory of finite groups and associative algebras”, Pure and Applied Mathematics, Vol. XI Interscience Publishers, a division of John Wiley & Sons, New York-London 1962. [D1] Dornhoff L., “Group representation theory. Part A: Ordinary representation theory”, Pure and Applied Mathematics, 7. Marcel Dekker, Inc., New York, 1971. [D2] Dornhoff, L., “Group representation theory. Part B: Modular representation theory”. Pure and Applied Mathematics, 7. Marcel Dekker, Inc., New York, 1972. [FH] Fulton; Harris, “Representation theory: A first course” Graduate Texts in Mathematics, 129, Readings in Mathematics, Springer-Verlag, New York, 1991. [Se] Serre J.P., “Linear representations of finite groups” Translated from the second French edition by Leonard L. Scott, Graduate Texts in Mathematics, Vol. 42, Springer-Verlag, New York-Heidelberg, 1977.. [Si] Simon, B., “Representations of finite and compact groups” Graduate Studies in Mathematics, 10, American Mathematical Society, Providence, RI, 1996. [M] Musili C. S., “Representations of finite groups” Texts and Readings in Mathematics, Hindustan Book Agency, Delhi, 1993. [DF] Dummit D. S.; Foote R. M. “Abstract algebra”, Third edition, John Wiley & Sons, Inc., Hoboken, NJ, 2004. [L] Lang, “Algebra”, Second edition, Addison-Wesley Publishing Company, Advanced Book Program, Reading, MA, 1984. [JL] James, Gordon; Liebeck, Martin, “Representations and characters of groups”, Second edition, Cambridge University Press, New York, 2001.

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