MA 553: Homework 1 Yingwei Wang ∗ Department of Mathematics, Purdue University, West Lafayette, IN, USA 1 Composition series 1.1 First part of Jordan Holder Theorem Theorem 1.1 (Jordan Holder Theorem: Part One). Every finite group has a composition series. Proof. Let G1 be a maximal normal subgroup of G; then G/G1 is simple. Let G2 be a maximal normal subgroup of Gl, and so on. Since G is finite, this process must end with Gn = 1. Thus G>G1 > ··· >Gn = 1 is a composition series. 1.2 Composition series Proposition 1.1. If G is a finite group and H ⊳ G. Then there is a composition series of G, one of whose term is H. Proof. On one hand, since H is also a finite group, then by Theorem 1.1, H has a composition series H>H1 > ··· >Hn = 1. On the other hand, consider the quotient group G/H. If G/H is not simple, then it has a nontrivial normal subgroup, which gives a group K1 between G and H extending the normal series H ⊳ K1 ⊳ G. Similarly we can consider G/K1. It indicates that we can find a normal series H = K0 ⊳ K1 ⊳ ··· ⊳ Km = G such that for each j ∈ [0, m], the quotient group Hj+1/Hj is simple. Now we find a composition series as 1= Hn < ··· <H1 <H<K1 ··· <Km−1 < G. ∗E-mail address: [email protected]; Tel: 765 237 7149 1 Yingwei Wang Abstract Algebra 2 Normal subgroup Proposition 2.1. Let A,B,C be subgroups of a group G with A ⊳ B and C ⊳ G. Then CA ⊳ CB. Proof. ∀ c1a ∈ CA, c2b ∈ CB, where c1,c2 ∈ C and a ∈ A, b ∈ B, then −1 −1 −1 (c2b)(c1a)(c2b) = c2bc1ab c2 −1 −1 −1 = c2(bc1b )(bab )c2 −1 −1 −1 = c2c˜1ac˜ 2 , wherec ˜1 = bc1b ∈ C, a˜ = bab ∈ A, −1 −1 = c2c˜1(˜ac2 a˜ )˜a −1 −1 = c2c˜1c˜2a,˜ wherec ˜2 =˜ac2 a˜ ∈ C, =˜ca˜ ∈ CA, wherec ˜ = c2c˜1c˜2 ∈ C. It implies that CA ⊳ CB. 3 Application of Isomorphism Theorem Proposition 3.1. Let G be a finite group with G =6 1. Suppose G has two composition series 1= N0 < N1 < ··· < Nr = G, 1= M0 < M1 < ··· < Ms = G. (I) Show for i,j > 0 that Ni−1(Mj−1 ∩ Ni) ⊳ Ni−1(Mj ∩ Ni). (II) Use the Second Isomorphism Theorem to establish isomorphisms Ni− (Mj ∩ Ni) Mj ∩ Ni (Mj ∩ Ni)Mj− 1 ∼= ∼= 1 . (3.1) Ni−1(Mj−1 ∩ Ni) (Mj−1 ∩ Ni)(Mj ∩ Ni−1) (Mj ∩ Ni−1)Mj−1 Proof. (I) For each i, j, we know that Mj <G and Ni <G. Thus it is obvious that Mj−1 ∩ Ni < Ni, Mj ∩ Ni < Ni. Besides, since Mj−1 ⊳ Mj, then Mj−1 ∩ Ni ⊳ Mj ∩ Ni. 2 Yingwei Wang Abstract Algebra Furthermore, we also know that Ni−1 ⊳ Ni. By the Proposition 2.1, we can conclude that Ni−1(Mj−1 ∩ Ni) ⊳ Ni−1(Mj ∩ Ni). (II) Define the map (Mj ∩ Ni)Mj−1 φ : Mj ∩ Ni → , (Mj ∩ Ni−1)Mj−1 by φ(y)= y(Mj ∩ Ni−1)Mj−1. It is easily seen to be a group homomorphism since Mj−1 ⊳Mj and Nj−1 ⊳Nj. Moreover, (Mj ∩Ni)Mj−1 since Mj ∩ Ni− 6 Mj ∩ Ni, it is easy to see that the image of φ is . 1 (Mj ∩Ni−1)Mj−1 Now we need to prove that ker φ =(Mj−1 ∩ Ni)(Mj ∩ Ni−1). On the one hand, if y ∈ Mj ∩ Ni−1, then φy is the identity coset (Mj ∩ Ni−1)Mj−1; similarly, if y ∈ Mj−1 ∩ Ni, then φy = y(Mj ∩ Ni−1)Mj−1 ⊆ Mj−1(Mj ∩ Ni−1)Mj−1 ⊆ (Mj ∩ Ni−1)Mj−1 since Mj−1 ⊳ Mj. Therefore, (Mj−1 ∩ Ni)(Mj ∩ Ni−1) ⊆ ker φ. On the other hand, if y ∈ ker φ then y ∈ (Mj ∩ Ni−1)Mj−1, so we can write y = ax with a ∈ Mj ∩ Ni−1 and x ∈ Mj−1. Now y ∈ Ni, x ∈ Ni, so x ∈ Ni and therefore y ∈ (Mj−1 ∩ Ni)(Mj ∩ Ni−1), which indicates that ker φ ⊆ (Mj−1 ∩ Ni)(Mj ∩ Ni−1). Then by the first isomorphism theorem, we can know that Mj ∩ Ni (Mj ∩ Ni)Mj− ∼= 1 . (Mj−1 ∩ Ni)(Mj ∩ Ni−1) (Mj ∩ Ni−1)Mj−1 In the same way we can prove another part of Eq.(3.1). 4 Second part of Jordan-Holder Theorem Theorem 4.1 (Jordan Holder Theorem: Part Two). Let G be a finite group with G =6 1. Then the composition factors in a composition series are unique, namely, if 1= N0 < N1 < ··· < Nr = G, 1= M0 < M1 < ··· < Ms = G. are two composition series for G, then r = s and there is some permutation π of {1, 2, ··· ,r} such that ∼ Mπ(i)/Mπi−1 = Ni/Ni−1. (4.1) 3 Yingwei Wang Abstract Algebra Proof. We refine the first series by inserting between each of the r pairs Ni−1 < Ni the following: Ni−1 = Ni−1(Ni ∩ M0) ⊳ Ni−1(Ni ∩ M1) ⊳ ··· ⊳ Ni−1(Ni ∩ Ms) ⊳ Ni. Note that the subquotient Ni/Ni−1 is replaced by the s subquotients N − (N ∩ M ) i 1 i j . (4.2) Ni−1(Ni ∩ Mj−1) Similarly, the second series can be refined to replace each of the s subquotient Mj/Mj−1 by the n subquotients (N ∩ M )M − i j j 1 . (4.3) (Ni−1 ∩ Mj)Mj−1 By Eq.(3.1), we know that (4.2) is isomorphic to (4.3). So if we delete any repetition in the refinements, we know that the sequences of subquotient are isomorphic after some permutation. 5 Unique factorization Theorem 5.1 (Fundamental Theorem of Arithmetic). Let n> 1 be a positive integer. Then there exist unique primes p1 >p2 < ··· <pk and unique positive integers r1,r2 ··· ,rk such r1 rk that n = p1 ··· pk . Proof. Let G = Z/nZ, then G is a cyclic group of order n. Let d be the largest proper divisor of n and let G1 be the normal subgroup of G of size d. Then G/G1 is simple and cyclic, hence of prime order. Repeating this construction, we obtain a composition series of G, G = G0 ⊲ G1 ⊲ ··· ⊲ Gm =1, where for each i, Gi/Gi+1 is of prime order pi. Thus n = |G| = |G/G1||G1/G2|···|Gm−1/Gm||Gm| = p1p2 ··· pm−1(1). By Jordan-Holder theorem (Thm 4.1), we can get the uniqueness of the prime decom- position of n. 4 MA 553: Homework 2 Yingwei Wang ∗ Department of Mathematics, Purdue University, West Lafayette, IN, USA 1 About S4 and D8 Question: Exhibit explicitly a Sylow 2-subgroup of the symmetric group S4, and show that it is (up to isomorphism) the dihedral group D8. Solution: Let H be a Sylow 2-subgroup of the symmetric group S4, then we can choose H as follows H = {(1), (1234), (13)(24), (1432), (24), (13), (12)(34), (14)(23)}. Let D8 be the dihedral group of order 8, then 2 3 2 3 D8 = {e, r, r ,r ,s,sr,sr ,sr }. Define the isomorphism π : D8 → H by π(e) = (1), π(r) = (1234), π(r2) = (13)(24), π(r3) = (1432), π(s) = (24), π(sr) = (14)(23), π(sr2) = (13), π(sr3) = (12)(43). ∼ Then it is obvious that D8 = H. 2 Non-abelian group of order 8 Let G be a non-abelian group of order 8. ∗E-mail address: [email protected]; Tel: 765 237 7149 1 Yingwei Wang Abstract Algebra 2.1 (a) Question: Show that the center of G has order 2. Denote its generator by c. Solution: Let C be the center of G, then C ⊳ G and C is abelian. Since |G| = 23, by Theorem 8 (on DF, p.125), C is nontrivial. Besides, G is non-abelian. Then we know that the only possibility for |C| is |C| =2, 22. If |C| = 22, then G/C = 2. Hence G/C is cyclic. Let G/C =<aC >, then ∀x, y ∈ G, m n ∃m, n ∈ N such that x = a c1,y = a c2. Then m n m+n n m xy = a c1a c2 = a c1c2 = a c2a c1 = yx. It implies that G is abelian, which is a contradiction. So |C| = 2. Denote C = {1,c}. 2.2 (b) Question: If G contains an element z =6 c of order 2, then G is isomorphic to a subgroup of S4. Thus by Section 2.1, we know that G = D8. Solution: Let H = {1, z}, S = {all of the left cosets of H}. Then |H| = 2, |S| = 4. Consider the action of G on S: G × S → S, g × xH → gxH ∈ S. Then we find a group homomorphism σ : G → Permutation on S defined by [σ(g)](xH)= gxH. ∼ It indicates that G is a subgroup of S4. Since |G| = 8, we know that G = D8. 2.3 (c) Question: If c is the only order 2 element in G, then G is generated by two elements j and k of order 4 such that j2 = k2 and jk = kj−1. Hence G is isomorphic to the quaternion group, i.e. the multiplicative group of complex matrices generated by 0 1 0 i & , −1 0 −i 0 2 Yingwei Wang Abstract Algebra where i2 = −1. Solution: For ∀g ∈ G,g =6 c, the possible order for g is 2, 4, 8. But in this case c is the only order 2 element, and G is non-abelian, so we can conclude that the possible order for g is 4.