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8 1 BASIC DEFINITIONS Cosets Let H be a subgroup of G. Aleft coset of H in G is a set of the form aH = {ah | h ∈ H}, some fixed a ∈ G; aright coset is a set of the form some fixed a ∈ G. Ha = {ha | h ∈ H}, Example 1.13. Let G = R 2 , regarded as a group under addition, and let H be a subspace of dimension 1 (line through the origin). Then the cosets (left or right) of H are the lines parallel to H. Proposition 1.14. (a) If C is a left coset of H, anda ∈ C, thenC = aH. (b) Two left cosets are either disjoint or equal. (c) aH = bH if and only if a −1 b ∈ H. (d) Any two left cosets have the same number of elements (possibly infinite). Proof. (a) Because C is a left coset, C = bH some b ∈ G, and because a ∈ C, a = bh for some h ∈ H. Now b = ah −1 ∈ aH, and for any other element c of C, c = bh ′ = ah −1 h ′ ∈ aH. Thus,C ⊂ aH. Conversely,ifc ∈ aH, thenc = ah ′ = bhh ′ ∈ bH. (b) If C and C ′ are not disjoint, then there is an element a ∈ C ∩ C ′ ,andC = aH and C ′ = aH. (c) We have aH = bH ⇐⇒ b ∈ aH ⇐⇒ b = ah, forsomeh ∈ H, i.e., ⇐⇒ a −1 b ∈ H. (d) The map (ba −1 ) L : ah ↦→ bh is a bijection aH → bH. In particular, the left cosets of H in G partition G, and the condition “a and b lie in the same left coset” is an equivalence relation on G. The index (G : H) ofH in G is defined to be the number of left cosets of H in G. In particular, (G :1)istheorderofG. Eachleft coset of H has (H :1)elementsandG is a disjoint union of the left cosets. When G is finite, we can conclude: Theorem 1.15 (Lagrange). If G is finite, then (G :1)=(G : H)(H :1). In particular, the order of H divides the order of G. Corollary 1.16. The order of every element of a finite group divides the order of the group. Proof. Apply Lagrange’s theorem to H = 〈g〉, recalling that (H :1)=order(g).
Normal subgroups 9 Example 1.17. If G has order p, a prime, then every element of G has order 1 or p. But only e has order 1, and so G is generated by any element g ≠ e. In particular, G is cyclic, G ≈ C p . Hence, up to isomorphism, there is only one group of order 1, 000, 000, 007; in fact there are only two groups of order 1, 000, 000, 014, 000, 000, 049. Remark 1.18. (a) There is a one-to-one correspondence between the set of left cosets and the set of right cosets, viz, aH ↔ Ha −1 . Hence (G : H) is also the number of right cosets of H in G. But, in general, a left coset will not be a right coset (see 1.22 below). (b) Lagrange’s theorem has a partial converse: if a prime p divides m =(G :1), then G has an element of order p; ifp n divides m, thenG has a subgroup of order p n (Sylow’s theorem 5.2). However, note that C 2 × C 2 has order 4, but has no element of order 4, and A 4 has order 12, but it has no subgroup of order 6 (see Exercise 31). More generally, we have the following result. Proposition 1.19. Let G be a finite group. If G ⊃ H ⊃ K with H and K subgroups of G, then (G : K) =(G : H)(H : K). Proof. Write G = ⋃ g i H (disjoint union), and H = ⋃ h j K (disjoint union). On multiplying the second equality by g i , we find that g i H = ⋃ j g ih j K (disjoint union), and so G = ⋃ g i h j K (disjoint union). Normal subgroups When S and T are two subsets of a group G, welet ST = {st | s ∈ S, t ∈ T }. A subgroup N of G is normal, written N ⊳ G, ifgNg −1 = N for all g ∈ G. An intersection of normal subgroups of a group is normal (cf. 1.6). Remark 1.20. To show N normal, it suffices to check that gNg −1 ⊂ N for all g, because gNg −1 ⊂ N ⇒ g −1 gNg −1 g ⊂ g −1 Ng (multiply left and right with g −1 and g); hence N ⊂ g −1 Ng for all g, and, on rewriting this with g −1 for g, we find that N ⊂ gNg −1 for all g. The next example shows however that there can exist an N and a g suchthat gNg −1 ⊂ N, gNg −1 ≠ N (famous exercise in Herstein, Topics in Algebra, 2nd Edition, Wiley, 1975, 2.6, Exercise 8). Let G = GL 2 (Q), and let H = {( 1 01 n ) | n ∈ Z}. Th en H is a Example 1.21. subgroup of G; in fact it is isomorphic to Z. Letg =( 01). 50 Then ( ) ( )( ) ( 1 n 5 5n 5 g g −1 −1 0 1 5n = = 0 1 0 1 0 1 0 1 Hence gHg −1 ⊂ H, but gHg −1 ≠ H. ) .
Page 1 and 2: GROUP THEORY J.S. Milne Abstract Th
Page 3 and 4: ii Notations. We use the standard (
Page 5 and 6: 1 1 Basic Definitions Definitions D
Page 7 and 8: Subgroups 3 group. For example GL n
Page 9 and 10: Groups of small order 5 and G can b
Page 11: Homomorphisms 7 Homomorphisms Defin
Page 15 and 16: Quotients 11 Quotients The kernel o
Page 17 and 18: 13 2 Free Groups and Presentations
Page 19 and 20: Free groups 15 We say two words w,
Page 21 and 22: Generators and relations 17 Lemma 2
Page 23 and 24: Finitely presented groups 19 group
Page 25 and 26: 21 3 Isomorphism Theorems. Extensio
Page 27 and 28: Direct products 23 (b) H 1 ∩ H 2
Page 29 and 30: Automorphisms of groups 25 and so w
Page 31 and 32: Semidirect products 27 Remark 3.12.
Page 33 and 34: Semidirect products 29 Write G = N
Page 35 and 36: Extensions of groups 31 Extensions
Page 37 and 38: Exercises 13-19 33 Part A has been
Page 39 and 40: General definitions and results 35
Page 45 and 46: Permutation groups 41 Permutation g
Page 47 and 48: Permutation groups 43 Note that the
Page 49 and 50: Permutation groups 45 Note that A 4
Page 51 and 52: The Todd-Coxeter algorithm. 47 The
Page 53 and 54: Exercises 20-33 49 Proof. =⇒: The
Page 55 and 56: 51 5 The Sylow Theorems; Applicatio
Page 57 and 58: The Sylow theorems 53 the highest p
Page 59 and 60: Applications 55 (a) α(V i ) ⊂ V
Page 61 and 62: Applications 57 Now assume that P i
Page 63 and 64:
59 6 Normal Series; Solvable and Ni
Page 65 and 66:
Solvable groups 61 Similarly G/H 1
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Solvable groups 63 (b) Let N be a n
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Nilpotent groups 65 Nilpotent group
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Nilpotent groups 67 Corollary 6.16.
Page 73 and 74:
Groups with operators 69 Example 6.
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Further reading 71 Theorem 6.29 (Kr
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73 10. The hint gives ab 3 a −1 =
Page 79 and 80:
75 Deduce that #C G (α) = 3, and s
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77 B Review Problems 34. Prove that
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79 (e) Describe all subgroups of G
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81 73. (a) Prove that subgroups A a
Page 87 and 88:
Solutions 83 Solutions 1. (a) False
Page 89 and 90:
Index action doubly transitive, 36
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