# Group Theory. Hwan Yup Jung. Department of Mathematics Education, Chungbuk National University

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1 Group Theory Hwan Yup Jung Department of Mathematics Education, Chungbuk National University Hwan Yup Jung (CBNU) Group Theory March 1, / 111

2 Groups Definition A group is a set G with a binary operation : G G G, (a, b) a b which satisfies the following properties: 1 (Closed) For any a, b G, a b G. 2 (Associative) For any a, b, c G, (a b) c = a (b c). 3 (Identity element) There exists an element e G such that e a = a e = a for all a G. Such element e is called an identity element of G. 4 (Inverse) For any a G, there exists an element a G such that a a = a a = e. In this case, a is called an inverse of a with respect to. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

3 Groups Definition A group (G, ) is called an abelian if is commutative, i.e. a b = b a for all a, b G. For a set X, X := the cardinality of X. Definition 1 A group G is called a finite group if G < and an infinite group otherwise. 2 When G is a finite group, G is called the order of G. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

4 Groups Theorem Let (G, ) be a group. 1 There exists a unique identity element in G. 2 Each element of G has a unique inverse in G. Proof. 1 If e and e are two identity elements of (G, ), e ( ) = e e ( ) = e. We view e as identity element in ( ). We view e as identity element in ( ). 2 If a and a are two inverses of a in G, a = a e = a (a a ) = (a a) a = e a = a. Notation. From now on we denote by a 1 the inverse of a in a group G. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

5 Groups Corollary Let (G, ) be a group. For all a, b G, we have 1 (a b) 1 = b 1 a 1, 2 (a 1 ) 1 = a. Proof. 1 For all a, b G, (a b) (b 1 a 1 ) = a (b b 1 ) a = a a 1 = e b 1 a 1 = (a b) 1. 2 For all a G, a a 1 = a 1 a = e a is the inverse of a 1. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

6 Groups Theorem The left and right cancelation laws hold in a group (G, ), i.e., for any a, b, c G 1 If a b = a c, then b = c (Left cancelation law). 2 If b a = c a, then b = c. (Right cancelation law). Proof. Suppose that a b = a c. a 1 (a b) = a 1 (a c) (a 1 a) b = (a 1 a) c e b = e c b = c. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

7 Groups Theorem In a group (G, ), 1 a x = b has unique solution x = a 1 b. 2 y a = b has unique solution y = ba 1. Proof. a x = b a 1 (a x) = a 1 b (a 1 a) x = a 1 b e x = a 1 b x = a 1 b. x = a 1 b is the unique solution of a x = b. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

8 Groups Example. 1 (N, +) is not a group. 2 (Z 0, +) is not a group. 3 (Z, +) is an infinite abelian group. 4 (Q, +), (R, +) and (C, +) are infinite abelian groups. 5 (Q, ), (R, ) and (C, ) are infinite abelian groups. 6 (U = {z C : z = 1}, ) is an infinite abelian group. 7 (U n = {z C : z n = 1}, ) is a finite abelian group of order n. 8 Let V be a vector space. Then (V, +) is an abelian group. 9 Let M m n (R) be the set of all m n matrices with real entries. Then (M m n (R), +) is an abelian group. 10 (M n n (R), ) is not group. 11 Let GL n (R) be the set of all invertible n n matrices with real entries. Then (GL n (R), ) is a group, but not abelian. The group GL n (R) is called the general linear group of degree n. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

9 Groups Example. For any a, b Z, a b mod n n (a b). This is an equivalence relation on Z. The equivalence class of a is ā = a + nz = {a + kn : k Z}. Let Z n = {ā : 0 a n 1} be the set of all equivalence classes. We define addition + on Z n as follows: for any a, b Z n, a + b = a + b. Then (Z n, +) is an abelian group of order n. (Well defined) Suppose a 1 = a 2 and b 1 = b 2. a 1 a 2 mod n, b 1 b 2 mod n a 1 + b 1 a 2 + b 2 mod n (Closed), (Associative) OK (Identity) 0 is the identity. (Inverse) ( a) is the inverse of a. a 1 + b 1 = a 2 + b 2. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

10 Subgroups Definition Let (G, ) be a group and H G. H is called a subgroup of G if (H, ) forms a group. In this case we write H < G. Remark. H < G if and only if (H, ) satisfies the followings: 1 x y H for all x, y H. 2 The identity element e H. 3 x 1 H for all x H. Theorem Let (G, ) be a group and H G. Then H < G if and only if x y 1 H for all x, y H. Proof. ( ) We will show that H satisfies the conditions in above Remark. For any x, y H, x x 1 = e H; e x 1 = x 1 H; x (y 1 ) 1 = x y H. ( ) For all x, y H; x, y 1 H x y 1 H. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

11 Subgroups Definition Let G be a group. 1 Any subgroup H G is called a proper subgroup and G itself the improper subgroup of G. 2 The subgroup {e} is called the trivial subgroup and all the other subgroups are called nontrivial. Example. 1 H = {(x 1, x 2,..., x n ) R n : x 1 = 0} is a subgroup of (R n, +). 2 U n is a subgroup of (C, ). 3 Let SL n (C) = {A GL n (C) : det(a) = 1}. Then SL n (C) < GL n (C). 4 nz < (Z, +) for any n Z; in fact, any subgroup of Z is of this form. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

12 Subgroups Definition Let (G, ) be a group. For any g G, we define g 0 = e and for n > 0, g n = g g g; }{{} g n = g 1 g 1 g 1. }{{} n-times n-times It is easy to check that for any integers m and n, we have g n g m = g n+m ; (g n ) 1 = g n. Theorem Let G be a group and g G. Then H = {g n : n Z} is the smallest subgroup of G containing g. (It is called the cyclic subgroup of G generated by g and denoted by g.) Proof. For any x, y H, x = g m, y = g n ( m, n Z) x y 1 = g m g n = g m n H Hence, H < G. For K < G with g K; g, g 1 K g n K ( n Z). Hence, H < K. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

13 Subgroups Definition 1 We say that g generates G (or g is a generator of G) if g = G. 2 A group G is called a cyclic group if G = g for some g G. Example. 1 Z is a cyclic group: Z = 1 = 1. 2 Z n = 1 is a cyclic group of order n. Definition Let G be a group with the identity element e. 1 g G is an element of finite order if g m = e for some positive integer m. In this case, the smallest positive integer m such that g m = e is called the order of g and denoted by (g). 2 Otherwise g is called an element of infinite order. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

14 Cyclic Groups Lemma Suppose that g G is an element of finite order m. Then 1 g n 1 = g n 2 n 1 n 2 mod m. 2 g n = e m n. 3 g = {e, g, g 2,..., g m 1 }. Proof. HW Lemma Suppose that g G is an element of infinite order. Then 1 g n 1 = g n 2 n 1 = n 2. 2 g = {, g 2, g 1, e, g, g 2, }. Proof. HW Hwan Yup Jung (CBNU) Group Theory March 1, / 111

15 Cyclic Groups Theorem Every cyclic group is abelian. Proof. Let G be a cyclic group with a generator g. For any x, y G, x = g m and y = g n for some integers m, n xy = g m g n = g m+n = g n g m = yx. Therefore, G is an abelian group. Theorem Any subgroup of a cyclic group is cyclic. Proof. Let G be a cyclic group with a generator g, and H < G. 1 If H = {e}, then H = e is a cyclic. 2 Assume that H {e}. Let m be the smallest positive integer such that g m H. Claim : H = g m. Corollary The subgroups of (Z, +) are precisely the groups nz for n Z. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

16 Cyclic Groups Theorem Let G = g be a finite cyclic group of order m. Then (g s ) = Proof. m gcd(m, s). (g s ) = min{k N : (g s ) k = g sk = e} = min{k N : m sk} { ( m ) ( s ) } m = min k N : k = gcd(m, s) gcd(m, s) gcd(m, s). Corollary If g is a generator of a finite cyclic group G of order m, then the other generators of G are the elements of the form g r, where r is relatively prime to m. Proof. For any g r G, g r is a generator of G g r = g gcd(m, r) = gcd(m, 1) = 1. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

17 Cyclic Groups Corollary Let G = g be a finite cyclic group of order m. Then g s = g t gcd(m, s) = gcd(m, t). Proof. ( ) Suppose that g s = g t. g s = g t m gcd(m,s) = m gcd(m,t) gcd(m, s) = gcd(m, t). ( ) Suppose that gcd(m, s) = gcd(m, t) = d. m = m 1 d, s = s 1 d, t = t 1 d with gcd(m 1, s 1 ) = gcd(m 1, t 1 ) = 1. am 1 + bs 1 = 1 and em 1 + ft 1 = 1 for some integers a, b, e, f. g s = g s(em1+ft1) = g ms1e g ts1f = (g t ) s1f g t g t = g t(am1+bs1) = g mt1a g st1a = (g s ) t1a g s. Hence, g t = g s. Corollary Let G be a finite cyclic group. For any subgroups H, K < G, H = K H = K. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

18 Cyclic Groups Corollary Let G = g be a finite cyclic group of order m. Let S G be the set of all subgroups of G and D m be the set of all positive divisors of m. Then there is a bijection between S G and D m. Explicitly, Ψ : S G D m, H Ψ(H) = H is a bijection with the inverse map Ψ 1 : D m S G by Ψ 1 (d) = g m/d. Proof. Let Φ : D m S G be defined by Φ(d) = g m/d. 1 For any positive divisor d of m, Ψ(Φ(d)) = Ψ( g m/d ) = g m/d = m gcd(m,m/d) = d. 2 For any subgroup H of G with H = d, since g m/d = d = H, g m/d = H. Φ(Ψ(H)) = Φ(d) = g m/d = H. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

19 Cyclic Groups Example. Consider the cyclic group Z 12 = 1. Z 12 has exactly one subgroup H d of order d, where d = 1, 2, 3, 4, 6 or 12. H 1 = 0 = {0}. H 2 = 6 = {0, 6}. H 3 = 4 = {0, 4, 8} = 8. H 4 = 3 = {0, 3, 6, 9} = 9. H 6 = 2 = {0, 2, 4, 6, 8, 10} = 10. H 12 = 1 = 5 = 7 = 11 = Z 12. H 12 = 1 H 6 = 2 H 4 = 3 H 3 = 4 H 2 = 6 H 1 = 0 Hwan Yup Jung (CBNU) Group Theory March 1, / 111

20 Groups of permutations Definition Any bijective map φ : A A is called a permutation of a set A. We denote by S A the set of all permutations of A. Theorem Let A be a nonempty set. Then S A forms a group with composition. Proof. We have to show that (S A, ) satisfies the group axioms. (Closed), (Associative) OK. (Identity) The identity map id A : A A is the identity element of S A. (Inverse) σ 1 S A for any σ S A. Definition When A = {1, 2, 3,..., n}, S A is called the symmetric group on n letters, and is denoted by S n. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

21 Groups of permutations For any σ S n, we write ( ) 1 2 n 1 n σ =. σ(1) σ(2) σ(n 1) σ(n) Example. In S 5, let ( ) σ =, τ = ( ) Then στ = = ( ) Hwan Yup Jung (CBNU) Group Theory March 1, / 111

22 Groups of permutations Example. S 3 = {ρ 0, ρ 1, ρ 2, µ 1, µ 2, µ 3 }, where ρ 0 := ( ) = id and ρ 1 := ( ), 120 -Counterclockwise rotation. ρ 2 := ( ), 240 -Counterclockwise rotation. µ 1 := ( ), Reflection with respect to 1. µ 2 := ( ), Reflection with respect to 2. µ 3 := ( ), Reflection with respect to 3. ρ 0 ρ 1 ρ 2 µ 1 µ 2 µ 3 ρ 0 ρ 0 ρ 1 ρ 2 µ 1 µ 2 µ 3 ρ 1 ρ 1 ρ 2 ρ 2 µ 1 µ 1 µ 2 µ 2 µ 3 µ 3 S 3 {ρ 0, µ 1 } {ρ 0, µ 2 } {ρ 0, µ 3 } {ρ 0, ρ 1, ρ 2 } {ρ 0 } Hwan Yup Jung (CBNU) Group Theory March 1, / 111

23 Groups of permutations Example. The 4-th dihedral group D 4 = {ρ 0, ρ 1, ρ 2, ρ 3, µ 1, µ 2, δ 1, δ 2 }. ρ 0 := ( ). ρ 1 := ( ), 90 -Counterclockwise rotation. ρ 2 := ( ), 180 -Counterclockwise rotation. ρ 3 := ( ), 270 -Counterclockwise rotation. µ 1 := ( ), Reflection µ 2 := ( ), Reflection δ 1 := ( ), Reflection δ 2 := ( ), Reflection Hwan Yup Jung (CBNU) Group Theory March 1, / 111

24 Groups of permutations ρ 0 ρ 1 ρ 2 ρ 3 µ 1 µ 2 δ 1 δ 2 ρ 0 ρ 0 ρ 1 ρ 2 ρ 3 µ 1 µ 2 δ 1 δ 2 ρ 1 ρ 1 ρ 2 ρ 2 ρ 3 ρ 3 µ 1 µ 1 µ 2 µ 2 δ 1 δ 1 δ 2 δ 2 D 4 {ρ 0, ρ 2, µ 1, µ 2 } {ρ 0, ρ 1, ρ 2, ρ 3 } {ρ 0, ρ 2, δ 1, δ 2 } {ρ 0, µ 1 } {ρ 0, µ 2 } {ρ 0, ρ 2 } {ρ 0, δ 1 } {ρ 0, δ 2 } {ρ 0 } Hwan Yup Jung (CBNU) Group Theory March 1, / 111

25 Groups of permutations Definition A function φ : G G is called a homomorphism if φ(ab) = φ(a)φ(b) for all a, b G. Lemma Let φ : G G be a homomorphism. 1 φ(e) = e. 2 φ(x 1 ) = φ(x) 1 for any x G. 3 φ(g) is a subgroup of G. Proof. (i) φ(e) = φ(ee) = φ(e)φ(e) φ(e) = e. (ii) For any x G, e = φ(e) = φ(xx 1 ) = φ(x)φ(x 1 ) φ(x 1 ) = φ(x) 1. (iii) For any x, y φ(g), x = φ(x) and y = φ(y) for some x, y G x y 1 = φ(x)φ(y) 1 = φ(x)φ(y 1 ) = φ(xy 1 ) φ(g) Hence, φ(g) is a subgroup of G. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

26 Groups of permutations Definition 1 Any bijective homomorphism φ : G G is called an isomorphism. 2 Two groups G and G are isomorphic if there exists an isomorphism of G onto G. Corollary Let φ : G G be a one-to-one homomorphism. Then φ : G φ(g) is an isomorphism of G onto φ(g). Hwan Yup Jung (CBNU) Group Theory March 1, / 111

27 Groups of permutations Theorem (Cayley) Every group is isomorphic to a group of permutations. Proof. Let G be any group and S G be the group of all permutations of G. We claim that G is isomorphic to a subgroup of S G. For each g G, we associate λ g S G defined by λ g (x) = gx. (In fact, λ 1 g = λ g 1 is the inverse of λ g ) φ : G S G, g φ(g) = λ g is an one-to-one homomorphism. Hence, G is isomorphic to φ(g), which is a subgroup of S G. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

28 Orbits, Cycles, and the Alternating groups Definition For anyσ S A, we define an equivalence relation σ on A as follows: a σ b def b = σ n (a) for some n Z. The equivalence class of a under σ is which is called an orbit of σ. Example. {b A : b σ a} = {σ n (a) : n Z}, 1 For each a A, the orbit of a under 1A is {1 n A (a) : n Z} = {a}. 2 Find the orbits of the permutation σ = ( ). Solution. 1 σ 3 σ 6 σ 1 σ 2 σ 8 σ 2 σ 4 σ 7 σ 5 σ 4 σ Thus, the orbits of σ are given by {1, 3, 6}, {2, 8}, {4, 7, 5}. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

29 Orbits, Cycles, and the Alternating groups Definition 1 A permutation σ S n is called a cycle if it has at most one orbit containing more than one element. 2 The length of a cycle is the number of elements in its largest orbit. Example. 1 In S 8, consider δ = ( ). The orbits of δ are {1, 3, 6}, {2}, {8}, {4}, {7}, {5}; δ is a cycle of length 3; In this case we write δ = (1, 3, 6). 2 Consider τ = ( ). The orbits of σ are {2, 8}, {1}, {3}, {4}, {5}, {6}, {7}; τ is a cycle of length 2; In this case we write τ = (2, 8). 3 Consider µ = ( ). The orbits of µ are {4, 7, 5}, {1}, {2}, {3}, {6}, {8}; µ is a cycle of length 3; In this case we write µ = (4, 7, 5). Hwan Yup Jung (CBNU) Group Theory March 1, / 111

30 Orbits, Cycles, and the Alternating groups Definition Two cycles (a 1, a 2,... a s ) and (b 1, b 2,..., b t ) in S n are said to be disjoint if a i b j for all i, j. Example. In above Examples, δ, τ and µ are mutually disjoint. Lemma Let σ = (a 1, a 2,..., a s ) be a cycle in S n. Then for any k {1, 2,..., n}, we have a i+1, if k = a i and i < s, σ(k) = a 1, if k = a s, k, if k a i for all i. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

31 Orbits, Cycles, and the Alternating groups Lemma If σ and τ are disjoint cycles in S n, then στ = τσ. Proof. Let σ = (a 1, a 2,... a s ) and τ = (b 1, b 2,..., b t ). By hypothesis, a i b j for all i, j. We have to show that στ(k) = τσ(k) for all 1 k n. If k = a i for some i, στ(k) = στ(a i ) = σ(a i ), τσ(k) = τσ(a i ) = σ(a i ). If k = b j for some j, στ(k) = στ(b j ) = τ(b j ), τσ(k) = τσ(b j ) = τ(b j ). If k a i, b j for all i, j, στ(k) = k = τσ(k). Hwan Yup Jung (CBNU) Group Theory March 1, / 111

32 Orbits, Cycles, and the Alternating groups Theorem Every permutation σ of a finite set is a product of disjoint cycles. Proof. Suppose that σ S n. Let B 1, B 2,..., B r be the orbits of σ. For each 1 i r, let µ i be the cycle defined by { σ(x), if x B i µ i (x) = x, if x B i. µ i and µ j are disjoint for i j (since B i B j = ). For any x {1, 2,..., n}, x B i (!i) µ j (x) = x, µ j (µ i (x)) = µ i (x) ( j i) µ 1 µ 2 µ r (x) = µ i (x) = σ(x). Hence, σ = µ 1 µ 2 µ r. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

33 Orbits, Cycles, and the Alternating groups Example. 1 Consider σ = ( ) σ 3 σ 6 σ 1, 2 σ 2, 4 σ 7 σ 4, 5 σ 5, 8 σ 8 The orbits of σ are {1, 3, 6}, {2}, {4, 7}, {5}, {8} and 2 Consider σ = σ = (1, 3, 6)(2)(4, 7)(5)(8) = (1, 3, 6)(4, 7). ( ) σ 6 σ 1, 2 σ 5 σ 3 σ 2, 4 σ 4 The orbits of σ are {1, 6}, {2, 5, 3}, {4} and σ = (1, 6)(2, 5, 3)(4) = (1, 6)(2, 5, 3). Hwan Yup Jung (CBNU) Group Theory March 1, / 111

34 Orbits, Cycles, and the Alternating groups Definition A cycle of length 2 is called a transposition. Lemma Any cycle can be written as a product of transpositions. Proof. It follows from the fact that (a 1, a 2,..., a r ) = (a 1, a r )(a 1, a r 1 ) (a 1, a 2 ). Corollary Any permutation of a finite set of at least two elements is a product of transpositions. Example. 1 (1, 6)(2, 5, 3) = (1, 6)(2, 3)(2, 5). 2 In S n, for n 2, the identity permutation is the product (1, 2)(1, 2) of transpositions. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

35 Orbits, Cycles, and the Alternating groups Theorem No permutations in S n can be written both as a product of an even number of transpositions and as a product of an odd number of transpositions. Proof. We remark that S A S B if A = B. We work with S A, where A = {e 1, e 2,..., e n } is the set of the n rows of the n n identity matrix I n. Let σ S A and e 1 σe 1 e 2 σ σe 2 I n = C =... e n σe n If σ = (i, j) is a transposition, then C is the matrix obtained from I n by interchange the i-th row and j-th row. det(c) = det(in ) = 1. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

36 Orbits, Cycles, and the Alternating groups If σ = τ r τ r 1 τ 1, where τ i is a transposition, then τ I 1 τ n 2 τ C1 3 τ C2 r Cr = C. det(c) = ( 1) r det(i ) = ( 1) r. Assume σ = τ r τ r 1 τ 1 = µ s µ s 1 µ 1, where τ i, µ j are a transposition. det(c) = ( 1) r = ( 1) s r s mod 2. Definition A permutation of a finite set is even (resp. odd) if it can be written as a product of an even (resp. odd) number of transpositions. Example. Since (1, 4, 5, 6)(2, 1, 5) = (1, 6)(1, 5)(1, 4)(2, 5)(2, 1), the permutation (1, 4, 5, 6)(2, 1, 5) in S 6 is odd. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

37 Orbits, Cycles, and the Alternating groups Theorem If n 2, then the collection of all even permutations of {1, 2, 3,..., n} forms a subgroup of order n!/2 of the symmetric group S n. Proof. Let A n be the subset of all even permutations in S n, and B n be the set of all odd permutations in S n. A n is a subgroup of S n. Let τ be a fixed transposition in S n. Then ρ τ : A n B n, τ τσ is a bijection with the inverse map ρ τ 1 : B n A n, σ τ 1 σ. An = B n, A n = Sn 2 = n! 2. Definition The subgroup of S n consisting of the even permutations of n letters is called the alternating group A n on n letters. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

38 Costs and The Theorem of Lagrange Definition Let H < G. We define two relations L and R on G as follows: for any a, b G, 1 a L b a 1 b H b = ah ( h H). 2 a R b ba 1 H b = ha ( h H). Theorem 1 The relations L and R on G are both equivalence relations on G. 2 For each a G, the equivalence class of a with respect to L and R are given by {b G : a L b} = {ah : h H} = ah. {b G : a R b} = {ha : h H} = Ha. Here, ah := {ah : h H} is called the left coset of H containing a and Ha := {ha : h H} is called the right coset of H containing a. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

39 Costs and The Theorem of Lagrange Example. Exhibit the left cosets (right cosets) of the subgroup 3Z of Z. a L b a + b 3Z b a mod 3 ā = b. There are three left cosets of the subgroup 3Z of Z. 0 = 0 + 3Z = {..., 9, 6, 3, 0, 3, 6, 9,...} 1 = 1 + 3Z = {..., 8, 5, 2, 1, 4, 7, 10,...} 2 = 2 + 3Z = {..., 7, 4, 1, 2, 5, 8, 11,...}. Example. The group Z 6 is an abelian group, and H = {0, 3} < Z H = {0, 3} = 3 + H = H 1 + H = {1, 4} = 4 + H 2 + H = {2, 5} = 5 + H. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

40 Costs and The Theorem of Lagrange Example. Let H = µ 1 = {ρ 0, µ 1 } < S 3. ρ 0 H = {ρ 0, µ 1 } = µ 1 H ρ 1 H = {ρ 1, ρ 1 µ 1 } = {ρ 1, µ 3 } = µ 3 H ρ 2 H = {ρ 2, ρ 2 µ 1 } = {ρ 2, µ 2 } = µ 2 H, Hρ 0 = {ρ 0, µ 1 } = Hµ 1 Hρ 1 = {ρ 1, µ 1 ρ 1 } = {ρ 1, µ 2 } = Hµ 2 Hρ 2 = {ρ 2, µ 1 ρ 2 } = {ρ 2, µ 3 } = Hµ 3. Lemma All left, right cosets of H in G have the same cardinality. Proof. For any g G, we define two maps φ g : H gh, h gh, ψ g : H Hg, h hg. φ g and ψ g are both bijective, so gh = H = Hg. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

41 Costs and The Theorem of Lagrange Theorem (Lagrange) Let H be a subgroup of a finite group G. Then H is a divisor of G. Moreover, G H is equal to the number of distinct left (right) cosets of H. Proof. Let g 1 H,..., g r H be all the distinct left cosets of H. r r r G = g i H (disjoint union) G = g i H = H = r H Corollary i=1 Every group of prime order is a cyclic. i=1 i=1 H is a divisor of G. Proof. Let G be a finite group of prime order p. We claim that G = a for any a G, a e. a is a divisor of p = G and a 2 a = p = G a = G Hwan Yup Jung (CBNU) Group Theory March 1, / 111

42 Costs and The Theorem of Lagrange Theorem Let G be a finite group and g G. Then (g) is a divisor of G, in particular, g G = e. Proof. By Lagrange Theorem, (g) = g is a divisor of G g G = ( g (g)) G / (g) = e. Lemma Let H < G. Let L and R be the set of all left cosets and right cosets of H in G, respectively. Then L and R have the same cardinality. Proof. The map φ : L R defined by φ(ah) = Ha 1 is a bijection. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

43 Costs and The Theorem of Lagrange Definition Let H < G. The cardinality of L (or R)is called the index of H in G and denoted by (G : H) Remark. 1 The index (G : H) may be finite or infinite. 2 If G is finite, then (G : H) is finite and (G : H) = G H. Theorem Let K H G. Suppose that (H : K) and (G : H) are both finite. Then (G : K) is also finite and (G : K) = (G : H)(H : K). G (G:H) (G:K) H K Hwan Yup Jung (CBNU) Group Theory March 1, / 111 (H:K)

44 Direct Products Theorem Let G 1, G 2,..., G n be groups. Then n i=1 G i = G 1 G 2 G n forms a group with the binary operation (a 1,..., a n )(b 1,..., b n ) = (a 1 b 1,..., a n b n ), and is called the direct product of the groups G i. Example. 1 If each G i is abelian group, then n i=1 G i is also an abelian group. 2 Z 2 Z 3 is a cyclic group, in fact, Z 2 Z 3 = ( 1, 1). ( 1, 1) = ( 1, 1), 2( 1, 1) = ( 0, 2), 3( 1, 1) = ( 1, 0), 4( 1, 1) = ( 0, 1), 5( 1, 1) = ( 1, 2), 6( 1, 1) = ( 0, 0). 3 Z 3 Z 3 is not a cyclic group. For any (a, b) Z3 Z 3, 3(a, b) = (0, 0). There are no element of order 9 in Z3 Z 3. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

45 Direct Products Theorem Z m Z n is cyclic group if and only if gcd(m, n) = 1. In this case, Z m Z n = Zmn. Proof. ( ) Suppose that gcd(m, n) = 1. We claim Z m Z n = ( 1, 1). d = the order of ( 1, 1) in Z m Z n = the smallest positive integer satisfying f ( 1, 1) = ( f, f ) = ( 0, 0) = the smallest positive integer satisfying m d and n d = lcm(m, n) = mn. ( ) Suppose that d = gcd(m, n) > 1, and let f = lcm(m, n) = mn/d < mn. For any (ā, b) Z m Z n, f (ā, b) = (fa, fb) = ( 0, 0). Any element of Z m Z n has order < mn. Hence, Z m Z n is not a cyclic group. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

46 Direct Products Corollary n i=1 Z m i is a cyclic if and only if gcd(m i, m j ) = 1 for all i j. In this case, n i=1 Z m i = Zm1 m 2 m n. In particular, if n = p e 1 1 per r, we have Z n = Zp e 1 Z 1 p er. r Theorem Let (a 1, a 2,..., a n ) n i=1 G i. If each a i is of finite order r i in G i, then the order of (a 1, a 2,..., a n ) in n i=1 G i is lcm(r 1, r 2,..., r n ). Proof. d = the order of (a 1,..., a n ) in G i = the smallest positive integer such that (a d 1,..., a d n) = (e 1,..., e n ) = the smallest positive integer such that a d i = e i for 1 i n = the smallest positive integer such that r i d for 1 i n = lcm(r 1, r 2,..., r n ) Hwan Yup Jung (CBNU) Group Theory March 1, / 111

47 Direct Products Example. Consider the element ( 8, 4, 10) in Z 12 Z 60 Z 24. The order of 8 in Z 12 is 12 gcd(12,8) = 3. The order of 4 in Z 60 is 60 gcd(60,4) = 15. The order of 10 in Z 24 is 24 gcd(24,10) = 12. Hence, the order of ( 8, 4, 10) in Z 12 Z 60 Z 24 is lcm(3, 15, 12) = 60. Lemma Let G be an abelian group and x 1, x 2,..., x r G. Then H = {x n 1 1 x n 2 2 x nr r : n 1, n 2,..., n r Z} is a subgroup of G, called the subgroup of G generated by x 1, x 2,..., x r and denoted by x 1, x 2,..., x r. Proof. For x, y H, x = x n 1 1 x r nr and y = x m 1 1 xr mr xy 1 = x n 1 m 1 nr mr 1 xr H. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

48 Homomorphisms Definition Let G and G be groups. A map φ : G G is called a homomorphism if for all a, b G. Remark. φ(ab) = φ(a)φ(b) ( ) 1 In ( ), the product ab on the left-hand side takes place in G, while the product φ(a)φ(b) on the right-hand side takes place in G. 2 For any groups G and G, there is always at least one homomorphism φ : G G, g e, called the trivial homomorphism, where e is the identity element of G. 3 If φ : G G and γ : G G are both homomorphisms, then their composition γ φ : G G is also a homomorphism. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

49 Homomorphisms Theorem Let φ : G G be a homomorphism. Then 1 φ(e) = e. 2 If a G, then φ(a 1 ) = φ(a) 1. 3 If H < G, then φ(h) < G. 4 If K < G, then φ 1 (K ) < G. Proof. (iv) For a, b φ 1 (K ), φ(ab 1 ) = φ(a)φ(b 1 ) = φ(a)φ(b) 1 K ab 1 φ 1 (K ). Definition For a homomorphism φ : G G, the subgroup φ 1 ({e }) = {a G : φ(a) = e } < G is called the Kernel of φ, and denoted by Ker(φ). Hwan Yup Jung (CBNU) Group Theory March 1, / 111

50 Homomorphisms Theorem Let φ : G G be a homomorphism with H = Ker(φ). Then for any a G, we have φ 1 ({φ(a)}) = ah = Ha. proof. We only show that φ 1 ({φ(a)}) = ah. ( ) For x φ 1 ({φ(a)}), φ(x) = φ(a) φ(a) 1 φ(x) = φ(a 1 x) = e a 1 x H = Ker(φ) x = ah ( h H). x ah. ( ) For y ah, y = ah ( h H) φ(y) = φ(ah) = φ(a)φ(h) = φ(a)e = φ(a) y φ 1 ({φ(a)}). Hwan Yup Jung (CBNU) Group Theory March 1, / 111

51 Homomorphisms Corollary A homomorphism φ : G G is injective if and only if Ker(φ) = {e}. Proof. ( ) φ is injective + φ(e) = e Ker(φ) = {e}. ( ) Suppose that Ker(φ) = {e}. For a, b G with φ(a) = φ(b), φ(a) = φ(b) φ(a)φ(b) 1 = φ(ab 1 ) = e ab 1 Ker(φ) = {e} ab 1 = e a = b. Definition A subgroup H < G is called a normal subgroup if ah = Ha for all a G, and we write H G. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

52 Homomorphisms Theorem For H < G, the following conditions are equivalent: 1 H G. 2 ghg 1 H for all g G and h H. 3 ghg 1 = H for all g G (ghg 1 = {ghg 1 : h H}). 4 gh = Hg for all g G. Proof. (i) (iv) by definition. (i) (ii) Let H G. For any g G and h H, gh gh = Hg gh = h g ( h H) ghg 1 = h H. (ii) (iii) Assume that ghg 1 H for all g G. H = g 1 (ghg 1 )g g 1 Hg ( g G) H ghg 1 ( g G) ghg 1 = H ( g G). Hwan Yup Jung (CBNU) Group Theory March 1, / 111

53 Homomorphisms (iii) (iv) Assume that gh = Hg for all g G. ghg 1 = H ( g G) (ghg 1 )g = Hg ( g G) Corollary gh = Hg ( g G). If φ : G G is a homomorphism, then Ker(φ) is a normal subgroup of G. Corollary Every subgroup of an abelian group is normal. Example. Let φ : G G be a surjective homomorphism. If G is abelian, then G must be also abelian. Proof. For any a, b G, φ(a) = a and φ(b) = b for some a, b G a b = φ(a)φ(b) = φ(ab) = φ(ba) = φ(b)φ(a) = b a. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

54 Homomorphisms Example. Let φ : S n Z 2 be defined by { 0 if σ A n, φ(σ) = 1 if σ S n A n. Then φ is a surjective homomorphism with Ker(φ) = A n. Proof. We have to show that φ(στ) = φ(σ) + φ(τ) for all σ, τ S n. σ τ στ φ(στ) φ(σ) + φ(τ) even even even = 0 even odd odd = 1 odd even odd = 1 odd odd even = 0 By definition, we have Ker(φ) = A n. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

55 Factor Groups Let H < G and G/H := {gh : g G}. 1 We want to define an operation on G/H by (ah)(bh) = (ab)h for all ah, bh G/H. 2 To be well defined, if ah = a H and bh = b H, then it must be satisfied (ab)h = (a b )H. Theorem Let H < G. Then the operation on G/H is well defined if and only if H G. (ah)(bh) = (ab)h Hwan Yup Jung (CBNU) Group Theory March 1, / 111

56 Factor Groups Proof. ( ) Assume that the operation (ah)(bh) = (ab)h is well defined. It means that if ah = a H and bh = b H, then (ab)h = (a b )H. We want to show that ah = Ha for all a G. ah Ha ẋ ah xh = ah (xa 1 )H = (xh)(a 1 H) = (ah)(a 1 H) = H xa 1 H x Ha Similarly, Ha ah. ah = Ha for all a G. ( ) Assume that H G. ah = a H and bh = b H a = ah, b = bk ( h, k H) a b = (ah)(bk) = a(hb)k hb = bh ( h H) (since Hb = bh) a b = a(bh )k = (ab)h k (a b )H = (ab)h Hwan Yup Jung (CBNU) Group Theory March 1, / 111

57 Factor Groups Corollary Let H G. Then G/H forms a group under the binary operation (ah)(bh) = (ab)h. Proof. We have to show that G/H satisfies the group axioms with respect to the operation (ah)(bh) = (ab)h. (Closed), (Associative) OK. (Identity) H = eh = He is the identity element. (Inverse) For any ah G/H, a 1 H is the inverse of ah. Definition The group G/H in the preceding Corollary is called the factor group (or quotient group) of G by H. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

58 Factor Groups Lemma Let H G. Then the map γ : G G/H, is a surjective homomorphism with kernel H. Theorem g gh Let φ : G G be a homomorphism with H = Ker(φ). Then the map φ : G/H G, ah φ(a) is an injective homomorphism satisfying φ γ = φ. G φ G γ G/H φ Hwan Yup Jung (CBNU) Group Theory March 1, / 111

59 Factor Groups Proof. Let φ : G/H G be defined by φ(ah) = φ(a). 1 (well defined) If ah = bh, ab 1 H = Ker(φ) φ(ab 1 ) = e φ(a) = φ(b). 2 (homomorphism) For any ah, bh G/H, φ((ah)(bh)) = φ((ab)h) = φ(ab) = φ(a)φ(b) = φ(ah) φ(bh). 3 (injective) If ah Ker( φ), φ(ah) = φ(a) = e a H = Ker(φ) ah = H 4 ( φ γ = φ) For any a G, φ γ(a) = φ(ah) = φ(a). Hwan Yup Jung (CBNU) Group Theory March 1, / 111

60 Factor Groups Theorem (Fundamental Homomorphism Theorem) Let φ : G G be a homomorphism with H = Ker(φ). Then there exists a unique injective homomorphism φ : G/H G satisfying φ γ = φ. In particular, φ : G/H φ(g) is an isomorphism. γ G G/H Proof. We only need to prove the uniqueness. Let ψ : G/H G be any homomorphism such that ψ γ = φ. For any ah G/H, Hence, ψ = φ. φ φ G ψ(ah) = ψ(γ(a)) = φ(a) = φ(γ(a)) = φ(ah). Hwan Yup Jung (CBNU) Group Theory March 1, / 111

61 Factor Groups Corollary Let φ : G G be a surjective homomorphism with H = ker(φ). Then it induces an isomorphism φ : G/H G satisfying φ(xh) = φ(x) for all x G. φ G G γ G/H φ Hwan Yup Jung (CBNU) Group Theory March 1, / 111

62 Factor Groups Theorem (The Structure of Cyclic Groups) 1 Any infinite cyclic group G is isomorphic to Z, +. 2 Any finite cyclic group G of finite order n is isomorphic to Z n, +. Proof. Let G be a cyclic group with a generator g and let φ : Z G, n g n with H = Ker(φ). H = mz for some integer m 0, and φ induces an isomorphism φ : Z/mZ G such that φ(n + mz) = g n for all n + mz Z/mZ. If G is an infinite cyclic group, Z/mZ = m = 0, so φ : Z G. If G is a finite cyclic group of order n 1, Z/mZ = G = n m = n, so φ : Z/nZ G. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

63 Factor Groups Definition An abelian group G is said to be finitely generated if there exists x 1, x 2,..., x r G such that x 1, x 2,..., x r = G. Theorem (Fundamental Theorem of Finitely Generated Abelian Groups) Every finitely generated abelian group G is isomorphic to a direct product of cyclic groups in the form Z p r 1 1 Z p r 2 2 Z p rn n Z Z Z, where the p i are primes, not necessary distinct, and the r i are positive integers. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

64 Factor Groups Example. 1 Let G and H be cyclic groups. Then G H is a finitely generated abelian groups. 2 (Z 4 Z 2 )/({0} Z 2 ) is a cyclic group of order 4. Proof. Define a map φ : Z 4 Z 2 Z 4, (ā, b) ā. φ is a surjective homomorphism with kernel Ker(φ) = {0} Z2. φ induces an isomorphism φ : (Z 4 Z 2 )/({0} Z 2 ) Z 4. 3 Find all finite abelian groups of order p 3 up to isomorphisms, where p is a prime. Solution. p 3 can be written as p 3, p 2 p and p p p. Thus all finite Abelian groups of order p 3 (up to isomorphisms) are Z p 3 p 3 Z p 2 Z p p 2 p Z p Z p Z p p p p. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

65 Factor Groups Example. Find all finite abelian groups of order p 3 q 2 up to isomorphisms, where p and q distinct are primes. Solution. All finite Abelian groups of order p 3 (up to isomorphisms) are Z p 3, Z p 2 Z p, Z p Z p Z p. All finite Abelian groups of order q 2 (up to isomorphisms) are Z q 2, Z q Z q. Thus all finite abelian groups of order p 3 q 2 (up to isomorphisms) are Z p 3 Z q 2, Z p 3 Z q Z q, Z p 2 Z p Z q 2, Z p 2 Z p Z q Z q, Z p Z p Z p Z q 2, Z p Z p Z p Z q Z q. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

66 Factor Groups Example. Find all abelian groups of order 360 up to isomorphisms. Solution. Note that 360 = All finite Abelian groups of order 2 3 (up to isomorphisms) are Z 2 3, Z 2 2 Z 2, Z 2 Z 2 Z 2. All finite Abelian groups of order 3 2 (up to isomorphisms) are Z 3 2, Z 3 Z 3. All finite Abelian groups of order 5 (up to isomorphisms) are Z 5. Thus all finite Abelian groups of order (up to isomorphisms) are Z 2 3 Z 3 2 Z 5, Z 2 3 Z 3 Z 3 Z 5, Z 2 2 Z 2 Z 3 2 Z 5, Z 2 2 Z 2 Z 3 Z 3 Z 5, Z 2 Z 2 Z 2 Z 2 Z 3 2 Z 5, Z 2 Z 2 Z 2 Z 2 Z 3 Z 3 Z 5. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

67 Factor Groups Definition 1 An isomorphism φ : G G of a group G with itself is called an automorphism of G. 2 For all g G, the map i g : G G, x gxg 1 is an automorphism of G, called the inner automorphism of G by g. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

68 Factor-Group Computations And Simple Groups Theorem Let H G and H G. Then H H G G and (G G )/(H H ) = G/H G /H. Proof.Consider φ : G G G/H G /H, (g, g ) (gh, g H ). Corollary φ is a surjective homomorphism with Ker(φ) = H H. φ induces an isomorphism φ : (G G )/(H H ) = G/H G /H. Let G = H K, H = H {e K } and K = {e H } K. Then H G, K G and G/ H = K, G/ K = H. Theorem A factor of a cyclic group is cyclic. Proof. HW Hwan Yup Jung (CBNU) Group Theory March 1, / 111

69 Factor-Group Computations And Simple Groups Example Compute the factor group (Z 4 Z 6 )/ ( 0, 1). 1 Solution 1. ( 0, 1) = { 0} Z 6. (Z 4 Z 6 )/ ( 0, 1) = (Z 4 Z 6 )/{ 0} Z 6 = Z4. 2 Solution 2. (Z 4 Z 6 )/ ( 0, 1) is an abelian group of order 4. There are two finite abelian group of order 4 (up to isomorphism) Z 2 Z 2 and Z 4. Z 2 Z 2 has no element of order 4. Z 4 has an element of order 4. ( 1, 0) + ( 0, 1) has order 4 in (Z 4 Z 6 )/ ( 0, 1). Hence, (Z 4 Z 6 )/ ( 0, 1) is isomorphic to Z 4. 3 Solution 3. Define φ : Z 4 Z 6 Z 4, (a, b) a. φ is a surjective homomorphism with kernel Ker(φ) = ( 0, 1). φ induces an isomorphism φ : (Z 4 Z 6 )/ ( 0, 1) Z 4. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

70 Factor-Group Computations And Simple Groups Example Compute the factor group (Z 4 Z 6 )/ ( 0, 2). Solution 1. ( 0, 2) = {( 0, 0), ( 0, 2), ( 0, 4)}; (Z 4 Z 6 )/ ( 0, 2) is a finite abelian group of order 8. There are three finite abelian group of order 8 (up to isomorphism) Z 8, Z 2 Z 4, Z 2 Z 2 Z 2. Z 8 has an element of order 8. Z 2 Z 4 has no element of order 8 and has an element of order 4. Z2 Z 2 Z 2 has no element of order 8 and 4. There is no element of order 8 in (Z 4 Z 6 )/ ( 0, 2). ( 1, 0) + ( 0, 2) has order 4 in (Z 4 Z 6 )/ ( 0, 2). Hence, (Z 4 Z 6 )/ ( 0, 2) is isomorphic to Z 4 Z 2. Solution 2. Consider φ : Z 4 Z 6 Z 4 Z 2, (a + 4Z, b + 6Z) (a + 4Z, 3b + 2Z). φ is well defined surjective homomorphism with kernel Ker(φ) = {( 0, 0), ( 0, 2), ( 0, 4)} = ( 0, 2). φ induces an isomorphism φ : (Z 4 Z 6 )/ ( 0, 2) Z 4 Z 2. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

71 Factor-Group Computations And Simple Groups Example Compute the factor group (Z 4 Z 6 )/ ( 2, 3). Solution 1. ( 2, 3) = {( 2, 3), ( 0, 0)}; (Z 4 Z 6 )/ ( 2, 3) is a finite abelian group of order 12. There are two finite abelian groups of order 12 (up to isomorphisms) Z 4 Z 3 Z 12, Z 2 Z 2 Z 3 Z 2 Z 6. Z4 Z 3 Z 12 has an element of order 12. Z2 Z 2 Z 3 Z 2 Z 6 has no element of order 12. ( 1, 2) + ( 2, 3) has order 12 in (Z 4 Z 6 )/ ( 2, 3). Hence, (Z 4 Z 6 )/ ( 2, 3) is isomorphic to Z 4 Z 3 Z 12. Solution 2. Can you find a surjective homomorphism φ : (Z 4 Z 6 ) Z 4 Z 3 with kernel ( 2, 3), which will induce an isomorphism φ : (Z 4 Z 6 )/ ( 2, 3) Z 4 Z 3? Hwan Yup Jung (CBNU) Group Theory March 1, / 111

72 Factor-Group Computations And Simple Groups Example Compute the factor group (Z Z)/ (1, 1). Solution. Note that (1, 1) = {(a, a) : a Z}. Consider φ : Z Z Z, (a, b) a b. φ is a surjective homomorphism with kernel Ker(φ) = (1, 1). φ induces an isomorphism φ : (Z Z)/ (1, 1) Z. Definition A group is called a simple if it is non-trivial and has no proper normal subgroups. Theorem A n is simple if n 5. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

73 Factor-Group Computations And Simple Groups Theorem Let φ : G G be a homomorphism. 1 If N G, then φ(n) φ(g). 2 If N G, then φ 1 (N ) G containing Ker(φ). Proof. 1 We know that φ(n) < φ(g). For any g φ(g) and x φ(n), g = φ(g) and x = φ(x) for some g G, x N g x (g ) 1 = φ(g)φ(x)φ(g) 1 = φ(gxg 1 ) φ(n). Hence, φ(n) is a normal subgroup of φ[g]. 2 We know that φ 1 (N ) < G containing Ker(φ). For any g G and x φ 1 (N ), φ(gxg 1 ) = φ(g)φ(x)φ(g) 1 N gxg 1 φ 1 (N ). Hence, φ 1 (N ) is a normal subgroup of G. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

74 Factor-Group Computations And Simple Groups Definition A normal subgroup M of G is called a maximal if M G and there is no proper normal subgroup N of G properly containing M. Theorem M is a maximal normal subgroup of G if and only if G/M is a simple. Proof. ( ) Let M be a maximal normal subgroup of G. For any normal subgroup N of G/M, Ker(γ) = M < γ 1 (N ) G γ 1 (N ) = M or γ 1 (N ) = G N = γ(m) = {M} or N = G/M. Hence, G/M has only two normal subgroups {M} and G/M itself, i.e., G/M is a simple. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

75 Factor-Group Computations And Simple Groups ( ) Assume that G/M is a simple group. For any normal subgroup N of G such that M N G, γ(n) γ(g) = G/M γ(n) = {M} or γ(n) = G/M N = M or N = G. Hence, M is a maximal normal subgroup of G. Example 1 In the abelian group Z, nz mz m n. 2 Thus for any positive integer n, nz is a maximal normal subgroup of Z if and only if n is a prime. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

76 Factor-Group Computations And Simple Groups Lemma Let G be a group and Z(G) = {z G : zg = gz for all g G}. 1 Z(G) G. 2 Z(G) = G if and only if G is abelian group. Definition The normal subgroup Z(G) of G is called the center of G. Example 1 S 3 has the trivial center, i.e., Z(S 3 ) = {ρ 0 }. 2 In general, S n has the trivial center if n 3. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

77 Group Action On A Set Definition Let X be a set and G a group. An action of G on X is a map such that 1 e x = x for all x X. : G X X, (g, x) g x 2 (g 1 g 2 ) x = g 1 (g 2 x) for all x X and g 1, g 2 G. In this case, X is called a G-set. Theorem Let X be a G-set. 1 For each g G, σ g : X X, x g x is a permutation of X. 2 The map φ : G S X, g σ g is a group homomorphism. Proof. HW Hwan Yup Jung (CBNU) Group Theory March 1, / 111

78 Group Action On A Set Lemma Let X be a G-set. Then N = {g G : g x = x for all x X } is a normal subgroup of G. Definition Let X be a G-set. 1 We say that G acts faithfully on X if N = {e}. 2 We say that G acts transitively on X if for each x 1, x 2, X, there exists g G such that g x 1 = x 2. Remark. We can regard X as a G/N-set, where the action G/N on X is given by (gn) x = g x. Then G/N acts faithfully on X. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

79 Group Action On A Set Example 1 Let G be a group and H < G. Then G is a H-set with the action : H G G, (h, g) hg. 2 Let H < G, and let G/H be the set of all left cosets of H. Then G/H is a G-set with the G-action Theorem : G G/H G/H, (g, xh) (gx)h. Let X be a G-set. For each x X, let G x = {g G : gx = x}. Then G x < G, called the isotropy subgroup (or stabilizer) of x. Proof. For any g G x, g x = x g 1 x = x g 1 G x. For any g 1, g 2 G x, we have (g 1 g2 1 ) x = g 1 (g2 1 x) = g 1 x = x g 1 g2 1 G x. Hence, G x < G. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

80 Group Action On A Set Let X be a G-set. We define a relation on X as follow: for x 1, x 2 X, Theorem is an equivalence relation on X. x 1 x 2 g x 1 = x 2 for some g G. Proof. (Reflexive) For any x X, x = e x x x. (Symmetric) If x y, g x = y ( g G) g 1 y = x y x. (Transitive) If x y and y z, g 1 x = y, g 2 y = z ( g 1, g 2 G) (g 2 g 1 ) x = z x z. Definition Let X be a G-set. For each x X, the equivalence class of x with respect to given by {y X : x y} = {gx : g G}, called the orbit of x and we denote it by Gx. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

81 Group Action On A Set Theorem Let X be a G-set. For any x X, we have Gx = (G : G x ). Especially, if G is finite, then Gx is a divisor of G. Proof. Let G/G x be the set of all left cosets of G x. Consider ψ : G/G x Gx, gg x g x. (well defined) If gg x = g G x, g = g h ( h G x ) gx = (g h)x = g (hx) = g x. (injective) If ψ(gg x ) = ψ(g G x ), gx = g x g 1 g G x gg x = g G x. (surjective) For any gx Gx, gx = ψ(gg x ). Hence, ψ is a bijection, and so Gx = (G : G x ). If G is a finite group, then Gx = (G : G x ) = G G x. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

82 Isomorphism Theorems Theorem (First Isomorphism Theorem) Let φ : G G be a homomorphism with H = Ker(φ) and let γ : G G/H be the canonical homomorphism. Then there exists a unique isomorphism φ : G/H φ(g) such that φ(x) = φ γ(x) for all x G. γ G G/H φ φ φ(g) Hwan Yup Jung (CBNU) Group Theory March 1, / 111

83 Isomorphism Theorems Lemma For N G, let X N (G) be the set of normal subgroups of G containing N and X (G/N) for the set of normal subgroups of G/N. Then we have the following one-to-one correspondence X N (G) X (G/N), whose inverse mapping is H φ 1 (H). L φ(l), Proof. Note that if L < G, then φ 1 (φ(l)) = NL < G. In particular, if N < L, then φ 1 (φ(l)) = L. (Injectivity) For L 1, L 2 X N (G), if φ(l 1 ) = φ(l 2 ), then L 1 = φ 1 (φ(l 1 )) = φ 1 (φ(l 2 )) = L 2. (Surjectivity) For H X (G/N), we have φ 1 (H) G and N = Ker(φ) < φ 1 (H) (that is, φ 1 (H) X N (G)). H = φ(φ 1 (H)). Hwan Yup Jung (CBNU) Group Theory March 1, / 111

84 Isomorphism Theorems Definition For any H, N < G, Lemma H N := the intersection of all subgroups of G that contain HN = the smallest subgroup of G containing both H and N (H N is called the join of H and N.) 1 If N G, H < G, then H N = HN = NH. 2 If N G, H G, then HN G. Proof. If N G, H < G, then HN = NH < G HN = NH = H N. Assume that H, N G. For any g G, g(hn)g 1 = (ghg 1 )(gng 1 ) = HN. Hence, HN G. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

85 Isomorphism Theorems Theorem (Second Isomorphism Theorem) If H < G, N G, then H N H and HN/N = H/(H N). G HN H N H N Hwan Yup Jung (CBNU) Group Theory March 1, / 111

86 Isomorphism Theorems Proof. Note that N HN, H N H and NH/N = {xn : x H}. Consider φ : H NH/N, x xn. φ is a homomorphism. φ is a surjective (H/N = {xn : x H}). Ker(φ) = H N. Hence, φ induces an isomorphism φ : H/H N NH/N. Example Let G = Z Z Z, H = Z Z {0} and N = {0} Z Z. HN = Z Z Z = G. H N = {0} Z {0}. HN/N = Z and H/H N = Z. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

87 Isomorphism Theorems Theorem (Third Isomorphism Theorem) Let H, K G with K H. Then H/K G/K and G/H = (G/K)/(H/K). Proof. It is easy to see that H/K G/K. Consider φ : G/K G/H, xk xh. φ is a surjective homomorphism. Ker(φ) = H/K. Hence, φ induces an isomorphism φ : (G/K)/(H/K) G/H. Example Consider K = 6Z < H = 2Z < G = Z. G/K = {6Z, 1 + 6Z, 2 + 6Z, 3 + 6Z, 4 + 6Z, 5 + 6Z}. H/K = 2Z/6Z = {6Z, 2 + 6Z, 4 + 6Z}. (G/K)/(H/K) = Z 2 = G/H. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

88 Series of Groups Definition 1 A subnormal (or subinvariant) series of a group G is a finite sequence H 0 = {e} < H 1 < < H n = G such that H i H i+1 for all 0 i n 1. 2 Moreover, if each H i G, then it is called a normal (or invariant) series of a group G. Examples 1 {0} < 8Z < 4Z < Z and {0} < 9Z < Z are normal series of Z. 2 {ρ 0 } < {ρ 0, µ 1 } < {ρ 0, ρ 2, µ 1, µ 2 } < D 4 is a subnormal series of D 4 (which is not normal series of D 4 ). Hwan Yup Jung (CBNU) Group Theory March 1, / 111

89 Series of Groups Definition A subnormal(normal) series {K j } is a refinement of a subnormal(normal) series {H i } of a group G if {H i } {K j }, that is, if each H i is one of the K j. Definition Two subnormal(normal) series {H i } and {K i } of G are isomorphic if there is a one-to-one correspondence between {H i+1 /H i } and {K i+1 /K i } such that corresponding factors are isomorphic. Theorem (Schreier Theorem) Any two subnormal (normal) series of a group G have isomorphic refinements. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

90 Series of Groups Example.Consider two series {0} < 8Z < 4Z < Z, {0} < 9Z < Z of Z {0} < 72Z < 8Z < 4Z < Z is a refinement of {0} < 8Z < 4Z < Z; 72Z/{0} = 72Z, 8Z/72Z = Z 9, 4Z/8Z = Z 2, Z/4Z = Z 4 {0} < 72Z < 18Z < 9Z < Z is a refinement of {0} < 9Z < Z; 72Z/{0} = 72Z, 18Z/72Z = Z 4, 9Z/18Z = Z 2, Z/9Z = Z 9 Two series {0} < 72Z < 8Z < 4Z < Z and {0} < 72Z < 18Z < 9Z < Z are isomorphic. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

91 Series of Groups Definition 1 A subnormal series {H i } of a group G is a composition series if all the factor groups H i+1 /H i are simple. 2 A normal series {H i } of G is called a principal or chief series if all the factor groups H i+1 /H i are simple. Example. 1 Z has no composition series. 2 The series {e} < A n < S n for n 5 is a composition series of S n. Theorem (Jordan-Hölder Theorem) Any two composition (principal) series of a group G are isomorphic. Proof. Let {H i } and {K j } be any two composition series of G. They have isomorphic refinements; Since all factor groups H i /H i+1 and K j /K j+1 are already simple, neither {H i } or {K j } has any further refinement. Hence, {H i } and {K j } are isomorphic. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

92 Series of Groups Theorem If G has a composition (principal) series, and if N is a proper normal subgroup of G, then there exists a composition (principal) series containing N. Proof. The series {e} < N < G is both a subnormal and normal series of G. Let {H i } be a composition series of G. The series {e} < N < G has a refinement which isomorphic to a refinement of {H i }; {H i } has no furthermore refinement (since it is a composition series) {e} < N < G has a refinement which isomorphic to {H i } {e} < N < G can be refined to a composition series {K j : 0 j n}. If K i = N, then {e} = K 0 < < K i 1 < K i = N is a composition series of N. Example. A composition series of Z 4 Z 9 containing (0, 1) is {(0, 0)} < (0, 3) < (0, 1) < 2 1 < 1 1 = Z 4 Z 9. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

93 Series of Groups Definition A group G is solvable if it has a composition series {H i } such that all factor groups H i+1 /H i are abelian. Remark. For a solvable group G, every composition series {H i } of G must have abelian factor groups H i+1 /H i (Jordan-Hölder Theorem). Example. 1 The group S 3 is a solvable group. {e} < A3 < S 3 is a composition series of S 3. A3 /{e} = A 3, S 3 /A 3 = Z2 are all abelian groups. 2 The group S 5 is not a solvable group. {e} < A5 < S 5 is a composition series of S 5 (since A 5 is simple). A3 /{e} = A 5 is not abelian. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

94 Sylow Theorems Let X be a finite G-set. Let Gx = {gx : g G} is the orbit of x X under G. X G = {x X : gx = x for all g G}. x XG Gx = {x}. X G is the union of the one-element orbits in X. Assume that X = r i=1 Gx i (disjoint union). X = r i=1 Gx i. Suppose there are s one-element orbits (0 s r); X G = s. Reordering the x i if necessary, X = X G + r i=s+1 Gx i. Throughout this section p will always be a prime number. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

95 Sylow Theorems Theorem Let G be a group of order p n and let X be a finite G-set. Then we have X X G mod p. Proof. From the equation X = X G + r i=s+1 Gx i, Gx i = G / G xi > 1, so Gx i is divisible by p for s + 1 i s. Hence, X X G mod p. Definition Let p be a prime number. 1 A group G is called a p-group if every element in G has order a power of p. 2 A subgroup of a group G is a p-subgroup of G if the subgroup is itself a p-group. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

96 Sylow Theorems Theorem (Cauchy s Theorem) Let p be a prime number. Let G be a finite group and p divide G. Then G has an element of order p and, consequently, a subgroup of order p. Proof. Let X = {(g 1,..., g p ) : g i G and g 1 g 2 g p = e}; X = G p 1. Let σ = (1, 2, 3,..., p) S p. We let σ act on X by σ(g 1, g 2,..., g p ) = (g σ(1), g σ(2),..., g σ(p), ) = (g 2, g 3,..., g p, g 1 ). g 2 g 3 g p g 1 = g 2 g 3 g p (g 2 g p ) 1 = e. We consider the subgroup σ of S p to act on X. Note that (g 1,..., g p ) X σ g 1 = = g p X σ since (e, e,..., e) X σ. X σ X = G p 1 0 mod p p divides X σ X σ p a G(a e) such that (a, a,..., a) X σ a p = e, so a has order p; a is a subgroup of G of order p. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

97 Sylow Theorems Corollary Let G be a finite group. Then G is a p-group G is a power of p. Proof. ( ) Lagrange s Theorem. ( ) Suppose that G is not a power of p. a prime divisor q (q p) of G g G of order q (Cauchy), which is a contradiction to the definition of p-group. Definition For H < G, N[H] := {g G : ghg 1 = H} is called the normalizer of H in G. Remark. 1 N[H] is a subgroup of G containing H. 2 H is a normal subgroup of N[H]. 3 N[H] is the largest subgroup of G having H as a normal subgroup. Hwan Yup Jung (CBNU) Group Theory March 1, / 111

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