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

Group Theory Hwan Yup Jung Department of Mathematics Education, Chungbuk National University Hwan Yup Jung (CBNU) Group Theory March 1, 2013 1 / 111

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, 2013 2 / 111

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, 2013 3 / 111

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, 2013 4 / 111

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, 2013 5 / 111

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, 2013 6 / 111

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, 2013 7 / 111

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, 2013 8 / 111

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, 2013 9 / 111

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, 2013 10 / 111

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, 2013 11 / 111

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, 2013 12 / 111

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, 2013 13 / 111

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, 2013 14 / 111

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, 2013 15 / 111

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, 2013 16 / 111

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, 2013 17 / 111

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, 2013 18 / 111

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, 2013 19 / 111

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, 2013 20 / 111

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 ( ) 1 2 3 4 5 σ =, τ = 4 2 5 3 1 ( ) 1 2 3 4 5. 3 5 4 2 1 Then 1 2 3 4 5 στ = 3 5 4 2 1 = 5 1 3 2 4 ( ) 1 2 3 4 5. 5 1 3 2 4 Hwan Yup Jung (CBNU) Group Theory March 1, 2013 21 / 111

Groups of permutations Example. S 3 = {ρ 0, ρ 1, ρ 2, µ 1, µ 2, µ 3 }, where ρ 0 := ( 1 1 2 2 3 3 ) = id and ρ 1 := ( 1 2 2 3 3 1 ), 120 -Counterclockwise rotation. ρ 2 := ( 1 3 2 1 3 2 ), 240 -Counterclockwise rotation. µ 1 := ( 1 1 2 3 3 2 ), Reflection with respect to 1. µ 2 := ( 1 3 2 2 3 1 ), Reflection with respect to 2. µ 3 := ( 1 2 2 1 3 2 ), 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, 2013 22 / 111

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 := ( 1 1 2 2 3 3 4 4 ). ρ 1 := ( 1 2 2 3 3 4 4 1 ), 90 -Counterclockwise rotation. ρ 2 := ( 1 3 2 4 3 1 4 2 ), 180 -Counterclockwise rotation. ρ 3 := ( 1 4 2 1 3 2 4 3 ), 270 -Counterclockwise rotation. µ 1 := ( 1 2 2 1 3 4 4 3 ), Reflection µ 2 := ( 1 4 2 3 3 2 4 1 ), Reflection δ 1 := ( 1 3 2 2 3 1 4 4 ), Reflection δ 2 := ( 1 1 2 4 3 3 4 2 ), Reflection Hwan Yup Jung (CBNU) Group Theory March 1, 2013 23 / 111

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, 2013 24 / 111

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, 2013 25 / 111

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, 2013 26 / 111

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, 2013 27 / 111

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 = ( 1 3 2 8 3 6 4 7 5 4 6 1 7 5 8 2 ). 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, 2013 28 / 111

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 = ( 1 2 3 4 5 6 7 8 3 2 6 4 5 1 7 8 ). 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 = ( 1 2 3 4 5 6 7 8 1 8 3 4 5 6 7 2 ). 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 = ( 1 1 2 2 3 3 4 7 5 4 6 6 7 5 8 8 ). 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, 2013 29 / 111

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, 2013 30 / 111

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, 2013 31 / 111

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, 2013 32 / 111

Orbits, Cycles, and the Alternating groups Example. 1 Consider σ = ( ) 1 2 3 4 5 6 7 8. 3 2 6 7 5 1 4 8 1 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). ( ) 1 2 3 4 5 6. 6 5 2 4 3 1 1 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, 2013 33 / 111

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, 2013 34 / 111

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, 2013 35 / 111

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, 2013 36 / 111

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, 2013 37 / 111

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, 2013 38 / 111

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 6. 0 + 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, 2013 39 / 111

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, 2013 40 / 111

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, 2013 41 / 111

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, 2013 42 / 111

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, 2013 43 / 111 (H:K)

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, 2013 44 / 111

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, 2013 45 / 111

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, 2013 46 / 111

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, 2013 47 / 111

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, 2013 48 / 111

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, 2013 49 / 111

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, 2013 50 / 111

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, 2013 51 / 111

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, 2013 52 / 111

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, 2013 53 / 111

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 0 + 0 = 0 even odd odd 1 0 + 1 = 1 odd even odd 1 1 + 0 = 1 odd odd even 0 1 + 1 = 0 By definition, we have Ker(φ) = A n. Hwan Yup Jung (CBNU) Group Theory March 1, 2013 54 / 111

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, 2013 55 / 111

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, 2013 56 / 111

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, 2013 57 / 111

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, 2013 58 / 111

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, 2013 59 / 111

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, 2013 60 / 111

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, 2013 61 / 111

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, 2013 62 / 111

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, 2013 63 / 111

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, 2013 64 / 111

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, 2013 65 / 111

Factor Groups Example. Find all abelian groups of order 360 up to isomorphisms. Solution. Note that 360 = 2 3 3 2 5. 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 2 3 3 2 5 (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, 2013 66 / 111

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, 2013 67 / 111

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, 2013 68 / 111

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, 2013 69 / 111

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, 2013 70 / 111

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, 2013 71 / 111

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, 2013 72 / 111

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, 2013 73 / 111

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, 2013 74 / 111

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, 2013 75 / 111

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, 2013 76 / 111

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, 2013 77 / 111

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, 2013 78 / 111

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, 2013 79 / 111

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, 2013 80 / 111

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, 2013 81 / 111

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, 2013 82 / 111

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, 2013 83 / 111

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, 2013 84 / 111

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, 2013 85 / 111

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, 2013 86 / 111

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, 2013 87 / 111

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, 2013 88 / 111

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, 2013 89 / 111

Series of Groups Example.Consider two series {0} < 8Z < 4Z < Z, {0} < 9Z < Z of Z. 1 1 4 9 4 9 4 9 2 2 2 8 8 8 18 9 9 4 72 72 {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, 2013 90 / 111

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, 2013 91 / 111

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, 2013 92 / 111

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, 2013 93 / 111

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, 2013 94 / 111

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, 2013 95 / 111

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, 2013 96 / 111

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, 2013 97 / 111