# I. GROUPS: BASIC DEFINITIONS AND EXAMPLES

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1 I GROUPS: BASIC DEFINITIONS AND EXAMPLES Definition 1: An operation on a set G is a function : G G G Definition 2: A group is a set G which is equipped with an operation and a special element e G, called the identity, such that (i) the associative law holds: for every x, y, z G, x (y z) = (x y) z; (ii) e x = x = x e for all x G; (iii) for every x G, there is x G (so-called, inverse) with x x = e = x x Set Operation + Operation Additional Condition N no no Z yes no Q yes no for Q \ {0} R yes no for R \ {0} R \ Q no no Set Operation + Operation Z >0 no no Z 0 no no Q >0 no yes Q 0 no no R >0 no yes R 0 no no 1

2 Set Operation + Operation {2n : n Z} yes no {2n + 1 : n Z} no no {3n : n Z} yes no {kn : n Z}, where k N is some fixed number yes no {a n : n Z}, where a R is some fixed number no yes { p 2 n : p Z, n Z 0} yes no Set Operation: a b = a 2 b 2 Operation: a b = a b R >0 no no Definition 3: A group is called abelian if x y = y x for any x, y G The parity group P has two elements, the words even and odd, with operation even even = even = odd odd and even odd = odd = odd even It is clear that: 1 even is the identity element; 2 The inverse of even is even and the inverse of odd is odd 2

3 The group Z 2 has two elements: [0] is the set of all even numbers and [1] is the set of all odd numbers Operation: [0] [0] = [1] [1] = [0] and [0] [1] = [1] [0] = [1] It is clear that: 1 [0] is the identity element; 2 The inverse of [0] is [0] and the inverse of [1] is [1] The group Z 3 has three elements: [0] is the set of numbers which are congruent to 0 mod 3; [1] is the set of numbers which are congruent to 1 mod 3; [2] is the set of numbers which are congruent to 2 mod 3 Operation: [0] [0] = [1] [2] = [2] [1] = [0], [0] [1] = [1] [0] = [2] [2] = [1], and [0] [2] = [2] [0] = [1] [1] = [2] It is clear that: 1 [0] is the identity element; 2 The inverse of [0] is [0], the inverse of [1] is [2], and the inverse of [2] is [1] The group Z 3 has two elements: [1] is the set of numbers which are congruent to 1 mod 3; [2] is the set of numbers which are congruent to 2 mod 3 Operation: and [1] [1] = [2] [2] = [1] [1] [2] = [2] [1] = [2] It is clear that: 1 [1] is the identity element; 2 The inverse of [1] is [1] and the inverse of [2] is [2] 3

4 Definition 4: II THREE EXAMPLES OF SPECIAL GROUPS The family of all the permutations of the set X = {1, 2,, n} is called the symmetric group It is denoted by S n Theorem 1: S n is a nonabelian group under operation of composition Proof (Sketch): It is obvious that S n is closed under operation of composition One can show that this operation is associative The identity element is (1) Finally, we know that every permutation α is either a cycle or a product of disjoint (with no common elements) cycles and the inverse of the cycle α = (i 1 i 2 i r ) is the cycle α 1 = (i r i r 1 i 1 ) Therefore, every element of S n is invertible To show that S n is nonabelian, we note that, for example, (123)(13) (13)(123) Definition 5: The set of the following four permutations is called the four-group Definition 6: V = {(1), (12)(34), (13)(24), (14)(23)} The set of all 2 2 nonsingular (determinant is nonzero) matrices with real entries and with operation matrix multiplication is called the general linear group It is denoted by GL(2, R) Theorem 2: GL(2, R) is a nonabelian group Proof (Sketch): It is obvious that GL(2, R) is closed under operation of multiplication One can show that this operation is associative The identity element is [ ] [ ] a b Finally, for every element A = there exists the inverse c d [ ] A 1 1 d b = ad bc c a To show that GL(2, R) is nonabelian, we note that, for example, [ ] [ ] [ ] [ ]

5 III MAIN THEOREM ABOUT GROUPS Theorem 3: Let G be a group (i) If x a = x b or a x = b x, then a = b (ii) The identity element e is unique (iii) For all x G, the inverse element x 1 is unique (iv) For all x G we have (x 1 ) 1 = x (v) For all a, b G we have (a b) 1 = b 1 a 1 (i) Let x a = x b, then x 1 (x a) = x 1 (x b), therefore by the associative law we get ( x 1 x ) a = ( x 1 x ) b, so e a = e b, and the result follows In the same way one can deduce a = b from a x = b x (ii) Assume to the contrary that there are two identity elements e 1 and e 2 Then which is a contradiction e 1 = e 1 e 2 = e 2, (iii) Assume to the contrary that for some x G there are two inverse elements x 1 1 and x 1 2 Then x 1 2 = e x 1 2 = ( x 1 1 x ) x 1 2 = x 1 1 ( ) x x 1 2 = x 1 1 e = x 1 1, which is a contradiction (iv) We have ( x 1 ) 1 x 1 = e Multiplying both sides by x, we get ( x 1 ) 1 ( x 1 x ) = e x, hence ( x 1 ) 1 e = x, and the result follows (v) We have (a b) ( b 1 a 1) = [ a ( b b 1)] a 1 = (a e) a 1 = a a 1 = e, and the result follows 5

6 IV SUBGROUPS: BASIC DEFINITIONS AND EXAMPLES Definition 7: A subset H of a group G is a subgroup if (i) e H; (ii) if x, y H, then x y H; (iii) if x H, then x 1 H Notation: If H is a subgroup of G, we write H G It is obvious that {e} and G are always subgroups of a group G Definition 8: We call a subgroup H proper, and we write H < G, if H G We call a subgroup H of G nontrivial if H {e} 1 Z + < Q + < R + 2 Q 0 < R 0 3 Q >0 < R >0 < R 0 4 A group of even numbers is a subgroup of Z + 5 V < S 4 6

7 V TWO THEOREMS ABOUT SUBGROUPS Theorem 4: A subset H of a group G is a subgroup H is nonempty and, whenever x, y H, then xy 1 H = ) Suppose H is a subgroup of G We should prove that H is nonempty and, whenever x, y H, then xy 1 H We first note that H is nonempty, because 1 H by part (i) of definition 7 Finally, if x, y H, then y 1 H by part (iii) of definition 7, and so xy 1 H, by part (ii) of definition 7 =) Suppose H is a nonempty subset of G and, whenever x, y H, then xy 1 H We should prove that H is a subgroup of G Since H is nonempty, it contains some element, say, h Taking x = h = y, we see that 1 = hh 1 H, and so part (i) of definition 7 holds If y H, then set x = 1 (which we can do because 1 H), giving y 1 = 1y 1 H, and so part (iii) holds Finally, we know that (y 1 ) 1 = y, by (iv) of Theorem 3 Hence, if x, y H, then y 1 H, and so xy = x(y 1 ) 1 H Therefore, H is a subgroup of G Theorem 5: A nonempty subset H of a finite group G is a subgroup H is closed = ) Suppose H is a subgroup of G Then it is closed by part (ii) of definition 7 =) Suppose H is a nonempty closed subset of a finite group G We should prove that H is a subgroup We first note that since H is closed, it follows that part (ii) of definition 7 holds This, in particularly means, that H contains all the powers of its elements Let us pick some element a H (we can do that, since H is nonempty) Then a n H for all integers n 1 Lemma: If G is a finite group and a G, then a k = 1 for some integers k 1 Consider the subset {1, a, a 2,, a n, } Since G is finite, there must be a repetition occurring on this infinite list So, there are integers m > n with a m = a n, hence 1 = a m a n = a m n So, we have shown that there is some positive power of a equal to 1 By this Lemma for any a G there is an integer m with a m = 1, hence 1 H and part (i) of definition 7 holds Finally, if h H and h m = 1, then h 1 = h m 1 (for hh m 1 = 1 = h m 1 h), so that h 1 H and part (iii) of definition 7 holds Therefore, H is a subgroup of G 7

8 VI CYCLIC GROUPS Definition 9: If G is a group and a G, write a = {a n : n Z} = {all powers of a}; a is called the cyclic subgroup of G generated by a G = {0, ±1, ±2, ±3, }, H = {0, ±2, ±4, ±6, } = 2 Definition 10: A group G is called cyclic if G = a In this case a is called a generator of G (a) {e} = e (b) Z + = {0, ±1, ±2, ±3, } = 1 (c) Q +, Q >0, R +, R >0 are not cyclic (d) S 2 = {(1), (12)} = (12) (1) (e) S m, m > 2, is not cyclic (f) Z + m = {[0], [1], [2],, [m 1]} = [1] (g) Z 3 = {[1], [2]} = [2] [1] (h) Z 5 = {[1], [2], [3], [4]} = [2] = [3] [1], [4] (i) Z 7 = {[1], [2],, [6]} = [3] = [5] [1], [2], [4], [6] (j) Z m is cyclic m is a prime (Lagrange, 1769) Remark: Recall, that if m is composite, Z m is not a group 8

9 Definition 11: VII ORDER Let G be a group and let a G If a k = 1 for some k 1, then the smallest such exponent k 1 is called the order of a; if no such power exists, then one says that a has infinite order (a) Let G = S 2 = {(1), (12)}, then the order of (1) is 1 and the order of (12) is 2 (b) Let G = Z + 4 = {[0], [1], [2], [3]}, then the order of [0] is 1, order of [1] is 4, order of [2] is 2, order of [3] is 4 Definition 12: If G is a finite group, then the number of elements in G, denoted by G, is called the order of G S n = n!, Z + n = n, Z p = p 1 Theorem 6: Let G be a finite group and let a G Then the order of a is a Part I: We first prove that if a = k, then the order of a is k In fact, the sequence 1, a, a 2,, a k 1 has k distinct elements, while has a repetition Hence, 1, a, a 2,, a k 1, a k a k {1, a, a 2,, a k 1 }, that is, a k = a i for some i with 0 i < k If i 1, then a k i = 1, contradicting the original list having no repetitions Therefore i = 1, so a k = a 0 = 1, and k is the order of a (being smallest positive such k) Part II: We now prove that if the order of a is k, then a = k If H = {1, a, a 2,, a k 1 }, then H = k; It suffices to show that H = a Clearly, H a For the reverse inclusion, take a i a By the division algorithm, i = qk + r, where 0 r < k Hence a i = a qk+r = a qk a r = ( a k) q a r = a r H; this gives a H, and so a = H Theorem 7: Let G be a group and let a G has finite order k If a n = 1, then k n 9

10 VIII LAGRANGE S THEOREM Definition 13: If H is a subgroup of a group G and a G, then the coset ah is the following subset of G : ah = {ah : h H} Remark: Cosets are usually not subgroups In fact, if a H, then 1 ah, for otherwise 1 = ah = a = h 1 H, which is a contradiction Let G = S 3 and H = {(1), (12)} Then there are 3 cosets: (12)H = {(1), (12)} = H, (13)H = {(13), (123)} = (123)H, (23)H = {(23), (132)} = (132)H Lemma: Let H be a subgroup of a group G, and let a, b G Then (i) ah = bh b 1 a H (ii) If ah bh, then ah = bh (iii) ah = H for all a G (i) ) Let ah = bh, then for any h 1 H there is h 2 H with ah 1 = bh 2 This gives since h 2 H and h 1 1 H ) Let b 1 a H Put b 1 a = h 0 Then b 1 a = h 2 h 1 1 = b 1 a H, ah bh, since if x ah, then x = ah = x = b(b 1 a)h = b h 0 h = bh }{{} 1 bh; h 1 bh ah, since if x bh, then x = bh = x = a(b 1 a) 1 h = a h 1 0 h = ah } {{ } 2 ah h 2 So, ah bh and bh ah, which gives ah = bh 10

11 (ii) Let ah bh, then there exists an element x with therefore ah = bh by (i) x ah bh = ah 1 = x = bh 2 = b 1 a = h 2 h 1 1 H, (iii) Note that if h 1 and h 2 are two distinct elements from H, then ah 1 and ah 2 are also distinct, since otherwise ah 1 = ah 2 = a 1 ah 1 = a 1 ah 2 = h 1 = h 2, which is a contradiction So, if we multiply all elements of H by a, we obtain the same number of elements, which means that ah = H Theorem 8 (Lagrange): If H is a subgroup of a finite group G, then Let G = t and be the family of all cosets of H in G Then H divides G {a 1 H, a 2 H,, a t H} G = a 1 H a 2 H a t H, because G = {a 1, a 2,, a t } and 1 H By (ii) of the Lemma above for any two cosets a i H and a j H we have only two possibilities: a i H a j H = or a i H = a j H Moreover, from (iii) of the Lemma above it follows that all cosets have exactly H number of elements Therefore G = H + H + + H = G = d H, and the result follows Corollary 1: If G is a finite group and a G, then the order of a is a divisor of G By Theorem 6, the order of the element a is equal to the order of the subgroup H = a By Lagrange s Theorem, H divides G, therefore the order a divides G Corollary 2: If a finite group G has order m, then a m = 1 for all a G Let d be the order of a By Corollary 1, d m; that is, m = dk for some integer k Thus, a m = a dk = (a d ) k = 1 11

12 Corollary 3: If p is a prime, then every group G of order p is cyclic Choose a G with a 1, and let H = a be the cyclic subgroup generated by a By Lagrange s Theorem, H is a divisor of G = p Since p is a prime and H > 1, it follows that and so H = G, as desired H = p = G, Theorem 9 (Fermat s Little Theorem): Let p be a prime Then n p n mod p for any integer n 1 We distinguish two cases Case A: Let p n, then, obviously, p n p n, and we are done Case B: Let p n Consider the group Z p and pick any [a] Z p Let k be the order of [a] We know that [a] is a subgroup of Z p and by Theorem 6 we obtain By Lagrange s Theorem we get which gives since [a] = k and Z p = p 1 So [a] = k [a] divides Z p, k p 1, p 1 = kd for some integer d On the other hand, since k is the order of [a], it follows that for any n [a] we have n k 1 mod p, hence and the result follows, since kd = p 1 n kd 1 d 1 mod p, 12

13 Definition 14: IX HOMOMORPHISMS AND ISOMORPHISMS If (G, ) and (H, ) are groups, then a function f : G H is a homomorphism if for all x, y G f(x y) = f(x) f(y) Let (G, ) be an arbitrary group and H = {e}, then the function f : G H such that f(x) = e for any x G is a homomorphism In fact, f(x y) = e = e e = f(x) f(y) Let (G, ) be an arbitrary group, then the function f : G G such that f(x) = x for any x G is a homomorphism In fact, f(x y) = x y = f(x) f(y) { [0] if x is even Let f : Z + Z + 2 be a function such that f(x) = [1] if x is odd In fact, if x + y is even, then f(x + y) = [0] = f(x) + f(y) Similarly, if x + y is odd, then Then f is a homomorphism f(x + y) = [1] = f(x) + f(y) Let f : GL(2, R) R 0 be a function such that Then f is a homomorphism In fact, f(m) = det M for any M GL(2, R) f(m 1 M 2 ) = det(m 1 M 2 ) = det(m 1 ) det(m 2 ) = f(m 1 )f(m 2 ) 13

14 Definition 15: Let a function f : G H be a homomorphism If f is also a one-one correspondence, then f is called an isomorphism Two groups G and H are called isomorphic, denoted by if there exists an isomorphism between them G = H, We show that R + = R >0 In fact, let f(x) = e x To prove that this is an isomorphism, we should check that is one-one correspondence and that f : R + R >0 f(x + y) = f(x)f(y) for all x, y R The first part is trivial, since f(x) = e x is defined for all x R and its inverse g(x) = ln x is also defined for all x R >0 The second part is also true, since f(x + y) = e x+y = e x e y = f(x)f(y) Definition 16: Let G = {a 1, a 2,, a n } be a finite group A multiplication table for G is an n n matrix whose ij entry is a i a j : G a 1 a 2 a j a n a 1 a 1 a 2 a 1 a 2 a 1 a j a 1 a n a 2 a 2 a 1 a 2 a 2 a 2 a j a 2 a n a i a i a 1 a i a 2 a i a j a i a n We will also agree that a 1 = 1 a n a n a 1 a n a 2 a n a j a n a n 14

15 Multiplicative table for Z 5 is Z 5 [1] [2] [3] [4] [1] [1] [2] [3] [4] [2] [2] [4] [1] [3] [3] [3] [1] [4] [2] [4] [4] [3] [2] [1] Remark: It is clear that two groups G = {a 1, a 2,, a n } and H = {b 1, b 2,, b n } of the same order n are isomorphic if and only if it is possible to match elements a 1, a 2,, a n with elements b 1, b 2,, b n such that this one-one correspondence remains also for corresponding entries a i a j and b i b j of their multiplication tables 1 The multiplication tables below show that P = Z + 2 = Z 3 : P even odd even even odd odd odd even Z + 2 [0] [1] [0] [0] [1] [1] [1] [0] Z 3 [1] [2] [1] [1] [2] [2] [2] [1] 2 The multiplication tables below show that Z + 4 = Z 5 : Z + 4 [0] [1] [2] [3] [0] [0] [1] [2] [3] [1] [1] [2] [3] [0] [2] [2] [3] [0] [1] [3] [3] [0] [1] [2] Z 5 [1] [2] [4] [3] [1] [1] [2] [4] [3] [2] [2] [4] [3] [1] [4] [4] [3] [1] [2] [3] [3] [1] [2] [4] 3 The multiplication tables below show that Z + 6 = Z 7 : Z + 6 [0] [1] [2] [3] [4] [5] [0] [0] [1] [2] [3] [4] [5] [1] [1] [2] [3] [4] [5] [0] [2] [2] [3] [4] [5] [0] [1] [3] [3] [4] [5] [0] [1] [2] [4] [4] [5] [0] [1] [2] [3] [5] [5] [0] [1] [2] [3] [4] Z 7 [1] [3] [2] [6] [4] [5] [1] [1] [3] [2] [6] [4] [5] [3] [3] [2] [6] [4] [5] [1] [2] [2] [6] [4] [5] [1] [3] [6] [6] [4] [5] [1] [3] [2] [4] [4] [5] [1] [3] [2] [6] [5] [5] [1] [3] [2] [6] [4] 15

16 Theorem 10: Let f : G H is a homomorphism of groups Then (i) f(e) = e; (ii) f(x 1 ) = f(x) 1 ; (iii) f(x n ) = [f(x)] n for all n Z (i) We have e e = e = f(e e) = f(e) = f(e)f(e) = f(e) Multiplying both sides by [f(e)] 1, we get [f(e)] 1 f(e)f(e) = [f(e)] 1 f(e) = e f(e) = e = f(e) = e (ii) We have x x 1 = e = f(x x 1 ) = f(e) = f(x)f(x 1 ) = f(e) Since f(e) = e by (i), we get f(x)f(x 1 ) = e Similarly, from x 1 x = e one can deduce that f(x 1 )f(x) = e So, f(x)f(x 1 ) = f(x 1 )f(x) = e, which means that f(x 1 ) = f(x) 1 (iii) If n 1, one can prove f(x n ) = [f(x)] n by induction If n < 0, then f(x n ) = f((x 1 ) n ) = [f(x 1 )] n, which is equal to [f(x)] n by (ii) Theorem 11: Any two cyclic groups G and H of the same order are isomorphic Proof (Sketch): Suppose that G = a = {1, a, a 2,, a m 1 } and H = b = {1, b, b 2,, b m 1 } Then f : G H with f(a i ) = b i, 0 i m 1, is an isomorphism and G = H 16

17 We know that any group of the prime order is cyclic Therefore by Theorem 11 any two groups of the same prime order are isomorphic Let p be a prime number We know that the group Z + p 1 is cyclic, since Z + p 1 = [1] It is possible to prove that Z p is also cyclic Also, Z + p 1 = Z p = p 1 Therefore from Theorem 11 it follows that Z + p 1 = Z p Problem: Show that V = Z + 4 Solution: Assume to the contrary that V = Z + 4 Then there is a one-one correspondence f : V Z + 4 From this, in particularly, follows that there exists x V such that This and (iii) of Theorem 10 give f(x) = [1] f(x 2 ) = [f(x)] 2 = [1] 2 = [1] + [1] = [2] We now recall that for any element x V we have By this and (i) of Theorem 10 we get This is a contradiction x 2 = e f(x 2 ) = f(e) = e = [0] Remark: One can show that any group of order 4 is isomorphic to either Z + 4 or V 17

18 X KERNEL Definition 17: If f : G H is a homomorphism, define ker f = {x G : f(x) = 1} Theorem 12: Let f : G H be a homomorphism Then ker f is a subgroup of G Proof 1: From (i) of Theorem 10 it follows that 1 ker f, since f(1) = 1 Next, if x, y ker f, then hence f(x) = 1 = f(y), f(xy) = f(x)f(y) = 1 1 = 1, so xy ker f Finally, if x ker f, then f(x) = 1 and so f(x 1 ) = [f(x)] 1 = 1 1 = 1, hence x 1 ker f Therefore ker f is a subgroup of G Proof 2: From (i) of Theorem 10 it follows that 1 ker f, since f(1) = 1 Therefore ker G is a nonempty set Next, by the definition of a homomorphism and Theorem 10 we have f(xy 1 ) = f(x)f(y 1 ) = f(x)[f(y)] 1 = = 1 for any x, y ker f This means that if x, y ker f, then xy 1 ker f This gives the desired result thanks to Theorem 4 18

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