# Definitions, Theorems and Exercises. Abstract Algebra Math 332. Ethan D. Bloch

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1 Definitions, Theorems and Exercises Abstract Algebra Math 332 Ethan D. Bloch December 26, 2013

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3 Contents 1 Binary Operations Binary Operations Isomorphic Binary Operations Groups Groups Isomorphic Groups Basic Properties of Groups Subgroups Various Types of Groups Cyclic Groups Finitely Generated Groups Dihedral Groups Permutations and Permutation Groups Permutations Part II and Alternating Groups Basic Constructions Direct Products Finitely Generated Abelian Groups Infinite Products of Groups Cosets Quotient Groups

4 2 Contents 5 Homomorphisms Homomorphisms Kernel and Image Applications of Groups Group Actions Rings and Fields Rings Polynomials Vector Spaces Fields Vector Spaces Subspaces Linear Combinations Linear Independence Bases and Dimension Bases for Arbitrary Vector Spaces Linear Maps Linear Maps Kernel and Image Rank-Nullity Theorem Isomorphisms Spaces of Linear Maps Linear Maps and Matrices Review of Matrices Multiplication Linear Maps Given by Matrix Multiplication All Linear Maps F n F m Coordinate Vectors with respect to a Basis Matrix Representation of Linear Maps Basics Matrix Representation of Linear Maps Composition Matrix Representation of Linear Maps Isomorphisms Matrix Representation of Linear Maps The Big Picture Matrix Representation of Linear Maps Change of Basis

5 1 Binary Operations

6 4 1. Binary Operations 1.1 Binary Operations Fraleigh, 7th ed. Section 2 Gallian, 8th ed. Section 2 Judson, 2013 Section 3.2 Definition Let A be a set. A binary operation on A is a function A A A. A unary operation on A is a function A A. Definition Let A be a set, let be a binary operation on A and let H A. The subset H is closed under if a b H for all a,b H. Definition Let A be a set, and let be a binary operation on A. The binary operation satisfies the Commutative Law (an alternative expression is that is commutative) if a b = b a for all a,b A. Definition Let A be a set, and let be a binary operation on A. The binary operation satisfies the Associative Law (an alternative expression is that is associative) if (a b) c = a (b c) for all a,b,c A. Definition Let A be a set, and let be a binary operation on A. 1. Let e A. The element e is an identity element for if a e = a = e a for all a A. 2. If has an identity element, the binary operation satisfies the Identity Law. Lemma Let A be a set, and let be a binary operation on A. If has an identity element, the identity element is unique. Proof. Let e,ê A. Suppose that e and ê are both identity elements for. Then e = e ê = ê, where in the first equality we are thinking of ê as an identity element, and in the second equality we are thinking of e as an identity element. Therefore the identity element is unique. Definition Let A be a set, and let be a binary operation of A. Let e A. Suppose that e is an identity element for. 1. Let a A. An inverse for a is an element a A such that a a = e and a a = e. 2. If every element in A has an inverse, the binary operation satisfies the Inverses Law. Exercises Exercise Which of the following formulas defines a binary operation on the given set?

7 1.1 Binary Operations 5 (1) Let be defined by x y = xy for all x,y { 1, 2, 3,...}. (2) Let be defined by x y = xy for all x,y [2, ). (3) Let be defined by x y = x y for all x,y Q. (4) Let be defined by (x,y) (z,w) = (x + z,y + w) for all (x,y),(z,w) R 2 {(0,0)}. (5) Let be defined by x y = x + y for all x,y N. (6) Let be defined by x y = ln( xy e) for all x,y N. Exercise For each of the following binary operations, state whether the binary operation is associative, whether it is commutative, whether there is an identity element and, if there is an identity element, which elements have inverses. (1) The binary operation on Z defined by x y = xy for all x,y Z. (2) The binary operation on R defined by x y = x + 2y for all x,y R. (3) The binary operation on R defined by x y = x + y 7 for all x,y R. (4) The binary operation on Q defined by x y = 3(x + y) for all x,y Q. (5) The binary operation on R defined by x y = x for all x,y R. (6) The binary operation on Q defined by x y = x + y + xy for all x,y Q. (7) The binary operation on R 2 defined by (x,y) (z,w) = (4xz,y + w) for all (x,y),(z,w) R 2. Exercise For each of the following binary operations given by operation tables, state whether the binary operation is commutative, whether there is an identity element and, if there is an identity element, which elements have inverses. (Do not check for associativity.) (1) (3) x y z w x x z w y y z w y x z w y x z w y x z w. (2) j k l m j k j m j k j k l m l k l j l m j m l m.

8 6 1. Binary Operations (4) a b c d e a d e a b b b e a b a d c a b c d e d b a d e c e b d e c a. (5) i r s a b c i i r s a b c r r s i c a b s s i r b c a a a b c i s r b b c a r i s c c a b s r i. Exercise Find an example of a set and a binary operation on the set such that the binary operation satisfies the Identity Law and Inverses Law, but not the Associative Law, and for which at least one element of the set has more than one inverse. The simplest way to solve this problem is by constructing an appropriate operation table. Exercise Let n N. Recall the definition of the set Z n and the binary operation on Z n. Observe that [1] is the identity element for Z n with respect to multiplication. Let a Z. Prove that the following are equivalent. a. The element [a] Z n has an inverse with respect to multiplication. b. The equation ax 1 (mod n) has a solution. c. There exist p,q Z such that ap + nq = 1. (It turns out that the three conditions listed above are equivalent to the fact that a and n are relatively prime.) Exercise Let A be a set. A ternary operation on A is a function A A A A. A ternary operation : A A A A is left-induced by a binary operation : A A A if ((a,b,c)) = (a b) c for all a,b,c A. Is every ternary operation on a set left-induced by a binary operation? Give a proof or a counterexample. Exercise Let A be a set, and let be a binary operation on A. Suppose that satisfies the Associative Law and the Commutative Law. Prove that (a b) (c d) = b [(d a) c] for all a,b,c,d A. Exercise Let B be a set, and let be a binary operation on B. Suppose that satisfies the Associative Law. Let Prove that P is closed under. P = {b B b w = w b for all w B}. Exercise Let C be a set, and let be a binary operation on C. Suppose that satisfies the Associative Law and the Commutative Law. Let Q = {c C c c = c}.

9 1.1 Binary Operations 7 Prove that Q is closed under. Exercise Let A be a set, and let be a binary operation on A. An element c A is a left identity element for if c a = a for all a A. An element d A is a right identity element for if a d = a for all a A. (1) If A has a left identity element, is it unique? Give a proof or a counterexample. (2) If A has a right identity element, is it unique? Give a proof or a counterexample. (3) If A has a left identity element and a right identity element, do these elements have to be equal? Give a proof or a counterexample.

10 8 1. Binary Operations 1.2 Isomorphic Binary Operations Fraleigh, 7th ed. Section 3 Gallian, 8th ed. Section 6 Definition Let (G, ) and (H, ) be sets with binary operations, and let f : G H be a function. The function f is an isomorphism of the binary operations if f is bijective and if f (a b) = f (a) f (b) for all a,b G. Definition Let (G, ) and (H, ) be sets with binary operations. The binary operations and are isomorphic if there is an isomorphism G H. Theorem Let (G, ) and (H, ) be sets with binary operations. Suppose that (G, ) and (H, ) are isomorphic. 1. (G, ) satisfies the Commutative Law if and only if (H, ) satisfies the Commutative Law. 2. (G, ) satisfies the Associative Law if and only if (H, ) satisfies the Associative Law. 3. (G, ) satisfies the Identity Law if and only if (H, ) satisfies the Identity Law. If f : G H is an isomorphism, then f (e G ) = e H. 4. (G, ) satisfies the Inverses Law if and only if (H, ) satisfies the Inverses Law. Exercises Exercise Prove that the two sets with binary operations in each of the following pairs are isomorphic. (1) (Z,+) and (5Z,+), where 5Z = {5n n Z}. (2) (R {0}, ) and (R { 1}, ), where x y = x + y + xy for all x,y R { 1}. (3) (R 4,+) and (M 2 2 (R),+), where M 2 2 (R) is the set of all 2 2 matrices with real entries. Exercise Let f : Z Z be defined by f (n) = n + 1 for all n Z. (1) Define a binary operation on Z so that f is an isomorphism of (Z,+) and (Z, ), in that order. (2) Define a binary operation on Z so that f is an isomorphism of (Z, ) and (Z,+), in that order. Exercise Prove Theorem (2). Exercise Prove Theorem (3). Exercise Prove Theorem (4).

11 2 Groups

12 10 2. Groups 2.1 Groups Fraleigh, 7th ed. Section 4 Gallian, 8th ed. Section 2 Judson, 2013 Section 3.2 Definition Let G be a non-empty set, and let be a binary operation on G. The pair (G, ) is a group if satisfies the Associative Law, the Identity Law and the Inverses Law. Definition Let (G, ) be a group. The group (G, ) is abelian if satisfies the Commutative Law. Lemma Let G be a group. If g G, then g has a unique inverse. Definition Let G be a group. If G is a finite set, then the order of the group, denoted G, is the cardinality of the set G. Definition Let n N, and let a,b Z. The number a is congruent to the number b modulo n, denoted a b (mod n), if a b = kn for some k Z. Theorem Let n N, and let a Z. Then there is a unique r {0,...,n 1} such that a r (mod n). Theorem Let n N. 1. Let a,b Z. If a b (mod n), then [a] = [b]. If a b (mod n), then [a] [b] = /0. 2. [0] [1]... [n 1] = Z. Definition Let n N. The set of integers modulo n, denoted Z n, is the set defined by Z n = {[0],[1],...,[n 1]}, where the relation classes are for congruence modulo n. Definition Let n N. Let + and be the binary operations on Z n defined by [a] + [b] = [a + b] and [a] [b] = [ab] for all [a],[b] Z n. Lemma Let n N, and let a,b,c,d Z. Suppose that a c (mod n) and b d (mod n). Then a + b c + d (mod n) and ab cd (mod n). Proof. There exist k, j Z such that a c = kn and b d = jn. Then a = c + kn and b = d + jn, and therefore The desired result now follows. a + b = (c + kn) + (d + jn) = c + d + (k + j)n, ab = (c + kn)(d + jn) = cd + (c j + dk + k jn)n. Corollary Let n N, and let [a],[b],[c],[d] Z n. Suppose that [a] = [c] and [b] = [d]. Then [a + b] = [c + d] and [ab] = [cd].

13 2.1 Groups 11 Lemma Let n N. Then (Z n,+) is an abelian group. Example We wish to list all possible symmetries of an equilateral triangle. Such a triangle is shown in Figure The letters A, B and C are not part of the triangle, but are added for our convenience. Mathematically, a symmetry of an object in the plane is an isometry of the plane (that is, a motion that does not change lengths between points) that take the object onto itself. In other words, a symmetry of an object in the plane is an isometry of the plane that leaves the appearance of the object unchanged. Because the letters A, B and C in Figure are not part of the triangle, a symmetry of the triangle may interchange these letters; we use the letters to keep track of what the isometry did. There are only two types of isometries that will leave the triangle looking unchanged: reflections (that is, flips) of the plane in certain lines, and rotations of the plane by certain angles about the center of the triangle. (This fact takes a proof, but is beyond the scope of this course.) In Figure we see the three possible lines in which the plane can be reflected without changing the appearance of the triangle. Let M 1, M 2 and M 3 denote the reflections of the planes in these lines. For example, if we apply reflection M 2 to the plane, we see that the vertex labeled B is unmoved, and that the vertices labeled A and C are interchanged, as seen in Figure The only two possible non-trivial rotations of the plane about the center of the triangle that leave the appearance of the triangle unchanged are rotation by 120 clockwise and rotation by 240 clockwise, denoted R 120 and R 240. We do not need rotation by 120 counterclockwise and rotation by 240 counterclockwise, even though they also leave the appearance of the triangle unchanged, because they have the same net effect as rotation by 240 clockwise and rotation by 120 clockwise, respectively, and it is only the net effect of isometries that is relevant to the study of symmetry. Let I denote the identity map of the plane, which is an isometry, and which can be thought of as rotation by 0. A C B Figure The set G = {I,R 120,R 240,M 1,M 2,M 3 } is the collection of all isometries of the plane that take the equilateral triangle onto itself. Each of these isometries can be thought of as a function R 2 R 2, and as such we can combine these isometries by composition of functions. It can be proved that the composition of isometries is an isometry, and therefore

14 12 2. Groups L 3 L 2 L 1 Figure M 2 A C C B A B Figure

15 2.1 Groups 13 composition becomes a binary operation on the set G; the details are omitted. We can then form the operation table for this binary operation: I R 120 R 240 M 1 M 2 M 3 I I R 120 R 240 M 1 M 2 M 3 R 120 R 120 R 240 I M 3 M 1 M 2 R 240 R 240 I R 120 M 2 M 3 M 1 M 1 M 1 M 2 M 3 I R 120 R 240 M 2 M 2 M 3 M 1 R 240 I R 120 M 3 M 3 M 1 M 2 R 120 R 240 I. Composition of functions is associative, and hence this binary operation on G is associative. Observe that I is an identity element. It is seen that I, M 1, M 2 and M 3 are their own inverses, and that R 120 and R 240 are inverses of each other. Therefore (G, ) is a group. This group is not abelian, however. For example, we see that R 120 M 1 M 1 R 120. The group G is called the symmetry group of the equilateral triangle. Exercises Exercise For each of the following sets with binary operations, state whether the set with binary operation is a group, and whether it is an abelian group. (1) The set (0,1], and the binary operation multiplication. (2) The set of positive rational numbers, and the binary operation multiplication. (3) The set of even integers, and the binary operation addition. (4) The set of even integers, and the binary operation multiplication. (5) The set Z, and the binary operation on Z defined by a b = a b for all a,b Z. (6) The set Z, and the binary operation on Z defined by a b = ab + a for all a,b Z. (7) The set Z, and the binary operation on Z defined by a b = a+b+1 for all a,b Z. (8) The set R { 1}, and the binary operation on R { 1} defined by a b = a + b + ab for all a,b R { 1}. Exercise Let P = {a,b,c,d,e}. Find a binary operation on P given by an operation table such that (P, ) is a group. Exercise Find an example of a set and a binary operation on the set given by an operation table such that each element of the set appears once and only once in each row of the operation table and once and only once in each column, but the set together with this binary operation is not a group.

16 14 2. Groups Exercise Let A be a set. Let P(A) denote the power set of A. Define the binary operation on P(A) by X Y = (X Y ) (Y X) for all X,Y P(A). (This binary operation is called symmetric difference. Prove that (P(A),) is an abelian group. Exercise Let (G, ) be a group. Prove that if x = x for all x G, then G is abelian. Is the converse to this statement true? Exercise Let (H, ) be a group. Suppose that H is finite, and has an even number of elements. Prove that there is some h H such that h e H and h h = e H.

17 2.2 Isomorphic Groups Isomorphic Groups Fraleigh, 7th ed. Section 3 Gallian, 8th ed. Section 6 Judson, 2013 Section 9.1 Definition Let (G, ) and (H, ) be groups, and let f : G H be a function. The function f is an isomorphism (sometimes called a group isomorphism) if f is bijective and if f (a b) = f (a) f (b) for all a,b G. Definition Let (G, ) and (H, ) be groups. The groups G and H are isomorphic if there is an isomorphism G H. If G and H are isomorphic, it is denoted G = H. Theorem Let G, H be groups, and let f : G H be an isomorphism. 1. f (e G ) = e H. 2. If a G, then f (a ) = [ f (a)], where the first inverse is in G, and the second is in H. Proof. We will prove Part (2), leaving the rest to the reader. Let and be the binary operations of G and H, respectively. (2). Let a G. Then f (a) f (a ) = f (a a ) = f (e G ) = e H, where the last equality uses Part (1) of this theorem, and the other two equalities use the fact that f is a isomorphism and that G is a group. A similar calculation shows that f (a ) f (a) = e H. By Lemma 2.1.3, it follows that [ f (a)] = f (a ). Theorem Let G, H and K be groups, and let f : G H and j : H K be isomorphisms. 1. The identity map 1 G : G G is an isomorphism. 2. The function f 1 is an isomorphism. 3. The function j f is an isomorphism. Lemma Let G and H be groups. Suppose that G and H are isomorphic. Then G is abelian if and only if H is abelian. Proof. This lemma follows immediately from Theorem (1). Lemma Let (G, ) be a group, let A be a set, and let f : A G be a bijective map. Then there is a unique binary operation on A such that (A, ) is a group and f is an isomorphism.

18 16 2. Groups Exercises Exercise Which of the following functions are isomorphisms? The groups under consideration are (R,+), and (Q,+), and ((0, ), ). (1) Let f : Q (0, ) be defined by f (x) = 5 x for all x Q. (2) Let k : (0, ) (0, ) be defined by k(x) = x 7 for all x (0, ). (3) Let m: R R be defined by m(x) = x + 3 for all x R. (4) Let g: (0, ) R be defined by g(x) = lnx for all x (0, ). Exercise Prove Theorem (1). Exercise Prove Theorem (1) and (2). Exercise Prove that up to isomorphism, the only two groups with four elements are Z 4 and the Klein 4-group K. Consider all possible operation tables for the binary operation of a group with four elements; use the fact that each element of a group appears once in each row and once in each column of the operation table for the binary operation of the group, as stated in Remark Exercise Let G be a group, and let g G. Let i g : G G be defined by i g (x) = gxg 1 for all x G. Prove that i g is an isomorphism.

19 2.3 Basic Properties of Groups Basic Properties of Groups Theorem Let G be a group, and let a,b,c G. 1. e 1 = e. 2. If ac = bc, then a = b (Cancellation Law). 3. If ca = cb, then a = b (Cancellation Law). 4. (a 1 ) 1 = a. 5. (ab) 1 = b 1 a If ba = e, then b = a If ab = e, then b = a 1. Proof. We prove Part (5), leaving the rest to the reader. Fraleigh, 7th ed. Section 4 Gallian, 8th ed. Section 2 Judson, 2013 Section 3.2 (5). By Lemma we know that ab has a unique inverse. If we can show that (ab)(b 1 a 1 ) = e and (b 1 a 1 )(ab) = e, then it will follow that a 1 b 1 is the unique inverse for ab, which means that (ab) 1 = b 1 a 1. Using the definition of a group we see that (ab)(b 1 a 1 ) = [(ab)b 1 ]a 1 = [a(bb 1 )]a 1 = [ae]a 1 = aa 1 = e. A similar computation shows that (b 1 a 1 )(ab) = e. Remark A useful consequence of Theorem (2) and (3) is that if the binary operation of a group with finitely many elements is given by an operation table, then each element of the group appears once and only once in each row of the operation table and once and only once in each column (consider what would happen otherwise). On the other hand, just because an operation table does have each element once and only once in each row and once and only once in each column does not guarantee that the operation yields a group; the reader is asked to find such an operation table in Exercise Theorem Let G be a group. The following are equivalent. a. G is abelian. b. (ab) 1 = a 1 b 1 for all a,b G.

20 18 2. Groups c. aba 1 b 1 = e for all a,b G. d. (ab) 2 = a 2 b 2 for all a,b G. Theorem (Definition by Recursion). Let H be a set, let e H and let k : H H be a function. Then there is a unique function f : N H such that f (1) = e, and that f (n + 1) = k( f (n)) for all n N. Definition Let G be a group and let a G. 1. The element a n G is defined for all n N by letting a 1 = a, and a n+1 = a a n for all n N. 2. The element a 0 G is defined by a 0 = e. For each n N, the element a n is defined by a n = (a n ) 1. Lemma Let G be a group, let a G and let n,m Z. 1. a n a m = a n+m. 2. (a n ) = a n. Lemma Let G be a group, let a G and let n,m Z. 1. a n a m = a n+m. 2. (a n ) 1 = a n. Proof. First, we prove Part (1) in the case where n N and m = n. Using the definition of a n we see that a n a m = a n a n = a n (a n ) 1 = e = a 0 = a n+m. We now prove Part (2). There are three cases. First, suppose n > 0. Then n N. Then the result is true by definition. Second, suppose n = 0. Then (a n ) 1 = (a 0 ) 1 = e 1 = e = a 0 = a 0 = a n. Third, suppose n < 0. Then n N. We then have e = e 1 = (a 0 ) 1 = (a n a n ) 1 = (a n ) 1 (a n ) 1, and similarly e = (a n ) 1 (a n ) 1. By the uniqueness of inverses we deduce that [(a n ) 1 ] 1 = (a n ) 1, and hence a n = (a n ) 1. We now prove Part (1) in general. There are five cases. First, suppose that n > 0 and m > 0. Then n,m N. This part of the proof is by induction. We will use induction on k to prove that for each k N, the formula a k a p = a k+p holds for all p N. Let k = 1. Let p N. Then by the definition of a p we see that a k a p = a 1 a p = a a p = a p+1 = a 1+p = a k+p. Hence the result is true for k = 1. Now let k N. Suppose that the result is true for k. Let p N. Then by the definition of a k we see that a k+1 a p = (a a k ) a p = a (a k a p ) = a a k+p = a (k+p)+1 = a (k+1)+p. Hence the result is true for k + 1. It follows by induction that a k a p = a k+p for all p N, for all k N. Second, suppose that n = 0 or m = 0. Without loss of generality, assume that n = 0. Then by the definition of a m we see that a n a m = a 0 a m = e a m = a m = a 0+m = a n+m.

21 2.3 Basic Properties of Groups 19 Third, suppose that n > 0 and m < 0. Then m > 0, and hence n N and m N. There are now three subcases. For the first subcase, suppose that n + m > 0. Therefore n + m N. Let r = n + m. Then n = r + ( m). Because r > 0 and m > 0, then by the first case in this proof, together with the definition of a ( m), we see that a n a m = a r+( m) a m = (a r a m )a ( m) = a r (a m a ( m) ) = a r (a m (a m ) 1 ) = a r e = a r = a n+m. For the second subcase, suppose that n + m = 0. Then m = n. Using the definition of a n we see that a n a m = a n a n = a n (a n ) 1 = e = a 0 = a n+m. For the third subcase, suppose that n + m < 0. Then ( n) + ( m) = (n + m) > 0. Because m > 0 and n < 0, we can use the first of our three subcases, together with the definition of a n, to see that a n a m = ((a n ) 1 ) 1 ((a m ) 1 ) 1 = (a n ) 1 (a m ) 1 = [a m a n ] 1 = [a ( m)+( n) ] 1 = [a (n+m) ] 1 = [(a n+m ) 1 ] 1 = a n+m. We have now completed the proof in the case where n > 0 and m < 0. Fourth, suppose that n < 0 and m > 0. This case is similar to the previous case, and we omit the details. Fifth, suppose that n < 0 and m < 0. Then n > 0 and m > 0, and we can proceed similarly to the third subcase of the third case of this proof; we omit the details. Definition Let A be a set, and let be a binary operation on A. An element e A is a left identity element for if e a = a for all a A. If has a left identity element, the binary operation satisfies the Left Identity Law. Definition Let A be a set, and let be a binary operation of A. Let e A. Suppose that e is a left identity element for. If a A, a left inverse for a is an element a A such that a a = e. If every element in A has a left inverse, the binary operation satisfies the Left Inverses Law. Theorem Let G be a set, and let be a binary operation of A. If the pair (G, ) satisfies the Associative Law, the Left Identity Law and the Left Inverses Law, then (G, ) is a group. Exercises Exercise Let H be a group, and let a,b,c H. Prove that if abc = e H, then bca = e H. Exercise Let G be a group. An element g G is idempotent if g 2 = g. Prove that G has precisely one idempotent element. Exercise Let H be a group. Suppose that h 2 = e H for all h H. Prove that H is abelian. Exercise Let G be a group, and let a,b G. Prove that (ab) 2 = a 2 b 2 if and only if ab = ba. Exercise Let G be a group, and let a,b G. Prove that (ab) 1 = a 1 b 1 if and only if ab = ba. (Do not use Theorem )

22 20 2. Groups Exercise Find an example of a group G, and elements a,b G, such that (ab) 1 a 1 b 1. Exercise Let (H, ) be a group. Let be the binary operation on H defined by a b = b a for all a,b H. (1) Prove that (H, ) is a group. (2) Prove that (H, ) and (H, ) are isomorphic. Exercise Let G be a group, and let g G. Let i g : G G be defined by i g (x) = gxg for all x G. Prove that i g is an isomorphism. Exercise Let G be a group, and let g G. Suppose that G is finite. Prove that there is some n N such that g n = e G.

23 2.4 Subgroups Subgroups Fraleigh, 7th ed. Section 5 Gallian, 8th ed. Section 3 Judson, 2013 Section 3.3 Definition Let G be a group, and let H G be a subset. The subset H is a subgroup of G if the following two conditions hold. (a) H is closed under. (b) (H, ) is a group. If H is a subgroup of G, it is denoted H G. Lemma Let G be a group, and let H G. Proof. 1. The identity element of G is in H, and it is the identity element of H. 2. The inverse operation in H is the same as the inverse operation in G. (1). Let e G be the identity element of G. Because (H, ) is a group, it has an identity element, say e H. (We cannot assume, until we prove it, that e H is the same as e G.) Then e H e H = e H thinking of e H as being in H, and e G e H = e H thinking of e H as being in G. Hence e H e H = e G e H. Because both e H and e G are in G, we can use Theorem (2) to deduce that e H = e G. Hence e G H, and e G is the identity element of H. (2). Now let a H. Because (G, ) is a group, the element a has an inverse a 1 G. We will show that a 1 H. Because (H, ) is a group, then a has an inverse â H. (Again, we cannot assume, until we prove it, that â is the same as a 1.) Using the definition of inverses, and what we saw in the previous paragraph, we know that a 1 a = e G and â a = e G. Hence â a = a 1 a. Using Theorem (2) again, we deduce that a 1 = â. Theorem Let G be a group, and let H G. Then H G if and only if the following three conditions hold. (i) e H. (ii) If a,b H, then a b H. (iii) If a H, then a 1 H.

24 22 2. Groups Proof. First suppose that H is a subgroup. Then Property (ii) holds by the definition of a subgroup, and Properties (i) and (iii) hold by Lemma Now suppose that Properties (i), (ii) and (iii) hold. To show that H is a subgroup, we need to show that (H, ) is a group. We know that is associative with respect to all the elements of G, so it certainly is associative with respect to the elements of H. The other properties of a group follow from what is being assumed. Theorem Let G be a group, and let H G. Then H G if and only if the following three conditions hold. (i) H /0. (ii) If a,b H, then a b H. (iii) If a H, then a 1 H. Proof. First, suppose that H is a subgroup. Then Properties (i), (ii) and (iii) hold by Lemma Second, suppose that Properties (i), (ii) and (iii) hold. Because H /0, there is some b H. By Property (iii) we know that b 1 H. By Property (ii) we deduce that b 1 b H, and hence e H. We now use Lemma to deduce that H is a subgroup. Lemma Let G be a group, and let K H G. If K H and H G, then K G. Lemma Let G be a group, and let {H i } i I be a family of subgroups of G indexed by I. Then i I H i G. Theorem Let G, H be groups, and let f : G H be an isomorphism. 1. If A G, then f (A) H. 2. If B H, then f 1 (B) G. Proof. We will prove Part (1), leaving the rest to the reader. Let and be the binary operations of G and H, respectively. (1). Let A G. By Lemma (1) we know that e G A, and by Theorem (1) we know that e H = f (e G ) f (A). Hence f (A) is non-empty. We can therefore use Theorem to show that f (A) is a subgroup of H. Let x,y f (A). Then there are a,b A such that x = f (a) and y = f (b). Hence x y = f (a) f (b) = f (a b), because f is a isomorphism. Because A is a subgroup of G we know that a b A, and hence x y f (A). Using Theorem (2) we see that x 1 = [ f (a)] 1 = f (a 1 ). Because A is a subgroup of G, it follows from Theorem (iii) that a 1 A. We now use Theorem to deduce that f (A) is a subgroup of H.

25 2.4 Subgroups 23 Lemma Let (G, ) and (H, ) be groups, and let f : G H be a function. Suppose that f is injective, and that f (a b) = f (a) f (b) for all a,b G. 1. f (G) H. 2. The map f : G f (G) is an isomorphism. Proof. Part (1) is proved using the same proof as Theorem (1), and Part (2) is trivial. Exercises Exercise Let GL 2 (R) denote the set of invertible 2 2 matrices with real number entries, and let SL 2 (R) denote the set of all 2 2 matrices with real number entries that have determinant 1. Prove that SL 2 (R) is a subgroup of GL 2 (R). (This exercise requires familiarity with basic properties of determinants.) Exercise Let n N. (1) Prove that (Z n,+) is an abelian group. (2) Suppose that n is not a prime number. Then n = ab for some a,b N such that 1 < a < n and 1 < b < n. Prove that the set {[0],[a],[2a],...,[(b 1)a]} is a subgroup of Z n. (3) Is (Z n {[0]}, ) a group for all n? If not, can you find any conditions on n that would guarantee that (Z n {[0]}, ) is a group? Exercise Find all the subgroups of the symmetry group of the square. Exercise Let G be a group, and let A,B G. Suppose that G is abelian. Let AB denote the subset AB = {ab a A and b B}. Prove that if A,B G, then AB G. Exercise Let G be a group, and let H G. Prove that H G if and only if the following two conditions hold. (i) H /0. (ii) If a,b H, then a b 1 H. Exercise Let G be a group. Suppose that G is abelian. Let I denote the subset Prove that I G. I = {g G g 2 = e G }.

26 24 2. Groups Exercise Let G be a group, and let H G. Suppose that the following three conditions hold. (i) H /0. (ii) H is finite. (iii) H is closed under. Prove that H G. Exercise Let G be a group, and let s G. Let C s denote the subset Prove that C s G. C s = {g G gs = sg}. Exercise Let G be a group, and let A G. Let C A denote the subset Prove that C A G. C A = {g G ga = ag for all a A}.

27 3 Various Types of Groups

28 26 3. Various Types of Groups 3.1 Cyclic Groups Lemma Let G be a group and let a G. Then Fraleigh, 7th ed. Section 6 Gallian, 8th ed. Section 4 Judson, 2013 Section 4.1 {a n n Z} = {H G a H}. (3.1.1) Definition Let G be a group and let a G. The cyclic subgroup of G generated by a, denoted a, is the set in Equation Definition Let G be a group. Then G is a cyclic group if G = a for some a G; the element a is a generator of G. Definition Let G be a group and let H G. Then H is a cyclic subgroup of G if H = a for some a G; the element a is a generator of H. Definition Let G be a group and let a G. If a is finite, the order of a, denoted a, is the cardinality of a. If a is infinite, then a has infinite order. Theorem (Well-Ordering Principle). Let A N be a set. If A is non-empty, then there is a unique m A such that m a for all a A. Theorem (Division Algorithm). Let a,b Z. Suppose that b 0. Then there are unique q,r Z such that a = qb + r and 0 r < b. Theorem Let G be a group, let a G and let m N. Then a = m if and only if a m = e and a i e for all i {1,...,m 1}. Proof. First, suppose that a = m. Then a = {a n n Z} has m elements. Therefore the elements a 0,a,a 2,...,a m cannot be all distinct. Hence there are i, j {0,...,m} such that i j and a i = a j. WLOG assume i < j. Then a j i = e. Observe that j i N, and j i m. Let M = {p N a p = e}. Then M N, and M /0 because j i M. By the Well-Ordering Principle (Theorem 3.1.6), there is a unique k M such that k x for all x M. Hence a k = e and a q e for all q {1,...,k 1}. Because j i M, it follows that k j i m. Let y Z. By the Division Algorithm (Theorem 3.1.7), there are unique q,r Z such that y = qk + r and 0 r < k. Then a y = a qk+r = (a k ) q a r = e q a r = a r. It follows that a {a n n {0,...,k 1}}. Therefore a k. Because a = m, we deduce that m k. It follows that k = m. Hence a m = e and a i e for all i {1,...,m 1}. Second, suppose a m = e and a i e for all i {1,...,m 1}. We claim that a {a n n {0,...,m 1}}, using the Division Algorithm similarly to an argument seen previously in this proof. Additionally, we claim that all the elements a 0,a,a 2,...,a m 1 are distinct. If not, then an argument similar to the one above would show that there are s,t {0,...,m 1}

29 3.1 Cyclic Groups 27 such that s < t and a t s = e; because t s m 1 < m, we would have a contradiction. It follows that a = {a n n {0,...,m 1}}, and hence a = m. Lemma Let G be a group. If G is cyclic, then G is abelian. Theorem Let G be a group and let H G. If G is cyclic, then H is cyclic. Corollary Every subgroup of Z has the form nz for some n N {0}. Theorem Let G be a group. Suppose that G is cyclic. 1. If G is infinite, then G = Z. 2. If G = n for some n N, then G = Z n. Definition Let a,b Z. If at least one of a or b is not zero, the greatest common divisor of a and b, denoted (a,b), is the largest integer that divides both a and b. Let (0,0) = 0. Lemma Let a, b Z. Then (a, b) exists and (a, b) 0. Definition Let a,b Z. The numbers a and b are relatively prime if (a,b) = 1. Lemma Let a,b Z. Then there are m,n Z such that (a,b) = ma + nb. Corollary Let a,b Z. If r is a common divisor of a and b, then r (a,b). Corollary Let a,b Z. Then (a,b) = 1 if and only if there are m,n Z such that ma + nb = 1. Corollary Let a,b,r Z. Suppose that (a,b) = 1. If a br then a r. Theorem Let G be a group. Suppose that G = a for some a G, and that G = n for some n N. Let s,r N. 1. a s = n (n,s). 2. a s = a r if and only if (n,s) = (n,r). Corollary Let G be a group. Suppose that G = a for some a G, and that G = n for some n N. Let s N. Then G = a s if and only if (n,s) = 1. Exercises Exercise Let C be a cyclic group of order 60. How many generators does C have? Exercise List all orders of subgroups of Z 20. Exercise Find all the subgroups of Z 36. Exercise Let p,q N be prime numbers. How many generators does the group Z pq have?

30 28 3. Various Types of Groups Exercise Let G be a group, and let g,h G. Prove that if gh has order p for some p N, then hg has order p. Exercise Let G, H be groups, and let f,k : G H be isomorphisms. Suppose that G = a for some a G. Prove that if f (a) = k(a), then f = k. Exercise Let G be a group. Prove that if G has a finite number of subgroups, then G is finite. Exercise Let G be a group, and let A,B G. Suppose that G is abelian, and that A and B are cyclic and finite. Suppose that A and B are relatively prime. Prove that G has a cyclic subgroup of order A B

31 3.2 Finitely Generated Groups Finitely Generated Groups Lemma Let G be a group and let S G. 1. {H G S H} /0. 2. {H G S H} G. 3. If K G and S K, then {H G S H} K. Fraleigh, 7th ed. Section 7 Proof. Part (1) holds because G {H G S H}, Part (2) holds by Lemma 2.4.6, and Part (3) is straightforward. Definition Let G be a group and let S G. The subgroup generated by S, denoted S, is defined by S = {H G S H}. Theorem Let G be a group and let S G. Then S = {(a 1 ) n 1 (a 2 ) n2 (a k ) n k k N, and a 1,...,a k S, and n 1,...,n k Z}. Remark In an abelian group, using additive notation, we would write the result of Theorem as S = {n 1 a 1 + n 2 a n k a k k N, and a 1,...,a k S, and n 1,...,n k Z}. Definition Let G be a group and let S G. Suppose S = G. The set S generates G, and the elements of S are generators of G. Definition Let G be a group. If G = S for some finite subset S G, then G is finitely generated.

32 30 3. Various Types of Groups 3.3 Dihedral Groups Fraleigh, 7th ed. Section 8 Gallian, 8th ed. Section 1 Judson, 2013 Section 5.2 Theorem Let n N. Suppose n 3. For a regular n-gon, let 1 denote the identity symmetry, let r denote the smallest possible clockwise rotation symmetry, and let m denote a reflection symmetry. 1. r n = 1, and r k 1 for k {1,...,n 1}. 2. m 2 = 1 and m rm = mr If p Z, then r p m = mr p. 5. If p N, then r p = r k for a unique k {0,1,...,n 1}. 6. If p,s {0,1,...,n 1}, then r p = r s and mr p = mr s if and only if p = s. 7. If p {0,1,...,n 1}, then m r p. 8. If p {0,1,...,n 1}, then (r p ) 1 = r n p and (mr p ) 1 = mr p. Proof. Parts (1) and (2) follow immediately from the definition of r and m. Part (3) can be verified geometrically by doing both sides to a polygon. For Part (4), prove it for positive p by induction, for p = 0 trivially, and for negative p by using the fact that it has been proved for positive p and then for p negative looking at mr p m = mr ( p) m = r p mm = r p. Part (5) uses the Division Algorithm (Theorem 3.1.7). Part (6) is proved similarly to proofs we saw for cyclic groups. Part (7) is geometric, because t reverses orientation and powers of r preserve orientation. Part (8) is also geometric, using basic ideas about rotations and reflections. Definition Let n N. Suppose n 3. For a regular n-gon, let 1 denote the identity symmetry, let r denote the smallest possible clockwise rotation symmetry, and let m denote a reflection symmetry. The n-th dihedral group, denoted D n, is the group D n = {1,r,r 2,...,r n 1,m,mr,mr 2,...,mr n 1 }. Theorem Let n N. Suppose n 3. Then there is a unique group with generators a and b that satisfy the following three conditions. (i) a n = 1, and a k 1 for all k {1,...,n 1}. (ii) b 2 = 1 and b 1. (iii) ab = ba 1.

33 3.4 Permutations and Permutation Groups Permutations and Permutation Groups Fraleigh, 7th ed. Section 8 Gallian, 8th ed. Section 5, 6 Judson, 2013 Section 5.1 Definition Let A be a non-empty set. A permutation of A is a bijective map A A. The set of all permutations of A is denoted S A. The identity permutation A A is denoted ι. Definition Let A be a non-empty set. The composition of two permutations of A is called permutation multiplication. Lemma Let A be a non-empty set. The pair (S A, ) is a group. Remark In the group S A, we will usually write στ as an abbreviation of σ τ. Lemma Let A and B be non-empty sets. Suppose that A and B have the same cardinality. Then S A = SB. Definition Let n N, and let A = {1,...,n}. The group S A is denoted S n. The group S n is the symmetric group on n letters. Proposition Let A be a non-empty set. Then S A is abelian if and only if A is finite and A 2. Proof. It is easy to see that S 1 and S 2 are abelian (look at the groups explicitly). We see that S 3 is not abelian by looking at the two permutations with second rows (231) and (132). However, those two permutations can be thought of as belonging to S A for any A that has three or more elements (including the case when A is infinite), and hence all such groups are not abelian. Theorem (Cayley s Theorem). Let G be a group. Then there is a set A such that G is isomorphic to a subgroup of S A. If G is finite, a finite set A can be found. Exercises Exercise Let σ,τ S 5 be defined by ( ) σ = and τ = Compute each of the following. (1) σ 2. (2) τ 1. ( )

34 32 3. Various Types of Groups (3) τσ. (4) στ. Exercise Find all subgroups of S 3. Exercise Let n N. Suppose n 3. Let σ S n. Suppose that στ = τσ for all τ S n. Prove that σ = ι. Exercise Let A be a non-empty set, and let P S A. The subgroup P is transitive on A if for each x,y A, there is some σ P such that σ(x) = y. Prove that if A is finite, then there is a subgroup Q S A such that Q is cyclic, that Q = A, and that Q is transitive on A.

35 3.5 Permutations Part II and Alternating Groups Permutations Part II and Alternating Groups Fraleigh, 7th ed. Section 9 Gallian, 8th ed. Section 5 Judson, 2013 Section 5.1 Definition Let A be a non-empty set, and let σ S A. Let be the relation on A defined by a b if and only if b = σ n (a) for some n Z, for all a,b A. Lemma Let A be a non-empty set, and let σ S A. The relation is an equivalence relation on A. Definition Let A be a set, and let σ S A. The equivalence classes of are called the orbits of σ. Definition Let n N, and let σ S n. The permutation σ is a cycle if it has at most one orbit with more than one element. The length of a cycle is the number of elements in its largest orbit. A cycle of length 2 is a transposition. Lemma Let n N, and let σ S n. Then σ is the product of disjoint cycles; the cycles of length greater than 1 are unique, though not their order. Corollary Let n N, and let σ S n. Suppose n 2. Then σ is the product of transpositions. Theorem Let n N, and let σ S n. Suppose n 2. Then either all representations of σ as a product of transpositions have an even number of transpositions, or all have an odd number of transpositions. Definition Let n N, and let σ S n. Suppose n 2. The permutation σ is even or odd, respectively, if it is the product of an even number or odd number, respectively, of transpositions. Definition Let n N. Suppose n 2. The set of all even permutations of A is denoted A n. Lemma Let n N. Suppose n The set A n is a subgroup of S n. 2. A n = n! 2. Definition Let n N. Suppose n 2. The group A n is the alternating group on n letters. Exercises

36 34 3. Various Types of Groups Exercise Compute the product of cycles (1,5,2)(3,4) as a single permutation in the following groups. (1) In S 5. (2) In S 6. Exercise Let σ S 7 be defined by ( ) σ = (1) Write σ as a product of cycles. (2) Write σ as a product of transpositions. Exercise Let n N. Suppose n 3. Let σ S n. (1) Prove that σ can be written as a product of at most n 1 transpositions. (2) Prove that if σ is not a cycle, it can be written as a product of at most n 2 transpositions. (3) Prove that if σ is odd, it can be written as a product of 2n + 3 transpositions. (4) Prove that if σ is even, it can be written as a product of 2n + 8 transpositions. Exercise Let n N. Suppose n 2. Let K S n. Prove that either all the permutations in K are even, or exactly half the permutations in K are even. Exercise Let n N. Suppose n 2. Let σ S n. Suppose that σ is odd. Prove that if τ S n is odd, then there is some η A n such that τ = ση. Exercise Let n N. Let σ S n. Prove that if σ is a cycle of odd length, then σ 2 is a cycle.

37 4 Basic Constructions

38 36 4. Basic Constructions 4.1 Direct Products Fraleigh, 7th ed. Section 11 Gallian, 8th ed. Section 8 Judson, 2013 Section 9.2 Definition Let H and K be groups. The product binary operation on H K is the binary operation defined by (h 1,k 1 )(h 2,k 2 ) = (h 1 h 2,k 1 k 2 ) for all (h 1,k 1 ),(h 2,k 2 ) H K. Lemma Let H and K be groups. The set H K with the product binary operation is a group. Definition Let H and K be groups. The set H K with the product binary operation is the direct product of the groups H and K. Lemma Let H and K be groups. Then H K = K H. Lemma Let H and K be groups. Suppose that H and K are abelian. Then H K is abelian. Theorem Let m,n N. The group Z m Z n is cyclic and is isomorphic to Z mn if and only if m and n are relatively prime. Definition Let G be a group, and let A,B G. Let AB denote the subset AB = {ab a A and b B}. Lemma Let H and K be groups. Let H = H {e K } and K = {e H } K. 1. H, K H K. 2. H K = H K. 3. H K = {(e H,e K )}. 4. hk = kh for all h H and k K. Lemma Let G be a group, and let H,K G. Suppose that the following properties hold. (i) HK = G. (ii) H K = {e}. (iii) hk = kh for all h H and k K. Then G = H K. Lemma Let G be a group, and let H,K G. Then HK = G and H K = {e} if and only if for every g G there are unique h H and k K such that g = hk.

39 4.1 Direct Products 37 Theorem Let m 1,...,m r N. The group r i=1 Z mi is cyclic and is isomorphic to Z m1 m 2 m r if and only if m i and m k are relatively prime for all i,k {1,...,r} such that i k. Exercises Exercise List all the elements of Z 3 Z 4, and find the order of each element. Exercise Find all the subgroups of Z 2 Z 2 Z 2. Exercise Prove Lemma Exercise Prove Lemma

40 38 4. Basic Constructions 4.2 Finitely Generated Abelian Groups Fraleigh, 7th ed. Section 11 Gallian, 8th ed. Section 11 Judson, 2013 Section 13.1 Theorem (Fundamental Theorem of Finitely Generated Abelian Groups). Let G be a finitely generated abelian group. Then G = Z (p1 ) n 1 Z (p2 ) n 2 Z (pk ) n k Z Z Z for some k N, and prime numbers p 1,..., p k N, and n 1,...,n k N. This direct product is unique up to the rearrangement of factors. Exercises Exercise Find, up to isomorphism, all abelian groups of order 16. Exercise Find, up to isomorphism, all abelian groups of order 720. Exercise How many abelian groups are there, up to isomorphism, of order 24. Exercise Let G be a group. Suppose that G is finite and abelian. Prove that G is not cyclic if and only if there is some prime number p N such that G has a subgroup isomorphic to Z p Z p.

41 4.3 Infinite Products of Groups Infinite Products of Groups Definition Let I be a non-empty set, and let {A i } i I be a family of sets indexed by I. The product of the family of sets, denoted i I A i, is the set defined by A i = { f F (I, A i ) f (i) A i for all i I}. i I i I If all the sets A i are equal to a single set A, the product i I A i is denoted by A I. Theorem Let I be a non-empty set, and let {A i } i I be a family of non-empty sets indexed by I. Then i I A i /0. Definition Let I be a non-empty set, and let {G i } i I be a family of non-empty groups indexed by I. Let f,g i I G i. 1. Let f g: I i I A i be defined by ( f g)(i) = f (i)g(i) for all i I. 2. Let f : I i I A i be defined by ( f )(i) = [ f (i)] 1 for all i I. 3. Let ē: I i I A i be defined by ē(i) = e Gi for all i I. Lemma Let I be a non-empty set, and let {G i } i I be a family of non-empty groups indexed by I. Let f,g i I G i. 1. f g i I G i. 2. f i I G i. 3. ē i I G i. Lemma Let I be a non-empty set, and let {G i } i I be a family of non-empty groups indexed by I. Then i I G i is group. Proof. We will show the Associative Law; the other properties are similar. Let f,g,h i I G i. Let i I. Then Hence ( f g)h = f (gh). [( f g)h](i) = [( f g)(i)]h(i) = [ f (i)g(i)]h(i) = f (i)[g(i)h(i)] = f (i)[(gh)(i)] = [ f (gh)](i).

42 40 4. Basic Constructions 4.4 Cosets Fraleigh, 7th ed. Section 10 Gallian, 8th ed. Section 7 Judson, 2013 Section 6.1, 6.2 Definition Let G be a group and let H G. Let L and R be the relations on G defined by a L b if and only if a 1 b H for all a,b G, and a R b if and only if ab 1 H for all a,b G. Lemma Let G be a group and let H G. The relations L and R are equivalence relations on G. Definition Let G be a group, let H G and let a G. Let ah and Ha be defined by ah = {ah h H} and Ha = {ha h H}. Lemma Let G be a group, let H G and let a G. 1. The equivalence class of a with respect to L is ah. 2. The equivalence class of a with respect to R is Ha. Proof. With respect to L, we have The proof for R is similar. [a] = {b G a L b} = {b G a 1 b H} = {b G a 1 b = h for some h H} = {b G b = ah for some h H} = {ah h H} = ah. Definition Let G be a group, let H G and let a G. The left coset of a (with respect to H) is the set ah. The right coset of a (with respect to H) is the set Ha. Lemma Let G be a group, let H G and let a,b G. 1. ah = bh if and only if a 1 b H. 2. Ha = Hb if and only if ab 1 H. 3. ah = H if and only if a H. 4. Ha = H if and only if a H. Proof. This lemma follows immediately from the definition of L and R and Lemma

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