3.3 Equivalence Relations and Partitions on Groups

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1 84 Chapter 3. Groups 3.3 Equivalence Relations and Partitions on Groups Definition Let (G, ) be a group and let H be a subgroup of G. Let H be the relation on G defined by a H b if and only if ab 1 2 H. (3.8) Theorem Let (G, ) be a group and let H be a subgroup of G. The relation H is an equivalence relation on G. Proof. See Exercise 2. Recall that when is an equivalence relation on a set A, weleta/ denote the partition {[a] :a 2 A} of A induced by (see Theorem and Definition 0.5.7). Definition Let (G, ) be a group and let H be a subgroup of G. For each a 2 G, the equivalence class of a is defined by [a] ={b 2 G : b H a}. We shall write G/H as the set of equivalence classes of the relation H, that is, G/H = G/ H = {[a] :a 2 G}. Thus, G/H = {[a] :a 2 G} is a partition of G by Theorem 0.5.6, the fundamental theorem on equivalence relations. The following notion of a right coset is very useful for computing each equivalence class in G/H, wheneverh is a subgroup of a group G. Definition (Right Cosets). Let (G, ) be a group and let H be a subgroup of G. A right coset of H in G is a subset of G of the form Ha = {ha : h 2 H} for some a 2 G. Theorem Let (G, ) be a group and let H be a subgroup of G. Consider the equivalence relation H, as defined by (3.8). For each a 2 G we have that [a] =Ha. Consequently, G/H = {[a] :a 2 G} = {Ha : a 2 G}. Proof. Let G and H be as stated in the theorem. Recall that the equivalence relation H on G is defined by a H b if and only if ab 1 2 H. (3.9) Now, let a 2 G be arbitrary. We shall prove that [a] =Ha, that is, we shall prove that these two sets are equal. Let x 2 G be arbitrary. We see that x 2 [a] i x H a i xa 1 2 H by (3.9) i (xa 1 )a 2 Ha by the definition of [a] by the definition of Ha i x 2 Ha since (xa 1 )a = x. Therefore, [a] =Ha. Consequently, G/H = {[a] :a 2 G} = {Ha : a 2 G}. Let (G, ) be a group. Our next theorem will show, for each a 2 G, that the sets H and Ha have the same number of elements. Theorem Let (G, ) be a group and let H be a subgroup of G. Let a 2 G be arbitrary and let Ha be the right coset of H; thatis,ha = {ha : h 2 H}. Consider the function f : H! Ha defined by f(x) =xa for all x 2 H. Then the function f is one-to-one and onto. Proof. See Exercise 3

2 3.3 Equivalence Relations and Partitions on Groups 85 Theorem Let (G, ) be a finite group and let H be a subgroup of G. Then for each a 2 G, we have that [a] =Ha and [a] = H (the set [a] and the set H have the same number of elements). Proof. Let (G, ) be a finite group and let H be a subgroup of G. [a] =Ha, and Theorem implies that [a] = H. Theorem asserts that Lagrange s Theorem Theorem (Lagrange s Theorem). Let (G, ) be a finite group and let H be a subgroup of G. Then the order of H divides the order of G. Proof. Let (G, ) be a finite group and let H be a subgroup of G. SinceG is finite, so is H. Recall that G denotes the number of elements in G, and H denotes the number of elements in H. We shall prove that H evenly divides G. Consider the equivalence relation H on G defined by a H b if and only if ab 1 2 H. Thus, G/H = {[a] :a 2 G} is a partition of G by Theorem 0.5.6, the fundamental theorem on equivalence relations. Since G is a finite set, there must be a finite list a 1,a 2,...,a k of elements in G such that all the distinct equivalence classes of H are in the list [a 1 ], [a 2 ],...,[a k ]. Thus, {[a] :a 2 G} = {[a 1 ], [a 2 ],...,[a k ]} is a partition of G as illustrated in the following figure: G = x y z a 1 a 2 a k " " " [a 1 ] [a 2 ] [a k ] Theorem states that [a i ] = H for each i =1, 2,...,k. Thus, each equivalence class [a i ] has H many elements. Since G =[a 1 ] [ [a 2 ] [ [[a k ] and the sets [a 1 ], [a 2 ],...,[a k ] are all mutually disjoint, it follows that G = [a 1 ] + [a 2 ] + + [a k ] = H + H + + H = k H. {z } k times So G = k H. Therefore, H divides G. Remark In the above proof we showed that G = k H where k is the number of elements in G/H. Consequently, the number of elements in G/H is equal to G H. These observations inspire the following corollary and definition. Corollary Let (G, ) be a finite group with subgroup H. Then G = G/H H. Definition Let (G, ) be a finite group and let H be a subgroup of G. Thentheindex of H in G is the natural number, denoted by [G : H], defined by [G : H] = G H.Thus,[G: H] denotes the number of elements in G/H.

3 86 Chapter 3. Groups Applications of Lagrange s Theorem Definition Let (G, ) be a group. G is called cyclic if there is an a 2 G such that G = hai = {a i : i 2 Z}. The element a is called a generator of the group G. Theorem Let (G, ) be a finite group and let p = G, the order of G. If p>1 is prime number, then G is a cyclic group. Proof. Let (G, ) be a finite group and let p = G the order of G. Assume that p>1isprime number. We shall prove that G is a cyclic group. Since p>1 then there is an element a 2 G such that a 6= e. Let hai = {a i : i 2 Z} be the cyclic subgroup of G generated by a. Note that a, e 2hai and so hai 2. By Lagrange s Theorem, we have that hai divides G = p. Butsincep is a prime number, it follows that hai = p. So, since hai G and hai has the same number of elements as G. We conclude that hai = G and thus, a is a generator for G. Therefore, G is a cyclic group. This completes the proof of the Theorem. Lemma Let (G, ) be a finite group. Let a 2 G and let m = o(a). Consider the set A = {e, a, a 2,...,a m 1 }. Then (A, ) is a subgroup of G. Proof. See Exercise 4. Theorem Let (G, ) be a finite group and let a 2 G. Then o(a), the order of a, divides G. Proof. Let (G, ) be a finite group. Let a 2 G and let m = o(a). Lemma states that the set A = {e, a, a 2,...,a m 1 } is a subgroup of G. Since A = o(a), Lagrange s Theorem implies that o(a) divides G. This completes the proof of the Theorem. Theorem Let (G, ) be a finite group and let n = G. Then a n = e, foralla 2 G. Proof. Let (G, ) be a finite group and let n = G. Let a 2 G be arbitrary and let m = o(a). So, by definition of o(a), we see that a m = e. By Theorem , we also see that m n, that is, n = km for some k 2 N. Thus, This completes the proof of the Theorem The Conjugacy Relation a n = a km =(a m ) k = e k = e. When G is a nonabelian group, the conjugacy relation on G is another equivalence relation on G which gives us alternative way to partition the group. This new partitioning of G o ers another tool that can be used to gain a better understanding of the structure of the group G. Definition Let (G, ) be a group. The conjugacy relation on G is defined for all a, b 2 G. a b if and only if a = g 1 bg for some g 2 G Theorem Let (G, ) be a group. Then the conjugacy relation on G is an equivalence relation on G. Proof. See Exercise 11.

4 3.3 Equivalence Relations and Partitions on Groups 87 Definition Let (G, ) be a group and let be the conjugacy relation on G. For each a 2 G, the equivalence class [a] = {b 2 G : a b} is called the conjugacy class of a. Thus, for any group G, we have that G/ = {[a] : a 2 G} is a partition of G. We now show how to evaluate each conjugacy class in this partition. Theorem Let (G, ) be a group, a 2 G, and let be the conjugacy relation on G. Then [a] = {g 1 ag : g 2 G}, where [a] is the conjugacy class of a. Proof. See Exercise 12. Corollary Let (G, ) be a group, a 2 G, and let be the conjugacy relation on G. Then [a] = {a} if and only if a 2 Z(G). Proof. See Exercise 13. Let (G, ) be a group with a 2 G. We know that C(a) is a subgroup of G and, by Theorem 3.3.2, we also know that C(a) is an equivalence relation on G. Theorem Let (G, ) be a group with a 2 G. Then x C(a) y if and only if x 1 ax = y 1 ay, for all x, y 2 G Proof. See Exercise The Class Equation Let (G, ) be a finite group with a 2 G. For each x 2 G, let[x] C(a) = {y 2 G : y C(a) x} be the C(a) equivalence class of x. Thus, G/C(a) is the resulting partition of G (see Definition 3.3.3). The class equation is a numerical equation describing this partitioning of G. Our next theorem will show that the set G/C(a) has the same number of elements as the conjugacy class [a]. Theorem Let (G, ) be a finite group with a 2 G. Then [a] = G/C(a), thatis,the sets [a] and G/C(a) have the same number of elements. Proof. Let (G, ) be a finite group with a 2 G. Let f :[a]! G/C(a) bedefinedbyf(g 1 ag) = [g] C(a) for each g 1 ag 2 [a]. Using Theorems and , one can prove that f is a welldefined function which is also one-to-one and onto. Thus, [a] = G/C(a). If G is a finite group, then for any a 2 G, the elements in the conjugacy class [a] are in one-to-one correspondence with the elements in G/C(a). Thus, the number of elements in the conjugacy class of a equals the index [G : C(a)] of the centralizer C(a) ing. We can now present the important and useful class equation for finite groups. Theorem (The Class Equation). Let (G, ) be a finite group. Then G = Z(G) + X [G : C(a)] (3.10) a/2z(g) where the sum runs through exactly one element from each conjugacy class [a] with a/2 Z(G). Proof. Let (G, ) be a finite group. We know that G/ = {[a] : a 2 G} is a partition of G. Furthermore, by Corollary , we know that a conjugacy class [a] = {a} if and only if a 2 Z(G). Thus, if we collect together all of the elements in these singleton classes, we obtain the center Z(G). We can now conclude that {Z(G)}[{[a] : a 2 G and a/2 Z(G)}

5 88 Chapter 3. Groups is also a partition of G. Therefore, G = Z(G) + X a/2z(g) [a] where the sum runs through exactly one element from each conjugacy class [a] with a/2 Z(G). Since the number of elements in each conjugacy class of [a] equals [G : C(a)], we conclude that G = Z(G) + X [G : C(a)] a/2z(g) where, again, the sum runs through exactly one element from each conjugacy class [a] with a/2 Z(G). This completes the proof. Remark Let G be a finite group. By Exercise 7 on page 83, we see that C(a) is a proper subgroup of G, when a/2 Z(G). Furthermore, Z(G), C(a), and [G : C(a)] all evenly divide G by Lagrange s Theorem. These are just some of the reasons that the class equation (3.10) has many important applications in group theory. Exercises Let (G, ) be a cyclic group. Prove that G is abelian. 2. Prove Theorem Prove Theorem Prove Lemma Suppose that (G, ) is a finite cyclic group of order m. Let G = {e, a, a 2,...,a m 1 }. Prove that o(a) = m. 6. Let (G, ) be a group and let N be a subgroup of G. Let be the equivalence relation on G defined by a N b if and only if ab 1 2 N. Prove the following: (a) [e] =N. (b) For all a 2 G, we have that a 2 N if and only if [a] =N. 7. Let (G, ) be a finite abelian group and let a, b 2 G. Suppose that o(a) =m and o(b) =n where m and n are relatively prime. Prove that (a) hai\hbi = {e}, (b) o(ab) =mn. 8. Let (G, ) be a group with subgroup H. Suppose that H K where K is also subgroup of G. Show that K/H G/H. 9. Let (G, ) be a group. Suppose that every subgroup of G is normal. Prove for all a, h 2 G, that aha 1 = h i for some integer i. 10. Let (G, ) be a cyclic group with generator a. Suppose that the order of G is n = jk where 1 <j,k<n. Let b = a j. (a) Show that o(a) = jk. (b) Show that o(b) = k. (c) Consider the subgroup hbi. Show that hbi = {e, a j,a 2j,...,a (k 1)j }. (d) Show that a/2hbi and thus hbi is a proper subgroup of G.

6 3.4 Permutation Groups Prove Theorem Prove Theorem Prove Corollary Prove Theorem Let (G, ) be a group and let be the conjugacy relation on G. For a, b 2 G, prove that if b a and a/2 Z(G), then b/2 Z(G). Exercise Notes: For Exercise 1, review Remark For Exercise 2, review Lemma For Exercise 4, apply Lemma and the Division Algorithm For Exercise 7, use Exercise 5 on page 44 and Theorem For Exercise 8, read Exercise 9 on page 83 and then use Theorem For Exercise 13, review Definition For Exercise 14, review Definition and Theorem Permutation Groups Permutations of a Set An important class of groups are called permutation groups. What is a permutation? Let S be a set, then a permutation of S is a one-to-one and onto function : S! S. In this section we shall be using lower case Greek letters to denote such functions. Example 1. Let Z be the set of integers and define : Z! Z by (x) =x+2. Since is one-to-one and onto, the function is a permutation of Z. Example 2. Let R be the set of real numbers and let a, b 2 R with a 6= 0. Define the function : R! R by (x) =ax + b. Then is a permutation of R. Definition Let S be any non-empty set. Then Per(S) is the set of all one-to-one and onto functions from S to S. Per(S) is called the set of all permutations of S. Given, 2 Per(S) define a binary operation on Per(S) by where is composition. A permutation group is a group whose elements are permutations of a given set with composition as its binary operation. We will soon show that the algebraic structure (Per(S), )is a permutation group. Remark Let and be permutations in Per(S) wheres is a nonempty set, and let n be a whole number. Then we shall write: 1. for the composition, whichisdefinedby( )(x) = ( (x)) for all x 2 S for the inverse function of and is defined by 1 (x) =y if and only if (y) =x. 3. for the identity function from S to S. [ is the Greek letter iota.] 4. n for the composition {z }. n times 5. n as shorthand for ( 1 ) n 1 1, that is, for the composition 1 {z }. n times 6. 0 =. Note that 1 2 Per(S) by Theorem and, since is one-to-one and onto, we see that 2 Per(S). It follows that i =, i j = i+j,( i ) j = ij and ( i ) 1 = i for all i, j 2 Z.

MA441: Algebraic Structures I. Lecture 18

MA441: Algebraic Structures I. Lecture 18 MA441: Algebraic Structures I Lecture 18 5 November 2003 1 Review from Lecture 17: Theorem 6.5: Aut(Z/nZ) U(n) For every positive integer n, Aut(Z/nZ) is isomorphic to U(n). The proof used the map T :